What is the voltage gain of the emitter follower in Fig. 33-29? If \beta = 150, what is the ac load voltage?
Chapter 33
Q. 33.11
Step-by-Step
Verified Solution
The dc base voltage is half the supply voltage:
V_B = 7.5 VThe dc emitter current is
I_E=\frac{6.8 V}{2.2 K\Omega}=3.09 mAand the ac resistance of the emitter diode is
r^\prime _e=\frac{25 mV}{3.09 mA}= 8.09 \OmegaThe external ac emitter resistance is
r_e = 2.2 k\Omega\parallel 6.8 k\Omega = 1.66 k\OmegaThe voltage gain equals
A_V=\frac{1.66 k\Omega }{1.66 k\Omega+8.09 \Omega }=0.995The input impedance of the base is
z_{in(base)} = 150(1.66 k\Omega + 8.09 \Omega) = 250 k\OmegaThis is much larger than the biasing resistors. Therefore, to a close approximation, the input impedance of the emitter follower is
z_{in(stage)} = 4.7 k\Omega \parallel 4.7 k\Omega = 2.35 k\OmegaThe ac input voltage is
v_{in}=\frac{2.35 k\Omega }{600\Omega+2.35 k \Omega } 1V= 0.797 VThe ac output voltage is
v_{out} = 0.995(0.797 V) = 0.793 V