Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: dbishopbliss on January 13, 2014, 04:54:36 pm
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I recently heard some demos of the Marshall 50th anniversary 1 Watt amps. I especially liked the JTM1, which got me thinking... there cannot be much to that amp. After some searching around, it seems that there are a number of existing designs out there. Then I started thinking more (which can be dangerous) and was hoping that a small amp like this could be a good way to learn how to actually design an amp as opposed to simply building one that someone else designed.
I am more of a hands-on interactive learner. Reading books, etc is OK, but I never know if I am really understanding. This seems like a friendly forum helping me on my earlier project (which I'm very close to finishing but I forgot a few resistors which are being shipped). So I'm going to start by posting my first questions and hopefully the members will chime in to help me out (or tell me to go away which is OK as well). As I go, I will continue to update my design on this thread and eventually build the amp. Sounds like fun to me, I hope you think so as well.
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There's some good info and a proven build here....
http://www.ax84.com/index.php/oldprojects.html?project_id=firefly (http://www.ax84.com/index.php/oldprojects.html?project_id=firefly)
diy tube guitar amp: ax84 firefly demo (http://www.youtube.com/watch?v=Ll9z6Qs5urY#)
This will at least give you an idea of a design to look at....
Good luck and let us know what your up to :thumbsup:
One of the the most important (and least glamorous) things you could learn is how to decipher the schematic.
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I'm not sure if this is the right place to start, but I know the amp I'm trying to emulate is a push pull configuration using a single 12AU7/ECC82 tube. I also plan on using a Hammond 125C output transformer because I already have one. Nice thing about this transformer is there are multiple secondaries so I can play around with different loads.
That said, I have seen other designs out there that suggest the 22.5K load sounds good with a 12AU7 push pull amp. I believe, since there are two tubes I should halve the load. So, I have drawn out a 11.25K loadline on the ECC82 grid curves in the attachment.
Next, I have chosen an operating point of 250V/8mA because it looks like a nice place to start. Based on the operating point and assuming I want the amp to be cathode biased, I would calculate the cathode resistor using the following formula.
9.5V / 8mA = 1,187.5R
Using standard resistor values, I would use a 1200 Ohm resistor (1K2) on the cathode of each triode.
Couple of questions now:
- How's my math?
- If I wanted the cathodes to share a resistor, would it be 600 Ohm or 2400 Ohm (or something completely different)?
- Is there a better way to choose a loadline?
- Is there a better way to choose an operating point?
Thanks for your help.
David
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There's some good info and a proven build here....
http://www.ax84.com/index.php/oldprojects.html?project_id=firefly (http://www.ax84.com/index.php/oldprojects.html?project_id=firefly)
I've seen the firefly and have referenced it in my post above. But I want to know why values were chosen rather than simply using an existing schematic. I supposed I could ask over there as well, but I have had good luck asking questions here before.
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I've seen the firefly and have referenced it in my post above. But I want to know why values were chosen rather than simply using an existing schematic. I supposed I could ask over there as well, but I have had good luck asking questions here before.
Oops....didn't catch the reference.....
You never know what someone knows and/or doesn't know until you know :think1:
Looks like you've got a good grip on it.......I'm just a big preacher of the schematic :icon_biggrin:
"Feed a hungry mind" is a quote someone threw at me when I came aboard.
Continue on sir.....
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> 1 Watt amps. ... there cannot be much to that amp.
Perhaps not much *less* than a big amp. Pretty near the same number of parts. You need a preamp tube, a power tube, an OT, a PT, a DC filter... either way.
Here we have a 2000 pound Miata MX5 and a 6000 pound Chevy K2500 pickup truck. The both have 4 wheels, they both have an engine. True, the truck wheels are 2X the weight; and 6-bolt instead of 4-bolt. The Chevy has more pistons but the Miata has more camshafts. Bolts on the Miata are easy to turn but often hard to get to and easy to break; the truck doesn't try to tuck everything in tight.
> a 11.25K loadline on the ECC82 grid curves in the attachment.
But you have assumed a Resistor load. That plate voltage will drop from 340V to 250V.
You have a transformer. It may be 11K for 100Hz-10KHz but is surely much lower at DC. Perhaps 1,000 Ohms. So your 8mA will only drop about 8V from 340V, giving 332V static plate voltage.
You have to check that the 8mA assumption will not violate Pdiss at 332V. And you have to find your G1 voltage based-on a 332V plate voltage.
The actual load-line PER TUBE is curved as shown in dash-line. The composite loadline for both tubes is still your purple line.
In happy cathode bias you probably want your idle current not much less than half your peak current. Your peak appears to be 18mA. So try 9mA.... OK, that's probably a hair over Pdiss rating. However 8mA is close-enough and safe. (This is a point best explored on a working prototype.)
With that correction.... yes, your loadline is functional and probably very near optimal. And triodes are quite un-fussy. You could load 6K/plate or 24K/plate with only small change in Watts and THD.
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> 1 Watt amps. ... there cannot be much to that amp.
Perhaps not much *less* than a big amp. Pretty near the same number of parts. You need a preamp tube, a power tube, an OT, a PT, a DC filter... either way.
Good point, but my comment was directed at the JTM1 as opposed to 1 watt amps in general. The JTM1 has a single input, a loudness knob and a tone knob. Compared to JVM1 which has two channels, bass, middle, treble, presence controls, effects loop, etc... there cannot be much to that amp.
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> a 11.25K loadline on the ECC82 grid curves in the attachment.
But you have assumed a Resistor load. That plate voltage will drop from 340V to 250V.
You have a transformer. It may be 11K for 100Hz-10KHz but is surely much lower at DC. Perhaps 1,000 Ohms. So your 8mA will only drop about 8V from 340V, giving 332V static plate voltage.
How quickly you learn what you don't know. :dontknow:
First, what do you mean by at DC?
Second, is there a way for me tell from the transformer spec what the impedance will be or is this something where taking a SWAG is probably good enough?
You have to check that the 8mA assumption will not violate Pdiss at 332V. And you have to find your G1 voltage based-on a 332V plate voltage.
I'm assuming that Pdiss is the Maximum Plate Dissipation found on this (http://www.wooaudio.com/docs/tube_data/12AU7.pdf) datasheet or Wa on this (http://www.jj-electronic.com/pdf/ECC%2082.pdf) datasheet. Since the red dot on your diagram appears below the curved line then I think it does not violate Pdiss (Wa). Let's see... P = I * V or 2.256 = 8mA * 332V. So, we are below the 2.7W rating.
I'm not sure what you mean by find your G1 voltage but I suspect you are jumping ahead.
The actual load-line PER TUBE is curved as shown in dash-line. The composite loadline for both tubes is still your purple line.
How did you determine the dashed line's slope/curve?
I have read that if the amp was running in ClassAB you need to draw the A loadline and the B loadline, then combine them where they intersect. I used 1/2 the anode to anode transformer impedance (Za-a) or 11.2K for Class A and 1/4 Za-a or 5.6K for Class B. However, I noticed that most of the time the amp would be running in Class A so I didn't include it in my diagram. I have the image with both lines on another computer so can post that later if you think it will help.
This is great. Thanks!
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... But I want to know why values were chosen rather than simply using an existing schematic. ...
Then let's reverse-engineer the existing schematic (because selecting from scratch is actually a much longer, iterative process that leads to the final answer; we''l start with the final answer).
The Firefly schematic (http://ax84.com/media/ax84_m276.gif) shows we're using Hammond 269EX PT (190-0-190), Hammond 125A OT (wired for 22.5kΩ:8Ω), a self-split 12AU7 push-pull output stage with 8.8vdc across a 440Ω shared cathode resistor.
Let's assume B+ equals 12AU7 plate voltage, so 190v * 1.414 = ~267vdc after diode drops in the full-wave rectifier. If we want to be exact, we can subtract the cathode voltage from the B+ to get resulting plate voltage; 267vdc - 8.8vdc = ~258vdc (call it 260v).
The idle current of both 12AU7 triodes is passing through the cathode resistor, so 8.8v / 440Ω = 20mA -> /2 = 10mA per triode. 260v * 10mA = 2.6w dissipation, so we're just below max dissipation (which is where you want to idle a class A output stage; anything less is just less power output).
The green line represents a guesstimate of the -9v grid bias curve (about halfway between -8v and -10v). The slope of this line at the idle point is also the internal plate resistance of the tube at that point. You find it by going a bit to either side of idle along the bias curve and dividing the change in voltage by the change in current. I think I see relevant coordinates at (255v, 10mA) and (275v, 12.5mA).
rp = (V2-V1)/(I2-I1) = (275v - 255v)/(12.5mA - 10mA) = 20v / 0.0025A = 8kΩ
In RDH4 (you can download a copy from the Library of Information), Ch. 13.5 iii (pg. 577) says that push-pull class A triodes develop maximum output power when the plate-to-plate load impedance is twice the plate resistance of one of the tubes at idle (the previous section went to great lengths to derive that information). RDH4 also says this value of load impedance ought to be considered a minimum value, as increasing the load will reduce odd harmonic distortion and peak plate currents.
So we need a OT with 8kΩ * 2 = 16kΩ plate-to-plate primary impedance. If the tubes operate in class A (never turn off), then the load that a single tube sees is half the primary impedance (it sees the half from its plate to B+). So that's an 8kΩ load to each tube.
The orange line is and 8kΩ load, while the blue line represents a 16kΩ load. 8kΩ ventures well above the dissipation limit and so seems inadvisable. The 16kΩ line looks good, but that represents a 32kΩ plate-to-plate load. High load impedances like that are not usually available, and the highest the 125A can manage with an 8Ω speaker load is 22.5kΩ (probably why the designers chose that impedance).
There's a 2nd rule for maximum output power for triodes (RDH4 p. 578): loadline for maximum power intersects the 0v gridline at 0.6*B+. The plate current at that point is taken as a maximum current. For our 260v case, the current at the point on the 0v gridline is 21.5mA, and:
Rl = 1.6 * B+ / Imax = 1.6*260v / 0.0215A = ~19kΩ
P = 0.2 * B+ * Imax = 0.2 * 260 * 0.0215A = ~1.1w
This is drawn on the 2nd set of curves as a blue line running from (155v, 21.5mA) down to B+ and zero current. This is a class AB loadline, which passes through B+ on the x-axis, and represents 1/4 of the plate-to-plate load impedance (which each tube sees once the other tube shuts off). Double-check: (260v - 155v) / 0.0215A = 4883Ω -> 4883Ω * 4 = 19.5kΩ. So the math and loadlines check.
You'll see this loadline (blue) strays into excessive dissipation at the peak current; that might not be a problem because the duration is so brief. A loadline represented by 22.5kΩ/4 is shown by the orange line, and the higher impedance reduces the peak current. The "perfect" greeen loadline represents (260v - 142.5v) / 0.019A = 6184Ω, 6184Ω * 4 = 24.7kΩ. Looking at the chart for 125A impedances, something approaching that may be an awkward speaker load for such a small amp.
So it looks like the Firefly design approaches the ideal load for maximum power if the tubes push somewhat into class AB (class A resulted in excessive dissipation), and we use the closest conveniently-available plate load with existing transformers.
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> a 11.25K loadline on the ECC82 grid curves in the attachment.
But you have assumed a Resistor load. That plate voltage will drop from 340V to 250V.
... I used 1/2 the anode to anode transformer impedance (Za-a) or 11.2K for Class A and 1/4 Za-a or 5.6K for Class B. However, I noticed that most of the time the amp would be running in Class A so I didn't include it in my diagram. ...
I think David mis-stated, and drew the class A loadline (11.25kΩ, for 22.5kΩ plate-to-plate) for a B+ of 250v (very close to the Firefly schematic voltage.
David, I think PRR saw your line as an 11.25kΩ class AB line through 340v B+. But that would be 11.25k * 4 = 45kΩ and not readily available.
That load and your operating point (different for Firefly's) will work. I think the Firefly idles hotter and stays slightly into excessive dissipation, but also switches to class AB at those (brief) peaks, all of which keeps the 12AU7 from overheating on average.
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I'm digesting this stuff now. I have downloaded and read the RDH4 before but I have always been overwhelmed by the information there. The description above is a great summary and I am re-reading the RDH4 chapter now with a better (but not complete) understanding.
I was feeling like I cheated a bit by using the load on the firefly schematic rather than knowing how to determine what would be the optimal load. I recall reading that the designer originally chose an 8K load but decided to use the 22.5K load because it sounded better. I'm not sure if he changed to 22.5K after doing an analysis like the one presented above or if it was just trying different things, but I thought I read it was trial and error.
Just for fun (and to confirm that I am understanding the information presented) I think I will see if I can figure out the optimal load lines for a 12BH7. Partly because I have heard it can be used in a Firefly and partly because I have a couple of them laying around.
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I recall reading that the designer originally chose an 8K load but decided to use the 22.5K load because it sounded better. I'm not sure if he changed to 22.5K after doing an analysis like the one presented above or if it was just trying different things, but I thought I read it was trial and error.
Because the internal plate impedance on a preamp tube is different then a power tube you need to use a different OT primary Z than the common >10K used for power tubes. PRR has said that winding a high Z primary OT has problems. Power tubes were made/designed with that in mind preamp tubes weren't because they were designed with a different goal/purpose in mind.
figure out the optimal load lines for a 12BH7. Partly because I have heard it can be used in a Firefly and partly because I have a couple of them laying around.
A lot of guys seem to favor the 12BH7 over 12AU7 in that amp.
Brad :icon_biggrin:
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On my post on this tread you can find some idea for a low power amp
http://www.el34world.com/Forum/index.php?topic=16504.msg162453#msg162453 (http://www.el34world.com/Forum/index.php?topic=16504.msg162453#msg162453)
K
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A lot of guys seem to favor the 12BH7 over 12AU7 in that amp.
Brad :icon_biggrin:
YES! A big "pleasurable" difference w/ a touch more output volume too. Don't forget you need to move the speaker cone even a little bit, if not what's the point?! No satisfaction and you may as well just make a simple LM386 SS toy amp instead. It'll be much easier and cheaper and play & sound about the same. This said, ECC99 tubes are the limit for the Firefly-type amps. You'll get about 3.5watts or so w/ these and about 2watts out of the 12BH7a tubes. I can play sitting in front of mine at full volume and when I walk out of the room my ears are just below the point where they would be ringing when I'm using the 12BH7a. Don't forget that you can run the 12BH7a & especially the ECC99 both hotter so a slight re-bias helps run them optimally. Good luck Bliss
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Thanks for all the links and advice about using other tubes. So, seems like my exercise in calculating loadlines and bias points for 12BH7 will not be wasted. So, here is my attempt...
For the purposes of this example I will assume I'm using a Hammond 269EX PT (190-0-190) and Hammond 125C OT. I will assume the B+ equals the plate voltage therefore:
190v * 1.414 = ~267vdc
I don't have a schematic to reference as in the example above, so I eyeballed the plate current close to the Pdiss and decided the cathode voltage will be somewhere around 12vdc. Therefore the resulting plate voltage will be 267vdc - 12vdc = ~255vdc. However, I'm going to use 250vdc because it makes the graphing and math easier.
Now I want to figure out the internal plate resistance at the idle point. Luckily for me, the 10vdc grid line is very close to Pdiss at 250V. The green line in my diagram is the guestimate of rp calculated using coordinates at (225v, 7.5mA) and (275v, 17.5mA).
rp = (V2-V1) / (I2-I1) = (275v - 225v) / (17.5mA - 7.5mA) = 50v / 0.01A = 5kΩ
This is pretty consistent with the 12BH7 datasheet which suggests they typical plate resistance is 5.3kΩ.
According to RDH4 pg 577, I need a OT with 5kΩ * 2 = 10kΩ plate-to-plate impedance for a push-pull class A triodes to develop maximum power. Even though the transformer is 10kΩ, each tube will only see 5kΩ.
The blue line in the diagram is the 10kΩ load while the orange line represents 5kΩ load. Just like the 12AU7 example above, the higher load exceeds the dissipation limit. The 10kΩ load is pretty good but it does exceed the limit a tiny bit. The 10kΩ represents a 20kΩ plate-to-plate load which is close to the values provided by the Hammond 125C (22k/4Ω, 22.5k/8Ω and 21.5k/15Ω).
Do I have this right so far?
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Keep in mind that using the formula (*1.41) to calculate voltage is the "unloaded/ideal" voltage. You're actual loaded B+ is actually going to be closer to the 250vdc that you're calculating for.
Another important thing to remember is that your component tolerance overall is going to be between +/- 10% to 20% anyway so figuring to "gnats ass" only has so much relevance & practicallity. So no need to be overly concerned getting everything down to 4 significant figures...exagerating here but you get the idea.
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... I have downloaded and read the RDH4 before but I have always been overwhelmed by the information there. ...
RDH4 is not written as a how-to book (which we'd really like it to be); it's written as a "you remember this formula ..." book. As in, it's supposed to be a reference of material you already learned.
I'm embarrassed to say how long it took for me to clue into very basic concepts, like selecting optimum load impedance, because it was never laid out as simply as it really is (or because I imagined there must be more to it).
... I was feeling like I cheated a bit by using the load on the firefly schematic rather than knowing how to determine what would be the optimal load. ...
Eventually, you'll learn to appreciate the ease of using someone else's work (i.e., shamelessly steal the operating condition they figured out).
... I recall reading that the designer originally chose an 8K load but decided to use the 22.5K load because it sounded better. I'm not sure if he changed to 22.5K after doing an analysis like the one presented above or if it was just trying different things, but I thought I read it was trial and error. ...
In general, people learn that maximum power is transferred from one network to another when the impedance of each (where they join) is the same. So if he also figured the internal plate resistance of his 12AU7 at the operating point was 8kΩ, then that may be why he chose an 8kΩ load.
Maybe he analyzed it, or maybe he re-jiggered the 125A wiring to get different loads. As long as you know the right formulas, how to apply them, and don't make any math errors, calculation is faster that try-n-see. But try-n-see is required to verify the circuit operates the way you calculated. And the theoretical results should jive with the practical results, or the theory is no use.
So analysis or not, the result seems to be valid.
More to follow on your 12BH7 example...
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Now let's see if the second rule values agree with values above. The second rule says the loadline for maximum power intersects the 0v gridline at 0.6 * B+ = 0.6 * 250v = 150v. The current at the point on 0v gridline is 40mA therefore:
Rl = 1.6 * B+ / Imax = 1.6 * 250v / 0.04A = 10kΩ
P = 0.2 * B+ * Imax = 0.2 * 250v * 0.04A = 2W
Now to double check... (250v - 150v) / 0.04 = 2500Ω -> 2500Ω * 4 = 10kΩ so the math and loadlines check.
However, the loadline (blue) exceeds the dissipation limit by quite a bit. So, I have drawn a "perfect" loadline in green that represents the following:
(250v - 113v) / 0.03A = ~4567Ω -> 4567Ω * 4 = ~18.3kΩ.
Based on this, there may be a better impedance to choose (but I'm not really sure) on the 125C. At 4Ω, there is a 15K impedance available. At 8Ω there is a 17.6K impedance available. Would any of these be "better" than choosing the values originally selected using the first method (22k/4Ω, 22.5k/8Ω)?
Did I get these numbers right?
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For the purposes of this example I will assume I'm using a Hammond 269EX PT (190-0-190) and Hammond 125C OT. I will assume the B+ equals the plate voltage therefore:
190v * 1.414 = ~267vdc
Keep in mind that using the formula (*1.41) to calculate voltage is the "unloaded/ideal" voltage. You're actual loaded B+ is actually going to be closer to the 250vdc that you're calculating for.
Turns out the 269EX is rated for 71mA from the high voltage winding, so if anything the B+ may be a bit higher. I'll call this "check" to our first step.
... I don't have a schematic to reference as in the example above, so I eyeballed the plate current close to the Pdiss and decided the cathode voltage will be somewhere around 12vdc. ... Luckily for me, the 10vdc grid line is very close to Pdiss at 250V.
No-Go. You have to pick a bias voltage by which the total output tube voltage will be reduced (by virtue of drop across a cathode resistor). This value can't change willy-nilly, or else you should go back and restart with the new value of bias voltage and see where that value leads you.
Yes, that means in a real design, if there are 8 steps to the process you might really do 43 steps, once you count all the times you go back and re-figure because your earlier assumption didn't pan out.
To keep moving forward, let's pretend your rectified supply voltage is 260vdc, and you'll be using a bias of 10v.
Now I want to figure out the internal plate resistance at the idle point. Luckily for me, the 10vdc grid line is very close to Pdiss at 250V. The green line in my diagram is the guestimate of rp calculated using coordinates at (225v, 7.5mA) and (275v, 17.5mA).
rp = (V2-V1) / (I2-I1) = (275v - 225v) / (17.5mA - 7.5mA) = 50v / 0.01A = 5kΩ
Well-drawn tangent, with good endpoints chosen. But be sure to update with the copy/paste with the new numbers you find... :icon_biggrin:
rp = (V2-V1) / (I2-I1) = (300v - 225v) / (22.5mA - 7.5mA) = 75v / 0.015A = 5kΩ (surprisingly, both land on the same result)
According to RDH4 pg 577, I need a OT with 5kΩ * 2 = 10kΩ plate-to-plate impedance for a push-pull class A triodes to develop maximum power. Even though the transformer is 10kΩ, each tube will only see 5kΩ.
Yes, that's correct if we can operate the tube in class A (neither triode is ever driven to cutoff).
The orange line in the diagram is the 10kΩ load while the blue line represents 5kΩ load.
No-Go.
These must be backwards, as the more-vertical loadline is always a smaller resistance/impedance.
Look at this another way: Decide a value of current-change (say, 5mA). Ohm's Law says 5mA through 5kΩ causes a 25v drop (or 25v change); 5mA through 10kΩ causes a 50v drop. Now, draw a line on the graph where the difference in current between the 2 endpoints is 5mA, and the plate-voltage change is 25v. Repeat for everything the same except the plate voltage changes 50v.
You'll see the higher resistance/impedance is a longer, flatter line.
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The orange line in the diagram is the 10kΩ load while the blue line represents 5kΩ load.
So let's figure how to check these. You already know 1 point for each line: 250v and ~12.5mA.
The orange line passes mighty close to 150v and 32.5mA, so I'll take that as a 2nd point for that line. rp = ΔV / ΔI = 100v / 20mA = 5kΩ (that checks with what I said about steeper = lower resistance).
... The 10kΩ load is pretty good but it does exceed the limit a tiny bit. The 10kΩ represents a 20kΩ plate-to-plate load which is close to the values provided by the Hammond 125C (22k/4Ω, 22.5k/8Ω and 21.5k/15Ω).
Check; I think you corrected yourself whether you realized it or not.
Notice that the 10kΩ line does go above the plate dissipation limit, but just barely and only for a very small percentage of the total signal cycle. Any designer I know would call that perfectly acceptable.
Now a final step: calculate cathode bias resistance (since you've gone this far). You operating point shows 12.5mA per tube, so for 2 tubes you'll have 25mA through the cathode resistor. You're figuring your bias will be 10v, so 10v / 0.025A = 400Ω. 400Ω is not a very common value, but is available in wirewound resistors. 10v * 25mA = 0.25w, so a 1w wirewound resistor wound be amply sufficient.
If you had to select a different, available value, you'd want to pick a higher resistance. That's because we have already noticed we're right at max dissipation and have acknowledged our supply voltage may be higher than calculated due to under-loading of the winding. higher resistance would give a bit more bias voltage and cool the tube off with high supply voltage.
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Now let's see if the second rule values agree with values above.
...
Did I get these numbers right?
Yes, that process looks correct.
... So, I have drawn a "perfect" loadline in green that represents the following:
(250v - 113v) / 0.03A = ~4567Ω -> 4567Ω * 4 = ~18.3kΩ.
Based on this, there may be a better impedance to choose (but I'm not really sure) on the 125C.
You did this properly.
However, what I'm seeing (that rules don't outright express) is that the composite loadline for the class AB condition (show in your latest example) seems to want slightly less supply voltage to stay within dissipation (maybe down to 225v or so). In practice, some straying above the dissipation curve may not be a problem (for the reason I mentioned before).
If you had to choose an available impedance, I'm sure you see that the higher impedance means a reduction of tube dissipation, and is probably the safer choice.
... I will assume I'm using a ... Hammond 125C OT.
I'm sure you realize this, but there's not a good reason for you to spend money on the bigger 125C, since all the same impedances are available on the 125A. I'm guessing you just wanted to change the initial conditions some.
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I'm sure you realize this, but there's not a good reason for you to spend money on the bigger 125C, since all the same impedances are available on the 125A. I'm guessing you just wanted to change the initial conditions some.
You missed the bit about my already owning a 125C. No reason to spend money on a 125A. :icon_biggrin:
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Pssshhhhh.... Trying to use parts already on-hand. Who ever heard of doing that!!?! :l2:
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> what do you mean by at DC?
> ...is there a way for me tell from the transformer spec what the impedance will be or is this something where taking a SWAG is probably good enough?
Like you poke a DC volt-meter at it.
The ideal transformer will be 11,500 ohms impedance (loaded) at all audio frequencies but ZERO impedance for DC current. DC impedance (resistance) is just pure waste of power. We wind as much fat copper as we can to get the DC resistance down. For many audio transformers we end up with DCR near 5%-10% of nominal audio impedance.
And you "know" that with 340V supply the plates will sit very-near 340V DC, not 250V.
(250V would be correct for a *resistance-loaded* amplifier. However the efficiency of a res-load amp is only 6%. We can tolerate that for the microWatt signals in preamps, but not when we want WATTs.)
What is a typical value for transformer DCR? Frankly for geetar amps you can assume "zero". It won't ever be *zero*, but it usually ends up "very very small". My assumption of ~~9% dropped 340v to 330V, which is just-the-same in practical terms.
> not sure what you mean by find your G1 voltage but I suspect
I liked idle to be Ipeak/2 or 18mA/2 or 9mA. But 9mA times 330V is 2.97W which sits just over the 2.75W Pdiss rating. 8mA is just as happy and gives 2.64W which is just under the line.
Then 8mA at 330V implies a G1 voltage between the -14V and -16V lines. Maybe it's -14.6V. However it is also +/-20% from tube to tube, so let's not be fussy. And noting that 16V and 8mA are a neat 2:1 ratio, I'd be thinking 200 ohms cathode resistor (per tube). Or even nearest-up standard value, 220 Ohms/tube. That's liable to end up a wee bit cold, but it can be trimmed after smoke-test.
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What is a "perfect" load-line?
There's no such thing.
And to get this you *really* need a FIRM grasp of Basic Electricity.
The power line to my house is (say) 120V and 1 Ohm. What is my "perfect" load?
For what? Maximum load on the utility company? Maximum power in my house? Reasonably constant voltage so lights don't flicker bright dim dead? Minimum electric bill?
The Max Power condition is easy. Load a 1 Ohm source with a 1 Ohm load, "matching". I get half the voltage, 60V. The other 60V is dropped in the line. The current is 60V/1r or 60 Amps. I have 60V*60A= 3600 Watts in my house. I also have 3600 Watts in the power line where it does me no good. I have 60V full-load but 120V if I ever turn-off most of the load, so the lamps will be hurting.
The voltage-sag, and paying for twice the power I get in the house (and the hot line) are distressing. Let's say I replace the line with one of 0.01 ohm. The same 1 Ohm load in the house is now just 1% loss. I get 119V at 119A or 14161 Watts in the house and lose just 142 Watts in the line. No-load to full load is just 120V-119V so the lamps are happy. Until the bill for that 0.01 Ohm line comes in. A 0.01 Ohm line obviously has to have 100 times more metal area than a 1 Ohm line. Same length, so it surely costs 100 times as much. The line I have is about $5K, so the "better" line is $500K. Which is many times more than the value of my house&land, many-many times more than the value of all the electric I will ever use. That's dumb.
Generally, utility power is wired for 2%-5% sag from street to house and 2% sag within the house. In fact I'm living with 10% sag. The lamps flicker but the PC doesn't reboot when the pump kicks in. Since a 5% sag would cost me $5K-$15K, I live with sag.
Look at the ratios of Load to Line resistance.
1% drop, 99% efficiency is Load/Line of 100/1 and is too darn expensive.
50% drop, 50% efficiency is Load/Line of 1/1 and is too darn inefficient.
10% drop, 90% efficiency is Load/Line of 10/1 and is acceptable (though 20/1 would be better IF the cost doesn't hurt).
You gotta understand this basic concept of source to load resistances.
Now tubes. Copper/Aluminum conducts pretty good per dollar. Thermionic devices conduct BAD. Where we often have wires 1 Ohm or less, a 12AU7 unit (in the negative grid range) conducts like a 6K resistor; a 6V6 like a 2K resistor. We can go lower but the price rises (bigger tubes or more in parallel).
For *general* use of triodes in transformer-coupled POWER amplifiers, the ratio Rl/Rp should be generally around 2. So a single unit of 12AU7 at 6K should be loaded in 12K. If you go to 6K load you have a wee bit more power but much more distortion. If you go to Rl/Rp=5 or 30K load the distortion is down but so is the power.
Much of the distortion cancels in push-pull. Then Rl/Rp can approach 1 (6K load per 12AU7 unit) with some increase of power.
Since 11.2K/6K is exactly 2/1 (for practical purpose) I think your dart is well-landed. If you were desperate for power you could try lower.... but if you were desperate for power you wouldn't be dinking with 12AU7 when bigger tubes exist at tolerable price. And while a higher load would be lower distortion, that's only true in the "clean" zone. Electric guitar amplifiers ARE pushed well past this, so the "clean" distortion is unimportant, while POWer is always important. (If the amp does have too much power, you can drop the B+ and use smaller parts all around for savings in cost and portability.)
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> The second rule says the loadline for maximum power intersects the 0v gridline at 0.6 * B+ = 0.6 * 250v = 150v.
What does "0.6" mean?
It means 60% dropped in load and 40% dropped in tube. So the load is 60/40 or 1.5 times the tube resistance, Rl/Rp= 1.5. This is very near the Rl/Rp ratios of 1/1 and 2/1 discussed above, so very near a "maximum power" condition.
FWIW, Western Electric went through this and found two interesting conditions:
Rl/Rp = 1 for maximum power output with insufficient input signal (telephone repeaters).
Rl/Rp = 2 for best output-to-distortion ratio single-ended).
RCA's "0.6" or Rl/Rp=1.5 just splits the difference.
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I just realized I have a pair of ECC99 in an amp I own (Audioromy 813). I'm going to play with those curves next. Sorry if this is redundant for folks, but it reinforces the concepts for me.
Next up with be calculating the cathode resistor for all of these tubes/operating points. I'm pretty sure I already know how to determine this but I'm sure there is plenty that I'm missing.
After that... Phase Inverter time. Uh oh!!! I hope my head doesn't explode.
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After that... Phase Inverter time. Uh oh!!! I hope my head doesn't explode.
The beauty of the Firefly is that since the triodes are ran in self-split there is no pi to worry about. How many tubes do you want or need for running simple triodes? Is this just mental "exercise time" again?
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The beauty of the Firefly is that since the triodes are ran in self-split there is no pi to worry about. How many tubes do you want or need for running simple triodes? Is this just mental "exercise time" again?
I would call it "learning time". Until now load lines and transformer impedances were mystery values that someone came up with and I followed. I want to know why the values were chosen. I don't expect to know as much as the moderators, but at least I will have a little more than a clue.
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I looked through the basement and found the PT I want to use for this project. Its an AnTek toroidal transformer model AN-05T240 (http://www.antekinc.com/as-05t240-50va-240v-transformer/). It has a 220V tap that I plan to use.
- The grid curves for the ECC99 don't show the Pdis so I plotted the curve in red.
- Assuming the B+ equals the plate voltage, so 220v * 1.414 = ~311vdc. The grid curve value close to Pdis at 311v is approximately 11vdc. Therefore the plate voltage will be 311vdc - 11vdc = 300vdc.
- I chose an operating point current of 16mA to use because it is close to Pdis at 300V (and easy to graph).
- Estimate the internal plate resistance at the operating point by calculating the slope of the -11v grid bias curve. The coordinates (280v, 10mA) and (320v, 24mA) are pretty close. Therefore, rp = (V2-V1)/(I2-I1) = (320v - 280v)/(24mA - 10mA) = 40v / 0.014A = 2857Ω. This is fairly close to the "typical" value given on the datasheet of 2.3kΩ so I don't think I'm way off.
The blue line represents a 2.8kΩ load, but that is WAY over Pdis so I won't be using this. The orange line represents a 5.6kΩ load. It exceeds Pdis for a little bit so it is probably OK, but I'm going to keep trying in the next post.
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Now I'm going to try the second rule where the loadline for maximum power intersects the 0v gridline at 0.6*B+.
- 0.6*300v = 180v. The current at the point on the 0v gridline is 83.5mA (had to get creative with the grid curves to figure this out).
- Rl = 1.6 * B+ / Imax = 1.6 * 300v / 0.0835A = ~ 5.75kΩ
- P = 0.2 * B+ * Imax = 0.2 * 300v * 0.0835 = 5W
- I have drawn the class AB loadline (blue) from (180v, 83.5mA) down to B+ and zero current. This represents 1/4 of the plate-to-plate load impedance. (300v - 180v) / 0.0835A = ~14375Ω -> 14375Ω * 4 = 5748Ω so the math and loadlines check.
- Since the loadline is way past Pdis, calculate the ideal load by drawing a line (orange) from Pdis and G0 intersect to B+ at zero current. The slope of this line can be calculated using (300v - 80v) / 0.048A = 4583Ω, 4583Ω * 4 = ~18kΩ.
So, looking at the Hammond 125C hookup chart I see the options are slightly different for the ECC99 tube. For 4 ohms the closest value is still 22kΩ. I could try 15kΩ but I think that would exceed Pdiss. For 8 ohms I could run 22.5kΩ to be safe or choose 17.6kΩ which I'm guessing would be ok. For 16 ohms, I could choose 21.5kΩ. Not huge differences from the 12AU7 and 12BH7 tubes, but slightly and a good exercise.
I think I will go back and play with the curves for the other tubes using my PT (300v), but I won't bother posting the results any more. I can hear the moderators cheering now.
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IMHO, This is one of the best threads, produced on this forum.
This thread could be chapter(s) a how to book.
Last year I was introduced to the Cornell Notes concept, as a substitute teacher. That format, and this info would go together so well.
Kudos, to all the contributors.
I'm glad someone else is finding this useful as well.
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The blue line represents a 2.8kΩ load, but that is WAY over Pdis so I won't be using this. The orange line represents a 5.6kΩ load. It exceeds Pdis for a little bit so it is probably OK ...
Both lines have slopes that reflect 2.8kΩ and 5.6kΩ, so that's good.
Now remember, the "First Rule" assumes the tube is running class A, so the load seen by a single tube is equal to plate-to-plate impedance divided by 2. That means the two notional transformers you plotted are 5.6kΩ (for the 2.8kΩ line) and 11.2kΩ (for the 5.6kΩ line).
Also, the first rule assumes you want a load for a single tube that is rp * 2, or your orange 5.6kΩ line. So you should be able to predict by the rule that the blue 2.8kΩ line would have exceeded dissipation limits.
The orange 5.6kΩ line is good, except a class A loadline must pass through the operating point, the same way the blue 2.8kΩ line did. So you need to slide that line bodily to the right (maintaining its slope) until it passes through the point at -11v grid, 300v plate and 16mA of plate current.
Yes, we didn't cover that before (because it confuses everyone), but the loadline for a class A load passes through the operating point, while the composite loadline for a class B (or class AB at the point when the other side shuts off) runs down to the supply voltage on the x-axis.
You previously did this correctly with your 12BH7 loadlines.
Now I'm going to try the second rule ...
Yep, all this looks good.
Now if you wanted to plot the class A portion of the loadline you settled on (orange ~18kΩ), you could draw a line with a slope corresponding to 4583Ω * 2 = ~9kΩ, which passes through the operating point at -11v grid, 16mA and 300v plate. Or the available 17.6kΩ / 2 = 8.8kΩ from your transformer.
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Not huge differences from the 12AU7 and 12BH7 tubes, but slightly and a good exercise.
I think I will go back and play with the curves for the other tubes using my PT (300v), but I won't bother posting the results any more. I can hear the moderators cheering now.
You'll find that if/when you build the amp those tubes can pretty much be swapped at will w/out doing anything and they sound/play well.
*don't worry about asking questions, it's what the forum and everyone are all here for. Plus there's many not participating actively either now or in the future that also benefit too that you may never know or realize.
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The orange 5.6kΩ line is good, except a class A loadline must pass through the operating point, the same way the blue 2.8kΩ line did. So you need to slide that line bodily to the right (maintaining its slope) until it passes through the point at -11v grid, 300v plate and 16mA of plate current.
I updated the diagram, but the 5.6kΩ line exceeds Pdis as well.
Now if you wanted to plot the class A portion of the loadline you settled on (orange ~18kΩ), you could draw a line with a slope corresponding to 4583Ω * 2 = ~9kΩ, which passes through the operating point at -11v grid, 16mA and 300v plate. Or the available 17.6kΩ / 2 = 8.8kΩ from your transformer.
I plotted the 8.8KΩ line and the 11.25kΩ line as well. As you can see 8.8kΩ exceeds Pdis whereas 11.25KΩ sits right on it. Looks like the 22.5kΩ value is still the best.
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I just plotted the 11.25KΩ load line for the 12AU7, 12BH7 and ECC99. Ends up, that load line exceeds Pdis at 300V for the 12AU7 and 12BH7.
I think my choices are either lower the idle current values or lower the idle voltage. What is the best thing to do in this situation?
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Now if you wanted to plot the class A portion of the loadline you settled on (orange ~18kΩ), you could draw a line with a slope corresponding to 4583Ω * 2 = ~9kΩ, which passes through the operating point at -11v grid, 16mA and 300v plate. Or the available 17.6kΩ / 2 = 8.8kΩ from your transformer.
I plotted the 8.8KΩ line and the 11.25kΩ line as well. As you can see 8.8kΩ exceeds Pdis whereas 11.25KΩ sits right on it. Looks like the 22.5kΩ value is still the best.
Well, it's the safest.
But look at the entirety of the 8.8kΩ line you drew. What percentage of the total line strays above the dissipation curve? And by how far? Maybe half the line, and just barely above it, right?
So in the actual built amp, you may not notice any redplating with that setup. If you did and (for whatever reason) had to keep the 17.6kΩ plate-to-plate load, you might idle a little cooler (a little more bias voltage, a little less idle current). That would shift that entire loadline bodily downward, and move it all under the dissipation curve.
I updated the diagram, but the 5.6kΩ line exceeds Pdis as well.
This all highlights the drawback of class A operation: to reduce your plate dissipation, you should reduce the plate voltage, shift the loadline bodily leftward to a new operating point at the lower plate voltage, and thereby keep the tube from exceeding its plate dissipation rating.
This is also where class AB winds up being used to get more output power: supply voltage is higher than class A (and similar to what you're trying to use now), load impedance is lower (to allow more peak plate current and, in conjunction with the higher plate voltage, more output power).
The loadline may appear to run above the plate dissipation curve in class AB, but the heating of the plate is offset by the period of time when the tube stops passing plate current. Additionally, the idle bias voltage may be raised to increase the amount of time the tube is shut off, and reduce average plate dissipation (which is where people come up with the "70% dissipation 'rule' ").
The drawback from a design standpoint is that to determine on paper whether the tube will overheat, you need to calculate average power input to the plate (an accurate calculation is involved) and subtract power output to the speaker; the difference is the power dissipated by the tube plate.
You should also be seeing that this process can be tedious, and if you have any flexibility on available supply voltages or load impedances, many calculations might be needed to figure out which works best. Which is why tube manufacturers already made data sheets with example conditions which they worked to determine were the best available (or at least looked "Wow!" on the sheet). They typically chose supply voltages that historically were the most likely to be encountered, so we may/may not find them terribly useful today (but you can apply convesion factors to estimate a good condition at your chosen supply voltage).
I just plotted the 11.25KΩ load line for the 12AU7, 12BH7 and ECC99. Ends up, that load line exceeds Pdis at 300V for the 12AU7 and 12BH7.
I think my choices are either lower the idle current values or lower the idle voltage. What is the best thing to do in this situation?
Or your 3rd option, increase load impedance.
More load impedance reduces plate current, which reduces dissipation. Except in this case, more load impedance is awkwardly high.
You could put 8Ω of speaker on a 4Ω wiring to double the reflected primary impedance, but the available impedances are already awkwardly high (you might have some restricted bandwidth, or not, if you try to reflect a higher load).
So either of the other options is good: choose a different PT to get a lower supply voltage; or keep high-voltage/low-load and increase bias voltage to run deeper in class AB. Different PT probably wins out because you need big-ish curent through a cathode resistor to allow the self-split push-pull operation of the original Firefly.
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Now that I understand (at a very basic level) how to determine the load line, I want to explore choosing an operating point a bit more and what Class A and Class B really means.
After going through the exercises above, I think I would be better off buying a new power transformer. I will probably buy a Hammond 269EX PT (190-0-190) and Hammond 125C OT (wired for 22.5kΩ:8Ω) just like the Firefly.
The target B+ is 250V. I realize that 190v * 1.414 = ~267vdc and that my cathode voltages will vary by the tube but I figure I can play with the filter caps, inductors, resistors etc to get pretty close to that.
I drew the class B load lines 5.6kΩ (22.5kΩ/4) starting at 250v, 0mA on the grid curves for the 12AU7, 12BH7 and ECC99 tubes.
Then I recorded the current where 250v intersects the Pdis curve for each of the tubes; 11mA (12AU7), 13mA (12BH7) and 20mA (ECC99). I read somewhere between 75% and 85% of the those values is a good idle current. I think that is conservative (safe) value to use but I am not sure if its really necessary.
For the 12AU7 I chose 8.5mA (77%) because a higher value would cause the Class A load line to exceed Pdis. For the 12BH7 I chose 11mA (84%) because that lets the class A load line sit right at Pdis. For the ECC99 I chose 17mA because that's basically 85%. If I increased the current, then I would exceed the 85% value.
Next I drew the class A load line 11.25kΩ (22.5kΩ/2) intersecting the idle points for each tube.
Now that I have that's set up... here are some questions:
- Do my lines and operating points seem ok?
- How does the operation of the tube go from Class A to Class AB?
- Does the ECC99 tube ever run as class AB? (I'm thinking "no", but the previous question will probably give me the answer)
- What does it mean to "run deeper in class AB"?
I know I said I was going to ask about cathode resistors, but I thought I should get this stuff answered first.
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- Do my lines and operating points seem ok?
Your class A operating points look fine.
Now that I understand (at a very basic level) how to determine the load line, I want to explore choosing an operating point a bit more and what Class A and Class B really means.
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- How does the operation of the tube go from Class A to Class AB?
You may not like the answer... if you stick to the Firefly model, it won't. The Firefly must run class A because of the self-split operation; class AB would require an actual phase inverter.
- How does the operation of the tube go from Class A to Class AB?
What is the definition of class A and class B?
- Class A = tube conducts all the time.
- Class B = tube conducts exactly half the time (half of the input signal cycle, or 180 degrees).
- Class AB = tube conducts less than all the time (not class A), but more than half the time (not class B); it is intermediate between the two.
Therefore, by definition the move from class A to class AB is when one side of the push-pull stage turns off, or stops conducting. Notice how the class B loadline runs down to the supply voltage and zero plate current? That point is equivalent to the tube being turned off.
Key point: We assume the input signal to the output tubes is a symmetrical sine wave; it swings as far negatively as it does positively.
Look at the class A loadline you drew for the 12AU7. The idle point is a bit more than -9v, so let's call it -9.5v.
An input signal could swing as far positive as 9.5v, such that the input signal voltage plus bias voltage momentarily equals zero. This happens at the 0v gridline; this line is considered a limit because when the grid is driven to a positive voltage it ceases looking like an infinite impedance, draws current from the stage driving it and drags down that driving voltage. In other words, you go from clean to severe distortion very fast.
If the input signal is a symmetrical sine wave, then after it swings to 9.5v peak, it will swing the opposite way back through zero volts and to -9.5v peak. -9.5v input signal plus -9.5v bias voltage puts the grid momentarily at -19v.
Look at your 12AU7 class A loadline again for the -19v gridline and where that would intersect your loadline. It's at ~317vdc and 2.5mA plate current. So your tube is not driven to cutoff on the negative half-cycle of the input signal. The tube never transitions out of class A.
But you should have known this would happen: previously, you selected loads that allowed class A operation without exceeding plate dissipation limits, and acknowledged that your supply voltage needed to be lower to enable you to use available load impedances (in class A). Class AB (or further, class B) is used because you wanted more power output than class A, and used both higher supply voltage and lower load impedance (along with a reduced idle dissipation) to get that power.
You can look at each of your class A loadlines and find the point which is 2x bias voltage, and see that this never equates to tube cutoff.
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I said self-split requires class A operation, right? How does self-split work? [EDIT: Drgonzonm's reply links an article with a more in-depth explanation. You may find the below easier to follow, and additionally, the article has a better version of a self-split circuit than what the Firefly uses (it will more closely approach the ideal I describe below, but also requires more supply voltage).]
In the case of the Firefly, the triodes share a cathode resistor. One triode is driven, the other has its grid grounded.
Imagine a positive-going signal is applied to the grid of the driven triode, such that its plate current increases. This pulls more current through the cathode resistor, which ohm's law says will result in more voltage across that resistor. The non-driven triode sees its cathode voltage increase, which is the same as applying more-negative bias voltage, and that turns off the non-driven triode somewhat. Its current decreases, which lowers the current drawn through the cathode resistor. This also lowers the voltage drop across the cathode resistor, in accordance with ohm's law.
During the other half-cycle, the opposite current relationships occur, with the same result that the voltage across the cathode resistor stays the same, and an equal but opposite current is induced in the non-driven triode.
If the non-driven tube mirrors the current change of the driven tube, then if the driven tube hits cutoff, the non-driven tube should stay at that one peak current until the driven tube starts conducting, at which time the non-driven tube will reduce its current. That implies the self-split design can't run in class AB and generate the additional clean output power.
Another problem is self-split doesn't work perfectly like I've described, and the non-driven triode may never get as far as perfectly mirroring the driven triode all the way down to zero current.
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Here's another self split example and info for a 6BQ5 power amp section.
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I want to be sure I understand how operation transitions from Class A to ClassAB. I think the small signal triodes may make it more difficult to illustrate because the tube is not driven to cutoff given the bias points I have chosen.
I've attached the grid curves for a triode connected 6V6 tube. I have included Pdis = 12W in red. The output transformer is assumed to be 8kΩ (got this from the datasheet). For illustration purposes I have selected an operating point of 320V, 20mA.
Based on the operating point lets say the input signal can swing as far positive as 30v (probaly more like 28v but I'm trying to make a point and 30v makes it easier). So, when the wave swings negative the grid will be at -60v. At that point, the plate current for the Class A load line will be 0.0mA; the tube is driven to cutoff. Since the tube is cut off in Class A, then Class B takes over.
I think I'm struggling to make the point here as well... maybe I should have plotted out the pentode grid curves. But, is that the general idea?
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I know that there is a lot of discussion regarding the firefly output.
In the case of the Firefly...
Lots of references to the Firefly on this thread, but in my original post I referenced the Marshall JTM1. On another board the designer of the JTM said the following:
The 60s has an input stage that has a cathode r/rc network designed to imitate the frequency response of a jumpered input with a little more bright than normal volume level, then gain then anode follower then the tone stack from a JTM45 - 56k slope and 220p treble. Only the treble control is fitted as a tone control. Then there is another anode follower to the PI. There is about 6dB of NFB into the grid of the PI.
So, I'm thinking I will want to use either a Cathodyne or Long Tail Pair PI (not that I know one from the other right now, but I know the names). I read somewhere that the Cathodyne is preferred with these low power output stages... I guess that is something to explore.
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Last question/point first, cause it's easier...
Lots of references to the Firefly on this thread, but in my original post I referenced the Marshall JTM1.
We can dissect the JTM1, as soon as you provide a link to a schematic. The Firefly was easier to talk about and use as an example, because the schematic linked gave voltages and OT primary impedance.
We have zero point of reference for the JTM1... unfortunately, the designer only gave rough details of preamp topology in that thread; I'm thinking he didn't want mass cloning of his design, especially since Marshall was charging stupid money for a 1w amp.
So, I'm thinking I will want to use either a Cathodyne or Long Tail Pair PI (not that I know one from the other right now, but I know the names). I read somewhere that the Cathodyne is preferred with these low power output stages... I guess that is something to explore.
We're still stuck on output stage operation questions. :icon_biggrin:
And so you know, power output has no bearing on the type of phase inverter used, but rather the size of the signal that has to drive the output tubes. Some of the highest wattage tube bass & guitar amps ever made use a cathodyne/split-load phase inverter.
On to the transition to class B, but I need to do some writing...
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I want to be sure I understand how operation transitions from Class A to ClassAB. ...
Based on the operating point lets say the input signal can swing as far positive as 30v ... So, when the wave swings negative the grid will be at -60v. At that point, the plate current for the Class A load line will be 0.0mA; the tube is driven to cutoff. Since the tube is cut off in Class A, then Class B takes over.
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If you idle at -30v, and apply a 30v peak signal such that the peak voltages land at 0v (positive input peak) and -60v (negative input peak, and cutoff bias for the tube), you will not have driven the tube into class AB. It has landed at limiting class A, which is the when all conditions exactly line up so the tube sits at a supply voltage, load and operating point such that the idle bias voltage is exactly 1/2 the voltage needed to cut off the tube.
To transition to the class B loadline (during a given tube's positive signal swing), the tube on the other side of the push-pull stage has to be cut off.
Why? The OT has a winding ratio such that when you attach a speaker load it reflects a given impedance on the primary (like plugging an 8Ω speaker into that 125C and getting a 22.5kΩ primary). If both tubes are conducting, each sees the portion of the primary from its plate to the center-tap (B+), which is half the total primary winding. But when the other tube cuts off, there's no current through that half-winding and the ratio between the secondary and the remaining half-winding reflects 1/4 the original plate-to-plate load impedance.
So you never get your half-of-half-of-the-primary-impedance, which is your class B loadline, until the other tube shuts off.
Look at the bottom graph on Valve Wizard's explanation of push-pull (http://www.valvewizard.co.uk/pp.html). Curves and loadlines for each tube are plotted on their own graph, then one side is turned 180 degrees. They are brought together so the 0mA axis are one line, and the plate voltage of the output tubes line up (300v in this case). If you take a straightedge, you can hold it along the class B portions of each line, and they make one continuous line from one graph to the other. The other portion not on this straight line is the class A area for each tube. You can see that one tube's load bends from class A to class B once its mate has reached cutoff at the zero current axis.
So revisit your lines with this knowledge: if the tube sits at -30v, and it takes -60v to cut off, then it also takes the opposing tube -60v to cut off. With your symmetrical input signal, by the time the "pull stage" has gone from idle to -60v to cut off, the "push stage" is just hitting 0v on the class A loadline. You never get out of class A before grid current clamps the signal swing, and big distortion.
So let's look at a different way to skin the cat and move to class AB operation (but now I need to write that section...)
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Alright, we'll step through a class AB example. NOTE: Like any real design, I had to go through a couple iterations, as well as make choices about trade-offs at every step. So this is not a mechanical process, but the following should help you see what "right looks like."
Class AB is mostly about making more power than is possible in Class A, and generally uses a higher supply voltage, lower load impedance and larger bias voltage than Class A. I'm going to stick with the 12AU7 as our output tube (no fair trying to change to the 6V6 in your last run-though; you won't see how things should change if you switch tubes).
... the PT I want to use for this project. Its an AnTek toroidal transformer model AN-05T240 (http://www.antekinc.com/as-05t240-50va-240v-transformer/). It has a 220V tap that I plan to use.
220v * 1.414 = 311vdc, and subtracting diode drops will put us around 310vdc. That seems like a usefully-high B+ to build a Class AB amp with, so we'll use that for our new amp.
Now it's time to look at a 12AU7 datasheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/093/1/12AU7A.pdf), and the maximum ratings on the first page. Our constraints are a plate dissipation of 2.75w and a peak cathode current of 60mA. We'll bend the rules a bit and use 310v instead of the data sheet maximum of 300v.
Now it's time to look at the plate characteristics on page 4. Look at 310v and 0mA plate current. If the -24v gridline continued all the way to the axis, it would land somewhere near 310v, so let's estimate the cutoff voltage at B+ as being -24v.
I said earlier that the limit of class A operation is half the cutoff bias for your plate voltage. If you run into no other limit first, the grid is able to swing up to 0v and down to 2x bias, which is just enough to momentarily cut off plate current. Since plate current cutoff is a requirement for class AB, we must bias more than half the cutoff voltage.
We need something between -12v and -24v... I'll pick -17v. It's far enough toward cutoff that when the class A loadline extends above B+, we're still likely to hit cutoff not far above -24v.
Now for our Rule #2:
The loadline for maximum power intersects the 0v gridline at 0.6*B+. That happens at 185v and ~27mA. You can calculate the loadline impedance manually (like I did at first) or with:
Rl = 1.6 * B+ / Imax = 1.6*310v / 0.027A = ~18.4kΩ plate-to-plate
Class A load = 18.4kΩ / 2 = 9.2kΩ
Class B load = 18.4kΩ / 4 = 4.6kΩ
Check: (310v - 185v) / 0.027A = 4.629kΩ, so our math makes sense. The attached graph shows the proposed operating point, the class A loadline (green) and the class B loadline (red).
When does the tube transition from class A to class B? When the other tube tuns off. We see the class A loadline intersect the axis between the -24v and -26v gridlines. Let's then say cutoff happens at about -26v. That is our idle bias minus 9v of signal. Since each tube is receiving equal and opposite signals, the opposing tube is at idle plus 9v at the same moment. Therefore, our tube switches from the class A to class B loadline at -17v + 9v = -8v.
In reality, there is a transition from the class A line to the class B line that forms a smooth curve between them, but it fully shifts to class B by -8v on the grid.
Both these lines look like they're well above the dissipation limit. What's up with that?
Well, the transition from class A to class B loadlines happens at a point on the class A loadline below the dissipation limit on the class A line. Our headache now is we need to calculate average plate current, power input, power output and see what the resulting plate dissipation is to see if this choice of operating condition will work.
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You'll need to go look up calculating average plate current on pages 579-580 of RDH4; I won't re-type it here, but suffice that we need to calculate the plate currents of both tubes over 90 degrees of a sine wave input signal and apply the formula at the top of page 580 to find the average. I tabulated the grid voltage to each tube during that time, and found the resulting plate currents for each from the plot of each loadline in the graph below (you have to pay attention to the moment when you switch from taking current numbers from one to taking numbers from the other). I have them rolled-up in a spreadsheet, but the forum doesn't allow excel attachments.
When all was said and done, the results were:
- Plate Voltage: 310v
- Minimum Plate Voltage: 185v
- Max Signal Average Current: 16.7mA
- Peak Plate Current: 27mA
- Power Output: 1/2 * 27mA * (310v - 185v) = 1.69w
- Power Input: Plate Volts * Average Current = 310v * 16.7mA = 5.18w
- Plate Dissipation: (Power In - Power Out) / # Tubes = (5.18w - 1.69w) / 2 = ~1.75w per tube
Since the dissipation rating is 2.75w, we will not risk redplating. Even though the lines ran above the dissipation limit curve, the low idle current and amount of off-time, as well as power transferred through the OT to the load, keeps the tubes cool on-average. You would design your power supply to deliver 16.7mA d.c., plus whatever current is drawn by the preamp and phase inverter.
Because of the very big change from idle to peak, cathode bias is not recommended (also why the calculated ~310vdc is what we work from the entire time). You will need a fixed-bias supply, and we will need a phase inverter (which you wanted anyway).
18.4kΩ plate-to-plate is not available on the Hammond 125-series transformer, but 17.6kΩ:8Ω is available, and will work fine. You'll get a hair less output power, and a hair more plate dissipation. You could use the 18kΩ:3.2Ω option, but that impedance is mostly available with cheesy small speakers.
Now you have some solid numbers to base a phase inverter design from. You probably could try an even lower load impedance for more peak current, because we haven't hit the 12AU7's peak plate current or dissipation limits, but you didn't ask for a world-beating design...
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- What does it mean to "run deeper in class AB"?
Class A is conducting all of the input signal cycle, or 100% of the time, or 360 degrees.
Class B is conducting exactly half the input signal cycle, or 50% of the time, or 180 degrees.
Class AB is something in between those extremes.
Imagine I'm your boss. I might make you work 24 hours a day (class A), or 12 hours a day (class B). If you're doing manual labor, you might be able to work a bit harder if I let you rest that 12 hours you get off in class B. I could work you 22 hours and maybe get a little more out of you than class A. Or maybe I could work you 14 hours a day (further into class AB, towards class B) and get much more out of you, though not as much as giving you the 12 hour rest.
Tubes are the same way. If you give them a little time off, you can push them harder when they're working. If you give them a lot of time off, you can work them very much harder.
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We can dissect the JTM1, as soon as you provide a link to a schematic. The Firefly was easier to talk about and use as an example, because the schematic linked gave voltages and OT primary impedance.
We have zero point of reference for the JTM1... unfortunately, the designer only gave rough details of preamp topology in that thread; I'm thinking he didn't want mass cloning of his design, especially since Marshall was charging stupid money for a 1w amp.
I don't expect to find an official JTM1 schematic any time soon. What I was hoping for (and is happening) was to learn how to go about designing something similar to the JTM1. I have built amps from schematics before but the mystery has always bugged me.
We're still stuck on output stage operation questions. :icon_biggrin:
And so you know, power output has no bearing on the type of phase inverter used, but rather the size of the signal that has to drive the output tubes. Some of the highest wattage tube bass & guitar amps ever made use a cathodyne/split-load phase inverter.
On to the transition to class B, but I need to do some writing...
Got it... yeah, still learning about power output.
BTW, this thread has been great and I really appreciate (more than you know) all the time the you have been committing to my education. Now, time to do some more reading.
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You're welcome.
I'm sorry that in the interest of brevity (and out of laziness), I skipped over some of the finer points that we've uncovered after the 2nd/3rd/etc practice of loadline drawing.
But I guess it's best to keep it very simple at first, then add the complications as we run into them.
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So are all the questions answered? :laugh:
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So are all the questions answered? :laugh:
I think so. There were several "ah ha" moments while reading. I'm going to do similar calculations for the 12BH7 to be sure but I haven't had a chance to sit behind the computer this weekend.
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Look at the bottom graph on Valve Wizard's explanation of push-pull (http://www.valvewizard.co.uk/pp.html). Curves and loadlines for each tube are plotted on their own graph, then one side is turned 180 degrees. They are brought together so the 0mA axis are one line, and the plate voltage of the output tubes line up (300v in this case). If you take a straightedge, you can hold it along the class B portions of each line, and they make one continuous line from one graph to the other. The other portion not on this straight line is the class A area for each tube. You can see that one tube's load bends from class A to class B once its mate has reached cutoff at the zero current axis.
First "ah ha" moment. I have read Valve Wizard's explanation before but didn't quite make the connection from the words to the diagram. Now I get it.
Now I will attempt to step through the class AB example for a 12BH7 tube.
Using the 220V tap on the AnTek AN-05T240 the B+ will be 220V * 1.414 = 311vdc or 310v for convenience.
From the 12BH7 datasheet:
Plate dissipation = 3.5W
Peak cathode current = 70mA
Maximum voltage = 300v (450v for a Vertical Deflection Amplfier whatever that is but I assume 310v will be fine.)
Looking at the gridcurves I estimate the -23v gridline intersects at 310v and 0mA plate current. So we need an operating point between -11.5v and -23v... I will choose -17v.
Rule #2. The loadline for maximum power intersects the 0v gridline at 0.6 * B+ = 0.6 * 310 = 185v. The current at 186v, grid line 0v is ~52mA (had to do some creative line drawing). Therefore:
Rl = 1.6 * B+ / Imax = 1.6 * 310v / 0.052A = ~9.5kΩ
Class A load = 9.5kΩ / 2 = 4.7kΩ
Class B load = 9.5kΩ / 4 = 2.4kΩ
Check: (310v - 185v) / 0.052A = 2.404kΩ so the math makes sense.
Based on the drawing below I can see that the class A load line intersects the axis around -25v. That is the idle bias minus 8v (-17v -8v = -25v). Therefore the opposite tube is idle plus 8v at the same moment. Therefore, the tube will switch from class A to Class B at -17v + 8v = -9v.
The problem that I see with this is that at -9v, both the class A and class B load lines are above the dissipation limit. I could go through the exercise of calculating the average plate current, power input and power output to see what the resulting plate dissipation is, but I believe that since the lines are exceeding the limit already that its an unnecessary exercise. In addition, I know that my transformer does not have a 9.5kΩ tap. Instead, I'm going to redraw the diagram using the available tap of 12.8kΩ in the next post.
Please let me know if this was the correct decision (i.e., try a different loadline when I see both class A and class B loadlines are above dissipation limit at transition point)? Other than choosing a bad point, do I basically have everything correct?
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... The problem that I see with this is that at -9v, both the class A and class B load lines are above the dissipation limit. I could go through the exercise of calculating the average plate current, power input and power output to see what the resulting plate dissipation is, but I believe that since the lines are exceeding the limit already that its an unnecessary exercise. ...
On the contrary! Since the lines exceed the plate dissipation curve, the only way you can know if this is workable is to go through the tedium of calculating average current, power input, power output and resulting dynamic dissipation.
There is a convenient fiction in class A that we can make use of to know we never exceed plate dissipation in use. But in Class AB, you can never be sure without running the numbers (or without a LOT of experience with the tube in question).
... In addition, I know that my transformer does not have a 9.5kΩ tap. Instead, I'm going to redraw the diagram using the available tap of 12.8kΩ in the next post.
Please let me know if this was the correct decision (i.e., try a different loadline when I see both class A and class B loadlines are above dissipation limit at transition point)? Other than choosing a bad point, do I basically have everything correct?
Yes, you're applying the approach correctly. The fact that the calculated load is not available is more of a deciding factor than whether the lines are above the dissipation curve. In fact, the relevant section of RDH4 will tell you to plot numerous loadlines radiating from your supply voltage for evaluation (yep, the tedium again).
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Well, I went ahead and calculated the values and came up with the following results:
- Plate Voltage: 310v
- Minimum Plate Voltage: 185v
- Max Signal Average Current: 30.5mA
- Peak Plate Current: 51mA
- Power Output: 1/2 * 51mA * (310v - 185v) = 3.19w
- Power Input: Plate Volts * Average Current = 310v * 30.5mA = 9.46w
- Plate Dissipation: (Power In - Power Out) / # Tubes = (9.46w - 3.19w) / 2 = ~3.15w per tube
Since the dissipation rating is 3.5w, it appears I was wrong and I will not be exceeding maximum dissipation. However, if I want to stick to the rule of choosing a point between 75% (2.6W) and 85% (2.9W) of Pdis then the value is still a little high. Doesn't really matter because the 125C doesn't have 9.5kΩ tap so time to try again with loadlines for the 12.8kΩ tap.
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This time I chose a 12.8kΩ tap with a 310v B+ and a 6.25mA operating point and came up with the following results:
- Plate Voltage: 310v
- Minimum Plate Voltage: 185v
- Max Signal Average Current: 23.1mA
- Peak Plate Current: 34.5mA
- Power Output: 1/2 * 34.5mA * (310v - 185v) = 2.16w
- Power Input: Plate Volts * Average Current = 310v * 23.1mA = 7.15w
- Plate Dissipation: (Power In - Power Out) / # Tubes = (7.15w - 2.16w) / 2 = ~2.5w per tube
This time the dissipation is slightly below 75% of Pdis so this appears to be a safe operating point. I could probably eek out a little more power, but then again I'm trying to build a One Watt amp so I don't think that is really necessary.
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... However, if I want to stick to the rule of choosing a point between 75% (2.6W) and 85% (2.9W) of Pdis ...
Note, this is not really a "rule". People have espoused a "70% rule" for class AB, but that doesn't originate in any of the classic texts that I can find. More, it has been an observed generally-safe value with production guitar amps. The original designer already figured out a safe load, so...
Even RDH4's "2nd Rule" (0.6*Supply voltage) is not much more than an intermediate between 2 conditions where maximum output power are likely to occur (which PRR pointed out much earlier in this thread).
So the real "rule" regarding class AB is this: it has a bias somewhere beyond the case of "limiting class A".
Limiting class A is an operating point such that during the positive input swing, the tube's grid is driven just to 0v momentarily, and during the (equally-big) negative swing the tube is driven just to the point of cutoff. In other words, any bigger signal only produces gross distortion on one half-cycle and cutoff on the other.
You could make a case that there is one exact load impedance for any given supply voltage which allows perfect "limiting class A" operation. Said another way, one load which is the dividing line between class A and class AB.
... the class AB example for a 12BH7 tube. ...
... I could probably eek out a little more power, but then again I'm trying to build a One Watt amp so I don't think that is really necessary.
True. What this shows then, is that the 12BH7 is more tube than you need for 1w. You could use it, but they're more expensive than 12AU7's, and some folks like me have gear which truly needs a 12BH7 (McIntosh MC-30). I know you chose other tubes to allow you to reach your own conclusions, but hopefully you'll see your needs are met with the 12AU7. Otherwise it's kinda like using two KT88's to get a 25w output.
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12AU7 results:
- Idle Current: 3.5mA
- Plate Voltage: 310v
- Minimum Plate Voltage: 185v
- Max Signal Average Current: 16.7mA
- Peak Plate Current: 27mA
- Power Output: 1/2 * 27mA * (310v - 185v) = 1.69w
- Power Input: Plate Volts * Average Current = 310v * 16.7mA = 5.18w
- Plate Dissipation: (Power In - Power Out) / # Tubes = (5.18w - 1.69w) / 2 = ~1.75w per tube
...
Because of the very big change from idle to peak, cathode bias is not recommended (also why the calculated ~310vdc is what we work from the entire time). You will need a fixed-bias supply, and we will need a phase inverter (which you wanted anyway).
Can you define very big and what would be an example of normal or average? I believe the Firefly (and most other 12AU7 power tube variants I have seen) are using a cathode bias.
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True. What this shows then, is that the 12BH7 is more tube than you need for 1w. You could use it, but they're more expensive than 12AU7's, and some folks like me have gear which truly needs a 12BH7 (McIntosh MC-30). I know you chose other tubes to allow you to reach your own conclusions, but hopefully you'll see your needs are met with the 12AU7. Otherwise it's kinda like using two KT88's to get a 25w output.
However, many people have reported that the 12BH7 sounds better than the 12AU7 and some say the ECC99 is even better than that. Of course, what this may really be saying is that people like having more power.
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Because of the very big change from idle to peak, cathode bias is not recommended ...
Can you define very big and what would be an example of normal or average? ...
I didn't mean "very big" as being equivalent to "abnormal".
Cathode bias works by creating a voltage drop across the cathode resistance, and that voltage is used to bias the tube. If you want your bias voltage steady, the current through the resistor should be steady, too. In fact, don't bias-vary tremolo systems work to wobble the volume by having an un-steady bias voltage?
While the positive and negative peaks of plate current in a class A stage can and do cause local negative feedback, on average the current barely changes from the idle value. Meaning that an ideal class A stage which runs from idle current, to a peak of 2x idle current, back to a negative peak of zero current, back to idle current has an average current over that whole time which exactly equals the idle value.
Real tubes have some amount of distortion, and even class A stages have an average current which is slightly greater than the idle value.
12AU7 results:
...
- Max Signal Average Current: 16.7mA
- Peak Plate Current: 27mA
...
[/list]
This time I chose a 12.8kΩ tap with a 310v B+ and a 6.25mA operating point
...
- Max Signal Average Current: 23.1mA
- Peak Plate Current: 34.5mA
...
[/list]
... class AB example for a 12BH7 tube.
... an operating point between -11.5v and -23v... I will choose -17v. [EDIT: 5mA idle current]
...
- Max Signal Average Current: 30.5mA
- Peak Plate Current: 51mA
...
[/list]
Above, I've pulled together your results and mine when selecting a class AB operating point, plotting loadlines and reading/calculating peak and average currents.
In each case, we mentioned an idle current which was for a single tube, but if we used a cathode resistor it would like be shared and pass the currents of both tubes.
Case 1
7mA Idle -> 27mA Peak -> 16.7mA Average
Average = 2.39*Idle
Case 2
12.5mA idle -> 34.5mA peak -> 23.1mA Average
Average = 1.85*Idle
Case 3
10mA idle -> 51mA peak -> 30.5mA Average
Average = 3.05*Idle
The rise of current above idle will tend to turn off the tube, which will fight it ever reaching the peak currents we found on the loadlines, and reduce the average current.
- If the current increase was relatively small (like 1.05-1.1*Idle), or if peak currents were relatively infrequent (music reproduction rather than distorted guitar), a bypass cap could store enough charge to make up for the peaks
So cathode bias is incompatible with running a tube "far into class AB" and spending more time on that class B loadline. Which also explains why all the big, powerful guitar and bass amps used fixed bias: because it stays stable during big peak current swings in class AB/class B.
That's why I only subtracted a volt or so from the calculated d.c. voltage of your Antek PT: I knew fixed-biased may be required by the peak currents encountered by the class AB condition I was working up, so I only subtracted enough voltage to account for that dropped across rectifier diodes, and not a cathode bias resistor.
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However, many people have reported that the 12BH7 sounds better than the 12AU7 and some say the ECC99 is even better than that. Of course, what this may really be saying is that people like having more power.
You should know from the loadlines they won't be getting more power output with the original 22.5kΩ primary impedance. Remember, they're just tube rolling (in all likelihood) rather than making a meaningful change in the output stage.
So they may be saying they like less distortion, or more distortion, or different-distortion with the alternate tubes than they got with the 12AU7. We'd have to plot the loadlines for the original Firefly output stage to see what difference would actually happen when they plug the bigger tube in the socket.
I have a 5E3 Deluxe copy which uses 6V6's. I've plugged 6K6's into it, and got (very, very) slightly less power output, but a very different flavor of distortion. If I plug KT88's into in, I bet the distortion character could change, but I won't get any real power output increase.
When you understand why the OT primary impedance, supply voltage and PT high voltage current output limit that power output, you'll understand how an output stage works.
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1. Will this amp work any better by having a filtering capacitor between the O/T and the power supply?
Define "better".
The Firefly shouldn't suffer from hum due to insufficient filtering because:
- The 1st filter is already 47uF and has very little drain from the output stage.
- The output is push-pull, which will cancel hum in the output stage, and
- Even if the output wasn't push-pull, triodes have a higher power supply rejection ratio (PSRR) which makes their output less sensitive to hum in the supply.
Maybe you're thinking frequency response, in which case skip to #3 below.
2. Aren't the typical nine pin tubes restricted to a maximum of about 4.5w combined, so those high wattage tubes like the ecc99, are total tube dissipation limited ...
I don't know about "typical". The 12AU7 (http://www.mif.pg.gda.pl/homepages/frank/sheets/093/1/12AU7A.pdf), 12BH7 (http://www.mif.pg.gda.pl/homepages/frank/sheets/093/1/12BH7A.pdf) and ECC99 (http://www.mif.pg.gda.pl/homepages/frank/sheets/163/e/ECC899.pdf)data sheets say 2.75w, 3.5w and 5w per triode, respectively. I'd read that as 5.5w, 7w and 10w per tube envelope, as the sheets don't state any restriction in that regard.
3. Does the high impedance chosen by the original builders help because the O/T also serves some functions as a filter see question 1?
There is little consideration given by the designers regarding any filter formed by the OT primary impedance, at least for initial choice of operating conditions. Rather, a higher primary impedance provides a load which keeps the output tube plate current from becoming excessive during operation. If Power = Current2 * Resistance (or Impedance), then more primary impedance reduces current for the same power.
When looking at the loadlines, an infinite impedance is represented by a horizontal line on the graph; voltage could increase infinitely with no increase of current. So also a vertical line is zero impedance. When our early class A loadlines tended to be too vertical and stray into excessive plate dissipation, we increased the load impedance the flatten the loadline, reduce plate current and bring plate dissipation back down to a safe level.
As PRR pointed out earlier, the choice of primary impedance loading a triode plate is generally related to the triode internal plate resistance, in a 1-to-1 or 2-to-1 ratio, which also sets how the load and the tube will divide voltage.
The drawback of small signal triodes as output tubes is that even the 12AU7 has a plate resistance high enough that we found 22kΩ+ to be good loads for class A operation at a convenient plate voltage. In order to wind a primary with enough inductance to have a reactance very much greater than 22kΩ at the OT's lowest rated frequency (and thereby meet the OT's specs), many, many, many turns of fine wire are needed. That will increase winding capacitance within the OT (unless it is extraordinarily well-designed and manufactured), which will cause treble roll-off in the audio range.
So the primary impedance required is set by the demands/requirements of the output stage, and any impact of frequency is an undesired byproduct of a practical device. This is the reason why you see so few transformers with wide frequency response and specified high-impedance primaries. And the (well-made & uncompromised) mic-to-grid & plate-to-line transformers you do see with high impedance windings will cost as much as guitar/bass amp OTs 20-50 times bigger.
4. Does anyone have a design discussion on the split phase inverters other than what I posted?
Self-split inverters can sometimes look like something-for-nothing on paper, but they don't deliver their full promise in practical use. That's why you won't see them outside of a few oddball Gibson amps, very cheap tabletop radios, and (surprisingly to me) the Firefly.
But David wanted to venture into class AB output stage design, and self-split inversion (as well as stable cathode bias) pretty much demands class A operation for a total plate current that doesn't vary much from idle to full-tilt. It seems likely at this point that whatever output tube he chooses, the amp will likely use a conventional phase splitter circuit.
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I don't have any more output stage questions (for now anyway). So, on to the Phase Inverter.
The first thing I need to decide is the probably the tube and operating point. Let's keep with the 12AU7, using the 22.5K load because it makes it simple to compare to the Firefly. I ran the numbers and a bias point of 310V, 5mA, -15.5V seems to be pretty good. Power Output is 1.5W and Plate Dissipation is 1.8W whereas 70% of Max Pdis is 1.9W.
The next thing I need to decide is whether I want the amp to run Class A or Class AB. Running Class AB means that I cannot use the Self Split PI, but I believe that running Class A does not mean that I must use it. Class A means I have to buy another transformer while Class AB means I can use what I have. So, I guess there are a few ways to approach the discussion.
- Start with Self Split and dissect the Firefly, then move to Cathodyne (one valve) then Long Tail Pair (two valves).
- Start with the Cathodyne PI then move to LTP since the self split has already been done.
- Start with the LTP since according to Valve Wizard, "It is the most common phase inverter found in push-pull guitar amps" and it is probably what is used in the JTM1.
- Dissect the Firefly then LTP.
I'm leaning towards the last option since I think it would be most helpful to me and others.
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Regarding choices of PI's on the firefly, I would suggest reading the posts found in the AX84 forum regarding the firefly. There is lot of discussion, and literally hundreds of posts on this amplifier. (I went through about 300 posts, and just touched the discussion, looking for the philosophy of the self-split PI and why the 12au7 was chosen. I did not find either).
I've been reading posts over there. However, this thread is more generically about learning how to determine the proper values to use when designing an amp as opposed to figuring out why values were used in the Firefly.
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> triodes have a higher power supply rejection ratio
Beware where the output is taken.
In resistance-coupled stages, the output is taken plate-to-ground. Low Rp means better PSRR.
In transformer coupled stages, the output is effectly taken plate-to-supply. Low Rp means more buzz across the OT. A pentode's high Rp means less buzz across the OT. (This is a secondary reason why pentodes took-over the radio market so decisively.)
> many, many, many turns of fine wire are needed. That will increase winding capacitance within the OT
Parasitic C has little to do with number of turns. (I admit that I've probably used this simplification.) It is pretty much about the overall size of the winding. And for a given bass-power level, the winding size is pretty much fixed.
But you do many-many turns to get the main inductance up (high impedance at low frequency). Not all that inductance couples to the secondary, and the proportion that doesn't is fairly fixed (for a given core material, and not sectioning the windings).
The stray inductance works against the stray C to form a 2-pole low-pass filter.
So for a fixed C, and stray L nearly proportional to main L and bass-limit, raising the impedance raises the stray L against a fixed C and the upper resonance comes down in frequency. About as square-root of L (and impedance). And the Q (against load resistance) rises.
In practice, impedances below 1K can usually span the entire audio band without trouble. But impedances like 100K are very prone to a top-end resonance and limited treble extension. A better core does help, but for "power" transformers (above a fraction of a Watt) the cost of Permalloy etc is enormous, and iron-losses can mount up.
Hence a 4K winding gives little trouble, small power-tubes are usually sized to support 8K-10K tops. At 22K you better have saved enough on the tiny tube to cover the added cost of fine/fragile winding and possible partitioning to support full treble.
________________
> suggest the 6n7, which is designed as both a class A and class B tube?
In class A, it's 400mW per side. Which might work, but the projected OT impedance for push-pull is 40K-80K, very awkwardly high. (10K-20K for parallel SE, but dbishopbliss wants push-pull.)
In class B the grids are VERY hungry. Need 40V into a 2K load, instead of 18V into 500K load for a 6V6-pair to make the same 10W output. So you need a second 6N7 and a transformer just to drive the final. This was hot stuff for about a year in the 1930s. 6F6 without a transformer killed 6N7 designs. (It lingered for a long time in minor uses. But note there is no Miniature version of 6N7, which means it was out-of-fashion by 1940s.)
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> triodes have a higher power supply rejection ratio
Beware where the output is taken.
Thanks for the correction; that was an oversight on my part.
And for the better detail on stray C and stray L rolling off high end.
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Not a JTM1 schematic but Traynors amp with switchable output:
http://traynoramps.com/downloads/servman/smdh15h.pdf (http://traynoramps.com/downloads/servman/smdh15h.pdf)
How "happy" do you guys think the 12au7 is in this circuit?
I'd mod this amp so that i would split the phase-inverter plate resistors and take the output for 12au7 from the middle and maybe use the screen voltage node as a plate voltage for 12au7. Thoughts?
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... but Traynors amp with switchable output:
Traynor Darkhorse (http://traynoramps.com/downloads/servman/smdh15h.pdf)
How "happy" do you guys think the 12au7 is in this circuit?
This question is more complex than you think. I believe I've also finally arrived at a definitive answer to the issue of buzzy distortion in this amp raised in this thread (http://www.el34world.com/Forum/index.php?topic=14877.0).
PRR said in that thread that the 12AU7 mode is seriously mis-loaded. I did a hack-n-slash loadline for an 8kΩ primary OT, which seemed a reasonable loading for the 6V6's.
The 12AU7 idles at a bias of -18.7v according to the schematic. This is across an unbypassed 1.8kΩ resistor. That implies ~5.2mA per tube and a resulting plate-to-cathode voltage of ~377vdc. When I plot these on a set of 12AU7 curves, it lands almost on the -20v gridline, but that's probably close enough.
The 4k & 2k loads for class A and class B operation are extremely steep, with the tube crossing the plate dissipation line by at least -17.5v (or ~1.2v of peak input signal). But crossing the dissipation line doesn't matter if the tube gets turned off a significant chunk of time, and this tube will cut off by ~-30v (about 10-11v peak input in the negative direction).
Calculating average current and plate dissipation was difficult, because the loadlines ran off the graph. But I came up with an average current of 29.9mA at full output.
Problem:
The rise of current above idle will tend to turn off the tube, which will fight it ever reaching the peak currents we found on the loadlines, and reduce the average current.
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So cathode bias is incompatible with running a tube "far into class AB" and spending more time on that class B loadline. Which also explains why all the big, powerful guitar and bass amps used fixed bias: because it stays stable during big peak current swings in class AB/class B.
The idle current for the full 12AU7 output stage in the Darkhorse is ~10.4mA, rising to 29.9mA at full output. But the bias is deived across an unbypassed cathode resistor.
29.9mA * 1800Ω = 53.82v
If the 12AU7 could ever actually reach the peak implied by the loadlines, the cathode resistor would have shifted the bias from -18.7v to -53.8v. Which is a good thing because my calculations for plate dissipation per 12AU7 (if it were not hamstrung by the cathode bias) was 4.52w per triode, which is approaching double the limit.
So the key in this case is to have the unbypassed cathode resistor dynamically reducing output. I can't tell you what that would do to the distortion, as you'd really have to breadboard and measure for different input levels... the graphical method is tedious enough without all conditions changing with each extra volt increment in input signal, which is what we have here.
So back to that thread... Given the cathode resistor slams the tube toward cutoff with a bigger input signal. Maybe the original poster got a 12AU7 not happy with this situation, or maybe they pushed a good thing too far and found a bad sound. The tube should be able to quickly recover from this type of overload/cutoff because it's not the same mechanism as grid-blocking, but maybe it didn't sound that way to them (or the input signal was big enough to add a layer of grid blocking over top of the cutoff bias situation, which probably would sound crappy).
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How "happy" do you guys think the 12au7 is in this circuit?
If the 12AU7 could ever actually reach the peak implied by the loadlines, the cathode resistor would have shifted the bias from -18.7v to -53.8v. Which is a good thing because my calculations for plate dissipation per 12AU7 (if it were not hamstrung by the cathode bias) was 4.52w per triode, which is approaching double the limit.
Seems like the 12AU7 wouldn't be very "happy" to me. The good news is that I was able to look at the grid curves and come up with a similar conclusion even though I didn't do the actual math.
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Thanks guys, I had a feeling that this might be the case.
Though I kind of like this concept, so is there a way to make this work better or does the compromising make one or the other power option just unusable?
I was thinking that something like hammond 1609 OT (10k) and 320V Plate voltage would suit the 12au7 better.
...one day I'm going to take the time and learn these loadlines.
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Based on the calculations we have been making on this thread, it seems to me that the 12AU7 wants a much higher impedance secondary so I'm not sure 10K vs 8K will make that much of a difference.
I'm really beyond my understanding now, but I wonder why they couldn't have two separate output transformers?
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but I wonder why they couldn't have two separate output transformers?
Money and chassis space.
Although there is the old "penny wise and dollar foolish" saying.
Brad :icon_biggrin:
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... I wonder why they couldn't have two separate output transformers?
I don't believe Traynor was trying to get "optimum performance" from the 12AU7 at all.
PRR pointed out in the other thread I referenced that if you mis-load an output tube, you get less-than-maximum (clean) output power. What I found drawing loadlines was if you use a severely-low load impedance for the tube & operating condition, you run the risk of redplating, unless you do something extra to keep it in check.
Traynor already had a "maximum power" setup in the 6V6's. The 12AU7 was added because you want quiet, real quiet, and other approaches (variable B+, lamp limiter, push-pull to SE switching) risk patent licensing issues or "me too!" accusations. So they went "Full Retard" (to mis-quote a movie).
Traynor used the 6V6 loading on the 12AU7, which should be a terrible idea and burn up your 12AU7. But they also did "wrong cathode bias application" by using a farily-large value unbypassed cathode resistor on a push-pull stage with big average current shifts. If the output stage was a "proper class AB design with ideal loading" then the unbypassed cathode resistor would wreck its performance. Instead, what it does it limit the 12AU7's dissipation, and cause distortion & compression. Exactly what the user wants in a low-output setting of a guitar amp to get tube distortion at bedroom levels.
Money and chassis space.
Although there is the old "penny wise and dollar foolish" saying.
So I wouldn't say Traynor is cheaping-out; in fact, I'd call it smart, non-obvious design. My only disclaimer is it's complicated to calculate on paper, and a sim would probably lie in several ways, so the real way to prove how it works (and how clever it really is) is to set it up on the breadboard and measure.
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Based on Valve Wizard's description of the Long Tail Pair Phase Inverter I have come up with following for my design. Start with a 12AX7 for the tube. I believe that is what Marshall uses. I assume the voltage will be around 300V after filter caps, choke, voltage dropping resistors. Wizard suggests making the tail voltage 25% to 30% of that so 100V is a nice round number so that leaves 200V at the plate.
Next, Wizard suggests 100K is a typical plate resistor so I will use that value as well. I drew the loadlines so I could choose a bias point. For now, I will assume center biasing which works out to be around -1.3V with a quiescent plate current of 0.6mA. Since both triodes share the tail resistor the total current will be 1.2mA. Therefore the bias resistor (Rb) will be 1.3V / 0.0012A = ~1083Ω or 1kΩ. Since the cathode voltage is 100V and Vgk is -1.3V there must be 98.7V across the tail resistor. Therefore the tail resistor will be 98.7V / 0.0012A = 82250Ω or 82kΩ.
Next, I chose 1MΩ for the grid-leak resistors. Wizard suggests 470kΩ to reduce the resistor noise but since I'm following Marshall's lead here I went with the traditional value. Would the Wizard's suggestion of 470kΩ for grid-leak resistors change the tone of the amp, or simply lower the noise?
Since the grid-leak resistors are 1MΩ then the input impedence can be estimated to be 2*Rg = 2MΩ. Based on that I used the following to determine the coupling capacitor:
C = 1 / (2 * pi * f * R) = 1 / (2 * 3.414 * 20Hz * 2M) = ~4nF or 0.0047uF
The traditional value seems to be 0.01uF. I could follow Marshall here, but this time I'm going to go with my calculations.
Wizard goes on to discuss balance and determining the gain of each stage of the Phase Inverter, but I'm going to look at that further in my next post.
I've attached a schematic of what I have so far. How am I doing?
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You may need some resistors on the 12AU7 grids.
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> change the tone of the amp, or simply lower the noise?
Probably neither. The noise resistance is shunted by the circuit *before* this, and will typically be under 500K over most of the audio band.
> grid-leak resistors are 1M then the input impedence can be estimated to be 2*Rg = 2M
It is not that simple.
The "input" resistor is boot-strapped on a large cathode resistor, and acts-like a much higher impedance.
The input cap can be much smaller than you'd expect for "1Meg".
The "back-side" resistor "leaks" input signal to the opposite input and cancels output signal.
The back-side cap needs to be much-much larger than you would think. Again, research and contemplation (or stealing) is a good plan.
PLAGIARIZE!! Look what is used on other popular amplifiers.
Your drawing omits any grid resistors on the 12AU7. Since it otherwise seems detailed, I call that an accident waiting to be built. (whoops- Steve hit that nit.)
As for your 12AX7 loadline calculations-- you could just steal values from the R-C-coupled amplifier suggestion tables.
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PLAGIARIZE!! Look what is used on other popular amplifiers.
I certainly will do that but I'm trying to understand why the values were chosen in the first place (at least at a high level). I can follow a schematic to build something but too often I ask myself, "Why is that there?" or "Why does Fender use one value and Marshall another?" This thread has really been helping me answer those questions.
> change the tone of the amp, or simply lower the noise?
Probably neither. The noise resistance is shunted by the circuit *before* this, and will typically be under 500K over most of the audio band.
I assume the circuit before this is the preamp (I don't plan only any effects loops, etc). I don't understand *what* will be under 500K. Will it be easier to explain this later when I add in the preamp circuit?
> grid-leak resistors are 1M then the input impedance can be estimated to be 2*Rg = 2M
It is not that simple.
The "input" resistor is boot-strapped on a large cathode resistor, and acts-like a much higher impedance.
The input cap can be much smaller than you'd expect for "1Meg".
The "back-side" resistor "leaks" input signal to the opposite input and cancels output signal.
The back-side cap needs to be much-much larger than you would think. Again, research and contemplation (or stealing) is a good plan.
I was trying to use the process described on the Valve Wizard (http://www.valvewizard.co.uk/acltp.html) site. I don't understand the point you are making. Are you saying the values I chose are incorrect or explaining how to determine the input impedance? Perhaps I should start labeling the components so I understand exactly which ones you are referring to.
Your drawing omits any grid resistors on the 12AU7. Since it otherwise seems detailed, I call that an accident waiting to be built. (whoops- Steve hit that nit.)
I was thinking the circuit was looking like it was missing some parts. I was about to say I can't find any examples of an indirect heated triode used in a push pull configuration with a LTP phase inverter, but then I remembered the link to the Traynor amp. I updated my circuit adding the resistors (not sure what to call them) based upon the Traynor values (see I plagiarized).
What are the 1500Ω resistors are called? What are the 1MΩ resistors called? And now for the big one... why were the values chosen? They seem to be common values but what happens if I were to use 1000Ω or 2200Ω instead?
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What are the 1500Ω resistors are called? What are the 1MΩ resistors called? And now for the big one... why were the values chosen? They seem to be common values but what happens if I were to use 1000Ω or 2200Ω instead?
The 1500Ω are called grid stoppers. The 1MΩ are called grid return or grid leak. The values are typical. The tube manuals may specify the 1MΩ as a typical or max grid resistance for a 12AU7. The 1.5KΩ is a typical Fender value. Marshall often uses 5.6KΩ. No big deal if you use a slightly different value for either of those resistors.
You have the 12AU7 cathodes connected to a -15.5vdc and the grid is connected to zero volts. That combination will melt that 12AU7.
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The tube manuals may specify the 1MΩ as a typical or max grid resistance for a 12AU7.
Are tube manuals different than the tube datasheet? If not... please tell me where in this datasheet (http://www.wooaudio.com/docs/tube_data/12AU7.pdf) I can find that value so I know where to look it up in the future if I design something with a different tube?
You have the 12AU7 cathodes connected to a -15.5vdc and the grid is connected to zero volts. That combination will melt that 12AU7.
My thought was the -15.5vdc is the cathode bias voltage that hasn't been designed yet so I just put the label there. Originally, I thought I would use a cathode resistor to ground, but HotBluePlates indicated that was not a good idea.
How should I change the circuit to avoid melt down?
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please tell me where in this datasheet I can find that value
Bottom of page 1. See attached pic.
How should I change the circuit to avoid melt down?
The cathode must be POSITIVE in respect to the grid. Where did you come up with a negative voltage?
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The cathode must be POSITIVE in respect to the grid. Where did you come up with a negative voltage?
I took the value from the grid curves I plotted (forgot to upload the image). I noticed Fender schematics showed the bias voltage as negative on their schematics, so that is what I copied. Will it be correct if I change the value to +15.5vdc? Or is this voltage something entirely different?
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Will it be correct if I change the value to +15.5vdc?
Yes. Those negative voltages on the load line chart refer to grid voltages, which must be negative in respect to the cathode.
That compares favorably with the Traynor schematic posted above (and here). That schematic shows +18.7 on the cathode and zero on the grid.
http://traynoramps.com/downloads/servman/smdh15h.pdf (http://traynoramps.com/downloads/servman/smdh15h.pdf)
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According the the 12AU7 tube datasheet, the Maximum Grid Circuit Resistance with Fixed Bias is 250kΩ and Cathode Bias is 1MΩ.
Since my design is Fixed Bias, does that mean that I should change the Grid Leak resistors to 250kΩ?
If not, should I be concerned about having the total value of the Grid Stopper and Grid Leak resistors being greater than 1MΩ? I'm guessing that since the value is less than one percent it doesn't really matter... that is less than the accuracy of most resistors.
On the other hand, if I were to use 5.6kΩ Grid Stopper and 250kΩ Grid Leak resistors, then it would be around 2%. I'm guessing that still isn't an issue.
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According the the 12AU7 tube datasheet, the Maximum Grid Circuit Resistance...
Those are design center values, not maximum values. And the tiny value of the stoppers is insignificant toward the purpose of the return resistors.
If I was gonna do fixed bias, I'd change the grid resistors to 220K and run a negative bias voltage to those 220Ks, rather than connect the 220Ks to ground. Then ground the cathodes. But I'd probably opt for a cathode biased amp like the Traynor. Much simpler and much, much more bullet proof.
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Because of the very big change from idle to peak, cathode bias is not recommended (also why the calculated ~310vdc is what we work from the entire time). You will need a fixed-bias supply, and we will need a phase inverter (which you wanted anyway).
But I'd probably opt for a cathode biased amp like the Traynor. Much simpler and much, much more bullet proof.
I'm all about simpler (especially for my first go at this) but I was under the impression that a fixed-bias supply was necessary. I suppose there are other ways to cathode bias with LED's or some other constant current sink that I might be able to use instead of a resistor.
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Have you looked at the Traynor schematic?
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Here's another example...
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Thanks for the references. The good news (to me) is that the values I have been proposing are very similar to the values in the schematics you have posted. As a reminder, I'm trying to understand why the values were chosen rather than simply copy them from another design. However, I think I will update my circuit to use a cathode bias for simplicity. Perhaps visit adding fixed bias later when I get to the power supply.
Next up will be a very simple preamp section - volume only no tone control. I want to be sure I understand how that works before I start adding in tone controls, etc.
I know I want to use a 12AX7 to mimic the Marshall (and most amps out there). Why do most amps use a 12AX7 as the first tube in a preamp?
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I've updated the schematic so the power section uses cathode bias. Based on a cathode voltage of +15.5V and a operating current of 5mA I used the following to calculate the value for the cathode resistor:
Total Current = 2 * 5mA = 10mA
Cathode Resistor (Rk) = 15.5V / 0.01A = 1550Ω or 1500Ω for standard value.
Do I have that right?
While doing this calculation I realized that I have messed up a few things along the way. For example, my power transformer has a secondary of 220V which will yield a rectified voltage around 310V. I have drawn my load lines based on that voltage. I need to go back and adjust the voltage remembering that I need to subtract the 15.5V so my operating voltage should have been closer to 295V. I'm not sure if this is going to make much of a difference or not but since I'm trying to be thorough about why values were chosen, I am going to do it.
Another thought... I don't necessarily care about running in Class AB over Class A. I wanted to understand how Class AB works, but I don't necessarily have a preference. There seems to be a mystique about Class A amps. Since I don't really care about getting the most power (after all this is supposed to be a one watt amp), would there be a benefit to moving my operating point?
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We're designing an amp 1 post at a time? If that's the case, we're tied to choices made earlier.
The first thing I need to decide is the probably the tube and operating point. Let's keep with the 12AU7, using the 22.5K load because it makes it simple to compare to the Firefly. I ran the numbers and a bias point of 310V, 5mA, -15.5V seems to be pretty good. Power Output is 1.5W and Plate Dissipation is 1.8W whereas 70% of Max Pdis is 1.9W.
The next thing I need to decide is whether I want the amp to run Class A or Class AB.
You don't pick an operating point and then decide if it's class A or class AB. The operating point dictates which you are running.
You ran the numbers, but didn't show your work. Plot the loadlines at your given operating point, then run the numbers for average plate current with maximum input signal. That will tell you if there is a small change from idle to full power (class A) or a big change from idle to full power (class AB). The change in average current will dictate your bias method.
How should I change the circuit to avoid melt down?
The cathode must be POSITIVE in respect to the grid. Where did you come up with a negative voltage?
The problem here is you don't know the questions to ask yet.
Working with the data sheet curves, you've been looking at negative voltages for Ec1, which is the control grid. Negative compared to what?
On the data sheet (and if you were inside the tube), all voltages are measured with respect to the cathode. So the plate voltage you set at your operating point is the plate-to-cathode voltage. The bias you settle on is your grid-to-cathode voltage.
If the bias is -15.5v, it does not matter if you connect the grid to 0v and the cathode to 15.5v (the cathode bias method), or you connect the cathode to 0v and the grid to -15.5v (the fixed bias method). In fact, the positive 15.5v at the cathode does not have to be derived across a resistor, but could come from a 15.5v supply if you had one handy (maybe part of a solid-state supply or the like).
What does have an impact, though, is if you figure out everything to a gnat's-ass for 310v plate (which we originally worked on when I presented an example I knew would be class AB and therefore fixed-bias), and then use a resistor to derive 15.5v positive at the cathode: Your tube's plate-to-cathode voltage is now 310v-15.5v = 294.5v.
Now this probably won't be a big deal in a 12AU7 output stage. Less current, less power. But you did go to all the trouble to calculate, and if 310v and -15.5v really did give the performance you were looking for, why get sloppy now?
Based on Valve Wizard's description of the Long Tail Pair Phase Inverter I have come up with following for my design. ...
Are we trying to replicate a specific amp, or are you seeking to design from scratch?
If you're designing from scratch, you still work from output-to-input and determine what type of phase inverter you need (or the amp's constraints will allow) to meet your needs. And that involves looking at the available supply voltage and the required gain from phase inverter input to output.
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We're designing an amp 1 post at a time? If that's the case, we're tied to choices made earlier.
Sorry for jumping around a bit and not being clear. I started over because so many of my initial examples were to be sure I was understanding the process.
The next thing I need to decide is whether I want the amp to run Class A or Class AB.
You don't pick an operating point and then decide if it's class A or class AB. The operating point dictates which you are running.
Again, I wasn't clear. The comment was tied to deciding what type of Phase Inverter to use. If I wanted to explore the self splitting phase inverter, then I would have to change my operating point to be Class A.
The problem here is you don't know the questions to ask yet.
Absolutely!!!
What does have an impact, though, is if you figure out everything to a gnat's-ass for 310v plate...why get sloppy now?
Yeah, I realized that my calculations were wrong after I had written the last post. I'm going to start over again to come up with operating points, loadlines, etc. This must be the iterative process you meantioned earlier in the thread. I will post new loadlines, operating points, and details about how I came up with average plate current, etc.
Are we trying to replicate a specific amp, or are you seeking to design from scratch?
Yes. :-) The inspiration is the JTM1 but since I don't have a schematic or any real details I'm taking inspiration from existing designs (e.g., Marhall typically uses LTP Phase Inverter) but trying to go through the design from scratch process even if I come up with exactly the same values.
If you're designing from scratch, you still work from output-to-input and determine what type of phase inverter you need (or the amp's constraints will allow) to meet your needs. And that involves looking at the available supply voltage and the required gain from phase inverter input to output.
Rather than assume I will be using an LTP, I would like to know why this choice is made so frequently and how you determine the type of phase inverter.
Be back soon with the power section details mentioned above.
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Starting over
Power Output Tube: 12AU7
Output Transformer: Hammond 125C
OT Secondary: 22.5kΩ
Class A Loadline: 11.25kΩ
Class B Loadline: 5.6kΩ (not used because cutoff happens a -26V)
Power transformer secondary: 220V
B+ after rectification: 220V * 1.414 = 310V
Grid Bias Voltage: -13V (this is a new operating point):
Plate Voltage: 310V - 13V = 297V
Minimum Plate Voltage: 150V
Bias Current: 7.4mA
Calculating Average Plate Current
I used the method described in RDH4 page 580 and the spreadsheet HotBluePlates sent me to help out. I used Visio to create a ruler that can be scaled so that I can more accurately determine the grid voltages and the current values when entering them into the spreadsheet (that's what all the pink lines are).
| V1 Volts | V1 Current | V2 Volts | V2 Current | Total Current |
| -13.0V | 7.4mA | -13.0V | 7.4mA | 14.8mA |
| -10.8V | 9.2mA | -15.2V | 5.8mA | 15.0mA |
| -8.6V | 11.0mA | -17.4V | 4.5mA | 15.5mA |
| -6.5V | 13.0mA | -19.5V | 3.3mA | 16.3mA |
| -4.7V | 15.0mA | -21.3V | 2.5mA | 17.5mA |
| -3.0V | 16.8mA | -23.0V | 1.8mA | 18.6mA |
| -1.7V | 18.4mA | -24.3V | 1.3mA | 19.7mA |
| -0.8V | 19.0mA | -25.2V | 1.0mA | 20.0mA |
| -0.3V | 20.3mA | -25.7V | 0.8mA | 21.0mA |
| 0.0V | 20.5mA | -26.0V | 0.7mA | 21.2mA |
Average Current = (0.5 * I0 + I12 + I34 + I50 + I64 + I77 + I87 + I94 + I98 + 0.5*I100) / 9 = 17.9mA
Power Input = Plate Voltage / Average Current = 297V / 0.179A = 5.33W
Power Output = Max Current * (Plate Voltage - Min. Plate Voltage) / 2 = 0.212A * (297V - 150V) / 2 = 1.55W
Plate Dissipation = (Power Input - Power Output) / 2 = (5.33W - 1.55W) / 2 = 1.89W
Percentage of Max Plate Dissipation: 1.89W / 2.75W = 69%
How's that for showing my work? Assuming my calculations are correct, how should I go about determining the phase inverter based on those values?
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Are we trying to replicate a specific amp, or are you seeking to design from scratch?
Yes. :-) The inspiration is the JTM1 but since I don't have a schematic or any real details I'm taking inspiration from existing designs (e.g., Marhall typically uses LTP Phase Inverter) but trying to go through the design from scratch process even if I come up with exactly the same values.
The thread linked earlier indicates the JTM1 was not a single amp, but a collection of 5 amps, one for each decade from '50's through 2000's. Which of those are you wanting to copy?
The little bit of detail we have says some models used a long-tail inverter, some used split-load. Some used a high load impedance, some used a low load impedance. Exact topology and tone controls differed by model.
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The JTM1 was one of five 50th Anniversary amps that Marshall released. The lineup was:
- 1960's JTM1
- 1970's JMP1
- 1980's JCM1
- 1990's DSL1
- 2000's JVM1
The JTM1 is supposed to mimic the sound of a JTM45 with jumpered inputs.
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The JTM1 was one of five 50th Anniversary amps that Marshall released. ...
The JTM1 is supposed to mimic the sound of a JTM45 with jumpered inputs.
Thanks for the correction! I must being losing the pulse of this thread, having forgotten that.
The inspiration is the JTM1 but since I don't have a schematic or any real details ...
Rather than assume I will be using an LTP, I would like to know why this choice is made so frequently and how you determine the type of phase inverter.
I couldn't find where we had a link to James Marchant's description of the JTM1, so I'm including it now (http://www.ax84.com/bbs/dm.php?thread=460461).
James says the 60's model JTM1 has a cathodyne phase inverter (same thing as the "concertina" and split-load" inverters). He also says the 12AU7 output is cathode biased, and that his designs had a lower OT primary impedance than one of the other designer's amps. So those are clues anyway; that said, you have a given PT, and we probably oughta proceed with a design which suits the transformer you'll actually use.
... I used Visio to create a ruler that can be scaled so that I can more accurately determine the grid voltages and the current values when entering them into the spreadsheet (that's what all the pink lines are).
...
How's that for showing my work?
Outstanding job showing your work!
Minor point:
For your pink ruler lines to determine the voltages between gridlines, the ruler will not be horizontal. Instead, it is along the loadline itself. In this case though, the error due to this oversight is likely small, but you should be aware of it when estimating grid voltages not on one of the curves drawn.
... Assuming my calculations are correct, how should I go about determining the phase inverter based on those values?
At this point, I'm going to trust your calculations.
Stating the obvious for those following along:
Your output stage idles at 2 * 7.4mA = 14.8mA, and has an average plate current at full output of 17.9mA. The small change from idle to full output suggests class A operation, as does the fact that on your class A loadline an input signal of 2x the bias voltage (2*-13v or 26v) doesn't cut the tube off. So cathode bias seems advisable (and you've already accounted for it, showing the iterative process)... -13v / 0.0148A = ~880Ω. That's a non-standard value, but there is a 880Ω 3w resistor out there.
I'll come back in a bit for the phase inverter discussion.
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James says the 60's model JTM1 has a cathodyne phase inverter.
I should have caught that. I thought he said it was a LTP.
That's a non-standard value, but there is a 880Ω 3w resistor out there.
-13v / 0.0148A = 878.4Ω
(1300 * 2700) / (1300 + 2700) = 877.5Ω
1K3 and 2K7 resistors are probably cheaper and easier to come by.
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I did some reading this weekend on the Cathodyne Phase Inverter - RDH4, Crowhurst and Valve Wizard. Conceptually the Cathodyne makes sense, but I'm not sure how to go about choosing the values I need. Valve Wizard (http://www.valvewizard.co.uk/cathodyne.html) states the following:
If the power valves are sensitive types like EL84s or 6V6s, then we don't need huge amounts of swing and a total load around 47k to 100k would probably do. If we need to overdrive bigger valves like EL34s then 200k is probably in order.
How do I know how much voltage I need to swing?
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Does the thread now turn to the preamp side?
Not quite... I will be using a cathodyne PI because that is what the JTM1 is using. I have actually drawn up the schematic but have not entered any values yet because I don't know what to choose. I have read descriptions of how to calculate the appropriate values to get the cathodyne to split the signal and reverse the phase, but I can't find a description that says this is how to choose values for your power section.
Once I understand that I will post the calculations, update the schematic with values as well as pictures of the loadlines.
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> How do I know how much voltage I need to swing?
I thought you had this all plotted out (http://www.el34world.com/Forum/index.php?topic=16536.msg164319#msg164319)?
Looks like 10V-11V bias, swing up to zero and down to twice that... say 11V peak.
The perfect cathodyne, 240V supply, could swing 240V/4= 60V peak.
Yours must be imperfect. You have grid load (probably 1Meg), and your driver tube has plate resistance. As a general thing, split the difference (geometric mean) of tube and load resistance. 12AX7, Rp near 60K? So 2Meg total (both sides) load, 60K. Square root of (2Meg*60K) is 346K. Half that is 173K in each of plate and cathode. Not fussy, use 150K standard value.
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> How do I know how much voltage I need to swing?
I thought you had this all plotted out (http://www.el34world.com/Forum/index.php?topic=16536.msg164319#msg164319)?
See... I don't know what I know. I can be a little thick with this stuff and things need to be spelled out for me the first time (but I learn quickly after that).
Looks like 10V-11V bias, swing up to zero and down to twice that... say 11V peak.
I'm confused. The bias point I chose was -13V. Were you eyeballing the diagram to come up with 11V or are you referring to something else?
The perfect cathodyne, 240V supply, could swing 240V/4= 60V peak.
Yours must be imperfect. You have grid load (probably 1Meg), and your driver tube has plate resistance. As a general thing, split the difference (geometric mean) of tube and load resistance. 12AX7, Rp near 60K? So 2Meg total (both sides) load, 60K. Square root of (2Meg*60K) is 346K. Half that is 173K in each of plate and cathode. Not fussy, use 150K standard value.
Again confused... are you making a general statement about the perfect cathodyne? That is, a perfect cathodyne can swing 1/4 of the supply voltage. For example, a 280V supply could swing 70V peak 300V could swing 75V, etc. Since B+ is around 310V for the power tubes, do I just choose a voltage for the PI supply and get there by using a voltage dropping resistor or is 240V some magic number that I should use for a cathodyne?
I understand that a 12AX7 generally has a plate resistance around 60K (but individual 12AX7's can vary). Please explain where the grid load (Rg) comes from and why you specified 1M.
In summary, use the following formula to determine the total load R = (Ra + Rk) to use for a cathodyne?
R = √(2 * Rg * Rp) = √(2 * 1,000,000 * 60,000) = 346kΩ
The Ra = Rk resistors are calculated using:
Ra = Rk = R / 2 = 173kΩ
Finally, use a standard value for Ra/Rk. You suggested 150kΩ but I could also use 180kΩ.
I think I get most of it but if you can explain the couple of things I was confused about I will plot my load lines, calculate values and draw out the schematic tomorrow. Thanks for your help.
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I was looking at the valve wizard, "The Cathodyne Phase Inverter", and saw the plot for the 12AX7, My 12ax7 plot doesn't do below -5 volts. The valve wizard show -4 volts.
Please confirm that you are designing using the 12ax7 and not a 12au7.
I will be designing the phase inverter to use a 12AX7. I noticed that the valve wizard's plot for 12AX7 is slightly different than the plot in the GE 12AX7 datasheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/093/1/12AX7.pdf).
I've been reading many pages, but I still cannot figure out how to choose the appropriate voltage, load, bias point, etc for a phase inverter/power section combination.
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the -13v bias point is what I have chosen for the 12AU7 tubes in the power section. I have not chosen a point for the Phase Inverter yet.
I'm guessing the grid voltage bias point for the 12AU7 phase inverter will be between -1.5 and -2.0V. Using the methods described by Valve Wizard with a B+ of 295V and total load of 54kΩ I think the PI will swing 165V (from 130V to 295V), but I my understanding of this is vague at best. That seems like WAY too much, but perhaps it just depends on the signal coming it.
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> is 240V some magic number
It's divisable by 4 without strain.
400V is even easier (if unlikely).
Imagine the perfect tube, two resistors, no load. The maximum output idles at K=100V, P=300V. Turn the tube off, K goes to 0V and P to 400V. Turn the tube FULL on (zero drop), K=200V and P=200V. That's a perfect cathodyne.
In real life you never get zero across the tube. And with-Load an OFF tube won't go all the way to zero and 400V. But 1/4 3/4 points are often a good starting point.
> B+ is around 310V for the power tubes
The cathodyne and its driver don't have great ripple rejection, like a push-pull output stage. So you want some B+ filtering. Usually an R-C with some drop. 80% is often a good trial value. That's 248V. So 240V is also a good pencil number.
I think the 1Meg is the power stage 12AU7 grid resistors?
So perfect 240V allows 60V peak. The assumed 173K+173K DC load against the 60K Rp means we lose 1/6th due to tube loss. 1Meg load on 173K is about another 1/6th loss. So we probably get 70% of 60V peak, or 42V peak.
> point I chose was -13V
I'm not even going to look back and see why 11V or 13V. We need about a dozen volts and we have over 3 dozen volts. Not even close. If it were slightly close to not-enough, you would probably want a different approach.
However most tubes are designed so they can be driven with a reasonable tube driver in front. 6L6 has Mu(g2) of 10 so that a resistance-coupled driver can eat Vg2 or a bit less and still give full drive. 50L6 has Mu(g2) near 7, because 105V operation needs HIGH current which means low Mu(g2) and it's only a cheap radio. Mu near 5 is about as low as you can do with R-C coupled drive. 2A3 or 300B has Mu near 4 because these tubes were intended for use with transformer driver (and because big triode power is a lot about current and thus low Mu).
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> We need about a dozen volts and we have over 3 dozen volts. Not even close. If it were slightly close to not-enough, you would probably want a different approach.
So, the Phase Inverter doesn't need to be designed to swing a specific voltage (e.g., 13V), it just needs to be able to swing at least the grid voltage of the following stage Therefore:
Voltage after Rectification: 310V
Phase Inverter B+ = 80% * 310V = 248V but use 240V for easier math.
Rtotal = √(2 * Rg * Rp) = √(2 * 1,000,000 * 62,500) = 354kΩ (I got Rp from 12AX7 datasheet, Rg from power section design earlier).
Ra = Rk = Rtotal/2 = 177kΩ but use standard value of 150kΩ
Plot the loadline R=Ra+Rk=300kΩ
Based on the loadline choose a bias point.
Grid Voltage (Vgq): -1.5V
Current (Iq): 0.35mA
Voltage (Vq): 135V
Cathode voltage = Rk * Iq = 150kΩ * 0.35mA = 52.5V.
Another way of calculating this is (B+ - Vq) / 2 = (240V - 135V) / 2 = 52.5V
Since the values are the same, then I probably chose the right values.
Valve Wizard goes on to calculate a cathode bias resistor. My calculations are as follows:
Rb = Vgq / Iq = -1.5V / 0.00035A = 4.3kΩ so use either a 3.9kΩ or 4.7kΩ standard value resistor.
Finally, I added in the arc protection and large grid stopper recommended by Valve Wizard. To be honest, I'm not sure if I have everything quite right on my schematic. Please review and let me know if I need to make some corrections. If everything is correct, then I will proceed to the preamp section.
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Since there has been no replies on the phase inverter, I'm going to assume that everything is ok or close enough that we can come back to it later. Now it's time to start examining the preamp.
As I mentioned before, I'm looking to emulate the JTM1 which is supposed to be a one watt version of the JTM45. Which, as I have learned is really just a Bassman 5F6-A with a 12AX7 instead of a 12AY7. There are a few other changes I'm sure but I've been comparing the schematics of both and its pretty remarkable how similar the circuits are. The Ampbooks (http://www.ampbooks.com/home/classic-circuits/fender-bassman-preamp/)website does a good analysis of the Bassman preamp but I still have some questions about how to apply the information to my design and how they came up with some of the values in the article.
For discussion purposes, I'm going to refer to the First Stage of the High Gain Input, Normal channel of the JTM45 (image from ampbooks):
Rgs = 34kΩ (two 68kΩ resistors in parallel)
Rg = 1MΩ
Rl = 100kΩ
Cg = 0.02uF
Rk = 1.64kΩ (cathode resistance per triode)
Ck = 250uF
Rv = 1MΩ
(http://www.ampbooks.com/home/classic-circuits/fender-bassman-preamp/bassman-preamp.gif)
If you look at an actual JTM45 schematic, you might notice that the cathode resistor is 820Ω. However, the resistor is shared between both triodes in the tube so the actual Rk value is twice that or 1640Ω.
Ampbooks goes on to plot the 100kΩ loadline and the 1.64kΩ lines for Vpp = 325.
(http://www.ampbooks.com/home/classic-circuits/fender-bassman-preamp/bassman-first-preamp-curves.png)
I have a couple of questions:
- How did they come up with the slope for the 1.64kΩ blue line?
- What voltage should I use for Vpp given 310V after rectification and 240V target for the Phase Inverter?
Thanks for the ongoing help!
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Since there has been no replies
Thanks for the ongoing help!
-Please don't take this wrongly- (and I get you're trying to understand every last infinitesimal detail) but - there sure is A LOT of "worry & planning info" here for "only" a one-watt amp! Just build the darn thing and play around with it. All this extra worry and planning is mostly wasting a lot time because you'll find that you're most likely going to be changing and modifying things once built anyway after it's up & running. So then what's the point after all is said & done?! You'll find out why if you just build it and play around with it. All this paralysis by over analysis isn't worth all the fuss if there's going to be changes anyway in the end to get it just how you want it. Also, since it's only a one-watt amp you're not going to even hear or experience the difference of every detail as it can barely move the speaker cone as it is. You will learn this quickly if you build enough amp designs that are not direct copies or clones.
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-Please don't take this wrongly- (and I get you're trying to understand every last infinitesimal detail) but - there sure is A LOT of "worry & planning info" here for "only" a one-watt amp! Just build the darn thing and play around with it. All this extra worry and planning is mostly wasting a lot time because you'll find that you're most likely going to be changing and modifying things once built anyway after it's up & running. So then what's the point after all is said & done?! You'll find out why if you just build it and play around with it. All this paralysis by over analysis isn't worth all the fuss if there's going to be changes anyway in the end to get it just how you want it. Also, since it's only a one-watt amp you're not going to even hear or experience the difference of every detail as it can barely move the speaker cone as it is. You will learn this quickly if you build enough amp designs that are not direct copies or clones.
You have no idea how much I'm enjoying learning the details. I've tried to get this dialog going on a number of boards for a long time but have never found the right board with people that have the knowledge and willingness to share.
I have built a number of amps over the last 15 years. Mostly stereo amps (2A3, 300B, 6V6, EL84), a few preamps and added tube output stages to some DACs as well. During that time, I have swapped caps, changed resistors, changed output transformers, etc. I even "designed" some things but I just copied and hacked things together hoping they worked (which amazingly they did). However, I never really knew what I was doing or why. I would do it because someone said I should or because I saw someone do something similar. This thread has opened my eyes tremendously. I have been re-reading books, sites, articles, etc. with a new understanding and a lot of "ah ha, now I know why this works..."
While I appreciate your perspective and completely agree that sometimes people can over analyze... my questions are about gaining a fundamental understanding that goes beyond being able to read a schematic and solder.
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"No harm, no foul" as the late great broadcaster Chick Hearn (of the LA Lakers) would say, I just "bunny-hopped through the pea patch" (another expression for traveling) through this thread. :icon_biggrin:
I fully appreciate & understand your perspective & glad you found the right place. Just out of curiosity are you actually planning on building one of these or is it just for the knowledge aspect alone? My comment before was coming from a perspective that some over analyze the gazoo out things & spend too much time fretting over the smaller stuff. Then in the end things chang away from the original plans anyway. Time can be better spent getting things done initially then trouble-shoot and/or perfect the build & design once things are up and running. Many times even the best designed plans fall by the wayside because of poor layout, grounding, component placements, bad soldering, and many other issues.
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> The Ampbooks website does a good analysis of the Bassman preamp but
Buy the darn book. Fender Bassman 5F6a. Kuehnel does an extended analysis with more numbers than you can shake a stick at.
There's another one, Guitar Amplifier Preamps, which compares different preamps.
The paths you are exploring, the level you are at, these books are excellent guides for thought.
http://www.ampbooks.com/home/books/preamps/ (http://www.ampbooks.com/home/books/preamps/)
No they are not free. But why should I/we type $38 words of words when you can get much more for your money in books?
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But why should I/we type $38 words of words when you can get much more for your money in books?
I just discovered the ampbooks site the other day. Since you recommend the books I have ordered them I was thinking of buying Valve Wizard's preamp book as well - John Broskie at tubecad.com gave his poweramp book a good review but its sold out.
The reason I hesitated to buy them is I own the Morgan Jones book as well as RDH4, Crowhurst and a few others. I have read them over and over but I get overwhelmed with the math and symbols. I learn better through interaction and repetition... that is why I drew loadlines for 3 different tubes and chose different operating points at the start of this thread. With that, I learned new things I didn't understand and new questions to ask. Reminds me of how I learned back in high school - I was the kid who asked all the questions.
Speaking of that, does Kuehnel explain how he plotted the 1.64kΩ blue line? I understand the 100kΩ line - 1mA for every 100V. The blue line looks like it has a similar but opposite slope, yet its 60 times less. What am I missing?
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I have found Merlin's Preamp book invaluable. I have his Power Supply book as well; for what I do it's not as useful, but it still has helped me out quite a lot.
Nothing beats this forum though.
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> how he plotted the 1.64k blue line? ... slope, yet its 60 times less
It's the cathode resistor, isn't it?
Plotted as cathode current versus _grid_ voltage (not plate voltage). By pure Mu this would be 100 times less. With 100K shoved in the plate circuit the effective Mu seen below the cathode is less, perhaps 60.
Actually.... 100K/1.64K is 60.9765
With the 325V and 100K we know the tube will settle somewhere on the red line. But where? The 1.64K line settles that.
In total ignorance, we would draw the 1.64K line all the way from zero to some absurd current, then see where it crosses the 325V 100K line. But since we already know the answer we want is the intersection with the 325V 100K line, he just drew that part. Since it isn't on a whole-number grid voltage, he drew from the next-lower to next-higher grid voltage line.
The 1.64K Rk line, plotted on plate current/voltage curves, is not a straight line. However for all practical purposes for any tube you would want to run audio through, it is straight-enough within the limits of ruler/graph plotting, and certainly in light of the +/-20% variation between two tubes of the same type. (This is perhaps due to Mu being very-nearly constant over the prime range of a tube.)
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This thread started out wanting to build a one watt amp in the Marshall flavor of amps.
That's what I thought too!? But instead it turned into something much different like a personal classroom. I never got an answer if bishop even intends to actually build this amp or not? Kind of rude to not answer questions and only ask them with such detail and expectations, IMHO.
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This thread started out wanting to build a one watt amp in the Marshall flavor of amps.
Yes, that is the idea.
In an early post I suggested using the Cornell note system. I would like to suggest that again.
I don't know about the Cornell note system (I'll look it up), but I do plan on creating a blog or something that details the steps involved with designing the amp. I have also been working on a spreadsheet with forumlas, where to find them and notes for what values to enter. Right now, I'm the only one who could decipher it, but I will clean it up for general consumption before posting online.
At the risk of being a butthead, hand calculations and lots of them is the best way to learn the tube technology.
Not butt-ish at all. That is exactly why I have been reposting my work with numbers and formulas each iteration. First it reinforces the the information and second it allows others to point out where I have messed up.
That's what I thought too!? But instead it turned into something much different like a personal classroom. I never got an answer if bishop even intends to actually build this amp or not?
Kind of rude to not answer questions and only ask them with such detail and expectations, IMHO.
Sorry, I missed your question. I ABSOLUTELY plan on building this amp.
I am more of a hands-on interactive learner. Reading books, etc is OK, but I never know if I am really understanding. ... So I'm going to start by posting my first questions and hopefully the members will chime in to help me out (or tell me to go away which is OK as well). As I go, I will continue to update my design on this thread and eventually build the amp. Sounds like fun to me, I hope you think so as well.
I also apologize for asking so many questions, but I was clear in my first post that is how I learn best. If folks would rather I stop posting, that is ok as well.
By the way, I ordered four books that were recommended. They should be here tomorrow so I will spend a few days reading through them to see if I can answer my own questions. That said, I'm sure I will still have more questions. Books can't tell you if you are doing things correctly.
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I also apologize for asking so many questions, but I was clear in my first post that is how I learn best. If folks would rather I stop posting, that is ok as well.
Don't stop posting. Most of us understand this is as much or more a design learning project as it is an actual nuts and bolts, get it built project. You're still among friends and kindred souls. Group participation will be less just because you want to know everything about every component and most folks just ain't interested in the nth degree of design.
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Awesome thread. Rather abruptly this thread took a turn from the pursuit of knowledge and a greater understanding of thermionics to the usual internet BS.
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Back to our regularly scheduled programming...
In total ignorance, we would draw the 1.64K line all the way from zero to some absurd current, then see where it crosses the 325V 100K line. But since we already know the answer we want is the intersection with the 325V 100K line, he just drew that part. Since it isn't on a whole-number grid voltage, he drew from the next-lower to next-higher grid voltage line.
Perhaps if I ask a different way. How do I determine the quiescent grid voltage, plate voltage and plate current from looking at the schematic?
For example, I have included the first stage of a Trace Elliot Velocette (my favorite amp I own).
RL = 220kΩ
Rk = 1500Ω
I have plotted the 220kΩ loadline starting at 345V as specified on the schematic. I included what I think is the correct plot of the 1500Ω line, but that is because the schematic showed the plate voltage as being 160V. What if that value was not on the schematic? I'm feeling a bit thick here... I still don't get how to plot the slope of Rk.
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Rather abruptly this thread took a turn from the pursuit of knowledge and a greater understanding of thermionics to the usual internet BS.
I don't agree and here's why:
It would better serve everyone if this thread was named appropriately from the beginning and therefore stayed on topic. One could then search the correct thread title to find, read, and learn about it. This would then be treated appropriately and straight forwardly. Instead, in the beginning it was only about making a one watt amp and segued from there. There is a distinct difference here.
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I have plotted the 220kΩ loadline starting at 345V as specified on the schematic. I included what I think is the correct plot of the 1500Ω line, but that is because the schematic showed the plate voltage as being 160V. What if that value was not on the schematic? I'm feeling a bit thick here... I still don't get how to plot the slope of Rk.
Turns out I was wrong!!!! (no surprize there)
The 1.64K Rk line, plotted on plate current/voltage curves, is not a straight line.
I get it now. Valve Wizard's Designing Tube Preamps... book (Merlin Blencowe) explains how to do it on page 14 (guess who got a book delivered). I have plotted four points to reinforce the concept when all I really needed was two. Since I know the cathode resistor is 1500Ω, I need to calculate the current for various grid voltages until I find a line that intersects the 220kΩ loadline.
Ia = Vk / Rk = 0.5V / 1500Ω = 0.3mA
Ia = Vk / Rk = 1.0V / 1500Ω = 0.7mA
Ia = Vk / Rk = 1.5V / 1500Ω = 1.0mA
Ia = Vk / Rk = 2.0V / 1500Ω = 1.3mA
Next plot the points on the graph and connect the dots. That is:
(-0.5, 0.3)
(-1.0, 0.7)
(-1.5, 1.0)
(-2.0, 1.3)
Once the points are connected, its easy to see that the computations for 0.5V and 2.0V were unnecessary (but as I said, good for illustrations purposes). By examining the graph you can see that the quiescent grid voltage, plate voltage and plate current are as follows:
Vgq = -1.25V
Vpq = 155V
Ipq = 8.5mA
The calculated voltage is slightly less than the 160V on the schematic, but within the 20% so I'm pretty sure I did that right. That's the operating point for the Trace Elliot and the earlier example was for the Bassman. What I really want to know is the operating point for the JTM45 so I'm going to go figure that out now.
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The JTM45 uses a 100kΩ load resistor and a 1.64kΩ cathode resistor (Actually it uses a common 820Ω resistor across 2 tubes but the effective resistance is 1.64kΩ). Therefore I can plot the 1.64kΩ loadline using the following points:
Ia = Vk / Rk = 1.5V / 1.64kΩ = 0.9mA
Ia = Vk / Rk = 2.0V / 1.64kΩ = 1.2mA
Therefore:
Vgq = -1.75V
Vpq = 220V
Ipq = 1.05mA
For this design, I'm thinking that it I will not be able to have the exact same operating points just because the B+ voltage will have already been dropped to around 240V for the phase inverter (see earlier calculations).
If I am correct, then which of these operating point targets should I try to achieve to emulate the Marshall sound or is the operating point irrelevant and I should look at something else? I'll do some more reading to see if I can answer my own question in the mean time.
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Related to this thread, check out RC30, ... (RC15 starts on page 28)... I will start another thread using the cornell notes format...
I'm not sure what RC30 and RC15 are. I look forward to seeing how you use cornell notes for this (so I can copy it).
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... If I am correct, then which of these operating point targets should I try to achieve to emulate the Marshall sound ...
We're not there yet. :icon_biggrin: Sorry for the absence, but I'm working on getting a job.
Regarding the loadlines/cathode lines:
Step back from it for a moment. The graph is just a visual way to describe all the possible outputs if you change a variable, just like graphing an equation in school. The plate loadline is easiest to grasp because it will be a straight line; this line describes a resistor, and how ohm's law spits various values of current (y-variable) with changing values of voltage (x-variable).
Now think about the resistor itself: one end is tied to the supply voltage. Pretend that the opposite end also gets attached to the supply voltage (let's say the voltage is 300vdc). If both ends are at 300v, there is zero difference in voltage between the ends, and zero current. You plot a point at 300v, 0mA.
Let's also say our resistor is 100kΩ. Let's say the side not attached to the supply voltage is at ground potential (like the tube is shorted from plate to ground). The value of current is solved by ohm's law, 300v / 100kΩ = 3mA. So you plot the 0v, 3mA point. Connect those 2 points together, and you get a 100kΩ loadline for a supply voltage of 300v.
Now the bottom-left corner of the graph is the origin or 0v, 0mA so why don't we have a point at that location? Because the non-varying end of the resistor is pegged to the supply voltage. The varying end is attached to the plate, and the values of voltage shown on the x-axis are tube plate voltages. So the plot of phm's law using this resistor move up and to the left.
The cathode resistor, however, has an end connected to ground. So 0v, 0mA will be a point on the curve for the cathode resistor. Notice I said curve. The graph is nice & square for plate voltages and plate currents. However, the cathode loadline uses plate current (which for a triode will be equal to cathode current, and therefore current through the cathode resistor) and voltage drop across the cathode resistor, which is the same as grid-to-cathode voltage. These voltages are not shown on the x-axis but by the grid voltage curves. As you found, you have to plot the value of current when a voltage equal to the grid voltage curve is dropped across the cathode resistance.
And where the 2 lines cross is your idle point of the stage.
The perfect cathodyne, 240V supply, could swing 240V/4= 60V peak.
Again confused... are you making a general statement about the perfect cathodyne? That is, a perfect cathodyne can swing 1/4 of the supply voltage.
Maybe you've moved beyond this, but I'm back-tracking a bit.
I don't have any memorized rules for phase inverters, so I looked in RDH4. Ch 12.6 ii (A), pages 522-523, it says it is normal practice to assume the maximum grid-to-grid output of the split-load/cathodyne/concertina inverter (which RDH4 simply calls "phase splitter") is 1/4 * supply voltage, peak-to-peak.
A run-of-the-mill common cathode gain stage can likely swing an output at least 60-75% of its supply voltage, depending on plate load and idle point. A paraphase inverter is basically one of these gain stages, and a 2nd one with a voltage divider in between to knock down the signal to the 2nd stage. The result is 2 outputs with similar size and opposite polarity.
I did not see any obvious rules for a long-tail inverter, or its root form in RDH4, the "cathode-coupled phase inverter" (Ch 7.2 viii (A), pages 347-348). What I do notice is that some amount of voltage must be dropped across the tail, and the bigger this voltage, the better the balance of the inverter. Within this guideline, Aiken (http://web.archive.org/web/20060101094749/http://www.aikenamps.com/LongTailPairDesign.htm) and Merlin (d.c. (http://www.freewebs.com/valvewizard/dcltp.html) and a.c. (http://www.freewebs.com/valvewizard/acltp.html) versions) basically calculate the stage as though is were a common cathode gain stage, except they go back to estimate the amount of imbalance and how far to trim plate loads to restore balance.
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The "needed peak-to-peak drive" is equal to the output tube bias voltage times 2, because the entire phase inverter has an output swinging to a peak in one direction equal to the bias voltage for one output tube, and also an output swinging to a peak in the opposite direction equal to the bias for the other output tube.
So we might arrive at some self-guidance that if the needed peak-to-peak drive to the output stage is less than 25% of the supply voltage for the phase inverter, we can use the split-load. If we need a drive voltage in excess of 25% of supply voltage, then we should look to the long-tail inverter. If we need a peak-to-peak drive something like 50-75% (exact amount needs to be determined experimentally) of supply voltage, then we probably need a paraphase that doesn't waste a lot of supply voltage (across a tail resistor).
You have 310vdc supplying the output stage, and that will be dropped by some amount before feeding the phase inverter. I might guess your entire preamp and phase inverter will draw 4mA (this will have to be revisited later), and a 18kΩ dropping resistor in the power supply reduces supply voltage by 72v to 238vdc (or call it "240v" like PRR). Your proposed bias is 13v (if I'm following correctly), so the needed peak-to-peak output is 26v. 26v * 4 = 104v, so you have plenty of supply voltage to use the split-load (or any other) inverter.
To design a split-load inverter, you simply plot loadlines until you find one you like which delivers the needed peak-to-peak output swing. The line you plot will be the plate load resistor plus the cathode load resistor; for a 5E3 Deluxe which uses a 56kΩ resistor for each of these, you draw a 112kΩ loadline. Or, you find a loadline that works for your desired output swing, divide that value by 2 and that is the value for plate load and cathode load.
You then pick a desired idle point on your loadline, and wherever that point is, determine the indicated grid-to-cathode voltage and tube plate current. Divide voltage by current, and you have your cathode bias resistor.
You seem to have this in-hand, but copied the diode and input resistor arrangement from Merlin... are you certain you're going to use this setup? Because I'm thinking you can't just plug-n-play those resistor values (I might be wrong, as I don't have his book, though I think I see what he's doing on his website).
I'm not sure what RC30 and RC15 are.
RCA Receiving Tube Manuals. RC30 is a later edition, but with maybe different tubes but otherwise much the same info as RC15 (or RC19, RC25, etc).
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> the exact same operating points
Did Jim Marshall work with plots that spilled off his monitor and compute to 3-digit precision?
We know he stole much from Leo. And Leo got it from the Westinghouse manual.
In general, you stick in resistors that may be right, then if unsure of happiness, try the values up and down from that.
You have a 200V possible swing. We rarely have more than 50V of audio. Imagine steering a 50 foot boat through a 200 foot canal. Mostly you aim for the center, but not too fussy. Maybe it's mucky on one side and hard/sharp on the other side, or the current is stronger to port, you lean a little to one side or the other. But this isn't like scraping the Caribbean Corgi through Panama. And you CAN tweak it on breadboard (beats tweaking heavy steel plate repairs in dry-dock).
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Great summary and explanation.
You seem to have this in-hand, but copied the diode and input resistor arrangement from Merlin... are you certain you're going to use this setup? Because I'm thinking you can't just plug-n-play those resistor values (I might be wrong, as I don't have his book, though I think I see what he's doing on his website).
I just got the book yesterday and will see if I am really going to use it after further reading. It sounds like a good idea based on his website description.
Did Jim Marshall work with plots that spilled off his monitor and compute to 3-digit precision?
In general, you stick in resistors that may be right, then if unsure of happiness, try the values up and down from that.
And you CAN tweak it on breadboard.
While Jim and Leo didn't have the level of precision we have today, I bet they had some background in electronics and I expect they understood loadlines, gain, input and output impedance, etc. So, while I can tweak on the breadboard, I still would like to know why a tone stack can present a difficult load and why it makes sense to use a cathode follower for a buffer (I'm just throwing out things that I have read but don't understand - I might not even be repeating them correctly but you get the idea). The help I'm getting on this thread and the books I got last night are getting me over that hurdle.
With regard to the loadline, you confirmed what I was thinking to myself last night. That is, as long as you are in the middle there somewhere it really doesn't matter.
Now I need to build me one of these (http://www.el34world.com/Forum/index.php?topic=16551.msg164456#new.).
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...as long as you are in the middle there somewhere it really doesn't matter.
PRR has a special way with explanation doesn't he? You are on the path & getting closer to snatching the pepple grasshopper. ;)
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Did Jim Marshall work with plots that spilled off his monitor and compute to 3-digit precision?
While Jim and Leo didn't have the level of precision we have today, I bet they had some background in electronics and I expect they understood loadlines, gain, input and output impedance, etc.
Jim Marshall was a drummer who owned a music store. The little I know is that Ken Bran (and possibly others) did the design work for him, but the first Marshall 50w amps were dead-nuts copies of the Fender 5F6-A Bassman schematic (I found out the hard way the actual as-built amps varied from the schematic) using Brit parts and available speakers.
Leo Fender was a radio repairman. His particular genius was to take common circuits and build them in an inexpensive manner for sale. Long-time employees at Fender said Leo "had an ear like an oak log," and that his other stroke of genius was to put his products in the hands of southern California professional musicians and solicit their feedback on what changes were needed. The more you learn about old tube electronics references, you'll see more & more that Leo copied standard circuits and the latest trend circuits rather than "tweaking for tone," and that his couple of tremolo circuits were among his few actual innovations in guitar amps. I don't know but suspect there were additional folks that assisted with amp design.
You seem to have this in-hand, but copied the diode and input resistor arrangement from Merlin... are you certain you're going to use this setup? Because I'm thinking you can't just plug-n-play those resistor values (I might be wrong, as I don't have his book, though I think I see what he's doing on his website).
I just got the book yesterday and will see if I am really going to use it after further reading. It sounds like a good idea based on his website description.
He's not very overt in explaining that the compression effect related to the diode is heavily loading the preceding stage so that its output is ham-strung. Exact resistor values should be determined on a breadboard (in my opinion) so that the effect happens just before the phase inverter can be driven to a nasty zone.
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Exact resistor values should be determined on a breadboard (in my opinion) so that the effect happens just before the phase inverter can be driven to a nasty zone.
I was originally thinking I could build this on the cheap because I already own the transformers. Now, I'm thinking the money I saved will go towards building a breadboard setup where I can easily swap components. That said, I still want to have a design in hand to start with so that when I swap parts I can understand what it is doing when I change it (at least have an idea).
As an aside... several years ago I designed a stereo power amp using 6V6 tubes, triode connected and single ended. It made sound and didn't sound half bad, but the operating points that I thought I would get where WAY off. If I recall, I was expecting the operating point to be around 250V and it ended up being more like 325V. I had no idea why at the time. I think I cold figure it out now but that amp has since been parted out for other projects.
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> I need to build me one of these.
Sorry; poor wording on my part.
The universal breadboard may be good when you spend all your time building systems you don't know what they will be.
In a guitar amp, you can generally have a good idea how many stages and knobs. You punch-up a chassis with enough holes, but generous on size, spacing, and tie-points. Build the amp but don't NASA-wrap the wires and leads. Stick em in the hole and solder. Play. with the loose construction you can quickly change a part and try some more.
Basically don't go all cramped and hay-wire like an old Fisher or some Bogens. The Fender lug-board is not bad, except you may wish to do the wires top-side. Turrets may be better.
MUCH amp design can be done with the tube manual. Power stages, just use the Suggestion, but perhaps with whatever voltage you can get with in-stock PTs; higher because guitarists favor MORE SOUND over long life. This also applies to radio RF/IF stages, except you like to cheat the voltage low until sound quality falls-off. For Resistance Coupled Audio Amplifiers, the data-sheet conditions are excessive, you turn to the R-C Amplifier tables.
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The universal breadboard may be good when you spend all your time building systems you don't know what they will be.
In a guitar amp, you can generally have a good idea how many stages and knobs. You punch-up a chassis with enough holes, but generous on size, spacing, and tie-points. Build the amp but don't NASA-wrap the wires and leads. Stick em in the hole and solder. Play. with the loose construction you can quickly change a part and try some more.
Great advice....
I've built 2 of those boards now and they have provided me with a great experimental platform....
Most of what I'm doing can be done around some fixed values.....and within the confines of an experimental amp chassis (as PRR has suggested), without having to build a fully functional "breadboard"...
I'm obsessed with pushing a bunch of envelopes before I'm done,,,and that's why I did it
I bailed very early in this thread because I knew you were going in over my head,,,BUT I'm glad I initially posted, because it has reminded me to stop back and read all of this GREAT THREAD!!
I will eventually print this out and try to digest more of the design principles that have been presented....
BUT, with that said,,,I have learned more about "designing" the sound in my head by swapping values and listening HARD,,,and then reading,,,,and swapping more values, and then listening harder...(read, swap, play, listen, repeat, repeat, repeat)
As I've read your thread, there have been times where I've thought to myself:
"Well, you CAN design an amp from scratch, on paper,,,that's very cool, but you're not going to have any idea what it's gonna SOUND like....etc., etc.."
If you aim for "perfection in design" and concentrate very hard on hitting all of the tubes right in the middle of their operating points,,,you could be building a very generic, possibly sterile sounding amp,,,,that lacks some of the personality of a really poorly designed stage or 2 :l2:
Thank you for pressing on,,,,,,and BIG THANKS! to all of the posters who have continued to answer ALL of your questions...it really has made for one "interesting" read.....these guys are GOOD!
Once you're done designing, the resulting amp "could" end up on that breadboard,,,at least for a couple minutes :icon_biggrin:
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Thank you for pressing on,,,,,,and BIG THANKS! to all of the posters who have continued to answer ALL of your questions...it really has made for one "interesting" read.....these guys are GOOD!
Once you're done designing, the resulting amp "could" end up on that breadboard,,,at least for a couple minutes :icon_biggrin:
Well put SG (as always). I especially like the last line. How true, how true it is! :l2:
ps- I've had all of ONE design that actually stayed entirely the same as after I first designed & built it and that I didn't touch or tweak a single time from begining to end! But, it was my smallest output one therefore one of the simpler ones w/out too much going on. Makes sense right?! I will say though on a few others that I thought it was fine/perfect a few times (from design to build)...until I later learned something new or better, then I went back and made the necessary changes (as I told you before already). :laugh:
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> onionskin ...over the plot, then drew lines with a sharp 6H pencil, a straight edge
Obviously possible. And done for some fields (pump curves come to mind).
What is interesting is that I can not off-hand recall any *complete* Old-Days write-up of such a process for Resistance Coupled Amplifiers. Vague incomplete pictures to decorate an article. Not a complete design.
OTOH, Class C amplifiers basically were designed this way. Complete process described in the literature and graphs were sometimes plenty big enough to plot on (and available in sheets rather than bound in a book).
I have to go look for snow, but later I may look for Pullen, who might have done a write-up to support his Conductance Curves approach.
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My current approach is to take pdf's of the tube curves and convert to tiffs (Black and white), import as raster drawings into CAD, store in model space. then over draw the plots in paperspace. I can rescale the rasters, make them transparent. Just a high tech way of placing tracing paper or onion skin over a plot.
Sounds interesting. Would love to see some of your work.
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Obvious question time... or really I need a confirmation that I am understanding something correctly.
People frequently use the word "hot" when describing how a tube is biased. I have never really understood what this meant. Sometimes, I thought it meant that the Plate Voltage was high. Sometimes I thought that it meant that the Plate Current was high. What I just realized is that it is a combination of the two (at least I think I'm right). Here is what I mean... P = I * V. And P is measured in Watts which is basically heat. Therefore, a tube with a plate voltage of 150V can be biased hotter than a tube with plate voltage of 300 depending upon how much current it is drawing. For example,
P = 300V * 0.5mA = 1.5W
P = 150V * 2.0mA = 3.0W
So, is my understanding of the term "hot" with regard to biasing a tube correct?
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If I put my hand in "hot water" what is its absolute temperature?
It's impossible to tell, and everyone's definition of "hot" could be different. Your definition of "hot" can change depending on environmental factors.
For most amps, you can't vary the plate voltage of the output tubes. Even when you can (VVR, powerscaling, etc), you set the plate voltage to a given (usually maximum) level and bias the tubes at that set voltage.
So when I say "biased hot" I mean a lot of plate current. That also implies more power dissipation, because more current * same voltage = more power.
Plate dissipation also strictly means the power being "wasted" by the tube as the plate radiates heats to the tube envelope which then heats the air. So more current yields more dissipation which yields more actual heat in the air.
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This also can be described as the biasing point of anything towards the side of grid current limiting/saturation from the center biasing point. The more, the "more hotter" if you will.
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Next obvious question... I'm confused by the symbols used for the input networks. The diagram is pretty standard for most Fender and Marshall schematics I have seen.
Which of the following is correct for Input 1?
- R2 is the Grid Stopper and R3 is the Grid Leak resistor.
- R1 and R2 in parallel is the Grid Stopper because Input 2's jack is shorted. R3 is the Grid Leak resistor.
- Neither of the above
Which of the following is correct for Input 2?
- R1 is the Grid Stopper, and R2 and R3 in series is the Grid Leak resistor
- R1 is the Grid Stopper and R2 is the Grid Leak resistor because the Input 1 jack is shorted to ground
- Neither of the above
Sometimes its the really simple things...
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Next obvious question... I'm confused by the symbols used for the input networks. The diagram is pretty standard for most Fender and Marshall schematics I have seen.
Which of the following is correct for Input 1?
- R2 is the Grid Stopper and R3 is the Grid Leak resistor.
- R1 and R2 in parallel is the Grid Stopper because Input 2's jack is shorted. R3 is the Grid Leak resistor. Correct
- Neither of the above
Which of the following is correct for Input 2?
- R1 is the Grid Stopper, and R2 and R3 in series is the Grid Leak resistor
- R1 is the Grid Stopper and R2 is the Grid Leak resistor because the Input 1 jack is shorted to ground Correct
- Neither of the above
Sometimes its the really simple things...
Look at page 6 of this pdf to see the jacks explained another way that reinforces what you just said...
http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf (http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf)
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Look at page 6 of this pdf to see the jacks explained another way that reinforces what you just said...
http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf (http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf)
Excellent! That clears it up quite nicely!
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Look at page 6 of this pdf to see the jacks explained another way that reinforces what you just said...
http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf (http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf)
Want more good advice...
While you're there,,,,download and save everything you see....before he starts up with his site host again and they give him the boot :icon_biggrin:
Beautiful schematics + immaculately documented build guides + visually stunning layouts = some of the best "amp porn" on the internet IMHO
I can't count how many times I have referenced something from that site.....because you can be 100% sure that it's 100% accurate.
He wouldn't have it any other way.
http://home.comcast.net/~seluckey/amps/index.htm (http://home.comcast.net/~seluckey/amps/index.htm)
Click on each amp for ALL the goods
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Now for the followup question... What about when the inputs are jumpered?
I've highlighted how I think the signal will flow. If I plug the guitar into the Hi input of the lower triode the grid stopper will be R2 and the grid leak resistor will be R3. This is because the jumper cable will open the Lo jack so the R1 and R2 cannot be in parallel.
Then, signal will flow through the R1 to the Hi input of the upper triode. The grid stopper for the upper triode will be R1 plus R4 and R5 in parallel and the grid leak will be R6. Or... will it be R6 in parallel with R1, R2 and R3 in series?
Or will it be some other combination. I'm trying to think of a way that might emulate a jumpered input using only one jack. I'm probably getting ahead of myself and should stick to a simple gain stage for now, but while I was thinking about it, I thought I would ask.
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What about when the inputs are jumpered?
You highlighted it correctly.
I'm trying to think of a way that might emulate a jumpered input using only one jack.
Here's how I did it. Look at this schematic. The preamp is a Marshall 1987 Plexi, probably the most jumpered input amp of all time...
http://home.comcast.net/~seluckey/amps/november/november.pdf (http://home.comcast.net/~seluckey/amps/november/november.pdf)
So now it's my turn. What do you think happens when you plug a guitar into the hi input and plug another guitar into the lo input? That's the original real reason for multiple inputs on an amp. Talking about a single channel here.
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So now it's my turn. What do you think happens when you plug a guitar into the hi input and plug another guitar into the lo input? That's the original real reason for multiple inputs on an amp. Talking about a single channel here.
Let's see (I don't have my computer with software to draw this out)... The Hi jack will have 68k grid stopper with 1M grid leak. For Lo jack there will be a 68k grid stopper and a 1,068k grid leak. But... how does the Lo jack get grounded? I guess it goes through the 1M resistor.
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The Hi jack will have 68k grid stopper with 1M grid leak. For Lo jack there will be a 68k grid stopper and a 1,068k grid leak. But... how does the Lo jack get grounded? I guess it goes through the 1M resistor.
That's correct. The hi/lo function is gone in this situation and each guitar will have essentially the same volume. The 68K resistors now serve as mixing/isolation resistors to prevent the volume/tone controls of one guitar from affecting the other guitar.
Sounds like you have a good handle on the versatile Switchcraft 12A jacks. For even more switching options take a look at the Cliff jacks. The have a switch for the tip and another switch for the sleeve.
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Yet another fundamental question that should be obvious but I'm not sure about...
The diagram below is the first two valves in the normal channel of a JTM45 including the volume control. Assuming the gain of the first valve is 60, does this mean that if the volume control was all the way up and the signal coming from the guitar was 0.1V that the voltage going into the second valve would be 6.0V or will this voltage be limited because of the valve cutoff? Is this clipping and one of thes cause of distortion?
Similarly, if the second valve has also has a gain of 60 (it probably wouldn't because its not bypassed) and it has an input voltage of 4.0V, does that mean that the output voltage will be 240V? That doesn't seem to make sense to me because a 12AX7 can't swing that much voltage.
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Yet another fundamental question that should be obvious but I'm not sure about...
The diagram below is the first two valves in the normal channel of a JTM45 including the volume control. Assuming the gain of the first valve is 60, does this mean that if the volume control was all the way up and the signal coming from the guitar was 0.1V that the voltage going into the second valve would be 6.0V or will this voltage be limited because of the valve cutoff? Is this clipping and one of thes cause of distortion?
Similarly, if the second valve has also has a gain of 60 (it probably wouldn't because its not bypassed) and it has an input voltage of 4.0V, does that mean that the output voltage will be 240V? That doesn't seem to make sense to me because a 12AX7 can't swing that much voltage.
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Assuming the gain of the first valve is 60, does this mean that if the volume control was all the way up and the signal coming from the guitar was 0.1V that the voltage going into the second valve would be 6.0V ...
Yes.
... or will this voltage be limited because of the valve cutoff? ...
Why would the output voltage of the 1st stage be limited? The bias is very likely more than 0.1vdc, implying it can handle much more than 0.1v of input signal. The supply voltage plus some allowance is very much greater than 6v, so the output is unlikely to be limited. This stage should amplify cleanly.
Similarly, if the second valve has also has a gain of 60 (it probably wouldn't because its not bypassed) and it has an input voltage of 4.0V, does that mean that the output voltage will be 240V? That doesn't seem to make sense to me because a 12AX7 can't swing that much voltage.
You might ask youself, "4v RMS or 4v peak?" 4v RMS is a bigger peak voltage.
Regardless, look at a 12AX7 data sheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/093/1/12AX7A.pdf). The gridlines generally only go up to -5v, and that's almost at cutoff with 400v on the plate. Therefore that size of input signal is "too big" and likely to guarantee distortion in that 2nd stage.
... does that mean that the output voltage will be 240V? That doesn't seem to make sense to me because a 12AX7 can't swing that much voltage.
The amount of output voltage a 12AX7 stage can swing depends on its loading and supply voltage. 240v of output is perhaps not impossible, but is improbable in most cases. You should plot the loadline to know for sure.
... the signal coming from the guitar was 0.1V that the voltage going into the second valve would be 6.0V or will this voltage be limited because of the valve cutoff? Is this clipping and one of thes cause of distortion? ...
You mean the 6v from the 1st stage causing valve cutoff in the 2nd stage? Then yes the result will be distortion... in the 2nd stage.
Usually, I see "clipping" referenced when a gain stage's output signal bumps into a supply voltage limit.
If you apply a 6v signal to the 2nd stage, when the input signal swings negative the output voltage will bump into the supply voltage, and clip. It will have been distorted before it even reaches clipping by virtue of the increasingly-close spacing of the grid lines in that direction.
When the input signal swings positive, it will distort more abruptly as the grid reaches an absolute voltage of 0v, as the grid stops being a roughly infinite impedance. The sudden drop in apparent resistance of the grid circuit then also loads the 1st stage more heavily, clamps its signal to a degree and may cause the 1st stage to distort.
That last bit is part of the subtlty of Merlin's diode & d.c. coupling of the split-load inverter, forcing the preceding stage to distort before being able to make the inverter crap out. But resistor selection is very dependent on the desired signal levels at the inverter, and why I said you cannot just copy parts values & also why it is probably best tweaked on a breadboard or in operation.
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Let me try again with a loadline to see if I'm understanding. Assuming the operating point for both valves is the same (I know the values don't match JTM45, but it shouldn't matter for my questions):
Vgq = -1.5V
Vpq = 170V
Ipq = 0.725mA
Based on the graph below, if a 0.2V peak signal is input to Valve 1, the grid voltage will swing from -1.3V to -1.7V. Or, do I have this wrong and grid voltage would swing from -1.4V to -1.6V for 0.2V input?
Regardless of the input voltage, say grid voltage swings from -1.3V to -1.7V. Looking at the graph I can see the anode voltage will swing from 160V to 178V. Therefore the output voltage is 18V (or is it 9V?).
I'm guessing the volume potentiometer is used to divide the voltage so that only a fraction of the 18V is input to the second tube. Assume the volume control is all the way up. Since the tube can only go 1.5V positive and 3.1V negative, then either the supply voltage or the grid voltage limit will be hit. Va will go from 75V to 240V. Does this mean that the second tube has 165V (240V - 75V) output?
Extending this to a third gain stage not on the diagram with the same operating points and no volume control. Does the same thing happen again... more distortion and the output voltage limited to 165V?
By the way, I do plan on breadboarding this stuff and tuning values, but I still think this is a good concept for me to understand before attempting to do that.
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Vgq = -1.5V
Vpq = 170V
Ipq = 0.725mA
Based on the graph below, if a 0.2V peak signal is input to Valve 1, the grid voltage will swing from -1.3V to -1.7V. Or, do I have this wrong and grid voltage would swing from -1.4V to -1.6V for 0.2V input?
Yes, from -1.3v to -1.7v.
You spec'd 0.2v peak. If you'd said 0.2v peak-to-peak, then the grid would swing from -1.4V to -1.6V. If you'd said 0.2v RMS, then the grid would swing from -1.217v to -1.783v.
... Looking at the graph I can see the anode voltage will swing from 160V to 178V. Therefore the output voltage is 18V (or is it 9V?). ...
Now you'e trying to spec peak-to-peak output. So it's 9v peak output.
As a check, calculate gain in like units: 9v peak / 0.2v peak = 45 which is probably reasonable for a 12AX7.
18v / 0.2v = 90, which you will almost never see without an active load for a 12AX7.
I'm guessing the volume potentiometer is used to divide the voltage so that only a fraction of the 18V is input to the second tube.
Yes; further, it is to allow a variety of pickups to drive the output section to full output power. Your 0.2v example is unrealistic, as generally hot pickups will put out something in the range of 100mV, and even that may consist of tall peaks and low average signal level.
A volume pot is almost universally a log pot (or audio taper) which has 10% of total resistance (and therefore 10% voltage) halfway up. So your 9v peak output from the 1st stage will be 0.9v peak input to the 2nd stage. Really 0.636v peak, because the input is likely more like 0.1v RMS for 6.36v peak output.
... Since the tube can only go 1.5V positive and 3.1V negative ...
Be careful... It can go from -1.5v to 0v (+1.5v) and from -1.5v to ~-3.1v (-1.6v). So you have equal range of input signal, and should call that "1.5v peak input."
... Since the tube can only go 1.5V positive and 3.1V negative, then either the supply voltage or the grid voltage limit will be hit. Va will go from 75V to 240V. Does this mean that the second tube has 165V (240V - 75V) output?
165v peak-to-peak. Really, +70v/-95 as the output swing is not symmetrical in either direction. Your meter will read that as something more than 49.5v RMS but less than 67v RMS. Exactly what it will say depends on how much of which distortion components are present.
Either way, you should see that the output signal is distorted even before you hit the hard limits.
Extending this to a third gain stage not on the diagram with the same operating points and no volume control. Does the same thing happen again... more distortion and the output voltage limited to 165V?
The volume control is some amount of loss between 1st & 2nd stages. You'll probably have a tone stack also between either 1st & 2nd stages or 2nd & 3rd stages. Any analysis of what a gain stage will do is dependent on the losses imposed and the resulting (reduced) input signal.
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The volume control is some amount of loss between 1st & 2nd stages. You'll probably have a tone stack also between either 1st & 2nd stages or 2nd & 3rd stages. Any analysis of what a gain stage will do is dependent on the losses imposed and the resulting (reduced) input signal.
In absence of a tone stack there will be somewhere between 49.5v RMS and 67v RMS hitting V3. Based on that, I can assume similar performance from V3. That is, the limits will be hit (HARD) and there will be between 49.5v RMS and 67v RMS output with even more distortion.
In reality, the JTM45 uses a cathode follower then a tone stack so gain will be very different but I wanted to be sure I was understanding the simple case first.
Your 0.2v example is unrealistic, as generally hot pickups will put out something in the range of 100mV, and even that may consist of tall peaks and low average signal level.
I thought I read somewhere that humbuckers will put out 500mV with some peaking at 1V. I was trying to figure out how you could get any clean tone at all.
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I thought I read somewhere that humbuckers will put out 500mV with some peaking at 1V. I was trying to figure out how you could get any clean tone at all.
Hook a meter to your guitar cable and bang out some chords. YOU tell ME what voltage you get (you may not know peaks unless your meter has a peak-hold function or you have an o'scope).
I put my money where my mouth is, and with my Tele (stock-ish Fralin pickups) and *banging* the strings with both pickups on, I got 99mV RMS max. This means over the measurement period, my meter recorded a maximum signal of 99mV RMS. When recording and playing normally, the maximum signal level recorded was 38.4mV RMS.
Sure, humbuckers will be hotter and some will be a lot hotter. But you also know from experience most amps will distort somewhere between 1/2 & 3/4 volume.
In absence of a tone stack there will be somewhere between 49.5v RMS and 67v RMS hitting V3. Based on that, I can assume similar performance from V3. That is, the limits will be hit (HARD) and there will be between 49.5v RMS and 67v RMS output with even more distortion.
In reality, the JTM45 uses a cathode follower then a tone stack so gain will be very different but I wanted to be sure I was understanding the simple case first.
In reality, the JTM45 did not use the supply voltage you assumed, and had a cathode resistor that may have resulted in a different bias voltage than your example. V2 also had no bypass cap, which causes local negative feedback and cuts gain by around half. V3 is a cathode follower, and its output is less than its input, and it then feeds a tone circuit which will introduce some loss.
You also do not calculate gains from input jack to output, but work from the output stage, determine needed drive signals and preceding stage gains, and figure the sensitivity at the input jack. You might also assume that you want your target sensitivity (the signal level smack full output power) to happen at half-volume, so you allow for turn-down for hot pickups and turn-up for weak pickups, so all can allow your amp to deliver all its output power.
You gotta design from speaker backwards to the input jack, then after all done perform your analysis on each stage and determine if there are bottlenecks (really, you should be fixing bottlenecks along the way). Maybe you'll decide that you want some stages to always run clean. Maybe you want every stage in the amp (including the output tubes) to break into distortion at approximately the same moment...
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I know my examples were not representative of the JTM45. I was just trying to understand how the input voltages are amplified and what happens when you present a gain stage with a very large input voltage. To be honest, I'm still not entirely sure my assumption is correct... that is, pretty much after the second gain stage you are going to get the same output voltage over and over again assuming all stages have the same operating points and there are no volume or tone controls in the way.
This probably wouldn't sound good (to me anyway) and would just be layers upon layers of distortion. But, I wanted to be sure I am understanding what would happen.
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You wouldn't hear it that way... because well before your volume (input signal) got to that point, the output stage would already be distorting, then the phase inverter, then...
You probably wouldn't have kept cranking it up to get to the point your example shows, and if you did you cranked up because you were enjoying the sound.
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I think the next thing for me to figure out is what James Marchant meant when he said,
The 60s (JTM1) has an input stage that has a cathode r/rc network designed to imitate the frequency response of a jumpered input with a little more bright than normal volume level, then gain then anode follower then the tone stack from a JTM45 - 56k slope and 220p treble. Only the treble control is fitted as a tone control. Then there is another anode follower to the PI. There is about 6dB of NFB into the grid of the PI.
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I know the JTM1 has only 3 tubes - two 12AX7's and one 12AU7. I know that it uses a cathodyne PI so that leaves 3 triodes for the gain stages. This is where the description diverges from the actual JTM45 (I think). The first stage is probably just a typical gain stage like the JTM45 with a volume pot between V1 and V2. However, the JTM45 has a second gain stage coupled to a cathode follower, then the tone stack feeding the phase inverter. So, even though the amp sounds like a mini-JTM, the topology of the preamp is different.
Either that or this is where I diverge from the JTM1. There are a number of interesting input networks documented in the books I have been reading. One of them might be fun to try... it has two inputs high gain and low gain. Instead of using a voltage divider for the low gain input, it adds a gain stage for the high gain input using some nifty input wiring. This would take care of one 12AX7.
Another idea is to have one input that would bridge both halves of the first 12AX7 (like a jumpered plexi). Each triode would have its own volume control that feed into a common second gain stage, cathode follower, tone stack then phase inverter.
Of course I'm just throwing out stuff that sounds like I know what I'm talking about. Perhaps I should examine why the JTM1 would have a final anode follower before the PI first.
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I think the next thing for me to figure out is what James Marchant meant when he said ...
He didn't build a JTM45 with a 12AU7 output stage. He built a 12AU7-output amp with a JTM45 flavor.
He says, " an input stage ... then gain then anode follower then the tone stack ... Then there is another anode follower to the PI."
I think you should read that as "common-cathode gain stage" (the kind you already know about), Volume control (cause it's the only knob aside from the single Tone control), then common-cathode gain stage, the tone circuit, then common-cathode gain stage, then split-load phase inverter. We already know from previous discussion that he used the split-load inverter.
WTF is an "anode follower"? Well, in this case, he almost certainly means a gain stage with an output at the plate. There is a circuit in tube lore called an "anode follower" or see-saw circuit, but it has nothing to do with what is going on in this amp (and usually implies the use of 2 triodes).
"The 60s (JTM1) has an input stage that has a cathode r/rc network designed to imitate the frequency response of a jumpered input with a little more bright than normal volume level ...
This is easier than you think. The normal input (with the 330uF cap) is just insane mud. I know from having an early 70's 50w Marshall. The Brilliant channel is too ice-picky for normal use. Jumpered input allows you to use the volumes to get a balanced sound.
So I'd interpret his comment as having less bass than the 330uF cap, but more bass than the 0.68uF/2.7kΩ setup. Subjectively tilted a bit closer to brighter than darker.
I'd read the tone circuit as being a full-up Marshall tone circuit, but with fixed resistors in place of the Bass and Midrange pots. That way you control the overall sound with the single (Treble) Tone pot.
I think the next thing for me to figure out is what James Marchant meant when he said,
... There is about 6dB of NFB into the grid of the PI.
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If you want an exact clone of the amp, you'll need some feedback from speaker output to split-load grid. 6dB worth.
... There are a number of interesting input networks documented in the books I have been reading. One of them might be fun to try... it has two inputs high gain and low gain. ...
If it were my amp, I'd just stick with the basic setup.
The folks I've known who had a JCM800 (the input setup that adds a gain stage to the High jack) never use the low gain setting. Of course, they also got the amp because it was master volume and they wanted distortion.
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This is easier than you think. The normal input (with the 330uF cap) is just insane mud. I know from having an early 70's 50w Marshall. The Brilliant channel is too ice-picky for normal use. Jumpered input allows you to use the volumes to get a balanced sound.
So I'd interpret his comment as having less bass than the 330uF cap, but more bass than the 0.68uF/2.7kΩ setup. Subjectively tilted a bit closer to brighter than darker.
That's true of an early 70's Marshall, plus the two input stages used different coupling caps. But in the JTM 45 the only difference between the channels is a treble bleed cap on the "High Treble" channels mix resistor, and a bright cap on its volume control. So the bass response is the same for the two channels, but some treble lift on one of them.
However I agree - he's using the cathode network of the first stage to shape the response a little, so it's brighter than "Normal" but not as much as the "High Treble". Maybe something a little more complex to give a stepped response, rather than just a smaller bypass cap ?
Edit - The designer says "... a cathode r/rc network...". So I think that's it - series resistor under the bypass cap to give a bit of hf shelving, to emulate the bleed/bypass caps. Don't know how you'd calculate that, personally I'd temporarily tack in a 10-20K trimmer, tune by ear then remove and measure!
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That's true of an early 70's Marshall, plus the two input stages used different coupling caps. But in the JTM 45 the only difference between the channels is a treble bleed cap on the "High Treble" channels mix resistor, and a bright cap on its volume control. ...
Good point. I mainly took the cue from the designer that it was accomplished with cathode cap/resistor tone shaping.
... I think that's it - series resistor under the bypass cap to give a bit of hf shelving, to emulate the bleed/bypass caps.
Good call. I'd tend to believe that, and agree that you'd have to tune by ear just like the designer did. You calculate to pick good starting values for the bypass cap, then tune the shelving resistance by ear.
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> Assuming the gain of the first valve is 60, does this mean that if the volume control was all the way up and the signal coming from the guitar was 0.1V that the voltage going into the second valve would be 6.0V
Yes. And the 2nd stage would distort and sound like %$#@! If this were a Hi-Fi, you would TURN IT DOWN. (Guitar tone-creation sometimes uses %$#@! for flavor.)
> if the second valve has also has a gain of 60 (it probably wouldn't because its not bypassed)
Pencil-check ordinary stages un-bypassed as "half".
The Fender 12AX7 100K 1.5K stage is more like Gv=50; unbypassed maybe 20-30.
> and it has an input voltage of 4.0V, does that mean that the output voltage will be 240V?
It will try. It can't (with normal 270V-350V B+ supply). Like a 7-foot truck in a 6-foot garage, it's gonna have damage. Actually for your *assumed* values, like a 6-foot garage swallowing a _60_foot truck. It's gonna be nearly all damage.
> That doesn't seem to make sense
With all knobs full-up, 20mV (5mV-50mV) at the Inst jack should just-touch 50 Watts (1W, whatever) at the speaker jack.
AND for nearly all "normal" plucking, for any nearly-clean sound, you will Turn Down one or more gain-knobs.
You should be able to put-in 500mV, turn-down, and get your Full Power out cleanly.
Or not, and take it dirty.
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I've been working on a schematic as a starting point. While I will have the schematic in hand, I plan on building this "bread board" style with lots of room so I can clip in different bypass caps and 10K pots for cathode resistors (should work fine for these tubes).
I will post it soon but work has been very busy.
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One curious point about this amp from the designers description is: "... There is about 6dB of NFB into the grid of the PI. "
I've never seen negative feedback taken to a cathodyne grid - it's usually (in Fenders at least) to the cathode of the previous gain stage.
So would he be forming a voltage divider from the OPT and then a mix resistor into the grid, along a mix resistor coming from the pervious stage?
Why do that, rather than go the Fender feedback route?
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One curious point about this amp from the designers description is: "... There is about 6dB of NFB into the grid of the PI. "
I've never seen negative feedback taken to a cathodyne grid - it's usually (in Fenders at least) to the cathode of the previous gain stage.
Yep, that's what I've seen, too.
FWIW, I don't know exactly how this amp is set up. So everything from here on is a guess...
So would he be forming a voltage divider from the OPT and then a mix resistor into the grid, along a mix resistor coming from the pervious stage?
You're always gonna have a series resistor coming from the OT secondary for any feedback loop. If the grid in question has a resistor from grid to ground (or from grid to the top of a cathode load resistor), then that existing resistor can be the 2nd half of a voltage divider network. So the only trick involved would be sizing the series resistor from the OT to a value that provides 6dB feedback when connected.
Or maybe he looks at this phase inverter the way I do, in that the gain stage ahead of the split-load is always needed, and whose grid is basically the input port to the power amp. And since the split-load has no gain of its own, any feedback applied would have to go to either this pre-gain stage, or would steal for the output tube's ability to amplify.
That said, 6dB of voltage feedback is a halving of gain. The 12AU7 might have a gain of 12-16 or so, depending on the exact operating point used. Maybe this was deemed a bit hot for the amount of preamp gain stages used, as typical power tubes might have gains more like 8-10. If the feedback was applied to the split-load input, it couldn't reduce the split-load's gain any further (which is already less than 1) so it's really allowing the 12AU7 to accept a bigger input signal.
So I guess there are arguments for either connection.
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I guess if negative feedback is going to the PI grid then it will reduce (halve, by his account) the signal into the PI. There's probably still enough signal from the pre-PI stage to fully drive the 12AU7. It's the kind of amp that will be using a certain amount of output stage distortion so he must have plenty of level there. Alternatively you could be right in that he's thinking of the pre-PI stage as an integral part of the PI and the feedback is returning there. Even so it's odd to take it to the grid.
I've been playing with low power amps, SE and PP, for some time now. My current PP low power amp is pretty much the same topology as this little Marshall - same number of gain stages, tone and volume controls in the same place, cathodyne PI. Major difference is I'm running a pair of EL91's for about 4W output, with a VVR circuit to drop the PI and output stage B+ for domestic friendly SPL when required. To me these little power pentodes sound better than running the triode halves of a 12AU7 as a PP output stage. I'm not using any negative feedback at the moment, might be interesting to try his arrangement - and also the usual Fender feedback method on their cathodyne PI amps.
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I guess if negative feedback is going to the PI grid then it will reduce (halve, by his account) the signal into the PI.
That's not what it means.
Let's pretend our output stage (with fictitious tube "X51") has a power output of 1w with an 8Ω load. That implies √(1w*8Ω) = ~2.83v RMS at the speaker output jack.
Let's also say our X51 output stage has a bias of -13v, and so a 13v peak signal (9.19v RMS) is needed to drive it to full output power. Our 12AX7 split-load inverter has a gain of 0.97, and the 12AX7 pre-gain stage has a gain of 50.
Normally, the X51 will convert the 9.19v RMS input signal it receives into a much larger voltage swing (amplification to the plate), while the OT convert the big-voltage small-current signal at its primary to a small-voltage big-current signal at its secondary. Without even knowing the gain of the X51 or the step-down ratio of the OT, we can see that the overall gain through this system is 9.19v RMS (X51 input) -> 2.83v RMS (speaker terminals, or a gain of 0.31.
The split load then is more loss, which tells us we need 9.19v RMS/0.97 = 9.47v RMS at the split load input to get full power output. If the feedback is returned to the split-load input grid, then 6dB of negative feedback means you need twice the input signal, or 18.9v RMS at the split load grid for full output.
The end result is the same if the feedback is returned to the pre-gain stage, but now this stage is influenced by the feedback and becomes the new point of reference for required input signal level. Originally, this stage had a gain of 50 and it would take 0.184v RMS to deliver the 9.19v RMS needed by the split-load stage. After closing the feedback loop, it would take 0.368v RMS at the pre-gain stage to deliver the full 2.83v RMS to the speaker terminals.
Shortening the feedback loop (reducing the number of stages inside the loop) reduces the number of reactive components in the loop, and allows more feedback before the loop turn the amp into an oscillator. Anyway, 6dB of feedback is petty modest, and unlikely to cause a problem.
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The normal input (with the 330uF cap) is just insane mud... So I'd interpret his comment as having less bass than the 330uF cap, but more bass than the 0.68uF/2.7kΩ setup. Subjectively tilted a bit closer to brighter than darker.
I'd read the tone circuit as being a full-up Marshall tone circuit, but with fixed resistors in place of the Bass and Midrange pots. That way you control the overall sound with the single (Treble) Tone pot.
I put in some place holder values for the bypass cap on the first gain stage and the fixed resistors that replace the pots in the tone controls. I looked at the JTM45 tone stack for the capaictor values. I'm sure these are great places to play with different values on the breadboard. I can clip in bypass caps but I may actually use pots on the breadboard, then measure values I like for the fixed mid and bass resistors.
If you want an exact clone of the amp, you'll need some feedback from speaker output to split-load grid. 6dB worth.
Just when I think I'm beginning to understand you throw feedback at me. :icon_biggrin: Time to read the chapter on Feedback.
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Thanks for the feedback thoughts HBP. Nicely written.
I have plenty more gain than necessary so I think some negative feedback wouldn't go amiss. Might also tighten the sound up a little which would be useful.
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Just to keep this thread going, Mojotone have a 1 watt amp on the market using a 12BH7.
http://www.mojotone.com/Marketplace/Mojotone-Studio-One-Amplifier-Head#.U2MOgp1-9Fo (http://www.mojotone.com/Marketplace/Mojotone-Studio-One-Amplifier-Head#.U2MOgp1-9Fo)
It looks like a tweed front end for the 'clean'channel', and a JCM800 style cascaded style preamp for the 'dirty' channel.
I took some screen shots from the video demo.
They also sell an OT for the 12BH7.
Rob