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## Q. 33.8

What is the voltage gain in Fig. 33-24? The output voltage across the load resistor?

## Verified Solution

The ac collector resistance is

$r_c = R_C \parallel R_L = (3.6 k\Omega \parallel 2.2 k\Omega) = 1.37 k\Omega$

The dc emitter current is approximately

$I_E=\frac{1.8 V-0.7 V}{1K\Omega }=1.1 mA$

The ac resistance of the emitter diode is

$r^\prime _c=\frac{25 mV}{1.1 mA} =22.7 \Omega$

The voltage gain is

$A_V=\frac{r_c}{r^\prime _c}=\frac{1.37 k\Omega}{22.7 \Omega}=60.35$

The output voltage is

$v_{out} = A_Vv_{in} = (60.35)(100) \mu V = 6.035 mV$