[See schematic image in post #3 below.]

I will probably regret this, but...

I'm having a debate with another hobbyist and I was wondering if someone would check my calculations. The image is a simplified schematic of part of a push-pull pentode amplifier. There is a 470K shunt feedback resistor from the plate of the output tube V2 to the input of the source follower. I would like to determine the effective load resistance seen by the plate of the driver tube V1.

1) I assume the gain of the source follower is unity.

2) I assume the gain of the output tube is 10.

3) The feedback resistor is bootstrapped by the gain from the MOSFET gate to the V2 plate, giving an effective resistance of 470k / (10 + 1) = 42.7K.

4) The effective plate load on V1 is the parallel combination of the plate resistor, the bias resistor, and the bootstrapped feedback resistor, 220K || 1M || 42.7K = 34.5K.

To throw some more numbers at this, assuming the amplifier is driven to full output with 20VRMS on the grid, giving 200VRMS on the plate:

Feedback resistor current = 220V / 470K = 0.47mA

Plate resistor current = 20V / 220K = 0.09mA

Bias resistor current = 20V / 1M = 0.02mA

So, the total AC current in V1 is 0.58mA. With a 20V swing, the load line resistance is 20V / 0.58mA = 34.5K, the same result.

The bottom diagram is a more basic model. The signal source is the Thevenin equivalent of V1's internal generator voltage, uVgk, and the parallel combination of its plate resistance (~44K assumed) the 220K plate resistor, and the 1M bias resistor. This gives Rs = ~35K.

Rs in series with Rf forms a voltage divider having a ratio of Rs / (Rs + Rf). This gives a feedback factor of 0.07, a little lower than Schade's recommended 10%. Using the feedback equation, the resulting closed-loop gain is 10 / (1 + 10 * 0.07) = 5.9. The resulting reduction in gain is -4.6dB.

In the real world there will be some cathode resistance on V1 which will increase its effective plate resistance, giving more feedback. Also, the value of plate resistance may be higher depending on the tube's operating point. So the actual gain reduction may easily be set to Schade's -6.95dB.

Note that the gain reduction is from the V1 internal voltage generator, not its plate. This is an abstraction, as uVgk cannot be measured directly in the circuit.

This example is based on a schematic floating around the forum. The real-world amplifier uses a 6SL7 differential driver with a tail current of 1.15mA, or 0.575mA per tube. Since we calculated above that we need an input current of 0.58mA RMS = 0.82mA peak, it follows that the circuit as designed will be unable to drive the amplifier to peak output. Nonetheless, the designer has built and tested the amplifier, hence the debate and the conundrum.

FWIW I have put the full amplifier schematic into SPICE and confirmed the values I obtained by calculation. I know SPICE sometimes lies or gives misleading results but the fact that the simulation yields the same result increases my confidence in my math.

Assuming the value of 34.5K for the plate load is correct, this would seem to be far too low a value for a high-mu tube like the 6SL7 to drive, even at levels below the clipping point.

Since a number of people have built the amplifier and it seems to work I'm at a loss to explain the apparent contradiction. It would seem that I've made a mistake. If so I would be very appreciative if someone would show me what I've done wrong.

Thanks.

-Henry

I will probably regret this, but...

I'm having a debate with another hobbyist and I was wondering if someone would check my calculations. The image is a simplified schematic of part of a push-pull pentode amplifier. There is a 470K shunt feedback resistor from the plate of the output tube V2 to the input of the source follower. I would like to determine the effective load resistance seen by the plate of the driver tube V1.

1) I assume the gain of the source follower is unity.

2) I assume the gain of the output tube is 10.

3) The feedback resistor is bootstrapped by the gain from the MOSFET gate to the V2 plate, giving an effective resistance of 470k / (10 + 1) = 42.7K.

4) The effective plate load on V1 is the parallel combination of the plate resistor, the bias resistor, and the bootstrapped feedback resistor, 220K || 1M || 42.7K = 34.5K.

To throw some more numbers at this, assuming the amplifier is driven to full output with 20VRMS on the grid, giving 200VRMS on the plate:

Feedback resistor current = 220V / 470K = 0.47mA

Plate resistor current = 20V / 220K = 0.09mA

Bias resistor current = 20V / 1M = 0.02mA

So, the total AC current in V1 is 0.58mA. With a 20V swing, the load line resistance is 20V / 0.58mA = 34.5K, the same result.

The bottom diagram is a more basic model. The signal source is the Thevenin equivalent of V1's internal generator voltage, uVgk, and the parallel combination of its plate resistance (~44K assumed) the 220K plate resistor, and the 1M bias resistor. This gives Rs = ~35K.

Rs in series with Rf forms a voltage divider having a ratio of Rs / (Rs + Rf). This gives a feedback factor of 0.07, a little lower than Schade's recommended 10%. Using the feedback equation, the resulting closed-loop gain is 10 / (1 + 10 * 0.07) = 5.9. The resulting reduction in gain is -4.6dB.

In the real world there will be some cathode resistance on V1 which will increase its effective plate resistance, giving more feedback. Also, the value of plate resistance may be higher depending on the tube's operating point. So the actual gain reduction may easily be set to Schade's -6.95dB.

Note that the gain reduction is from the V1 internal voltage generator, not its plate. This is an abstraction, as uVgk cannot be measured directly in the circuit.

This example is based on a schematic floating around the forum. The real-world amplifier uses a 6SL7 differential driver with a tail current of 1.15mA, or 0.575mA per tube. Since we calculated above that we need an input current of 0.58mA RMS = 0.82mA peak, it follows that the circuit as designed will be unable to drive the amplifier to peak output. Nonetheless, the designer has built and tested the amplifier, hence the debate and the conundrum.

FWIW I have put the full amplifier schematic into SPICE and confirmed the values I obtained by calculation. I know SPICE sometimes lies or gives misleading results but the fact that the simulation yields the same result increases my confidence in my math.

Assuming the value of 34.5K for the plate load is correct, this would seem to be far too low a value for a high-mu tube like the 6SL7 to drive, even at levels below the clipping point.

Since a number of people have built the amplifier and it seems to work I'm at a loss to explain the apparent contradiction. It would seem that I've made a mistake. If so I would be very appreciative if someone would show me what I've done wrong.

Thanks.

-Henry

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