Question 19.15: Find the h parameters of the network in Fig. 19.49....
Find the h parameters of the network in Fig. 19.49.
From Eq. (19.16),
\begin{matrix}h_{11} = \frac{V_{1}}{I_{1}} \mid_{V_{2} = 0} , h_{12} = \frac{V_{1}}{V_{2}} \mid_{I_{1} = 0} \\\\ h_{21} = \frac{I_{2}}{I_{1}} \mid_{V_{2} = 0} , h_{22} = \frac{I_{2}}{V_{2}} \mid_{I_{1} = 0}\end{matrix} \quad (19.16)
h_{11} = \frac{V_1}{I_1}|_{v_2=0} ,\quad h_{21} = \frac{I_2}{I_1}|_{v_2=0}showing that h_{11} and h_{21} can be found by setting V_{2} = 0 . Also by setting I_{1} = 1 A , h_{11} becomes V_{1} /1 while h_{21} becomes I_{2}/1 . With this in mind, we draw the schematic in Fig. 19.50(a). We insert a 1-A dc current source
IDC to take care of I_{1} = 1 A the pseudocomponent VIEWPOINT to display V_{1} and pseudocomponent IPROBE to display I_{2} . After saving the schematic, we run PSpice by selecting Analysis/Simulate and note the values displayed on the pseudocomponents. We obtain
h_{11} = \frac{V_{1}}{1} = 10 Ω , h_{21} = \frac{I_{2}}{1} = – 0.5
Similarly, from Eq. (19.16),
h_{12} = \frac{V_{1}}{V_{2}} \mid_{I_{1} = 0} , h_{22} = \frac{I_{2}}{V_{2}} \mid_{I_{1} = 0}
indicating that we obtain h_{12} and h_{22} by open-circuiting the input port (I_{1} = 0 ) . By making V_{2} = 1 V ,h_{12} becomes V_{1} /1 while h_{22} becomes I_{2}/1 . Thus, we use the schematic in Fig. 19.50(b) with a 1-V dc voltage source VDC inserted at the output terminal to take care of V_{2} = 1 V. The pseudocomponents VIEWPOINT and IPROBE are inserted to display the values of V_{1} and I_{2} respectively. (Notice that in Fig. 19.50(b), the 5-Ω resistor is ignored because the input port is opencircuited and PSpice will not allow such. We may include the 5-Ω resistor if we replace the open circuit with a very large resistor, say, 10 MΩ. ) After simulating the schematic, we obtain the values displayed on the pseudocomponents as shown in Fig. 19.50(b). Thus,
h_{12} = \frac{V_{1}}{1} = 0.8333 , h_{22} = \frac{I_{2}}{1} = 0.1833 S
