Chapter 19, Problem 98P

A two-stage amplifier in Fig. 19.134 contains two identical stages with

$[h]=\left[\begin{array}{cc}2\text{ kΩ}& 0.004\\ 200& 500\mu \text{S}\end{array}\right]$

If Z_L = 20 kΩ, find the required value of V_s to produce V_o = 16 V.

Figure 19.134

To determine

Calculate the required source voltage for the given two-stage amplifier.

Answer to Problem 98P

The required source voltage for the given two-stage amplifier is $\underset{\_}{\text{464}\text{.224}\mu \text{V}}$.

Explanation of Solution

Given Data:

The h parameters of each stage of the amplifier are given as follows:

$\left[h\right]=\left[\begin{array}{cc}2\text{k}\Omega & 0.004\\ 200& 500\mu \text{S}\end{array}\right]$

The load impedance and output voltage of the given two-stage amplifier are given as follows:

$\begin{array}{l}{Z}_L=20\text{k}\Omega \\ V_o=16\text{V}\end{array}$

Refer to Figure 19.134 in the textbook for the given two-stage amplifier circuit.

Formula used:

Refer to TABLE 19.1 in the textbook, write the expression for transmission parameters in terms of h parameters as follows:

$\left[T\right]=\left[\begin{array}{cc}-\frac{{\Delta }_h}{h_{21}}& -\frac{h_{11}}{h_{21}}\\ -\frac{h_{22}}{h_{21}}& -\frac{1}{h_{21}}\end{array}\right]$        (1)

Write the expression for ${\Delta }_h$ as follows:

${\Delta }_h=h_{11}{h}_{22}-h_{12}{h}_{21}$        (2)

Write the expression for ${\Delta }_T$ as follows:

${\Delta }_T=AD-BC$        (3)

From TABLE 19.1 in the textbook, write the expression for impedance parameters in terms of transmission parameters as follows:

$\left[z\right]=\left[\begin{array}{cc}\frac{A}{C}& \frac{{\Delta }_T}{C}\\ \frac{1}{C}& \frac{D}{C}\end{array}\right]$        (4)

Calculation:

From the given circuit, the two-stage amplifier is the cascade connection of two identical transistors. Obtain the transmission parameters for two-stage amplifier and then convert into impedance parameters for simple analysis.

Find the transmission parameters for one stage of the two-stage amplifier as follows:

Substitute $2\text{k}\Omega$ for $h_{11}$, 0.004 for $h_{12}$, 200 for $h_{21}$, and $500\mu \text{S}$ for $h_{22}$ in Equation (2) to obtain the value of ${\Delta }_h$.

$\begin{array}{c}{\Delta }_h=\left(2\text{k}\Omega \right)\left(500\mu \text{S}\right)-\left(0.004\right)\left(200\right)\\ =\left(2\times {10}^3\Omega \right)\left(500\times {10}^{-6}\text{S}\right)-0.8\\ =1--0.8\\ =0.2\end{array}$

Substitute $2\text{k}\Omega$ for $h_{11}$, 0.004 for $h_{12}$, 200 for $h_{21}$, $500\mu \text{S}$ for $h_{22}$, and 0.2 for ${\Delta }_h$ in Equation (1) to obtain the transmission parameters.

$\begin{array}{c}\left[T_1\right]=\left[\begin{array}{cc}-\frac{0.2}{200}& -\frac{2\text{k}\Omega }{200}\\ -\frac{500\mu \text{S}}{200}& -\frac{1}{200}\end{array}\right]\\ =\left[\begin{array}{cc}-1\times {10}^{-3}& -\frac{2\times {10}^3\Omega }{200}\\ -\frac{500\times {10}^{-6}\text{S}}{200}& -\frac{1}{200}\end{array}\right]\\ =\left[\begin{array}{cc}-0.001& -10\Omega \\ -2.5\times {10}^{-6}\text{S}& -0.005\end{array}\right]\end{array}$

As both the stages are identical, the transmission parameters of second stage of the amplifier are equal to the first stage.

$\begin{array}{c}\left[T_2\right]=\left[T_1\right]\\ =\left[\begin{array}{cc}-0.001& -10\Omega \\ -2.5\times {10}^{-6}\text{S}& -0.005\end{array}\right]\end{array}$

As the two identical stages are connected in cascade to form a two-stage amplifier, write the expression for overall transmission parameters for the amplifier as follows:

$\left[T\right]=\left[T_1\right]\left[T_2\right]$

Substitute $\left[\begin{array}{cc}-0.001& -10\Omega \\ -2.5\times {10}^{-6}\text{S}& -0.005\end{array}\right]$ for $\left[T_1\right]$ and $\left[\begin{array}{cc}-0.001& -10\Omega \\ -2.5\times {10}^{-6}\text{S}& -0.005\end{array}\right]$ for $\left[T_2\right]$ as follows:

$\begin{array}{c}\left[T\right]=\left[\begin{array}{cc}-0.001& -10\Omega \\ -2.5\times {10}^{-6}\text{S}& -0.005\end{array}\right]\left[\begin{array}{cc}-0.001& -10\Omega \\ -2.5\times {10}^{-6}\text{S}& -0.005\end{array}\right]\\ =\left[\begin{array}{cc}\left(-0.001\right)\left(-0.001\right)+\left(-10\Omega \right)\left(-2.5\times {10}^{-6}\text{S}\right)& \left(-0.001\right)\left(-10\Omega \right)+\left(-10\Omega \right)\left(-0.005\right)\\ \left(-2.5\times {10}^{-6}\text{S}\right)\left(-0.001\right)+\left(-0.005\right)\left(-2.5\times {10}^{-6}\text{S}\right)& \left(-2.5\times {10}^{-6}\text{S}\right)\left(-10\Omega \right)+\left(-0.005\right)\left(-0.005\right)\end{array}\right]\\ =\left[\begin{array}{cc}2.6\times {10}^{-5}& 0.06\Omega \\ 1.5\times {10}^{-8}\text{S}& 5\times {10}^{-5}\end{array}\right]\end{array}$

From the obtained transmission parameters of the two-stage amplifier, write the transmission parameters as follows:

$\begin{array}{c}A=2.6\times {10}^{-5}\\ B=0.06\Omega \\ C=1.5\times {10}^{-8}\text{S}\\ D=5\times {10}^{-5}\end{array}$

Convert the obtained transmission parameters into impedance parameters as follows:

Substitute $2.6\times {10}^{-5}$ for $A$, $0.06\Omega$ for $B$, $1.5\times {10}^{-8}\text{S}$ for $C$, and $5\times {10}^{-5}$ for $D$ in Equation (3) to obtain the value of ${\Delta }_T$.

$\begin{array}{c}{\Delta }_T=\left(2.6\times {10}^{-5}\right)\left(5\times {10}^{-5}\right)-\left(0.06\Omega \right)\left(1.5\times {10}^{-8}\text{S}\right)\\ =\left(13\times {10}^{-10}\right)-\left(0.09\times {10}^{-8}\right)\\ =0.04\times {10}^{-8}\end{array}$

Substitute $2.6\times {10}^{-5}$ for $A$, $0.06\Omega$ for $B$, $1.5\times {10}^{-8}\text{S}$ for $C$, $5\times {10}^{-5}$ for $D$, and $\left(0.04\times {10}^{-8}\right)$ for ${\Delta }_T$ in Equation (4) to obtain the impedance parameters for given network.

$\begin{array}{c}\left[z\right]=\left[\begin{array}{cc}\frac{2.6\times {10}^{-5}}{1.5\times {10}^{-8}\text{S}}& \frac{0.04\times {10}^{-8}}{1.5\times {10}^{-8}\text{S}}\\ \frac{1}{1.5\times {10}^{-8}\text{S}}& \frac{5\times {10}^{-5}}{1.5\times {10}^{-8}\text{S}}\end{array}\right]\\ =\left[\begin{array}{cc}1733.3\Omega & 0.0267\Omega \\ 0.6667\times {10}^8\Omega & 3333.3\Omega \end{array}\right]\end{array}$

Refer to Figure 19.5 (b) in the textbook for general equivalent circuit of impedance parameters and draw the circuit for given amplifier as shown in Figure 1.

From the circuit in Figure 1, apply KVL to the input loop as follows:

$-V_s+\left(1\Omega +z_{11}\right)I_1+z_{12}{I}_2=0$

Simplify the expression as follows:

$V_s=\left(1\Omega +z_{11}\right)I_1+z_{12}{I}_2$        (5)

From the circuit in Figure 1, apply KVL to the output loop as follows:

$-V_o+z_{22}{I}_2+z_{21}{I}_1=0$

Simplify the expression as follows:

$V_o=z_{22}{I}_2+z_{21}{I}_1$        (6)

From Figure 1, write the expression for $V_o$ as follows:

$V_o=-I_2{Z}_L$

Rearrange the expression as follows:

$I_2=-\frac{V_o}{Z_L}$        (7)

Substitute $\left(-\frac{V_o}{Z_L}\right)$ for $I_2$ in Equation (6) as follows:

$\begin{array}{l}{V}_o=z_{22}\left(-\frac{V_o}{Z_L}\right)+z_{21}{I}_1\\ z_{21}{I}_1=V_o+\frac{z_{22}}{Z_L}{V}_o\end{array}$

$I_1=\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)V_o$        (8)

From Equations (7) and (8), substitute $\left(-\frac{V_o}{Z_L}\right)$ for $I_2$ and $\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)V_o$ for $I_1$ in Equation (5) as follows:

$\begin{array}{c}{V}_s=\left(1\Omega +z_{11}\right)\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)V_o+z_{12}\left(-\frac{V_o}{Z_L}\right)\\ =\left[\left(1\Omega +z_{11}\right)\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)-\frac{z_{12}}{Z_L}\right]V_o\end{array}$

Substitute $1733.3\Omega$ for $z_{11}$, $0.0267\Omega$ for $z_{12}$, $\left(0.6667\times {10}^8\right)\Omega$ for $z_{21}$, $3333.3\Omega$ for $z_{22}$, $20\text{k}\Omega$ for $Z_L$, and 16 V for $V_o$ to obtain the required source voltage to the amplifier.

$\begin{array}{c}{V}_s=\left(\left(1\Omega +1733.3\Omega \right)\left\{\frac{1}{\left(0.6667\times {10}^8\right)\Omega }+\frac{3333.3\Omega }{\left[\left(0.6667\times {10}^8\right)\Omega \right]\left(20\text{k}\Omega \right)}\right\}-\frac{0.0267\Omega }{20\text{k}\Omega }\right)\left(16\text{V}\right)\\ =\left(\left(1734.3\Omega \right)\left\{\frac{\left(20000\Omega \right)+3333.3\Omega }{\left[\left(0.6667\times {10}^8\right)\Omega \right]\left(20000\Omega \right)}\right\}-\frac{0.0267\Omega }{20000\Omega }\right)\left(16\text{V}\right)\\ =\left[\left(3.0349\times {10}^{-5}\right)-\left(0.1335\times {10}^{-5}\right)\right]\left(16\text{V}\right)\\ =\left(2.9014\times {10}^{-5}\right)\left(16\text{V}\right)\end{array}$

Simplify the expression as follows:

$\begin{array}{c}{V}_s=464.224\times {10}^{-6}\text{V}\\ =\text{464}\text{.224}\mu \text{V}\end{array}$

Conclusion:

Thus, the required source voltage for the given two-stage amplifier is $\underset{\_}{\text{464}\text{.224}\mu \text{V}}$.