Chapter 18, Problem 1P

To determine

Find the $g$ and $h$ parameters for the given circuit.

Answer to Problem 1P

The value of $g$ parameters $g_{11}$, $g_{12}$, $g_{21}$, and $g_{22}$ for the given circuit are $0.1\text{S}$, $-0.75$, $0.75$, and $3.75\Omega$ respectively, and the value of $h$ parameters $h_{11}$, $h_{12}$, $h_{21}$, and $h_{22}$ for the given circuit are $4\Omega$, $0.8$, $-0.8$, and $0.1067\text{S}$ respectively.

Explanation of Solution

Given data:

Refer to given figure in the textbook.

Formula used:

Write the expression to find inverse hybrid or $g$ parameters for the circuit.

$g_{11}={\frac{I_1}{V_1}|}_{I_2=0}$        (1)

$g_{21}={\frac{V_2}{V_1}|}_{I_2=0}$        (2)

$g_{12}={\frac{I_1}{I_2}|}_{V_1=0}$        (3)

$g_{22}={\frac{V_2}{I_2}|}_{V_1=0}$        (4)

Here,

$g_{11}$ is the open circuit input admittance,

$g_{12}$ is the short circuit reverse current gain,

$g_{21}$ is the open circuit forward voltage gain, and

$g_{22}$ is the short circuit output impedance.

Write the expression to find hybrid or $h$ parameters for the circuit.

$h_{11}={\frac{V_1}{I_1}|}_{V_2=0}$        (5)

$h_{21}={\frac{I_2}{I_1}|}_{V_2=0}$        (6)

$h_{12}={\frac{V_1}{V_2}|}_{I_1=0}$        (7)

$h_{22}={\frac{I_2}{V_2}|}_{I_1=0}$        (8)

Here,

$h_{11}$ is the input admittance of the network,

$h_{12}$ is the reverse voltage gain,

$h_{21}$ is the forward current gain, and

$h_{22}$ is the output impedance of the network.

Calculation:

When port 2 is open $\left(I_2=0\right)$, we can apply a $1\text{A}$ current source at port 1 to find $V_1$ and $V_2$. The modified circuit is shown in Figure 1.

Find the equivalent impedance of the circuit in Figure 1.

$\begin{array}{c}{Z}_{\text{eq}}=\left(20\Omega \right)\parallel \left(5\Omega +15\Omega \right)\\ =\left(20\Omega \right)\parallel \left(20\Omega \right)\\ =\frac{20\times 20}{20+20}\Omega \\ =10\Omega \end{array}$

Write the expression to calculate voltage $V_1$.

$V_1=Z_{\text{eq}}\left(1\text{A}\right)$

Substitute $10\Omega$ for $Z_{\text{eq}}$.

$\begin{array}{c}{V}_1=\left(10\Omega \right)\left(1\text{A}\right)\\ =10\text{V}\end{array}$

Calculate the voltage $V_2$ using voltage division rule.

$\begin{array}{c}{V}_2=\left(\frac{15}{15+5}\right)V_1\\ =\frac{15}{20}\left(10\text{V}\right)\\ =7.5\text{V}\end{array}$

Substitute $1\text{A}$ for $I_1$ and $10\text{V}$ for $V_1$ in equation (1) to find $g_{11}$.

$\begin{array}{c}{g}_{11}=\frac{1\text{A}}{10\text{V}}\\ =0.1\text{S}\end{array}$

Substitute $7.5\text{V}$ for $V_2$ and $10\text{V}$ for $V_1$ in equation (2) to find $g_{21}$.

$\begin{array}{c}{g}_{21}=\frac{7.5\text{V}}{10\text{V}}\\ =0.75\end{array}$

When port 1 is short circuited $\left(V_1=0\right)$, the reduced circuit is shown in Figure 2.

In Figure 2, the short circuit path neglects the effect of $20\Omega$ resistor.

From Figure 2, write the expression for current $I_2$.

$\begin{array}{c}{I}_2=\frac{V_2}{15}+\frac{V_2}{5}\\ =\left(\frac{1}{15}+\frac{1}{5}\right)V_2\\ =\frac{4}{15}{V}_2\end{array}$

$V_2=\frac{15}{4}{I}_2$        (9)

Substitute equation (9) in (4) to find $g_{22}$.

$\begin{array}{c}{g}_{22}=\frac{\frac{15}{4}{I}_2}{I_2}\\ =3.75\Omega \end{array}$

From Figure 2, write the expression for current $I_1$.

$I_1=-\left(\frac{15}{15+5}\right)I_2$

$I_1=-\frac{15}{20}{I}_2$        (10)

Substitute equation (10) in (3) to find $g_{12}$.

$\begin{array}{c}{g}_{12}=\frac{-\frac{15}{20}{I}_2}{I_2}\\ =-0.75\end{array}$

When port 1 is open $\left(I_1=0\right)$, we can apply a $1\text{A}$ current source at port 2 to find $V_1$ and $V_2$. The modified circuit is shown in Figure 3.

Find the equivalent impedance of the circuit in Figure 3.

$\begin{array}{c}{Z}_{\text{eq}}=\left(15\Omega \right)\parallel \left(5\Omega +20\Omega \right)\\ =\left(15\Omega \right)\parallel \left(25\Omega \right)\\ =\frac{15\times 25}{15+25}\Omega \\ =9.375\Omega \end{array}$

Write the expression to calculate voltage $V_2$.

$V_2=Z_{\text{eq}}\left(1\text{A}\right)$

Substitute $9.375\Omega$ for $Z_{\text{eq}}$.

$\begin{array}{c}{V}_2=\left(9.375\Omega \right)\left(1\text{A}\right)\\ =9.375\text{V}\end{array}$

Calculate the voltage $V_1$ using voltage division rule.

$\begin{array}{c}{V}_1=\left(\frac{20}{20+5}\right)V_2\\ =\frac{20}{25}\left(9.375\text{V}\right)\\ =7.5\text{V}\end{array}$

Substitute $7.5\text{V}$ for $V_1$ and $9.375\text{V}$ for $V_2$ in equation (7) to find $h_{12}$.

$\begin{array}{c}{h}_{12}=\frac{7.5\text{V}}{9.375\text{V}}\\ =0.8\end{array}$

Substitute $1\text{A}$ for $I_2$ and $9.375\text{V}$ for $V_2$ in equation (8) to find $h_{22}$.

$\begin{array}{c}{h}_{22}=\frac{1\text{A}}{9.375\text{V}}\\ =0.1067\text{S}\end{array}$

When port 2 is short circuited $\left(V_2=0\right)$, the reduced circuit is shown in Figure 4.

In Figure 4, the short circuit path neglects the effect of $15\Omega$ resistor.

From Figure 1, write the expression for current $I_1$.

$\begin{array}{c}{I}_1=\frac{V_1}{20}+\frac{V_1}{5}\\ =\left(\frac{1}{20}+\frac{1}{5}\right)V_1\\ =\left(\frac{5}{20}\right)V_1\end{array}$

$V_1=4I_1$        (11)

Substitute equation (11) in (5) to find $h_{11}$.

$\begin{array}{c}{h}_{11}=\frac{4I_1}{I_1}\\ =4\Omega \end{array}$

From Figure 4, write the expression for current $I_2$.

$I_2=-\left(\frac{20}{20+5}\right)I_1$

$I_2=-\left(\frac{20}{25}\right)I_1$        (12)

Substitute equation (12) in (6) to find $h_{21}$.

$\begin{array}{c}{h}_{21}=\frac{-\left(\frac{20}{25}\right)I_1}{I_1}\\ =-0.8\end{array}$

Conclusion:

Thus, the value of $g$ parameters $g_{11}$, $g_{12}$, $g_{21}$, and $g_{22}$ for the given circuit are $0.1\text{S}$, $-0.75$, $0.75$, and $3.75\Omega$ respectively, and the value of $h$ parameters $h_{11}$, $h_{12}$, $h_{21}$, and $h_{22}$ for the given circuit are $4\Omega$, $0.8$, $-0.8$, and $0.1067\text{S}$ respectively.