When set is 0 or 1 I could find it. It either attenuates by 10 or amplifies by 10. However, the circuit becomes different when it is between 0 and 1.

# This circuit gain when set is in [0,1]

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# This circuit gain when set is in [0,1]

circuit analysisgain

When set is 0 or 1 I could find it. It either attenuates by 10 or amplifies by 10. However, the circuit becomes different when it is between 0 and 1.

## Best Answer

No, the pot has nothing to do with gain, at least not DC gain. Note the capacitor in series with the wiper.

To analyze this, first look at what happens to DC. Hint: All capacitors become open circuits then.

Now note that the pot adjustment only effects frequency response. With the wiper all the way to the top, Cs and Re2 are acting on the input signal after the 30 kΩ resistor. With the wiper all the way at the bottom, Cs and Re2 are action on the feedback signal, so the effect will be the inverse of what it was on the input signal.

Now remember that the rolloff frequency of a R-C filter is F = 1 / 2πRC. When R is in Ohms, C in Farads, then F is in Hz.

From the above you should be able to get a pretty good intuitive idea what this circuit is intended to do, and with the equation, get a good idea at what frequencies it does it.