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The circuit shown in Fig Ex 4.22 uses transistor $h_{fe} h_{fe}=1k\Omega$ and $h_{fb}=20\Omega$. Circuit lower cutoff frequency (f) is to be 100 Hz. Calculate the following quantities

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a) $C_{2}$ determination

$X_{C2}=h_{ib}=20\Omega$

$C_{2}=\frac{1}{2\pi f_{1}X_{C2}}=\frac{1}{2\times Hz\times 20\Omega}$

=79.6$\mu F$ (use 80$\mu F$ standard value) .......(i)

b) Voltage gain

$A_{v}$=$-\frac{-h_{fe}(R_{C}||R_{L})}{h_{ie}}$

$\frac{-50\times 3.3k\Omega}{1k\Omega}$=-165 ........(ii)

c)$A_{v}$ at 100 Hz

At f=100Hz, $X_{C2}=20\Omega$

$A_{v(1)}$=$\frac{-h_{fe(R_{C}||R_{L})}}{\sqrt {h_{ie}^{2}+[(1+h_{fe})X_{C2}]^{2}}}$

=$\frac{-50\times 3.3k\Omega}{\sqrt{(1k\Omega)^{2}+[(1+50)\times 20\Omega]^{2}}}$

=-116$\frac{A_{v}}{\sqrt 2} \ \ \ \ \ ........(iii)$

d) $A_{v}$ at 100 Hz when $X_{C2}=R_{E}/10$

$A_{Vx}=\frac{-h_{fe}(R_{C}||R_{L})}{\sqrt{(h_{ie}^2+[(1+h_{fe})(R_{E}/10)]^{2})}}$

=$\frac{-50\times 3.3k\Omega}{\sqrt{ [1k\Omega]^{2}+(1+50)\times(1k\Omega /10)]^{2}}}$

=-31.7 ........(iv)

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