The parameters of a two port network are designated b and Their values for any network are such that the following two network equations are always satisfied In other word for any possible combination of input and output voltage and current might ever exist in a given network, equations 9-1 and 9-2 will be satisfied if the parameter values of that network are used in the equations. Think and dwell for a moment on the implications of the previous statement: It should seem startling that such a powerful set of parameters can exist. However as we shall presently see. the two equations themselves provide us with the means for defining the parameters in a way that forces them to satisfy the equations .Note that the left side of equation 9-1 is a voltage. Therefore each term on the right-hand side must also be a voltage. Similarly each term in equation 9-2 current. It follows that the units of the h parameters are different. For example, “II must have the units of ohms, since it multiplies the current to produce a voltage term, while the units of must be Siemens, since it multiplies the v Oz current term. This mixture of units accounts for the name hybrid parameters. If we set Vz equal to 0 in equation 9-1, we obtain In words, is the ratio of input voltage to input current when the output voltage (vz) is 0, i.e., when output is short-circuited. See Figure 9-3. It is clear from Ohm’s law that 1111 must have the units of ohms, and it is called the short-circuit input v-c will consider all v and currents to be Quantities, so we will refer to more properly as the short-circuit input impedance In words, is the ratio of input voltage to output voltage when the input current is 0, i.e., when the input i~ open-circuited. Since is the ratio of two voltages, it is a dimensionless quality (It L sometimes said to have the units of volts per volt. V/V). Since we are accustomed to thinking of voltage gain as the ratio of output to input voltage, and is just the reverse, it is called the open-circuit reverse voltage ratio, (It is sometimes referred to as the reverse voltage transfer ratio.)Since the ratio of the output to input current with the output short circuited it is a dimensionless quantity called the short-circuit forward current ratio (sometimes called the forward current transfer ratio). Setting 0 in equation 9-2, we obtain Since “!2 is the ratio of output current to output voltage with the input it has the units of and is called the open-circuit output admittance (or conductance. in the case of circuits). Table 9-1 summarizes the definitions and names of the h parameters. Note that short-circuit always refers to the condition at the output port, and open-circuit always refers to the condition at the input port. The next two examples demonstrate procedures that can be used to find the values of the parameters of some simple networks.Find the” parameters of the network shown in Figure 9-4 Solution. Figure 9-5 shows the network with the output short-circuited (U2 = 0), which is the condition required in the definitions of 1111 and Sin is the resistance looking into the input and since the 40-0 resistor is Values of the can be found without assigning explicit values to V2, or However it is sometimes easier to perform the computations by assuming ,I value for one of those quantities and then solving for another. For example, in Figure 9-5 could have been found by assuming that VI equals (say) no V and then solving for It is obvious that ;1 = I A when VI = 60 V, so Si ice the -n resistor is shorted, all the current ;1 will flow in the short. Thus, the magnitudes of and ;2 arc equal. N current flows out of port 2. that is opposite to the assumed positive direction of Therefore i. = and Figure 9-6 shows the network with the input open-circuited (i, = 0), the con necessary for finding 12 and 22. To find 12, we must find the ratio of VI to V2 in Figure 9-6. Since = (l, there is no ‘voltage drop across the resistor. Therefore. the voltage VI must be the same as the voltage V2 developed across the 40-H resistor v = V2. It follows that .The value of “12 could have been found by imagining that a voltage source, say. 20 V. was applied across the output terminals. With the input open-circuited. it would again be apparent that VI = V2 = 20 V, leading to the same value for 1112: VI/V2 = 1. Note. however. that it would be 10 VI across the input. calculate. and then find If a voltage source VI were applied across the input, then current would now into the circuit, violating the condition that i, = O.