3.6.3    Gain of the voltage amplifier with self-bias

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_a. Let V_in 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 R_k opposite to the cathode. The voltage V_fb measured between the two terminals of R_k, corresponding to the voltage drop introduced by R_k itself, is the feedback voltage. The grid to cathode voltage , resulting from the combined action of the input and feedback voltages, is:

.

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_in, because of the voltage drop introduced by R_k. Given that the voltage drop is V_fb,  will be equal to V_in plus 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_k, as shown in Figure 13. The impedance of the capacitor depends on the signal frequency 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_k in the equation for A^fb. We obtain:

.

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.

Example 5: Gain of a voltage amplifier with self-bias Let us consider again the configuration of a voltage amplifier with the green loadline in Figure 8. In Example 1, we set the load RL to 150K Ohm. In Example 2 we determined that the anode resistance ra is 75K Ohm, and the gain of the amplifier at the operating point identified by the red spot, with no cathode resistor, is 66.6. In Example 4, we determined that the cathode resistor needed to set the operating point at the red spot, using self-bias, is 2K Ohm. Remember that he amplification factor of a 12AX7 vacuum tube is μ=100. Therefore, the gain of the voltage amplifier, using self-bias with this cathode resistor is . Considering that the full gain with no cathode resistor is 66.6, the new gain corresponds to 20∙log(35.12/66.6) = -5.5dB with respect to full gain. Suppose now we use a bypass capacitor of 150μF. At a frequency of f=1 Hz the capacitor impedance is . The impedance of the cathode resistor in parallel with the bypass capacitor is . The gain of the voltage amplifier at 1 Hz, with the bypass capacitor, is: . This corresponds to 20∙log(47/66.6) = -3dB with respect to full gain. Repeating the same process for f=10Hz, we obtain a gain of 63.6, corresponding to 20∙log(63.6/66.6) = -0.4dB. This is fairly acceptable, since no attenuation is realistically perceived at all audible frequencies. Figure 14 reports the output gain attenuation, with the above configuration, with frequency ranging up to 100 Hz. It can be seen that the attenuation is practically eliminated from 40Hz on. Figure 14: Output gain attenuation with self-bias and bypass capacitor. Using a capacitor of 150μF we have -3dB output gain attenuation at 1 Hz and -0.4dB at 10 Hz. Attenuation is practically eliminated at 40 Hz.