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
$$\begin{array}{c}\mathbf{V}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2} \\\mathbf{I}_{2}=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \mathbf{V}_{2}\end{array} \tag{1}$$
or in matrix form,
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{h}_{11} & \mathbf{h}_{12} \\\mathbf{h}_{21} & \mathbf{h}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right]=[\mathbf{h}]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right] \tag{2}$$
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.
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.
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
$$\bbox[10px,border:1px solid grey]{\begin{array}{ll}\mathbf{h}_{11}=\left.\frac{\mathbf{V}_{1}}{\mathbf{I}_{1}}\right|_{\mathbf{V}_{2}=0}, & \mathbf{h}_{12}=\left.\frac{\mathbf{V}_{1}}{\mathbf{V}_{2}}\right|_{\mathbf{I}_{1}=0} \\\mathbf{h}_{21}=\left.\frac{\mathbf{I}_{2}}{\mathbf{I}_{1}}\right|_{\mathbf{V}_{2}=0}, & \mathbf{h}_{22}=\left.\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}\right|_{\mathbf{I}_{1}=0}\end{array}} \tag{3}$$
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,
$$\begin{array}{l}\mathbf{h}_{11}=\text { Short-circuit input impedance } \\\mathbf{h}_{12}=\text { Open-circuit reverse voltage gain } \\\mathbf{h}_{21}=\text { Short-circuit forward current gain } \\\mathbf{h}_{22}=\text { Open-circuit output admittance }\end{array}$$
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.
The h-parameter equivalent
network of a two-port network.
Fig. 1: The h-parameter equivalent network 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
$$\begin{aligned}\mathbf{I}_{1} &=\mathbf{g}_{11} \mathbf{V}_{1}+\mathbf{g}_{12} \mathbf{I}_{2} \\\mathbf{V}_{2} &=\mathbf{g}_{21} \mathbf{V}_{1}+\mathbf{g}_{22} \mathbf{I}_{2}\end{aligned}$$
$$\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{g}_{11} & \mathbf{g}_{12} \\\mathbf{g}_{21} & \mathbf{g}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]=[\mathbf{g}]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]$$
The values of the $ g $ parameters are determined as Thus, the inverse hybrid parameters are specifically called
$$\begin{array}{l}\mathbf{g}_{11}=\text { Open-circuit input admittance } \\\mathbf{g}_{12}=\text { Short-circuit reverse current gain } \\\mathbf{g}_{21}=\text { Open-circuit forward voltage gain } \\\mathbf{g}_{22}=\text { Short-circuit output impedance }\end{array}$$
Figure $ 2 $ shows the inverse hybrid model of a two-port network.
The g-parameter model of a
two-port network.
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.
The g-parameter model of a
two-port network.
Fig. 3: For Example 1.
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).
Fig. 4: For Example 1: (a) computing $h_{11}$ and $h_{21}$, (b) computing $h_{12}$ and $h_{22}$.
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, $$\mathbf{h}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}=\frac{1}{9} \mathrm{~S}$$
Example 2: Determine the Thevenin equivalent at the output port of the circuit in Fig. 5.
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 $V_{Th}$.
From Eq. (1),
$$\begin{array}{l}\mathbf{V}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2} \quad (2.1) \\ \mathbf{I}_{2}=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \mathbf{V}_{2} \quad (2.2)\end{array}$$
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
$$\begin{array}{c}-40 \mathbf{I}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \Longrightarrow \quad \mathbf{I}_{1}=-\frac{\mathbf{h}_{12}}{40+\mathbf{h}_{11}} \quad \text { (2.3) } \\ \mathbf{I}_{2}=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \quad \text { (2.4) }
\end{array}$$ Substituting Eq. (2.3) into Eq. (2.4) gives
$$\mathbf{I}_{2}=\mathbf{h}_{22}-\frac{\mathbf{h}_{21} \mathbf{h}_{12}}{\mathbf{h}_{11}+40}=\frac{\mathbf{h}_{11} \mathbf{h}_{22}-\mathbf{h}_{21} \mathbf{h}_{12}+\mathbf{h}_{22} 40}{\mathbf{h}_{11}+40}$$
$$\mathbf{Z}_{\mathrm{Th}}=\frac{\mathbf{V}_{2}}{\mathbf{I}_{2}}=\frac{1}{\mathbf{I}_{2}}=\frac{\mathbf{h}_{11}+40}{\mathbf{h}_{11} \mathbf{h}_{22}-\mathbf{h}_{21} \mathbf{h}_{12}+\mathbf{h}_{22} 40}$$
Substituting the values of the $ h $ parameters,
$$\begin{aligned}\mathbf{Z}_{\mathrm{Th}} &=\frac{1000+40}{10^{3} \times 200 \times 10^{-6}+20+40 \times 200 \times 10^{-6}} \\&=\frac{1040}{20.21}=51.46 \Omega\end{aligned}$$
To get $ \mathbf{V}_{\mathrm{Th}} $, we find the open-circuit voltage $ \mathbf{V}_{2} $ in Fig. 6(b). At the input port,
$$-60+40 \mathbf{I}_{1}+\mathbf{V}_{1}=0 \quad \Longrightarrow \quad \mathbf{V}_{1}=60-40 \mathbf{I}_{1} \tag{2.5} $$
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
$$60-40 \mathbf{I}_{1}=\mathbf{h}_{11} \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2}$$
$$60=\left(\mathbf{h}_{11}+40\right) \mathbf{I}_{1}+\mathbf{h}_{12} \mathbf{V}_{2} \tag{2.7}$$
$$0=\mathbf{h}_{21} \mathbf{I}_{1}+\mathbf{h}_{22} \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{I}_{1}=-\frac{\mathbf{h}_{22}}{\mathbf{h}_{21}} \mathbf{V}_{2} \tag{2.8}$$
Now substituting Eq. (2.8) into Eq. (2.7) gives
$$60=\left[-\left(\mathbf{h}_{11}+40\right) \frac{\mathbf{h}_{22}}{\mathbf{h}_{21}}+\mathbf{h}_{12}\right] \mathbf{V}_{2}$$
$$\mathbf{V}_{\mathrm{Th}}=\mathbf{V}_{2}=\frac{60}{-\left(\mathbf{h}_{11}+40\right) \mathbf{h}_{22} / \mathbf{h}_{21}+\mathbf{h}_{12}}=\frac{60 \mathbf{h}_{21}}{\mathbf{h}_{12} \mathbf{h}_{21}-\mathbf{h}_{11} \mathbf{h}_{22}-40 \mathbf{h}_{22}}$$
Substituting the values of the $ h $ parameters,
$$\mathbf{V}_{\text {Th }}=\frac{60 \times 10}{-20.21}=-29.69 \mathrm{~V}$$

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