Power amplifier with driver

Question

Consider the circuit of Figure 1 using Vcc = 6 volts, Vee = −6 volts, R1 = 6.8 kΩ, R4 = 1 kΩ, R3 = 100 Ω, Rload = 100 Ω, Cin = 1 µF and Cout = 100 µF. R2 is an adjustable resistance (pot or decade box). For proper bias, the emitters of the output transistors should be at 0 volts DC. For this to be true there must be a voltage of Vcc − Vbe, or approximately 5.3 volts, across R4. Ignoring base currents, this establishes the ICQ of transistor 1 which in turn creates a potential drop across R3. From this the voltage across R2 may be determined. Knowing the value of R1 and the total supply presented, Ohm’s law or the voltage divider rule may be used to compute the required setting for R2. Compute the required value for R2 and record it in Table 1.

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Answer 1

Ignoring base currents, the current through R1 and R2 is the same, and is a function of Vcc, Vee, R1 and R2. The voltage at the base of Q1 is a function of this current, Vee and R2. Once the base voltage of Q1 is known R2 can be solved for.

Similarly, the current through R3 and R4 is the same. This current is (Vcc-be)/R4 as described in the problem. This current multiplied by R3 sets the voltage drop from the emitter of Q1 to Vee. There is an additional Vbe to get to the base of Q1. After calculating the base voltage of Q1 plug back into the equation with R2 to get the value of R2.