With pentodes, the combination of g2 voltage( setting the envelope the load line runs in), and loading sets the stage for the distortion spectrum. Drive the load line out under the knee and I suspect the h3 goes up. Drive it out to the right of the knee and it runs through an increasingly wider spacing of the g1 lines( asymetrical/even hd), yes?
Douglas
Douglas
Yes, that's the way it should work. But in this SP-10 circuit...
For the driver pentodes, if I flatten the load line by making Rp and Rg2 really large in value, the load line will run well under the knee, yielding a more symmetrical waveform from the pentode's plate (output), so I'd expect more H3 and not less.
Conversely, if I reduce the values of Rp and Rg2 so that the driver stage pentodes draw more Ip and Ig2, the load line will become more vertical and that should increase H2, as the load line is now running through more widely spaced g1 lines, which are more widely spaced towards the top of the plate curves, and more closely packed together down at the bottom. That would mean more asymmetrical waveforms in the output, indicating more even harmonics (H2). But...
Everything is working exactly the opposite way in this wacky SP-10 circuit.
When I increase the current through the driver pentodes (whether they be 6AU6, 6CB6, 6EW6 or whatever), the amount of H3 from the output goes up, not down. However, increasing the current through the output tubes does decrease H3. So that part is acting as expected.
Maybe it's distortion cancellation? If the drivers put out more H2, does that end up as more H3 from the output stage?
For the driver pentodes, if I flatten the load line by making Rp and Rg2 really large in value, the load line will run well under the knee, yielding a more symmetrical waveform from the pentode's plate (output), so I'd expect more H3 and not less.
Conversely, if I reduce the values of Rp and Rg2 so that the driver stage pentodes draw more Ip and Ig2, the load line will become more vertical and that should increase H2, as the load line is now running through more widely spaced g1 lines, which are more widely spaced towards the top of the plate curves, and more closely packed together down at the bottom. That would mean more asymmetrical waveforms in the output, indicating more even harmonics (H2). But...
Everything is working exactly the opposite way in this wacky SP-10 circuit.
When I increase the current through the driver pentodes (whether they be 6AU6, 6CB6, 6EW6 or whatever), the amount of H3 from the output goes up, not down. However, increasing the current through the output tubes does decrease H3. So that part is acting as expected.
Maybe it's distortion cancellation? If the drivers put out more H2, does that end up as more H3 from the output stage?
When you change the Rp are you checking the change in the cathode voltage? I think the right way to change the idle current is to change the Rk. At 6.6v/1.8KR I think you're running the driver too cold. I'm looking at 6AU6 curves and data. The low g2 voltage is making it hard to get some current throughput. Yes?
Just a thought about Push Pull drivers, such as those 6AU6s.
I am giving a generalization of the drivers, and of the output stages too.
For triodes and pentodes, a single tube has 2nd harmonic distortion (they have a unidirectional change of gain).
And we understand about the cancellation of the 2nd harmonic when the push pull drivers are in anti-phase to each other.
But think again about that nice 6AU6.
The 2nd harmonic distortion of a single tube is because the closer g1 gets to 0Volts, the more the gain is.
OK.
But that same extra gain is when the push pull drivers are at the + and - crests of the signal voltages.
Hmmm, that means there is not any compression at the peaks of the signal . . . no compression at the peaks of push pull = no 3rd harmonic distortion.
Even if the driver is perfect, there still might be some compression of the push pull Output stage, so we might still get some 3rd harmonic distortion.
Sometimes, the 2nd harmonic distortion and the 3rd harmonic distortion are the same % on a push pull amplifier.
Example: My 2 stage balanced amplifiers (special case push pull) . . . @ 1Watt, the 2nd harmonic distortion is -52dBc, and the 3rd harmonic distortion is Also -52dBc . . . Equal!
And, I did not use any global negative feedback!
The only negative feedback paths are from cathode to cathode of the drivers, and cathode to cathode of the output tubes.
Have fun playing with the relative Percent (%) of the 2nd harmonic distortion, versus the Percent (%) of the 3rd harmonic distortion.
$0.03
I am giving a generalization of the drivers, and of the output stages too.
For triodes and pentodes, a single tube has 2nd harmonic distortion (they have a unidirectional change of gain).
And we understand about the cancellation of the 2nd harmonic when the push pull drivers are in anti-phase to each other.
But think again about that nice 6AU6.
The 2nd harmonic distortion of a single tube is because the closer g1 gets to 0Volts, the more the gain is.
OK.
But that same extra gain is when the push pull drivers are at the + and - crests of the signal voltages.
Hmmm, that means there is not any compression at the peaks of the signal . . . no compression at the peaks of push pull = no 3rd harmonic distortion.
Even if the driver is perfect, there still might be some compression of the push pull Output stage, so we might still get some 3rd harmonic distortion.
Sometimes, the 2nd harmonic distortion and the 3rd harmonic distortion are the same % on a push pull amplifier.
Example: My 2 stage balanced amplifiers (special case push pull) . . . @ 1Watt, the 2nd harmonic distortion is -52dBc, and the 3rd harmonic distortion is Also -52dBc . . . Equal!
And, I did not use any global negative feedback!
The only negative feedback paths are from cathode to cathode of the drivers, and cathode to cathode of the output tubes.
Have fun playing with the relative Percent (%) of the 2nd harmonic distortion, versus the Percent (%) of the 3rd harmonic distortion.
$0.03
This RCA SP-10 circuit has some crazy stuff going on in it. The 6AU6 grids are lifted up +4.5VDC so that a larger than normal value of Rk can allow a reasonable value of feedback resistor (from the output tube plate to the 6AU6 cathode.
In the 6AU6 driver pentodes, the problem is that changing the value of Rk (R9 in the schematic below) changes the level of NFB, so you also have to change the value of that 100k resistor (R19) feeding inverted signal from the output tube plate, as well as the 330k resistor (R8) from the driver triode cathode back to its grid. They all interact at both DC and AC. The RCA engineers added R8 (330k) to raise the 6AU6 grid up a few volts DC so they could use a 1.8k resistor for Rk, allowing them to keep the value of the NFB resistor up at 100k ohms.
In the Sams Photofact schematic (below), that R19 NFB resistor drops about 320V - 6.5V = 313.5V, which means there's something like 3.1mA current drawn across that resistor. So the combined current across Rk (R9) is 3.6mA. Checking that, 1800 ohms * 0.0036A = 6.48V.
The 6AU6 (V2) sees a grid-to-cathode voltage of -2V, which we know because the cathode sits at +6.5V and the grid sits at +4.5V. Therefore, the 6AU6 operating points are Vp = 193.5V, Vg2 = 59.5V and Vg1 = -2V. The plate current is about 400uV and the screen grid current is only about 100uA (each 6AU6).
Increasing the B+ to 400V or so would allow the 6AU6 to be operated with a more normal 1.5mA or 2mA plate current, but now more voltage would need to be dropped across the R19 NFB resistor (400V - 6.5V = 393.5V, which means 393.5V/100k ohms = 3.935mA, which means that resistor would dissipate 1.5W at idle.
Chris Hornbeck suggested replacing the R8 330k cathode-to-grid resistor with a +4.5VDC supply to the grids. That could simplify calculation of the value of the R19 NFB feedback resistor. I can use a simple voltage source in the simulation to test that out. That might allow changing the values of Rk and the NFB resistor to allow more current through the 6AU6 driver pentodes. I hope... Maybe use three AAA rechargeable batteries in series to make +4.5V?
The goal is to keep the value of NFB resistor as high as possible to avoid loading down the driver pentode plate.
There's a lot going on in this circuit. I'm still puzzling it out...
In the 6AU6 driver pentodes, the problem is that changing the value of Rk (R9 in the schematic below) changes the level of NFB, so you also have to change the value of that 100k resistor (R19) feeding inverted signal from the output tube plate, as well as the 330k resistor (R8) from the driver triode cathode back to its grid. They all interact at both DC and AC. The RCA engineers added R8 (330k) to raise the 6AU6 grid up a few volts DC so they could use a 1.8k resistor for Rk, allowing them to keep the value of the NFB resistor up at 100k ohms.
In the Sams Photofact schematic (below), that R19 NFB resistor drops about 320V - 6.5V = 313.5V, which means there's something like 3.1mA current drawn across that resistor. So the combined current across Rk (R9) is 3.6mA. Checking that, 1800 ohms * 0.0036A = 6.48V.
The 6AU6 (V2) sees a grid-to-cathode voltage of -2V, which we know because the cathode sits at +6.5V and the grid sits at +4.5V. Therefore, the 6AU6 operating points are Vp = 193.5V, Vg2 = 59.5V and Vg1 = -2V. The plate current is about 400uV and the screen grid current is only about 100uA (each 6AU6).
Increasing the B+ to 400V or so would allow the 6AU6 to be operated with a more normal 1.5mA or 2mA plate current, but now more voltage would need to be dropped across the R19 NFB resistor (400V - 6.5V = 393.5V, which means 393.5V/100k ohms = 3.935mA, which means that resistor would dissipate 1.5W at idle.
Chris Hornbeck suggested replacing the R8 330k cathode-to-grid resistor with a +4.5VDC supply to the grids. That could simplify calculation of the value of the R19 NFB feedback resistor. I can use a simple voltage source in the simulation to test that out. That might allow changing the values of Rk and the NFB resistor to allow more current through the 6AU6 driver pentodes. I hope... Maybe use three AAA rechargeable batteries in series to make +4.5V?
The goal is to keep the value of NFB resistor as high as possible to avoid loading down the driver pentode plate.
There's a lot going on in this circuit. I'm still puzzling it out...
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OK, as I get a bit deeper into this, it looks like 95% of the R9 current is from the FB, so it means they are running the 6AU6 very near cutoff which is class B territory. I guess you can do that with a push-pull driver but geeesh, it's got to be borderline a distortion generator if the crossovers between both tubes aren't in good sync. Man, you have a handfull here to get it tuned. Or just go ahead and throw some music through it and not worry about a sine that looks wonky.
The RCA 50 Watt design in RC-30 has a 390 Ohm resistor under the driver cathode with a 120K Ohm N Fdbk resistor. 120V on the 6CB6A screen grid.
The SP-10 running starved in class B seems nonsensical. What could they be up to? Maybe playing complementary gm variation between the 6AU6s and the 6V6 outputs? The 6AU6 drivers would be limited in ability to turn on the 6V6s. You can only turn OFF the 6AU6s so much. Such a driver would only work with 6V6 and 6AU6 tubes of specific complementary matched gm conditions. Maybe RCA sold special tube sets for these SP-10/20 amplifiers. $$$
The SP-10 running starved in class B seems nonsensical. What could they be up to? Maybe playing complementary gm variation between the 6AU6s and the 6V6 outputs? The 6AU6 drivers would be limited in ability to turn on the 6V6s. You can only turn OFF the 6AU6s so much. Such a driver would only work with 6V6 and 6AU6 tubes of specific complementary matched gm conditions. Maybe RCA sold special tube sets for these SP-10/20 amplifiers. $$$
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I'm not sure we could even call this a push-pull driver stage. It's really just two seperate anti-phase drivers, after the splitter. The signals are not summed like in a PP OPT.
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The RCA 50 Watt design adds plate-grid feedback around its 7027A outputs, and a global NFB loop to the cathode of the 7199 pentode input voltage amplifier. Three different types of NFB, nested.
This SP-10/20 design (page 361 in the RCA RC-19 Receiving Tubes manual from 1959) only uses NFB from the 6V6 plates to the 6AU6 cathodes, and nothing else.
What's weird is that people who've heard the SP-10 or SP-20 amps state they sound really good.
I looked up what OPT they used and it was a garden variety Stancor 10k:VC rated for 25W from 60Hz to 20kHz. Nothing special at all, although it did weigh 4 lbs. I don't think there was anything special going on with the OPT.
Could it be that this circuit works better in real life than it does in theory?
Also strange is that LTspice simulations predict exceptional performance. Could it be that the models can't resolve what happens in real life to a small signal pentode when it's biased that cold? Why does LTspice predict THD of something like 0.02% at 1W out when 6CB6 is used in place of the 6AU6 and 6JC5 is used in place of the 6V6GT?
FWIW, simply swapping in 6CB6 for the 6AU6 increases the plate current by a mere 100uA, but halves the THD, even if the 6V6GTs are retained.
I wish I could say what the RCA engineers were up to. I would dismiss this as vintage weirdness if it didn't simulate so well, and if people had reported that the amp sounded like a typical PP 6V6 amp of its day.
Can you explain that in more detail? Are you saying the 6AU6s must be operating with a lot of crossover notch (zero current in both 6AU6s for a portion of the waveform)? I guess we need someone to find a working one of these amps and see what the driver stage looks like on a 'scope. LTspice seems to think it works just great. That's a mystery to me, because SPICE simulation seems pretty good at picking up gross errors in the AC signal domain. This is weird.
This SP-10/20 design (page 361 in the RCA RC-19 Receiving Tubes manual from 1959) only uses NFB from the 6V6 plates to the 6AU6 cathodes, and nothing else.
What's weird is that people who've heard the SP-10 or SP-20 amps state they sound really good.
I looked up what OPT they used and it was a garden variety Stancor 10k:VC rated for 25W from 60Hz to 20kHz. Nothing special at all, although it did weigh 4 lbs. I don't think there was anything special going on with the OPT.
Could it be that this circuit works better in real life than it does in theory?
Also strange is that LTspice simulations predict exceptional performance. Could it be that the models can't resolve what happens in real life to a small signal pentode when it's biased that cold? Why does LTspice predict THD of something like 0.02% at 1W out when 6CB6 is used in place of the 6AU6 and 6JC5 is used in place of the 6V6GT?
FWIW, simply swapping in 6CB6 for the 6AU6 increases the plate current by a mere 100uA, but halves the THD, even if the 6V6GTs are retained.
I wish I could say what the RCA engineers were up to. I would dismiss this as vintage weirdness if it didn't simulate so well, and if people had reported that the amp sounded like a typical PP 6V6 amp of its day.
Can you explain that in more detail? Are you saying the 6AU6s must be operating with a lot of crossover notch (zero current in both 6AU6s for a portion of the waveform)? I guess we need someone to find a working one of these amps and see what the driver stage looks like on a 'scope. LTspice seems to think it works just great. That's a mystery to me, because SPICE simulation seems pretty good at picking up gross errors in the AC signal domain. This is weird.
It is surprising the simulation thinks it's great too. But the gm variation between the drivers and the output tubes may be playing off against each other to give a constant gain combo. ( this is like the pentode version of the usual class A triode Harmonic Cancellation scheme ) Especially if the simulation tube models have the same math, leading to near miraculous cancellation. (needs both stages operating in the same amount of class AB )
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Maybe we should consider the scheme Pete Millett used in his DCPP amplifier driver stage. A CCS tail below the driver cathodes, BUT with two cathode resistors going out to the cathodes from the tail. The N Fdbks would then connect to the two cathodes directly and the CCS gets adjusted to cancel or set the DC there.
http://www.pmillett.com/file_downloads/dcpp_sch.pdf
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Otherwise, keeping with the original SP-10/20 plan, the driver stage would be set up for the typically optimum class aB, and the output tubes get a bias adjust pot that adjusts Both tubes (for the same class aB ) with a separate pot for DC balance. ( like Citation II has ) The user can tweak them for best sound, and maybe some range of output tubes would be allowed.
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Maybe we should consider the scheme Pete Millett used in his DCPP amplifier driver stage. A CCS tail below the driver cathodes, BUT with two cathode resistors going out to the cathodes from the tail. The N Fdbks would then connect to the two cathodes directly and the CCS gets adjusted to cancel or set the DC there.
http://www.pmillett.com/file_downloads/dcpp_sch.pdf
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Otherwise, keeping with the original SP-10/20 plan, the driver stage would be set up for the typically optimum class aB, and the output tubes get a bias adjust pot that adjusts Both tubes (for the same class aB ) with a separate pot for DC balance. ( like Citation II has ) The user can tweak them for best sound, and maybe some range of output tubes would be allowed.
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You can imagine if one of the two tubes has reduced emission as it ages it would go into cutoff before the other tube and there would be a 0 signal period as we wait for that tube to come back out of cutoff. Class B is a cutoff period of 180 degrees. It is the point of exact biasing at 0 idle current. It takes much precision to make that work or else we need to bias the tubes to stay out of that range for a small period of time to create an overlap to keep a signal flowing. What could go wrong..... plenty.
The idea of matching up the amount of class aB (crossover) for both the driver stage and the outputs is ingenious for smooth gain, but drift-wise tenuous, just like class B. The adjustment of crossover for both stages needs to be simple so they can be matched up regularly against ageing. The bias control ( a'la Cit II ) for the outputs, and a current adjust pot for the CCS driver tail could do that. (or even set up for class A in the driver stage with plenty more CCS current ) Would the operator be happy with that from just listening, or does some test method need to be designed that would give a meter indication? I think some bigger driver tubes are called for to give a bigger drive range to handle from 6V6s to KT88s in UL (high Miller capacitance ). So the front end pre-amp stage needs a gain adjust pot too. With these pot knobs, one can design their own Amp on the fly!
Hmm, well the driver load resistors need to drop in value as the driver stage current goes up. Can one get a triple pot with two different R values. Maybe have to design for the same R values in the triple pot. And possibly another pot for the screen V. Might take a while to tune this baby up! And a two channel O'scope!
Some simplification: Just a multi-pole switch to convert the driver mode from class aB to more usual class A. And a current adjust pot for the CCS to match up driver class aB overlap with the output tubes. The multi-pole switch could also change the CCS current adjustment range (and driver load resistors ) for the two modes. Pretty straight-forward then.
Hmm, well the driver load resistors need to drop in value as the driver stage current goes up. Can one get a triple pot with two different R values. Maybe have to design for the same R values in the triple pot. And possibly another pot for the screen V. Might take a while to tune this baby up! And a two channel O'scope!
Some simplification: Just a multi-pole switch to convert the driver mode from class aB to more usual class A. And a current adjust pot for the CCS to match up driver class aB overlap with the output tubes. The multi-pole switch could also change the CCS current adjustment range (and driver load resistors ) for the two modes. Pretty straight-forward then.
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Just thinking out loud here: Wouldn't it be sweet with a PCB for this design, preferably monoblocks and designed to allow different sockets (or off-board sockets) for the output tubes? I imagine this could be the next DCPP, a platform that could use a wide range of output tubes and OPTs, from 6V6-oids or even smaller up to KT88 or big sweep tubes.
Woops, I just realized that the CCS tail idea won't work for the SP-10/20 mode driver since those tubes have to start near zero current. I guess one could just add a current nulling resistor (if necessary, the DC Fdbks are doing the same ) from the CCS tail to B+. Then the CCS can keep balance with the DC Fdbks, but can be adjusted for class aB operation of the driver tubes above. Basically, a switchable Amp between the two modes (tracked class aB driver or usual class A driver ).
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One way I can simulate tube mis-matching effects is to plug different versions of tube models into the schematic, and run the simulation. See what we get.
I have a promising looking circuit mockup. using 6SN7 > 2x 6CB6 > 2x 6L6GC with B+ = 330V and a Dynaco A470 OPT (I have a pair of those too). The simulation says the circuit should make 18W output at 1% THD.
I have two 6AU6 models, one from Ayumi N. and the other from a member named Rob McLean.
I also have several 6L6 models, including the ones from Ayumi N. and Adrian Immler.
I ran that, and:
Still the circuit makes 1W into 8 ohms with only 0.049% THD.
The FFT looks like it's from an SE Triode amp at low levels. Could that be the allure of this design?
Let's see what things look like at the output with 1% THD.
That doesn't look crazy for a tube amp that's going into clipping, does it? What do you guys see here? One driver 6AU6 in cutoff? Or is that one 6L6GC in cutoff?
I remember at one point I was playing around with the RCA 50 Watt Amplifier circuit by removing the global NFB, leaving only the combination of output tube plate-grid NFB and output tube plate to driver tube cathode NFB. Why not take this SP-10 circuit and reduce the amount of NFB to the cathodes then add a small amount of plate-grid NFB? If the plate-grid FB resistor is a high value (perhaps 1M ohms) that shouldn't load down the driver pentode plates too badly. At the same time, if the output plate to driver cathode resistor is a high value (perhaps 470k ohms) that wouldn't draw so much current causing biasing problems for the driver stage.
I think I'll start from scratch with push-pull 6L6GC pentodes into a Dynaco A470 OPT (which we know should work fine), and a pair of 6CB6 for the driver LTP. Start open loop, then add 8dB of NFB from 6L6GC plate to 6CB6 cathode. Readjust bias to make sure the driver is still in a happy place. Then add 8dB of NFB from 6L6GC plate to grid, for a total of 16dB NFB. If the 6CB6 pentodes are biased at a reasonable operating point, the amp should perform very well, right? We shall see...
I have a promising looking circuit mockup. using 6SN7 > 2x 6CB6 > 2x 6L6GC with B+ = 330V and a Dynaco A470 OPT (I have a pair of those too). The simulation says the circuit should make 18W output at 1% THD.
I have two 6AU6 models, one from Ayumi N. and the other from a member named Rob McLean.
I also have several 6L6 models, including the ones from Ayumi N. and Adrian Immler.
I ran that, and:
- The simulator had to work really hard. Lots more processing to do.
- Gain remained exactly the same. That's what you'd expect from a circuit using 23dB of NFB.
- THD at 1W output is only a little bit higher. Still very low (0.05%).
- Waveforms at the 6AU6 plates (6L6GC grids) still look like sine waves, but the driver stage balance is not good:
Still the circuit makes 1W into 8 ohms with only 0.049% THD.
The FFT looks like it's from an SE Triode amp at low levels. Could that be the allure of this design?
Let's see what things look like at the output with 1% THD.
- Output is 19W rms into 8 ohms at 0.91% THD.
- The driver imbalance remains. Here are the 1kHz waveforms at the 6L6GC grids:
That doesn't look crazy for a tube amp that's going into clipping, does it? What do you guys see here? One driver 6AU6 in cutoff? Or is that one 6L6GC in cutoff?
I remember at one point I was playing around with the RCA 50 Watt Amplifier circuit by removing the global NFB, leaving only the combination of output tube plate-grid NFB and output tube plate to driver tube cathode NFB. Why not take this SP-10 circuit and reduce the amount of NFB to the cathodes then add a small amount of plate-grid NFB? If the plate-grid FB resistor is a high value (perhaps 1M ohms) that shouldn't load down the driver pentode plates too badly. At the same time, if the output plate to driver cathode resistor is a high value (perhaps 470k ohms) that wouldn't draw so much current causing biasing problems for the driver stage.
I think I'll start from scratch with push-pull 6L6GC pentodes into a Dynaco A470 OPT (which we know should work fine), and a pair of 6CB6 for the driver LTP. Start open loop, then add 8dB of NFB from 6L6GC plate to 6CB6 cathode. Readjust bias to make sure the driver is still in a happy place. Then add 8dB of NFB from 6L6GC plate to grid, for a total of 16dB NFB. If the 6CB6 pentodes are biased at a reasonable operating point, the amp should perform very well, right? We shall see...
Your last image shows the driver clipping in one direction and the output valve clipping in the other.
All good fortune,
Chris
All good fortune,
Chris
Thanks Chris. That makes sense to me. But I have a question:
Which is the side on which the driver is clipping?
Which is the side on which the output tube is clipping?
Even if I use the same models for all tubes, so that they should be perfectly (ideally) matched, I still see that same shape with one side flat-topped and the other side with a 'nipple'.
Which is the side on which the driver is clipping?
Which is the side on which the output tube is clipping?
Even if I use the same models for all tubes, so that they should be perfectly (ideally) matched, I still see that same shape with one side flat-topped and the other side with a 'nipple'.
And another question...
Here's those same two waveforms at the grids of the output tubes, when the driver tubes are 6CB6 and the output tubes are 6JC5, into an 8k ohm primary (plate-to-plate).
In all the above screenshots, the red trace is the 'top' output tube and the gold trace is the 'bottom' output tube, as they appear in this detail from the schematic:
I hope that makes sense.
Thanks for looking at this.
Here's those same two waveforms at the grids of the output tubes, when the driver tubes are 6CB6 and the output tubes are 6JC5, into an 8k ohm primary (plate-to-plate).
In all the above screenshots, the red trace is the 'top' output tube and the gold trace is the 'bottom' output tube, as they appear in this detail from the schematic:
I hope that makes sense.
Thanks for looking at this.
I just slapped together a simulation of Pete Millett's DCPP Engineer's Amplifier, with 6EW7 drivers, 6L6GC outputs and a Dynaco A470 OPT.
It looks excellent, but in a different way. THD is really low, but it's dominated by H3, while H2 is pretty much eliminated. That's acting like an ideal push-pull amp, right?
The RCA SP-10 is quite different, with H2 more prominent than H3.
It looks excellent, but in a different way. THD is really low, but it's dominated by H3, while H2 is pretty much eliminated. That's acting like an ideal push-pull amp, right?
The RCA SP-10 is quite different, with H2 more prominent than H3.