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Old 9th July 2018, 07:24 PM   #1
rongon is offline rongon  United States
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Default M. Kosterís "Schadeode" Push-Pull Version?

I was looking at Michael J Koster's 'Schadeode' circuit, as found here:
RE: hybrid - Michael Koster - Tube DIY Asylum

and here:
Tube DIY Asylum

Looks like a fantastic way to use 6L6s.

So I was thinking.... Why not push-pull? I threw it all into spice, and as long as you can match up a couple of DN2540 or 01N100D MOSFETs, the results are nothing short of spectacular. BUT...

How likely is it that matching these MOSFETs would result in a well-balanced push-pull circuit? Would using the two MOSFETs in a LTP with a good stiff CCS in the tail balance them well enough?

I'm wondering why nobody else has posted about this, since it seems an obvious thing to try, and looks so good in simulations.

What do you think?

(Sorry I can't post a schematic. That will have to wait until I get home.)
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Old 9th July 2018, 07:50 PM   #2
Wavebourn is offline Wavebourn  United States
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I breadboarded similar thingy, with 6J52P driving GU-50 with cathode bias. MOSFET has higher transconductance, and more linear V to I conversion. The drawback is, linear current swing drives non-linear resistance of parallel feedback. That's why I went by a different path, when a pentode drives more linear impedance of a feedback voltage divider.
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Old 10th July 2018, 02:43 AM   #3
rongon is offline rongon  United States
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Here's a schematic of what I was thinking of.
Attached Images
File Type: png _PP-Schadeode_EXPERIMENTAL_00.png (31.7 KB, 499 views)
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Old 10th July 2018, 06:32 PM   #4
rongon is offline rongon  United States
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Quote:
Originally Posted by Wavebourn View Post
The drawback is, linear current swing drives non-linear resistance of parallel feedback. That's why I went by a different path, when a pentode drives more linear impedance of a feedback voltage divider.
I'm trying to understand this.

Are you saying that a pentode input tube with its own separate plate resistor and screen resistor, with a 'plate-to-plate' feedback resistor from the output tube plate to the plate of the input pentode, results in the input tube working into a more linear impedance than if the feedback resistor is also the plate (or in this case, drain) load resistor for the input stage?

In other words, are you criticizing the 'Schadeode' idea in general, whether for SE or PP, or is this only an issue if trying to use the 'Schadeode' idea in a push-pull configuration?

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Old 11th July 2018, 04:43 AM   #5
leadbelly is online now leadbelly  Canada
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You might want to read some of the older Schade threads that discussed pentode vs triode drivers. There's a lot of BS on this topic, a lot of the anti-pentode anti-fdbk crowd act like local feedback came from their deity.
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Last edited by leadbelly; 11th July 2018 at 04:53 AM.
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Old 11th July 2018, 06:14 AM   #6
Wavebourn is offline Wavebourn  United States
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Yes. I mean, pentode loaded on the resistance equal to feedback resistor's value divided by amplification factor of the output stage. Since the amplification factor is non-linear, the driver is loaded on non-linear resistance, providing non-linear voltage swing.
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Old 11th July 2018, 08:49 AM   #7
Ketje is offline Ketje  Belgium
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Isn't it just the way feedback should behave ? Less amplification of the output stage gives less load and more amplification of the driver to compensate. After all every amp is non linear and the feedback is there to correct things (allmost).

Mona
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Old 11th July 2018, 02:12 PM   #8
rongon is offline rongon  United States
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I'd figure that the higher the transconductance and gain of the input device, the better it would be able to correct non-linearities in the output device.

Is the problem in this M. Koster-inspired push-pull circuit that the drain resistor and feedback resistor have been combined into one? (R4 and R5 in the schematic.)

Click the image to open in full size.

Is it a better idea to take the more traditional approach of RC-coupling the input stage to the output stage, and adding a separate feedback resistor from the output pentode plate to the drain (or plate) of the input device? (Actually from output pentode plate to output pentode control grid, but it amounts to the same thing in this circuit.)

In other words, is it better for the input device to work against a voltage divider composed of passive resistors than it is to use its internal resistance as one leg of the feedback voltage divider?
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Old 16th July 2018, 08:12 AM   #9
Sorento is offline Sorento  Europe
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Concerning the original Schadeode schematic, why is the cathode bypass cap connected to B+ rather than gnd? Makes PSR worse and requires a high voltage cap ... Cannot see why it should be beneficial .

Last edited by Sorento; 16th July 2018 at 08:14 AM.
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Old 16th July 2018, 09:58 AM   #10
Wavebourn is offline Wavebourn  United States
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Quote:
Originally Posted by Ketje View Post
Isn't it just the way feedback should behave ? Less amplification of the output stage gives less load and more amplification of the driver to compensate. After all every amp is non linear and the feedback is there to correct things (allmost).

Mona
Yes, if feedback divider's ratio is stable. But if is non-linear, the result is non-linear. Thst's why I always use the best resistors in feedback voltage divider.
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