If the parameters of a circuit are known then it is possible to construct another circuit that 1. has a current source and a voltage source (whose values are determined by the known h-parchment values and 2. is entirely equivalent to the original circuit. Figure 9-9 shows the hybrid equivalent circuit that can be substituted for any circuit whose h parameters are known. Note that it contains a controlled voltage source whose output is volts and a controlled current source whose output is amperes. An impedance equal to ohms is in series with the voltage source, and an impedance equal to Jilin ohms is in parallel it the current source. It is easy to verify that the hybrid equivalent circuit shown in Figure 9-9 has the same h parameters as the circuit to which it is equivalent. If we set Vz = 0 in the equivalent circuit, by shorting the output port, then the voltage source on the input side produces 0 V (i.e it is a short circuit). and the input resistance is then really ohms. Also. with the output shorted. it can be seen that 2 = H so hell = l as required. It is left as an exercise at the end of thee $ ch P Cr to verify that the reverse voltage ratio and the output admitted nee in f~p. re 9-9 are hi and Ian, respectively. It is, important to realize that even though the hybrid circuit contains active sources. it is a valid equivalent for a network that contains only passive com opponents. Of course for very simple networks there is no reason to use the hybrid seq equivalent circuit as an analysis tool. For complex devices such as transistors, the hybrid circuit is a very useful means for obtaining a relatively simple but accurate model, provided the values of the device’s parameters are known. The next example demonstratives the validity of using the hybrid circuit for a simple, restive network Using the h-parameter values found in Example 9-1 for the network shown in re 0_10. verify that the hybrid equivalent circuit can be used to determine to v1 across the 24- resistor when {he pupil, is V. Solution”. Let us first use simple network analysis to find V2 The parallel combination of the 40-0 and 24-0 resistors is equivalent to IS 0, so by the voltage-divider f sure 9-11 shows the hybrid equivalent circuit of v, = V and a 24-fl resistor connected across the output. (The controlled voltage source is shown as a source, since we are modeling a circuit.) Note that = IV2 = V2 and Although the computations are laborious, we have demonstrated that analysis~sing the hybrid equivalent circuit produces the same result (V2 = 3 V) as the direct analysis.As a memory aid. h parameters are given special subscripts n place of the double-number subscripts we have been using. Following is a list of those special designations:Figure 9-12 shows the hybrid equivalent circuit using these new designations. We will use this notation in all future discussions.

**Gain and Impedance Computations Using h Parameters t**

We will now derive h-parameter expressions for the voltage gain, current gain, input impedance, and output impedance of a circuit that is driven by a signal source having signal-source impedance and that has a load impedance connected across its output. Figure 9-13 shows the general hybrid equivalent circuit with signal source

and load impedance connected. We should note that the quantities we seek are not equal to the h parameters themselves, since the values of those parameters are derived only under the very special conditions V2 = 0 or i, = O.We begin our analysis by finding the voltage gain A. = VI/V” i.e., the gain between the input of the hybrid circuit and the load. omitting Vs and i impedance for the time being. The input current is the voltage difference across h, divided by hi:To Lind the current gain A; = Liz,. we apply the current-divider rule to desuetude the portion of the current in the current source that flows in the load:If Z,. = 00, then A since there is no load current in that case. If Z,- = 0 (output short-circuited), then A; reduces to hi. as would be expected from the definition of hI’ The input impedance Z; looking into Y”rid circuit is found by dividing v, by it. From equation 9-12. we have v, a g by li,Just as the load impedance Z” on the output side affects the input impedance, the source impedance Zs on the input side affects the output impedance. To include Zs in the analysis, we must set Us to O.In that case, the input current ;1 is the voltage -h V2 divided by the total impedance in the input loop: