Question

In the circuit shown below, the output impedance $\left(Z_{out}\right)$ is

• A. $h_{11}-\frac{h_{12}{h}_{21}}{h_{22}{Z}_s+1}$
• B. $\frac{1}{h_{22}-\frac{h_{12}{h}_{21}}{h_{11}+Z_s}}$
• C. $\frac{1}{h_{11}-\frac{h_{12}{h}_{21}}{h_{11}+Z_s}}$
• D. $h_{22}-\frac{h_{12}{h}_{21}}{h_{11}{Z}_x+1}$

Answer 1

The correct option is B $\frac{1}{h_{22}-\frac{h_{12}{h}_{21}}{h_{11}+Z_s}}$

To determine $Z_{out}$ the source voltage $V_1$ is made zero with $Z_s$ being maintained


Applying KVL in the input loop
$h_{12}{V}_2=-I_1{h}_{11}-Z_s{I}_1$
$-h_{12}{V}_2=I_1[h_{11}+Z_s]$ $...\left(i\right)$
From the output port KCL gives
$I_2-h_{21}{I}_1=h_{22}{V}_2$ $...\left(ii\right)$
From (i) and (ii), we get
$I_2-h_{21}\left[\frac{-h_{12}}{h_{11}+Z_x}\right]=h_{22}{V}_2$
$I_2+\frac{h_{12}{h}_{21}}{h_{11}+Z_s}.V_2=h_{22}{V}_2$
$V_2[h_{22}-\frac{h_{12}{h}_{21}}{h_{11}+Z_s}]=I_2$
Output impedance, $Z_{out}=\frac{V_2}{I_2}=\frac{1}{h_{22}-\frac{h_{12}{h}_{21}}{h_{11}+Z_s}}\Omega$