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19.4 Hybrid Parameters

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19.4 Hybrid Parameters

The z and y parameters of a two-port network do not always exist. So there is a need for developing another set of parameters. This third set of parameters is based on making V1 and I2 the dependent variables. Thus, we obtain

V1=h11I1+h12V2V_1 = h_{11}I_1 + h_{12}V_2

\n

I2=h21I1+h22V2I_2 = h_{21}I_1 + h_{22}V_2

(19.14)

or in matrix form,

[V1I2]=[h11h12h21h22][I1V2]=[h][I1V2]\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{h}_{11} & \mathbf{h}_{12} \\ \mathbf{h}_{21} & \mathbf{h}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{h}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix}

(19.15)

The h terms are known as the hybrid parameters (or, simply, h parameters) because they are a hybrid combination of ratios. They are very useful for describing electronic devices such as transistors (see Section 19.9); it is much easier to measure experimentally the h parameters of such devices than to measure their z or y parameters. In f act, we ha ve seen that the ideal transformer in Fig. 19.6, described by Eq. (19.7), does not ha ve z parameters. The ideal transformer can be described by the hybrid parameters, because Eq. (19.7) conforms with Eq. (19.14).

The values of the parameters are determined as

h11=V1I1∣V2=0,h12=V1V2∣I1=0\mathbf{h}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}, \qquad \mathbf{h}_{12} = \frac{\mathbf{V}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}

\n

h21=I2I1∣V2=0,h22=I2V2∣I1=0\mathbf{h}_{21} = \frac{\mathbf{I}_2}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}, \qquad \mathbf{h}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}

\n(19.16)

It is evident from Eq. (19.16) that the parameters h11, h12, h21, and h22 represent an impedance, a voltage gain, a current gain, and an admittance, respectively. This is wh y they are called the h ybrid parameters. To be specific,

h11=Short-circuit input impedance\mathbf{h}_{11} = \text{Short-circuit input impedance}

\n

h12=Open-circuit reverse voltage gain\mathbf{h}_{12} = \text{Open-circuit reverse voltage gain}

\n

h21=Short-circuit forward current gain\mathbf{h}_{21} = \text{Short-circuit forward current gain}

\n

h22=Open-circuit output admittance\mathbf{h}_{22} = \text{Open-circuit output admittance}

\n(19.17)

The procedure for calculating the h parameters is similar to that used for the z or y parameters. We apply a v oltage or current source to the appropriate port, short-circuit or open-circuit the other port, depending on the parameter of interest, and perform re gular circuit analysis. F or reciprocal networks, h12 = −h21. This can be proved in the same way as we proved that z12 = z21. Figure 19.20 shows the hybrid model of a twoport network.

A set of parameters closely related to the h parameters are the g parameters or inverse hybrid parameters. These are used to describe the terminal currents and voltages as

I1=g11V1+g12I2I_1 = g_{11}V_1 + g_{12}I_2

\n

V2=g21V1+g22I2V_2 = g_{21}V_1 + g_{22}I_2

\n(19.18)

Figure 19.20 The h-parameter equivalent network of a two-port network.

or

[

[I1V2]=[g11g12g21g22][V1I2]=[g][V1I2]\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{g}_{11} & \mathbf{g}_{12} \\ \mathbf{g}_{21} & \mathbf{g}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = [g] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix}

(19.19)

The values of the g parameters are determined as

g11=I1V1∣I2=0,g12=I1I2∣V1=0\mathbf{g}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} \Big|_{\mathbf{I}_2=0}, \qquad \mathbf{g}_{12} = \frac{\mathbf{I}_1}{\mathbf{I}_2} \Big|_{\mathbf{V}_1=0}

\n

g21=V2V1∣I2=0,g22=V2I2∣V1=0\mathbf{g}_{21} = \frac{\mathbf{V}_2}{\mathbf{V}_1} \Big|_{\mathbf{I}_2=0}, \qquad \mathbf{g}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} \Big|_{\mathbf{V}_1=0}

\n(19.20)

Thus, the inverse hybrid parameters are specifically called

g11 = Open-circuit input admittance **g**12 = Short-circuit reverse current gain (19.21) g21 = Open-circuit forward voltage gain g22 = Short-circuit output impedance

Figure 19.21 shows the inverse hybrid model of a tw o-port network. The g parameters are frequently used to model field-effect transistors.

Example 19.5 Find the hybrid parameters for the two-port network of Fig. 19.22.

Solution:

To find h11 and h21, we short-circuit the output port and connect a current source I1 to the input port as shown in Fig. 19.23(a). From Fig. 19.23(a),

V1=I1(2+3∥6)=4I1\mathbf{V}_1 = \mathbf{I}_1(2 + 3 \parallel 6) = 4\mathbf{I}_1

Hence,

For Example 19.5.

Figure 19.23

Figure 19.21

network.

For Example 19.5: (a) computing h11 and

h21, (b) computing h12 and h22.

6 Ω

2 Ω 3 Ω

h11 = ___ V1 I1 = 4 Ω

Also, from Fig. 19.23(a) we obtain, by current division,

−I2=66+3I1=23I1-\mathbf{I}_2 = \frac{6}{6+3} \mathbf{I}_1 = \frac{2}{3} \mathbf{I}_1

Hence,

h21=I2I1=−23\mathbf{h}_{21} = \frac{\mathbf{I}_2}{\mathbf{I}_1} = -\frac{2}{3}

To obtain h12 and h22, we open-circuit the input port and connect a voltage source V2 to the output port as in Fig. 19.23(b). By voltage division,

V1=66+3V2=23V2\mathbf{V}_1 = \frac{6}{6+3} \mathbf{V}_2 = \frac{2}{3} \mathbf{V}_2

Hence,

h12=V1V2=23\mathbf{h}_{12} = \frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{2}{3}

Also,

V2=(3+6)I2=9I2\mathbf{V}_2 = (3+6)\mathbf{I}_2 = 9\mathbf{I}_2

The g-parameter model of a two-port

Thus,

h22=I2V2=19S\mathbf{h}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} = \frac{1}{9} S

Answer:

h11=2.4 Ω\mathbf{h}_{11} = 2.4 \, \Omega

, h12=0.4\mathbf{h}_{12} = 0.4 , h21=−0.4\mathbf{h}_{21} = -0.4 , h22=200 mS\mathbf{h}_{22} = 200 \, \text{mS} .

Determine the Thevenin equivalent at the output port of the circuit in Example 19.6 Fig. 19.25.

Solution:

To find ZTh and VTh, we apply the normal procedure, keeping in mind the formulas relating the input and output ports of the h model. To obtain ZTh, remove the 60-V voltage source at the input port and apply a 1-V voltage source at the output port, as shown in Fig. 19.26(a). From Eq. (19.14),

V1=h11I1+h12V2V_1 = h_{11}I_1 + h_{12}V_2

(19.6.1)

I2=h21I1+h22V2I_2 = h_{21}I_1 + h_{22}V_2

(19.6.2)

But

V2=1V_2 = 1

, and V1=−40I1V_1 = -40I_1 . Substituting these into Eqs. (19.6.1) and

(19.6.2), we get

−40I1=h11I1+h12⇒I1=−h1240+h11-40I_1 = h_{11}I_1 + h_{12} \Rightarrow I_1 = -\frac{h_{12}}{40 + h_{11}}

(19.6.3)

I2=h21I1+h22I_2 = h_{21}I_1 + h_{22}

(19.6.4)

Substituting Eq. (19.6.3) into Eq. (19.6.4) gives

I2=h21I1+h22\mathbf{I}_2 = \mathbf{h}_{21}\mathbf{I}_1 + \mathbf{h}_{22}

(19.6.3) into Eq. (19.6.4) gives

I2=h22−h21h12h11+40=h11h22−h21h12+h2240h11+40\mathbf{I}_2 = \mathbf{h}_{22} - \frac{\mathbf{h}_{21}\mathbf{h}_{12}}{\mathbf{h}_{11} + 40} = \frac{\mathbf{h}_{11}\mathbf{h}_{22} - \mathbf{h}_{21}\mathbf{h}_{12} + \mathbf{h}_{22}40}{\mathbf{h}_{11} + 40}

Therefore,

L2=h22−L1⋅L2h11+40=1.72⋅L1⋅L2h11+40\mathbf{L}_2 = \mathbf{h}_{22} - \frac{\mathbf{L}_1 \cdot \mathbf{L}_2}{\mathbf{h}_{11} + 40} = \frac{1.72 \cdot \mathbf{L}_1 \cdot \mathbf{L}_2}{\mathbf{h}_{11} + 40} ZTh=V2I2=1I2=h11+40h11h22−h21h12+h2240\mathbf{Z}_{\text{Th}} = \frac{\mathbf{V}_2}{\mathbf{I}_2} = \frac{1}{\mathbf{I}_2} = \frac{\mathbf{h}_{11} + 40}{\mathbf{h}_{11}\mathbf{h}_{22} - \mathbf{h}_{21}\mathbf{h}_{12} + \mathbf{h}_{22}40}

Substituting the values of the h parameters,

Substituting the values of the h parameters,
\n

ZTh=1000+40103×200×10−6+20+40×200×10−6\mathbf{Z}_{\text{Th}} = \frac{1000 + 40}{10^3 \times 200 \times 10^{-6} + 20 + 40 \times 200 \times 10^{-6}}

\n

=104020.21=51.46 Ω= \frac{1040}{20.21} = 51.46 \ \Omega

To get VTh, we find the open-circuit voltage V2 in Fig. 19.26(b). At the input port,

−60+40I1+V1=0⇒V1=60−40I1(19.6.5)-60 + 40I_1 + V_1 = 0 \qquad \Rightarrow \qquad V_1 = 60 - 40I_1 \qquad (19.6.5)

Figure 19.26 For Example 19.6: (a) finding ZTh, (b) finding VTh.

(b)

Figure 19.25

For Example 19.6.

At the output,

I2=0(19.6.6)\mathbf{I}_2 = 0 \tag{19.6.6}

Substituting Eqs. (19.6.5) and (19.6.6) into Eqs. (19.6.1) and (19.6.2), we obtain

60−40I1=h11I1+h12V260 - 40\mathbf{I}_1 = \mathbf{h}_{11}\mathbf{I}_1 + \mathbf{h}_{12}\mathbf{V}_2

or

60=(h11+40)I1+h12V260 = (\mathbf{h}_{11} + 40)\mathbf{I}_1 + \mathbf{h}_{12}\mathbf{V}_2

(19.6.7)

and

0=h21I1+h22V2⇒I1=−h22h21V2(19.6.8)0 = \mathbf{h}_{21}\mathbf{I}_1 + \mathbf{h}_{22}\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{I}_1 = -\frac{\mathbf{h}_{22}}{\mathbf{h}_{21}}\mathbf{V}_2 \tag{19.6.8}

Now substituting Eq. (19.6.8) into Eq. (19.6.7) gives

60=[−(h11+40)h22h21+h12]V260 = \left[ -(\mathbf{h}_{11} + 40) \frac{\mathbf{h}_{22}}{\mathbf{h}_{21}} + \mathbf{h}_{12} \right] \mathbf{V}_2

or

60=[−(h11+40)22h21+h12]V260 = \left[ -(\mathbf{h}_{11} + 40) \frac{22}{\mathbf{h}_{21}} + \mathbf{h}_{12} \right] \mathbf{V}_2 VTh=V2=60−(h11+40)h22/h21+h12=60h21h12h21−h11h22−40h22\mathbf{V}_{\text{Th}} = \mathbf{V}_2 = \frac{60}{-(\mathbf{h}_{11} + 40)\mathbf{h}_{22}/\mathbf{h}_{21} + \mathbf{h}_{12}} = \frac{60\mathbf{h}_{21}}{\mathbf{h}_{12}\mathbf{h}_{21} - \mathbf{h}_{11}\mathbf{h}_{22} - 40\mathbf{h}_{22}}

Substituting the values of the h parameters,

VTh=60×10−20.21=−29.69 VV_{\text{Th}} = \frac{60 \times 10}{-20.21} = -29.69 \text{ V}

1 Ω 1 H 1 F

Figure 19.28 For Example 19.7.

Example 19.7 Find the g parameters as functions of s for the circuit in Fig. 19.28.

Solution:

In the s domain,

1 H⇒sL=s,1 F⇒1sC=1s1 \text{ H} \Rightarrow sL = s, \quad 1 \text{ F} \Rightarrow \frac{1}{sC} = \frac{1}{s}

To get g11 and g21, we open-circuit the output port and connect a voltage source V1 to the input port as in Fig. 19.29(a). From the figure,

I1=V1s+1\mathbf{I}_1 = \frac{\mathbf{V}_1}{s+1} g11=I1V1=1s+1\mathbf{g}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{1}{s+1}

By voltage division,

**V**2 = _____ 1 s + 1 V1

or

g21=V2V1=1s+1\mathbf{g}_{21} = \frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{1}{s+1}

To obtain g12 and g22, we short-circuit the input port and connect a current source I2 to the output port as in Fig. 19.29(b). By current division,

I1=−1s+1I2\mathbf{I}_1 = -\frac{1}{s+1} \mathbf{I}_2

g12 = __ I1 I2 = − _____ 1 s + 1

Also,

or

V2 = I2( __1 s + s ‖ 1)

or

g22=V2I2=1s+ss+1=s2+s+1s(s+1)\mathbf{g}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} = \frac{1}{s} + \frac{s}{s+1} = \frac{s^2 + s + 1}{s(s+1)}

Thus,

[g]=[1s+1−1s+11s+1s2+s+1s(s+1)][\mathbf{g}] = \begin{bmatrix} \frac{1}{s+1} & -\frac{1}{s+1} \\ \frac{1}{s+1} & \frac{s^2+s+1}{s(s+1)} \end{bmatrix}

For the ladder network in Fig. 19.30, determine the g parameters in the s domain.

Answer: [g]=[ __________ s + 2 s 2 + 3s + 1 −__________ 1 s 2 + 3s + 1 __________ 1 s 2 + 3s + 1 s(s + 2) __________ s 2 + 3s + 1 ].