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## Q. 33.11

What is the voltage gain of the emitter follower in Fig. 33-29? If $\beta$  = 150, what is the ac load voltage? ## Verified Solution

The dc base voltage is half the supply voltage:

$V_B = 7.5 V$

The dc emitter current is

$I_E=\frac{6.8 V}{2.2 K\Omega}=3.09 mA$

and the ac resistance of the emitter diode is

$r^\prime _e=\frac{25 mV}{3.09 mA}= 8.09 \Omega$

The external ac emitter resistance is

$r_e = 2.2 k\Omega\parallel 6.8 k\Omega = 1.66 k\Omega$

The voltage gain equals

$A_V=\frac{1.66 k\Omega }{1.66 k\Omega+8.09 \Omega }=0.995$

The input impedance of the base is

$z_{in(base)} = 150(1.66 k\Omega + 8.09 \Omega) = 250 k\Omega$

This is much larger than the biasing resistors. Therefore, to a close approximation, the input impedance of the emitter follower is

$z_{in(stage)} = 4.7 k\Omega \parallel 4.7 k\Omega = 2.35 k\Omega$

The ac input voltage is

$v_{in}=\frac{2.35 k\Omega }{600\Omega+2.35 k \Omega } 1V= 0.797 V$

The ac output voltage is

$v_{out} = 0.995(0.797 V) = 0.793 V$