We discussed that a grid voltage, negative with respect to the cathode voltage, controls the current that traverses a vacuum tube and that the grid bias voltage determines the quiescent vacuum tube operating point. The input voltage signal is added to the grid bias voltage and is amplified.

There are two largely used techniques to provide a vacuum tube with a grid bias voltage, negative with respect to the cathode. The fixed bias technique requires a separate power supply that provides the wanted negative voltage. The cathode bias or self-bias technique connects the grid to ground and elevate the cathode voltage above ground. In this way the grid voltage is negative with respect to the cathode voltage. Let us discuss these two biasing techniques more in details.

### 3.6.1    Fixed bias

The fixed bias schema is given in Figure 11. Two different power supplies are used. PS1 gives the high-tension V+ to the anode of the vacuum tube, through the load. The negative of PS1 is connected to ground. PS2 produces the needed grid bias voltage Vg. The negative grid bias -Vg is obtained by connecting the positive of PS2 to ground. In this way, the positive of PS2 is at ground level and the negative is at –Vg with respect to ground. The grid receives the -Vg bias voltage using a resistor Rl. Given that no current goes through the grid, in normal operations, the resistor Rl does not affect the voltage seen by the grid. The cathode is also connected to ground so that the grid is at –Vg with respect to the cathode, as needed.

The usage of resistor Rl will be better discussed in Section 4.1.1. For the moment, we just mention that one of the purposes of Rl (also called the grid leak) is to provide the input signal, received from previous stage, with a high impedance path to ground.

The capacitor Cd, from the grid leak resistor to ground, decouples the residual input signal, that traverses Rl, from the bias voltage supply. Consider that, generally the bias power supply provides bias voltage to several vacuum tubes in the amplifier. For instance, in a stereo amplifier, both left and right channels are sometimes biased by the same power supply. The residual input signal, which traverses Rl, is added to the bias voltage and goes also to the other channels, where it is amplified by the other vacuum tubes, creating problems of cross-talk. In order to avoid that, the capacitor Cdforms, with the resistor Rl, a low-pass filter that shorts to ground the residual input signal. The value of this capacitor should be large, so that even very low frequencies are shorted to ground and do not go to the grid of the other vacuum tubes.

 Example 3: Determining the decoupling capacitor for fixed bias Suppose, for simplicity, that Rl is the only resistance seen by the capacitor Cd, and suppose its value is 200K Ohm. In order to have a low-pass filter with a very low cut-off frequency, for instance 1 Hz, we use the low-pass filter equation and we obtain the following capacitance: .

### 3.6.2    Cathode bias or self-bias

Negative voltage between grid and cathode can also be obtained by connecting the grid to ground voltage and by elevating the cathode voltage. This technique is generally referred as cathode bias or self-bias. The cathode voltage is elevated by connecting it to ground through the resistor Rk, generally called the cathode resistor, as shown in Figure 12. Given that, generally, there is an anode current also at the quiescent state, the resistor Rk produces a voltage drop from the cathode to ground so that the cathode voltage is above ground. The grid, being at ground voltage, is negative with respect to the cathode.

Note that, also in this case, the grid is not directly connected to ground. Rather, a grid leak resistor Rl is used to provide the input signal with a high impedance path to ground, as we already discussed for the fixed bias. Since there is no current flowing through the grid, it is at ground voltage.

The value of Rk can be computed using the Ohm law by knowing the bias current, that is the cathode current at the operating point (quiescent state), as shown in next Example.

 Example 4: Determining the cathode resistor for self-bias Suppose we use a 12AX7 vacuum tube and we want to set the operating point at the red spot in Figure 8, using the green loadline. This corresponds to a bias current of 0.75 mA. Using the average anode characteristics graphs, we see that a grid voltage of -1.5V, with respect to the cathode, is needed to obtain a current of 0,75 mA. Using the Ohm law, we find that the resistance needed to elevate the cathode at 1.5V, when the current is 0.75 mA, is: .

It is important to mention that the cathode resistor introduces a form of local negative feedback. In fact, when the current increases, the cathode voltage increases as well. In this case, the grid becomes more negative, with respect to the cathode, and tends to reduce the vacuum tube conduction. When current decrease, we have the opposite effect and the grid becomes less negative, increasing vacuum tube conduction. In other words, the cathode resistor tends to oppose the amplification of the signal and reduces the gain of the vacuum tube. In order to mitigate and almost eliminate this effect, a bypass or decoupling capacitor Ck, is generally introduced in the circuit, as shown in Figure 12. The bypass capacitor compensates the cathode voltage variation trying to maintain it as stable as possible, when amplifying a signal. In this way, local negative feedback is significantly reduced and gain significantly increased, as discussed in next Section.

### 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 cathode resistor reduces the gain of the vacuum tube. This local negative feedback can be significantly neutralized 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 Ck 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 ra. Let Vin 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 Rk opposite to the cathode. The voltage Vfb measured between the two terminals of Rk, corresponding to the voltage drop introduced by Rk 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). Vin 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 Vin, because of the voltage drop introduced by Rk. Given that the voltage drop is Vfb,  will be equal to Vin plus Vfb.

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 RL and can be computed using the voltage divider equation as

.

Using the voltage divider equation again, the feedback voltage Vfb is

.

Replacing Vfb 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 Ck is used to bypass the cathode resistor Rk, as shown in Figure 13. The impedance of the capacitor depends on the signal frequency 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 Rk with Zk in the equation for Afb. 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. 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.