One watt amp

[dbishopbliss]

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. 

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.

[dbishopbliss]

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.

[PRR]

> 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).

[dbishopbliss]

> 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.

[dbishopbliss]

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 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



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.
  


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!

[jojokeo]

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.

[dbishopbliss]

-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. 

[jojokeo]

"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.
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.

[PRR]

> 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/

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?

[dbishopbliss]

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?

[John]

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.

[PRR]

> 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.)

[jojokeo]

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.

[dbishopbliss]

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. 

[sluckey]

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 n^th degree of design.

[clyde]

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.   

[dbishopbliss]

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).

R_L = 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.

[jojokeo]

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.

[dbishopbliss]

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.

[dbishopbliss]

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.

[dbishopbliss]

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).

[HotBluePlates]

... 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.   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 and Merlin (d.c. and a.c. 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.

[HotBluePlates]

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).

[PRR]

> 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).

[dbishopbliss]

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.

[jojokeo]

...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. ;)

[HotBluePlates]

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.

[dbishopbliss]

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.

[PRR]

> 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.

[SILVERGUN]

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 

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 

[jojokeo]

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 

Well put SG (as always). I especially like the last line. How true, how true it is! 
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). 

[PRR]

> 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.

[sluckey]

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.

[dbishopbliss]

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?

[HotBluePlates]

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.

[jojokeo]

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.

[dbishopbliss]

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...

[sluckey]

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

[dbishopbliss]

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

Excellent!  That clears it up quite nicely!

[SILVERGUN]

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

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  

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
Click on each amp for ALL the goods

[dbishopbliss]

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.

[sluckey]

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

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.

[dbishopbliss]

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.

[sluckey]

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.

[dbishopbliss]

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.

[dbishopbliss]

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.

[HotBluePlates]

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. 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.

[dbishopbliss]

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. 

[HotBluePlates]

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.

[dbishopbliss]

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.