The cathode resistor, used for self-bias, introduces a form of local negative feedback: when the current increases, the cathode voltage increases as well reducing the grid to cathode voltage, and vice versa. The result is that the gain of the voltage amplifier with self-bias is reduced. This local negative feedback can be significantly neutralised by using a bypass capacitor connected in parallel to the cathode resistor.

The gain of the voltage amplifier with self-bias can be determined using the equivalent circuit shown in Figure 13. Let us first suppose that no bypass capacitor *C _{k}* is used. Similarly, to what we discussed in section 3.4, when we estimated the gain of the voltage amplifier, the vacuum tube is represented by an AC power supply, in series with the anode resistance

*r*. Let

_{a}*V*be the voltage of the input signal measured from the grid to the ground, that is from the grid to the terminal of the cathode resistor

_{in}*R*opposite to the cathode. The voltage

_{k}*V*measured between the two terminals of

_{fb}*R*, corresponding to the voltage drop introduced by

_{k}*R*itself, is the feedback voltage. The grid to cathode voltage , resulting from the combined action of the input and feedback voltages, is:

_{k}.

The above equation can be better understood using the equivalent circuit in Figure 13, as we already did in Section 3.4. In the circuit, power is supplied by the AC power supply, replacing the vacuum tube. Voltage drops, through the resistors in the circuit, proceeding clockwise. Suppose highest voltage is at the top of the AC power supply (the anode of the vacuum tube), and lowest voltage is at the bottom of the AC power supply (the cathode of the vacuum tube). *V _{in}* is the voltage difference between the input

*in*and the cathode resistor end, opposite to the voltage source (opposite to the cathode). The voltage difference between the input

*in*and the other cathode resistor end (between the grid and the cathode) is higher than

*V*, because of the voltage drop introduced by

_{in}*R*. Given that the voltage drop is

_{k}*V*, will be equal to

_{fb}*V*plus

_{in}*V*.

_{fb}The AC power supply produces a voltage equal to . As we said in Section 3.4, the minus sign here indicates that the phase of the AC power supply is reversed with respect to that of . The output voltage is taken between the two ends of the load *R _{L}* and can be computed using the voltage divider equation as

.

Using the voltage divider equation again, the feedback voltage *V _{fb}* is

.

Replacing *V _{fb}* into the equation for and simplifying we have:

.

Now we can replace this into the equation for and we obtain:

.

Finally, we can express the gain of the voltage amplifier, with local feedback introduced by a non-bypassed cathode resistor, as:

.

As we said in Section 3.4, we do not consider the phase so we omit the minus sign.

Let us now consider the case where a capacitor *C _{k}* is used to bypass the cathode resistor

*R*, as shown in Figure 13. The impedance of the capacitor depends on the signal frequency

_{k}*f*and is

.

The impedance of the cathode resistor in parallel with the bypass capacitor is

.

The gain of the voltage amplifier, with local feedback introduced by a cathode resistor and bypassed by a capacitor, is obtained by replacing *R _{k}* with

*Z*in the equation for

_{k}*A*. We obtain:

^{fb}.

The value of the gain depends on the frequency and on the capacitor. It ranges between these two extremes:

.

Minimum gain, equal to the non-bypassed cathode resistor, occurs when the capacitor impedance is maximum (at very low frequencies and/or very low capacitances). Maximum gain, similar to the circuit without cathode resistor, occurs when the capacitor impedance is minimum (at high frequencies and/or large capacitance).

The value of the capacitor should be chosen so that gain is maximum even at very low audible frequencies, as discussed in next example.