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Figure 19.12

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This procedure provides us with a means of calculating or measuring the y parameters. The impedance and admittance parameters are collectively referred to as immittance parameters.

Figure 19.12

Determination of the y parameters: (a) finding y11 and y21, (b) finding y12 and y22.

For a two-port network that is linear and has no dependent sources, the transfer admittances are equal ( y12 = y21). This can be proved in the same way as for the z parameters. A reciprocal network (y12 = y21) can be modeled by the Π-equivalent circuit in Fig. 19.13(a). If the network is not reciprocal, a more general equivalent network is shown in Fig. 19.13(b).

Figure 19.13

(a) Π-equivalent circuit (for reciprocal case only), (b) general equivalent circuit.

Figure 19.14 For Example 19.3.

Example 19.3 Obtain the y parameters for the Π network shown in Fig. 19.14.

Solution:

■ METHOD 1 To find y11 and y21, short-circuit the output port and connect a current source I1 to the input port as in Fig. 19.15(a). Because the 8-Ω resistor is short-circuited, the 2-Ω resistor is in parallel with the 4-Ω resistor. Hence,

V1=I1(4∥2)=43I1\mathbf{V}_1 = \mathbf{I}_1(4 \parallel 2) = \frac{4}{3}\mathbf{I}_1

, y11=I1V1=I143I1=0.75 S\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{\mathbf{I}_1}{\frac{4}{3}\mathbf{I}_1} = 0.75 \text{ S}

By current division,

−I2=44+2I1=23I1,y21=I2V1=−23I143I1=−0.5 S-\mathbf{I}_2 = \frac{4}{4+2}\mathbf{I}_1 = \frac{2}{3}\mathbf{I}_1, \qquad \mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1} = \frac{-\frac{2}{3}\mathbf{I}_1}{\frac{4}{3}\mathbf{I}_1} = -0.5 \text{ S}

(a)

To get y12 and y22, short-circuit the input port and connect a current source I2 to the output port as in Fig. 19.15(b). The 4-Ω resistor is shortcircuited so that the 2- and 8-Ω resistors are in parallel.

V2=I2(8∥2)=85I2,y22=I2V2=I285I2=58=0.625 S\mathbf{V}_2 = \mathbf{I}_2(8 \parallel 2) = \frac{8}{5}\mathbf{I}_2, \qquad \mathbf{y}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} = \frac{\mathbf{I}_2}{\frac{8}{5}\mathbf{I}_2} = \frac{5}{8} = 0.625 \text{ S}

By current division,

−I1=88+2I2=45I2,y12=I1V2=−45I285I2=−0.5 S-\mathbf{I}_1 = \frac{8}{8+2} \mathbf{I}_2 = \frac{4}{5} \mathbf{I}_2, \qquad \mathbf{y}_{12} = \frac{\mathbf{I}_1}{\mathbf{V}_2} = \frac{-\frac{4}{5} \mathbf{I}_2}{\frac{8}{5} \mathbf{I}_2} = -0.5 \text{ S}

as obtained previously.

Figure 19.15

For Example 19.3: (a) finding y11 and y21, (b) finding y12 and y22.

Figure 19.16 For Practice Prob. 19.3.

Determine the y parameters for the two-port shown in Fig. 19.17. Example 19.4

Solution:

We follow the same procedure as in the previous example. To get y11 and y21, we use the circuit in Fig. 19.18(a), in which port 2 is short-circuited and a current source is applied to port 1. At node 1,

V1−Vo8=2I1+Vo2+Vo−04\frac{\mathbf{V}_1 - \mathbf{V}_o}{8} = 2\mathbf{I}_1 + \frac{\mathbf{V}_o}{2} + \frac{\mathbf{V}_o - 0}{4}

But I1 = V1 _______ − V*o* 8 ; therefore,

0=V1−Vo8+3Vo40 = \frac{V_1 - V_o}{8} + \frac{3V_o}{4} 0=V1−Vo+6Vo⇒V1=−5Vo0 = \mathbf{V}_1 - \mathbf{V}_o + 6\mathbf{V}_o \qquad \Rightarrow \qquad \mathbf{V}_1 = -5\mathbf{V}_o

2 Ω

Solution of Example 19.4: (a) finding y11 and y21, (b) finding y12 and y22.

Hence,

I1=−5Vo−Vo8=−0.75Vo\mathbf{I}_1 = \frac{-5\mathbf{V}_o - \mathbf{V}_o}{8} = -0.75\mathbf{V}_o

and

y11=I1V1=−0.75Vo−5Vo=0.15 S\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{-0.75 \mathbf{V}_o}{-5 \mathbf{V}_o} = 0.15 \text{ S}

At node 2,

Vo−04+2I1+I2=0\frac{\mathbf{V}_o - 0}{4} + 2\mathbf{I}_1 + \mathbf{I}_2 = 0 −I2=0.25Vo−1.5Vo=−1.25Vo-I_2 = 0.25V_o - 1.5V_o = -1.25V_o

Hence,

y21=I2V1=1.25Vo−5Vo=−0.25 S\mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1} = \frac{1.25\mathbf{V}_o}{-5\mathbf{V}_o} = -0.25 \text{ S}

Similarly, we get y12 and y22 using Fig. 19.18(b). At node 1,

0−Vo8=2I1+Vo2+Vo−V24\frac{0 - \mathbf{V}_o}{8} = 2\mathbf{I}_1 + \frac{\mathbf{V}_o}{2} + \frac{\mathbf{V}_o - \mathbf{V}_2}{4}

But I1=0−Vo8\mathbf{I}_1 = \frac{0 - \mathbf{V}_o}{8} ; therefore,

0=−Vo8+Vo2+Vo−V240 = -\frac{\mathbf{V}_o}{8} + \frac{\mathbf{V}_o}{2} + \frac{\mathbf{V}_o - \mathbf{V}_2}{4}

or

0=−Vo+4Vo+2Vo−2V2⇒V2=2.5Vo0 = -\mathbf{V}_o + 4\mathbf{V}_o + 2\mathbf{V}_o - 2\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{V}_2 = 2.5\mathbf{V}_o

Hence,

y12=I1V2=−Vo/82.5Vo=−0.05 S\mathbf{y}_{12} = \frac{\mathbf{I}_1}{\mathbf{V}_2} = \frac{-\mathbf{V}_o/8}{2.5\mathbf{V}_o} = -0.05 \text{ S}

At node 2,

Vo−V24+2I1+I2=0\frac{\mathbf{V}_o - \mathbf{V}_2}{4} + 2\mathbf{I}_1 + \mathbf{I}_2 = 0 −I2=0.25Vo−14(2.5Vo)−2Vo8=−0.625Vo-\mathbf{I}_2 = 0.25\mathbf{V}_o - \frac{1}{4}(2.5\mathbf{V}_o) - \frac{2\mathbf{V}_o}{8} = -0.625\mathbf{V}_o

Thus,

or

y22=I2V2=0.625Vo2.5Vo=0.25 S\mathbf{y}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} = \frac{0.625 \mathbf{V}_o}{2.5 \mathbf{V}_o} = 0.25 \text{ S}

Notice that y12 ≠ y21 in this case, given that the network is not reciprocal.

Practice Problem 19.4 Obtain the y parameters for the circuit in Fig. 19.19.

Figure 19.19 For Practice Prob. 19.4.

Answer: y11 = 312.5 mS, y12 = −62.5 mS, y21 = 187.5 mS, y22 = 62.5 mS.