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Offline HotBluePlates

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Re: One watt amp
« Reply #50 on: January 20, 2014, 09:11:02 am »
So are all the questions answered?  :laugh:

Offline dbishopbliss

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Re: One watt amp
« Reply #51 on: January 20, 2014, 09:44:53 am »
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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Offline dbishopbliss

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Re: One watt amp
« Reply #52 on: January 21, 2014, 02:38:26 pm »
Look at the bottom graph on Valve Wizard's explanation of push-pull. 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?

« Last Edit: January 21, 2014, 03:06:57 pm by dbishopbliss »
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Offline HotBluePlates

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Re: One watt amp
« Reply #53 on: January 21, 2014, 04:59:25 pm »
... 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).
« Last Edit: January 21, 2014, 10:31:22 pm by HotBluePlates »

Offline dbishopbliss

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Re: One watt amp
« Reply #54 on: January 21, 2014, 09:52:03 pm »
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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Offline dbishopbliss

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Re: One watt amp
« Reply #55 on: January 21, 2014, 10:43:54 pm »
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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Offline HotBluePlates

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Re: One watt amp
« Reply #56 on: January 21, 2014, 10:57:27 pm »
...  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.

Offline dbishopbliss

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Re: One watt amp
« Reply #57 on: January 22, 2014, 07:50:32 am »

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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Offline dbishopbliss

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Re: One watt amp
« Reply #58 on: January 22, 2014, 07:55:02 am »
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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Offline HotBluePlates

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Re: One watt amp
« Reply #59 on: January 22, 2014, 04:00:15 pm »
    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:
    • Idle Current: 3.5mA
    ...
    • 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.

    Offline HotBluePlates

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    Re: One watt amp
    « Reply #60 on: January 22, 2014, 04:05:34 pm »
    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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    Re: One watt amp
    « Reply #61 on: January 23, 2014, 11:34:11 pm »
    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, 12BH7 and ECC99 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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #62 on: January 24, 2014, 10:58:53 am »
    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.
    « Last Edit: January 24, 2014, 01:47:58 pm by dbishopbliss »
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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #63 on: January 24, 2014, 03:19:42 pm »
    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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    Re: One watt amp
    « Reply #64 on: January 24, 2014, 04:24:49 pm »
    > 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.)


    Offline HotBluePlates

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    Re: One watt amp
    « Reply #65 on: January 25, 2014, 09:38:35 am »
    > 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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    Re: One watt amp
    « Reply #66 on: January 26, 2014, 06:49:00 pm »
    Not a JTM1 schematic but Traynors amp with switchable output:

    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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    Re: One watt amp
    « Reply #67 on: January 27, 2014, 01:07:31 am »
    ... but Traynors amp with switchable output:

    Traynor Darkhorse

    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.

    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.
    ...
    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).

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #68 on: January 27, 2014, 07:42:46 am »
    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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    Offline VMS

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    Re: One watt amp
    « Reply #69 on: January 27, 2014, 03:34:40 pm »
    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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #70 on: January 27, 2014, 04:24:02 pm »
    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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    Offline Willabe

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    Re: One watt amp
    « Reply #71 on: January 27, 2014, 09:38:32 pm »
    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:

    Offline HotBluePlates

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    Re: One watt amp
    « Reply #72 on: January 27, 2014, 10:04:42 pm »
    ... 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.

    Offline dbishopbliss

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    Phase Inverter Time
    « Reply #73 on: January 28, 2014, 04:34:45 pm »
    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?
    « Last Edit: January 28, 2014, 04:40:47 pm by dbishopbliss »
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    Offline sluckey

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    Re: One watt amp
    « Reply #74 on: January 28, 2014, 05:16:17 pm »
    You may need some resistors on the 12AU7 grids.
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    Offline PRR

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    Re: One watt amp
    « Reply #75 on: January 28, 2014, 10:09:47 pm »
    > 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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #76 on: January 29, 2014, 10:09:12 am »
    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 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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    Offline sluckey

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    Re: One watt amp
    « Reply #77 on: January 29, 2014, 11:01:51 am »
    Quote
    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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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #78 on: January 29, 2014, 12:19:53 pm »
    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 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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    Offline sluckey

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    Re: One watt amp
    « Reply #79 on: January 29, 2014, 12:49:25 pm »
    Quote
    please tell me where in this datasheet I can find that value
    Bottom of page 1. See attached pic.

    Quote
    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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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #80 on: January 29, 2014, 01:48:56 pm »
    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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    Offline sluckey

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    Re: One watt amp
    « Reply #81 on: January 29, 2014, 02:11:34 pm »
    Quote
    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
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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #82 on: January 29, 2014, 02:33:47 pm »
    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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    Offline sluckey

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    Re: One watt amp
    « Reply #83 on: January 29, 2014, 02:47:08 pm »
    Quote
    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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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #84 on: January 29, 2014, 04:12:34 pm »
    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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    Offline sluckey

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    Re: One watt amp
    « Reply #85 on: January 29, 2014, 04:37:19 pm »
    Have you looked at the Traynor schematic?
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    Offline sluckey

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    Re: One watt amp
    « Reply #86 on: January 29, 2014, 09:25:37 pm »
    Here's another example...

    A schematic, layout, and hi-rez pics are very useful for troubleshooting your amp. Don't wait to be asked. JUST DO IT!

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #87 on: January 30, 2014, 05:03:14 am »
    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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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #88 on: January 30, 2014, 07:48:29 am »
    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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    Offline HotBluePlates

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    Re: One watt amp
    « Reply #89 on: January 30, 2014, 11:27:21 pm »
    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.

    Quote
    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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #90 on: January 31, 2014, 07:41:46 am »
    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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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #91 on: January 31, 2014, 09:48:44 am »
    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 VoltsV1 CurrentV2 VoltsV2 CurrentTotal Current
    -13.0V7.4mA-13.0V7.4mA14.8mA
    -10.8V9.2mA-15.2V5.8mA15.0mA
    -8.6V11.0mA-17.4V4.5mA15.5mA
    -6.5V13.0mA-19.5V3.3mA16.3mA
    -4.7V15.0mA-21.3V2.5mA17.5mA
    -3.0V16.8mA-23.0V1.8mA18.6mA
    -1.7V18.4mA-24.3V1.3mA19.7mA
    -0.8V19.0mA-25.2V1.0mA20.0mA
    -0.3V20.3mA-25.7V0.8mA21.0mA
    0.0V20.5mA-26.0V0.7mA21.2mA

    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?
    « Last Edit: January 31, 2014, 10:03:20 am by dbishopbliss »
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    Offline HotBluePlates

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    Re: One watt amp
    « Reply #92 on: January 31, 2014, 03:43:11 pm »
    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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #93 on: January 31, 2014, 03:48:52 pm »
    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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    Offline HotBluePlates

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    Re: One watt amp
    « Reply #94 on: January 31, 2014, 05:59:34 pm »
    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.

    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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #95 on: January 31, 2014, 06:35:11 pm »
    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.

    « Last Edit: January 31, 2014, 06:42:49 pm by dbishopbliss »
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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #96 on: February 03, 2014, 11:35:37 am »
    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 states the following:

    Quote
    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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    Offline dbishopbliss

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    Re: One watt amp
    « Reply #97 on: February 03, 2014, 01:49:10 pm »
    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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    Offline PRR

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    Re: One watt amp
    « Reply #98 on: February 03, 2014, 07:33:14 pm »
    > How do I know how much voltage I need to swing?

    I thought you had this all plotted out?

    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.

    Offline dbishopbliss

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    Re: One watt amp
    « Reply #99 on: February 03, 2014, 09:06:04 pm »
    > How do I know how much voltage I need to swing?

    I thought you had this all plotted out?


    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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