Question 17.5: The circuit in Figure 17–6 is a small-signal model of a stan......

The sum of currents at nodes A and B can be written as

NodeA:    I_{1} + g_{m}V_{1}  –  \frac{V_{1}}{R_{E}}= 0

NodeB:    I_{2}  –  g_{m}V_{1}  –  \frac{V_{2}}{R_{D}}= 0

Solving the node A equation for V_{1} yields

V_{1} = \left[\frac{R_{E}}{1  –  g_{m}R_{E}}\right]I_{1} + [0]V_{2}

= h_{11}I_{1}          +        h_{12}V_{2}

Comparing this result and h-parameter definitions we have h_{11} = R_{E}/ (1  –  g_{m}R_{E}) and h_{12} = 0. Inserting the expression for V_{1} from node A into the node B equation and solving for I_{2} yields

I_{2} = \left[\frac{g_{m}R_{E}}{1  –  g_{m}R_{E}}\right]I_{1} + \left[\frac{1}{R_{D}}\right] V_{2}

= h_{21}I_{1}           +           h_{22}V_{2}

which means that h_{21} = g_{m}R_{E}/ (1  –  g_{m}R_{E} ) and h_{22} = 1/R_{D}.