Chapter 12, Problem 12.49P

The circuit in Figure P 12.49 has transistor parameters: $h_{FE}=100$ $V_{BE}(\text{on})=0.7\text{V},$ and $V_A=\infty .(\text{a})$ From the quiescent values, determine.

the small-signal parameters for $Q_1$ and $Q_2$ . (b) Using nodal analysis, determine the small-signal closed-loop current gain $A_{if}=i_o/i_s.$ (c) Using nodal analysis, find the input resistance $R_{if}$ .

To determine

The value of the small signal parameters for $Q_1$ and $Q_2$ .

Answer to Problem 12.49P

Thevalue of small signal resistances are $r_{\pi 1}$ is $13.1\text{k}\Omega$ and $r_{\pi 2}$ is $1.9\text{k}\Omega$ . The value of the trans-conductance are $g_{m1}$ is $7.62\text{mA}/\text{V}$ and $g_{m2}$ is $52.7\text{mA}/\text{V}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Calculation:

The small signal circuit for the given circuit is shown in Figure 2.

  

Figure 2

The expression to determine the value of the current $I_{C1}$ is given by,

  $I_{C1}=\left(\frac{h_{FE}}{1+h_{FE}}\right)I_{E1}$

Substitute $100$ for $h_{FE}$ and $0.2\text{mA}$ for $I_{E1}$ in the above equation.

  $\begin{array}{c}{I}_{C1}=\left(\frac{100}{1+100}\right)\left(0.2\text{mA}\right)\\ =0.198\text{mA}\end{array}$

The expression to determine the value of the voltage $V_{C1}$ is given by,

  $V_{C1}=10\text{V}-I_{C1}\left(40\text{k}\Omega \right)$

Substitute $0.198\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{V}_{C1}=10\text{V}-\left(0.198\text{mA}\right)\left(40\text{k}\Omega \right)\\ =2.08\text{V}\end{array}$

The value of the current $I_{E2}$ is calculated as,

  $I_{E2}=\frac{V_{C1}-V_{BE}\left(\text{on}\right)}{1\text{k}\Omega }$

Substitute $2.08\text{V}$ for $V_{C1}$ and $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{c}{I}_{E2}=\frac{2.08\text{V}-0.7\text{V}}{1\text{k}\Omega }\\ =1.38\text{mA}\end{array}$

The expression to determine the value of the current $I_{C2}$ is given by,

  $I_{C2}=\left(\frac{h_{FE}}{1+h_{FE}}\right)I_{E2}$

Substitute $100$ for $h_{FE}$ and $1.38\text{mA}$ for $I_{E2}$ in the above equation.

  $\begin{array}{c}{I}_{C2}=\left(\frac{100}{1+100}\right)\left(1.38\text{mA}\right)\\ =1.37\text{mA}\end{array}$

The expression for the small signal input resistance is given by,

  $r_{\pi 1}=\frac{h_{FE}{V}_T}{I_{CQ1}}$

Substitute $100$ for $h_{FE}$ , $0.026$ for $V_T$ and $0.198\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 1}=\frac{\left(100\right)\left(0.026\text{V}\right)}{0.198\text{mA}}\\ =13.1\text{k}\Omega \end{array}$

The expression for the trans-conductance of the first transistor is given by,

  $g_{m1}=\frac{I_{CQ1}}{V_T}$

Substitute $0.198\text{mA}$ for $I_{C1}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m1}=\frac{0.198\text{mA}}{0.026\text{V}}\\ =7.62\text{mA}/\text{V}\end{array}$

The expression for the small signal input resistance is given by,

  $r_{\pi 2}=\frac{h_{FE}{V}_T}{I_{C2}}$

Substitute $100$ for $h_{FE}$ , $0.026$ for $V_T$ and $1.37\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{\left(100\right)\left(0.026\text{V}\right)}{1.37\text{mA}}\\ =1.9\text{k}\Omega \end{array}$

The expression for the trans-conductance of the first transistor is given by,

  $g_{m2}=\frac{I_{C2}}{V_T}$

Substitute $1.37\text{mA}$ for $I_{C2}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{1.37\text{mA}}{0.026\text{V}}\\ =52.7\text{mA}/\text{V}\end{array}$

Conclusion:

Therefore, the value of small signal resistances are $r_{\pi 1}$ is $13.1\text{k}\Omega$ and $r_{\pi 2}$ is $1.9\text{k}\Omega$ . The value of the trans-conductance are $g_{m1}$ is $7.62\text{mA}/\text{V}$ and $g_{m2}$ is $52.7\text{mA}/\text{V}$ .

To determine

The value of small signal closed loop current gain.

Answer to Problem 12.49P

The value of the small signal closed loop current gain is $8.6$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Calculation:

Apply KCL at the source.

  $i_S=\frac{V_{\pi 1}}{10\text{k}\Omega }+\frac{V_{\pi 1}}{r_{\pi 1}}+\frac{V_{\pi 1}-V_{CE}}{10\text{k}\Omega }$

Substitute $13.1\text{k}\Omega$ for $r_{\pi 1}$ in the above equation.

  $\begin{array}{l}{i}_S=\frac{V_{\pi 1}}{10\text{k}\Omega }+\frac{V_{\pi 1}}{13.1\text{k}\Omega }+\frac{V_{\pi 1}-V_{CE}}{10\text{k}\Omega }\\ i_S=0.2763V_{\pi 1}-0.1V_{CE}\end{array}$ …… (1)

Consider the KCL equation.

  $g_{m1}{V}_{\pi 1}+\frac{V_{\pi 2}+V_{CE}}{40\text{k}\Omega }+\frac{V_{\pi 2}}{r_{\pi 2}}=0$

Substitute $7.62\text{mA}/\text{V}$ for $g_{m1}$ and $1.9\text{k}\Omega$ for $r_{\pi 2}$ in the above equation.

  $\begin{array}{l}7.62\text{mA}/\text{V}{V}_{\pi 1}+\frac{V_{\pi 2}+V_{CE}}{40\text{k}\Omega }+\frac{V_{\pi 2}}{1.9\text{k}\Omega }=0\\ 7.62V_{\pi 1}+0.5513V_{\pi 2}+00.25V_{CE}=0\end{array}$ …… (2)

Apply KCL at $V_{CE}$ .

  $\frac{V_{\pi 2}}{r_{\pi 2}}+g_{m2}{V}_{\pi 2}=\frac{V_{CE}}{1\text{k}\Omega }+\frac{V_{CE}-V_{\pi 1}}{10\text{k}\Omega }$

Substitute $1.9\text{k}\Omega$ for $r_{\pi 2}$ and $52.7\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $\begin{array}{l}\frac{V_{\pi 2}}{1.9\text{k}\Omega }+\left(52.7\text{mA}/\text{V}\right)V_{\pi 2}=\frac{V_{CE}}{1\text{k}\Omega }+\frac{V_{CE}-V_{\pi 1}}{10\text{k}\Omega }\\ V_{CE}=48.39V_{\pi 2}+0.0909V_{\pi 1}\end{array}$

Substitute $48.39V_{\pi 2}+0.0909V_{\pi 1}$ for $V_{CE}$ in the above equation.

  $\begin{array}{l}{i}_S=0.2763V_{\pi 1}-0.1\left(48.39V_{\pi 2}+0.0909V_{\pi 1}\right)\\ i_S=0.2672V_{\pi 1}-4.839V_{\pi 2}\end{array}$ ……. (3)

Substitute $48.39V_{\pi 2}+0.0909V_{\pi 1}$ for $V_{CE}$ in the above equation.

  $\begin{array}{l}7.62V_{\pi 1}+0.5513V_{\pi 2}+00.25\left(48.39V_{\pi 2}+0.0909V_{\pi 1}\right)=0\\ V_{\pi 1}=-0.23V_{\pi 2}\end{array}$

Substitute $-0.23V_{\pi 2}$ for $V_{\pi 1}$ in the equation.

  $\begin{array}{c}{i}_S=0.2672\left(-0.23V_{\pi 2}\right)-4.839V_{\pi 2}\\ =-4.901V_{\pi 2}\end{array}$ ….. (4)

The expression to determine the value of the output current is given by,

  $i_o=-g_{m2}{V}_{\pi 2}\left(\frac{2\text{k}\Omega }{2\text{k}\Omega +0.5\text{k}\Omega }\right)$

Substitute $52.7\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $i_o=-\left(52.7\text{mA}/\text{V}\right)V_{\pi 2}\left(\frac{2\text{k}\Omega }{2\text{k}\Omega +0.5\text{k}\Omega }\right)$

Substitute $\frac{i_S}{-4.901}$ for $V_{\pi 2}$ in the above equation.

  $\begin{array}{l}{i}_o=-\left(52.7\text{mA}/\text{V}\right)\left(\frac{i_S}{-4.901}\right)\left(\frac{2\text{k}\Omega }{2\text{k}\Omega +0.5\text{k}\Omega }\right)\\ \frac{i_o}{i_s}=8.6\end{array}$

Conclusion:

Therefore, the value of the small signal closed loop current gain is $8.6$ .

To determine

The value of the resistance $R_{if}$ .

Answer to Problem 12.49P

The value of the resistance is $47.4\Omega$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Calculation:

Substitute $\frac{V_{\pi 1}}{-0.23}$ for $V_{\pi 2}$ in equation (4).

  $\begin{array}{c}{i}_S=-4.901\left(\frac{V_{\pi 1}}{-0.23}\right)\\ =21.22V_{\pi 1}\end{array}$

The expression to determine the value of the input resistance is given by,

  $R_i=\frac{V_{\pi 1}}{i_s}$

Substitute $21.22V_{\pi 1}$ for $i_S$ in the above equation.

  $\begin{array}{c}{R}_i=\frac{V_{\pi 1}}{21.22V_{\pi 1}}\\ =47.4\text{m}\Omega \end{array}$

The expression to determine the value of the input feedback resistance is given by,

  $R_i=\frac{R_S{R}_{if}}{R_{if}+R_S}$

Substitute $10\text{k}\Omega$ for $R_S$ and $47.4\text{m}\Omega$ for $R_i$ in the above equation.

  $\begin{array}{l}47.4\text{m}\Omega =\frac{\left(47.4\text{m}\Omega \right)R_{if}}{R_{if}+10\text{k}\Omega }\\ R_{if}=47.4\Omega \end{array}$

Conclusion:

Therefore, the value of the resistance is $47.4\Omega$ .