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
or in matrix form,
The $ \mathbf{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; it is much easier to measure experimentally the $ h $ parameters of such devices than to measure their $ z $ or $ y $ parameters.
In fact, we have seen that the ideal transformer in Fig. (i), described by Eq. (A), does not have $ z $ parameters. The ideal transformer can be described by the hybrid parameters, because Eq. (A) conforms with Eq. (1).
The values of the parameters are determined as
It is evident from Eq. (3) that the parameters $ \mathbf{h}_{11}, \mathbf{h}_{12}, \mathbf{h}_{21} $, and $ \mathbf{h}_{22} $ represent an impedance, a voltage gain, a current gain, and an admittance, respectively. This is why they are called the hybrid parameters. To be specific,
The procedure for calculating the $ h $ parameters is similar to that used for the $ z $ or $ y $ parameters. We apply a voltage or current source to the appropriate port, short-circuit or open-circuit the other port, depending on the parameter of interest, and perform regular circuit analysis. For reciprocal networks, $ \mathbf{h}_{12}=-\mathbf{h}_{21} $. This can be proved in the same way as we proved that $ \mathbf{z}_{12}=\mathbf{z}_{21} $. Figure $ 1$ shows the hybrid model of a two-port 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
or
The values of the $ g $ parameters are determined as
Thus, the inverse hybrid parameters are specifically called
Figure $ 2 $ shows the inverse hybrid model of a two-port network.
$$\begin{array}{c}\mathbf{V}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2} \\\mathbf{I}_{2}=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \mathbf{V}_{2}\end{array} \tag{1}$$
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{h}_{11} & \mathbf{h}_{12} \\\mathbf{h}_{21} & \mathbf{h}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right]=[\mathbf{h}]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right] \tag{2}$$
Equations from the other pages
$$\mathbf{V}_{1}=\frac{1}{n} \mathbf{V}_{2}, \quad \mathbf{I}_{1}=-n \mathbf{I}_{2}\tag{A}$$
Fig. (i): An ideal transformer has no z parameters.
The values of the parameters are determined as
$$\bbox[10px,border:1px solid grey]{\begin{array}{ll}\mathbf{h}_{11}=\left.\frac{\mathbf{V}_{1}}{\mathbf{I}_{1}}\right|_{\mathbf{V}_{2}=0}, & \mathbf{h}_{12}=\left.\frac{\mathbf{V}_{1}}{\mathbf{V}_{2}}\right|_{\mathbf{I}_{1}=0} \\\mathbf{h}_{21}=\left.\frac{\mathbf{I}_{2}}{\mathbf{I}_{1}}\right|_{\mathbf{V}_{2}=0}, & \mathbf{h}_{22}=\left.\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}\right|_{\mathbf{I}_{1}=0}\end{array}} \tag{3}$$
$$\begin{array}{l}\mathbf{h}_{11}=\text { Short-circuit input impedance } \\\mathbf{h}_{12}=\text { Open-circuit reverse voltage gain } \\\mathbf{h}_{21}=\text { Short-circuit forward current gain } \\\mathbf{h}_{22}=\text { Open-circuit output admittance }\end{array}$$
Fig. 1: The h-parameter equivalent
network of a two-port network.
$$\begin{aligned}\mathbf{I}_{1} &=\mathbf{g}_{11} \mathbf{V}_{1}+\mathbf{g}_{12} \mathbf{I}_{2} \\\mathbf{V}_{2} &=\mathbf{g}_{21} \mathbf{V}_{1}+\mathbf{g}_{22} \mathbf{I}_{2}\end{aligned}$$
$$\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{g}_{11} & \mathbf{g}_{12} \\\mathbf{g}_{21} & \mathbf{g}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]=[\mathbf{g}]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]$$
$$\begin{array}{l}\mathbf{g}_{11}=\text { Open-circuit input admittance } \\\mathbf{g}_{12}=\text { Short-circuit reverse current gain } \\\mathbf{g}_{21}=\text { Open-circuit forward voltage gain } \\\mathbf{g}_{22}=\text { Short-circuit output impedance }\end{array}$$
Fig. 2: The g-parameter model of a
two-port network.
Example 1: Find the hybrid parameters for the two-port network of Fig. 3.
Solution:
To find $ \mathbf{h}_{11} $ and $ \mathbf{h}_{21} $, we short-circuit the output port and connect a current source $ \mathbf{I}_{1} $ to the input port as shown in Fig. 4(a).

From Fig. 4(a),
$$\mathbf{V}_{1}=\mathbf{I}_{1}(2+3 \| 6)=4 \mathbf{I}_{1}$$
Hence,
$$\mathbf{h}_{11}=\frac{\mathbf{V}_{1}}{\mathbf{I}_{1}}=4 \Omega$$
Also, from Fig. 4(a) we obtain, by current division,
$$-\mathbf{I}_{2}=\frac{6}{6+3} \mathbf{I}_{1}=\frac{2}{3} \mathbf{I}_{1}$$
Hence,
$$\mathbf{h}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{I}_{1}}=-\frac{2}{3}$$
To obtain $ \mathbf{h}_{12} $ and $ \mathbf{h}_{22} $, we open-circuit the input port and connect a voltage source $ \mathbf{V}_{2} $ to the output port as in Fig. 4(b). By voltage division,
$$\mathbf{V}_{1}=\frac{6}{6+3} \mathbf{V}_{2}=\frac{2}{3} \mathbf{V}_{2}$$
Hence,
$$\mathbf{h}_{12}=\frac{\mathbf{V}_{1}}{\mathbf{V}_{2}}=\frac{2}{3}$$
Also,
$$\mathbf{V}_{2}=(3+6) \mathbf{I}_{2}=9 \mathbf{I}_{2}$$
Thus,
$$\mathbf{h}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}=\frac{1}{9} \mathrm{~S}$$
Fig. 3: For Example 1.

Fig. 4: For Example 1: (a) computing $h_{11}$ and $h_{21}$, (b) computing $h_{12}$ and $h_{22}$.
Example 2: Determine the Thevenin equivalent at the output port of the circuit in Fig. 5.
Solution:
To find $ \mathbf{Z}_{\mathrm{Th}} $ and $ \mathbf{V}_{\mathrm{Th}} $, we apply the normal procedure, keeping in mind the formulas relating the input and output ports of the $ h $ model. To obtain $ \mathbf{Z}_{\mathrm{Th}} $, remove the $ 60\mathrm{V} $ voltage source at the input port and apply a 1-V voltage source at the output port, as shown in Fig. 6(a).
From Eq. (1),
But $ \mathbf{V}_{2}=1 $, and $ \mathbf{V}_{1}=-40 \mathbf{I}_{1} $. Substituting these into Eqs. (18.6.1) and (18.6.2), we get
\end{array}$$
Substituting Eq. (2.3) into Eq. (2.4) gives
Therefore,
Substituting the values of the $ h $ parameters,
To get $ \mathbf{V}_{\mathrm{Th}} $, we find the open-circuit voltage $ \mathbf{V}_{2} $ in Fig. 6(b). At the input port,
At the output,
$$\mathbf{I}_{2}=0 \tag{2.6}$$
Substituting Eqs. (2.5) and (2.6) into Eqs. (2.1) and (2.2), we obtain
or
and
Now substituting Eq. (2.8) into Eq. (2.7) gives
or
Substituting the values of the $ h $ parameters,
Fig. 5: For Example 2.
To find $ \mathbf{Z}_{\mathrm{Th}} $ and $ \mathbf{V}_{\mathrm{Th}} $, we apply the normal procedure, keeping in mind the formulas relating the input and output ports of the $ h $ model. To obtain $ \mathbf{Z}_{\mathrm{Th}} $, remove the $ 60\mathrm{V} $ voltage source at the input port and apply a 1-V voltage source at the output port, as shown in Fig. 6(a).

Fig. 6: For Example 2: (a) finding
$Z_{Th}$, (b) finding $V_{Th}$.
$$\begin{array}{l}\mathbf{V}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2} \quad (2.1) \\
\mathbf{I}_{2}=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \mathbf{V}_{2} \quad (2.2)\end{array}$$
$$\begin{array}{c}-40 \mathbf{I}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \Longrightarrow \quad \mathbf{I}_{1}=-\frac{\mathbf{h}_{12}}{40+\mathbf{h}_{11}} \quad \text { (2.3) } \\
\mathbf{I}_{2}=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \quad \text { (2.4) }
$$\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}$$
$$\mathbf{Z}_{\mathrm{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}$$
$$\begin{aligned}\mathbf{Z}_{\mathrm{Th}} &=\frac{1000+40}{10^{3} \times 200 \times 10^{-6}+20+40 \times 200 \times 10^{-6}} \\&=\frac{1040}{20.21}=51.46 \Omega\end{aligned}$$
$$-60+40 \mathbf{I}_{1}+\mathbf{V}_{1}=0 \quad \Longrightarrow \quad \mathbf{V}_{1}=60-40 \mathbf{I}_{1} \tag{2.5} $$
$$60-40 \mathbf{I}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2}$$
$$60=\left(\mathbf{h}_{11}+40\right) \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2} \tag{2.7}$$
$$0=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{I}_{1}=-\frac{\mathbf{h}_{22}}{\mathbf{h}_{21}} \mathbf{V}_{2} \tag{2.8}$$
$$60=\left[-\left(\mathbf{h}_{11}+40\right) \frac{\mathbf{h}_{22}}{\mathbf{h}_{21}}+\mathbf{h}_{12}\right] \mathbf{V}_{2}$$
$$\mathbf{V}_{\mathrm{Th}}=\mathbf{V}_{2}=\frac{60}{-\left(\mathbf{h}_{11}+40\right) \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}}$$
$$\mathbf{V}_{\text {Th }}=\frac{60 \times 10}{-20.21}=-29.69 \mathrm{~V}$$
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