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Changing Gain in Two-Stage Tube Circuit

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Hi,

I have an amp with c3m driving AD1. This amp has very high gain - too high for my setup.

Now the c3m is capacitor-coupled to the AD1 so I wonder if I can reduce the gain by splitting the c3m's anode resistor, say x ohms, into two resistors in series, e.g., B+->0.1x->0.9x->c3m anode, and then tap the middle of the two resistors to capacitor-coupled to the next stage's tube grid.

Am I correct that the signal amplitude will then be reduced by 10x and also the source impedance (viewed by the next stage) will also be lowered by 10x?

Any issues in such a modification?

Thanks.
 
The Linearity of a common cathode stage with an un-bypassed self bias resistor is very good . . . as long as the loss of gain is not too large (otherwise, the first stage will clip before the next stage can be driven to its full dynamic range). This stage topology has Degeneration, a form of local negative feedback.

An LED has a low impedance. But it still has a non linear IV curve (not a perfectly straight line). That means as the grid signals changes, the cathode current changes, and so does the LED voltage. So, the LED voltage changes as the input stage signal changes, but in a non-linear fashion. How much does it change? Depends on the signal amplitude, the tube, the LED, the standing current, and the plate load resistance in parallel with the next stage input loading.

Negative feedback in the form of degeneration, when it is from a non-linear device does generate additional distortion. Unless it roughly matches the tube non-linearity amplitude, and is in the opposite direction, then it can not reduce 2nd harmonic distortion. The effect of the tube and the LED is that the distortions are in the same direction, which increases the 2nd harmonic distortion (no matter how small the effect is).

A Good bypass capacitor can be more linear than an LED.

But in the case where you are trying to reduce gain, use an un-bypassed self bias resistor. Make it so the cathode voltage and current are the same as it was for the LED.
If there is now not enough gain (reduced gain too much), then Series up a resistor and a capacitor, and connect that series network Across the self bias resistor. That will get the gain between the 2 extremes.

"All Generalizations Have Exceptions"

One thing I notice about this web site is that hardly anybody puts up a schematic initially when they have a question. And, so often, we do not initially get a full enough description of the circuit either. Answers (good ones/right ones) take longer to get when those 2 factors are the case.
 
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If you decide to split the plate load in order to use it as a signal divider, you have to make sure that there is NO ripple in that B+. Otherwise you will have more hum than when you had the higher gain. The power supply rejection is lower the closer you get to B+, when you use a resistive divider from B+ to the plate.

With a true pentode that action is not as dramatic because the plate resistance is higher than the plate load resistance, but with either a triode wired pentode or a triode, the plate resistance and the single resistor plate load form a reasonable voltage divider of the B+ ripple.
 
One thing I notice about this web site is that hardly anybody puts up a schematic initially when they have a question. And, so often, we do not initially get a full enough description of the circuit either. Answers (good ones/right ones) take longer to get when those 2 factors are the case.

How true. We don't even know if the C3M is being operated as a triode or pentode.
 
One thing I notice about this web site is that hardly anybody puts up a schematic initially when they have a question. And, so often, we do not initially get a full enough description of the circuit either. Answers (good ones/right ones) take longer to get when those 2 factors are the case.

Exactly. The amount of speculation generated when there is too little information is unnecessary. In this case I believe the forum is chasing its tail.

The problem appears to be that the user has too much gain in his signal chain; not in the amp. Apparently, the AD1 has the same operating point as a 2A3: Eg = -45 for Ep 250V and Ip of 60mA. The C3m has a gain of 19 in triode configuration so an input of 1.6Vrms is required just to drive the AD1 to clipping. The LED will give a bias of about 1.5V - 2.0V depending on the LED. The bottom line: the amp appears to be properly designed, so the user must be interposing a preamp between the source and the amp creating too much gain, noise, and distortion. This amp simply needs a volume control on its input.
 
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