Chapter 12, Problem 12.37P

Consider the series-shunt feedback circuit in Figure $\text{P}12.37$ , with transistor parameters: $h_{FE}=120,V_{BE}(\text{on})=0.7\text{V},$ and $V_A=\infty .$ (a) Determine the small-signal parameters for $Q_1,Q_2,$ and $Q_3.$ Using nodal analysis, determine: (b) the small-signal voltage gain $A_{vf}=v_o/v_i,$ (c) the input resistance $R_{if},$ and $(\text{d})$ the output resistance $R_{of}$

Figure P12.37

To determine

The small signal parameter for $Q_1,Q_2$ and $Q_3$ .

Answer to Problem 12.37P

The trans-conductance of the first transistor $Q_1$ is $32.81\text{mA}/\text{V}$ , transistor $Q_2$ is $19.12\text{mA}/\text{V}$ and for transistor $Q_3$ is $78.08\text{mA}/\text{V}$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Calculation:

The value of the Thevenin voltage is given by,

  $\begin{array}{c}{V}_{TH}=\left(\frac{75\text{k}\Omega }{400\text{k}\Omega +75\text{k}\Omega }\right)\left(10\text{V}\right)\\ =1.579\text{V}\end{array}$

The expression for the equivalent resistance is given by,

  $\begin{array}{c}{R}_{TH}=\frac{\left(400\text{k}\Omega \right)\left(75\text{k}\Omega \right)}{\left(400\text{k}\Omega \right)+\left(75\text{k}\Omega \right)}\\ =63.2\text{k}\Omega \end{array}$

The expression for the current $I_{BQ1}$ is given by,

  $I_{BQ1}=\frac{V_{TH}-V_{BE{Q}_{11}}}{R_{TH}+\left(h_{FE}+1\right)R_{E1}}$

Substitute $1.579\text{V}$ for $V_{TH}$ , $0.7\text{V}$ for $V_{BE{Q}_1}$ , $120$ for $h_{FE}$ , $63.2\text{k}\Omega$ for $R_{TH}$ and $0.5\text{k}\Omega$ for $R_{E1}$ in the above equation.

  $\begin{array}{c}{I}_{BQ1}=\frac{1.579\text{V}-0.7\text{V}}{63.2\text{k}\Omega +\left(120+1\right)0.5\text{k}\Omega }\\ =0.007106\text{k}\Omega \end{array}$

The expression to determine the value of the collector current $I_{C{Q}_1}$ is given by,

  $I_{C{Q}_1}=h_{FE}{I}_{B{Q}_1}$

Substitute $120$ for $h_{FE}$ and $0.007106\text{mA}$ for $I_{BQ1}$ in the above equation.

  $\begin{array}{c}{I}_{C{Q}_1}=\left(120\right)0.007106\text{mA}\\ =0.853\text{mA}\end{array}$

The expression for the collector voltage $V_{C1}$ is given by,

  $V_{C1}=10\text{V}-I_{CQ1}\left(8.8\text{k}\Omega \right)$

Substitute $0.853\text{mA}$ for $I_{CQ1}$ in the above equation.

  $\begin{array}{c}{V}_{C1}=10\text{V}-0.853\text{mA}\left(8.8\text{k}\Omega \right)\\ =2.49\text{V}\end{array}$

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

  $I_{C2}=\frac{V_{C1}-V_{BE2}}{3.6\text{k}\Omega }$

Substitute $0.7\text{V}$ for $V_{BE2}$ and $2.49\text{V}$ for $V_{CE1}$ in the above equation.

  $\begin{array}{c}{I}_{C2}=\frac{2.49\text{V}-0.7\text{V}}{3.6\text{k}\Omega }\\ =0.497\text{mA}\end{array}$

The expression for the collector voltage $V_{C2}$ is given by,

  $V_{C2}=10\text{V}-I_{C2}\left(13\text{k}\Omega \right)$

Substitute $0.497\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{V}_{C2}=10\text{V}-\left(0.497\text{mA}\right)\left(13\text{k}\Omega \right)\\ =3.54\text{V}\end{array}$

The expression for the value of the current $I_{C3}$ is given by,

  $I_{C3}=\frac{V_{C2}-V_{BE3}}{1.4\text{k}\Omega }$

Substitute $0.7\text{V}$ for $V_{BE3}$ and $3.54\text{V}$ for $V_{C2}$ in the above equation.

  $\begin{array}{c}{I}_{C3}=\frac{3.54\text{V}-0.7\text{V}}{1.4\text{k}\Omega }\\ =2.03\text{mA}\end{array}$

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

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

Substitute $120$ for $h_{FE}$ , $0.026$ for $V_T$ and $0.853\text{mA}$ for $I_{CQ1}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 1}=\frac{\left(120\right)\left(0.026\text{V}\right)}{0.853\text{mA}}\\ =3.66\text{k}\Omega \end{array}$

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

  $g_{m1}=\frac{I_{CQ1}}{V_T}$

Substitute $0.853\text{mA}$ for $I_{C{Q}_1}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m1}=\frac{0.853\text{mA}}{0.026\text{V}}\\ =32.81\text{mA}/\text{V}\end{array}$

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

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

Substitute $120$ for $h_{FE}$ , $0.026$ for $V_T$ and $0.497\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{\left(120\right)\left(0.026\text{V}\right)}{0.497\text{mA}}\\ =6.28\text{k}\Omega \end{array}$

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

  $g_{m2}=\frac{I_{C2}}{V_T}$

Substitute $0.497\text{mA}$ for $I_{C2}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{0.497\text{mA}}{0.026\text{V}}\\ =19.12\text{mA}/\text{V}\end{array}$

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

  $r_{\pi 3}=\frac{h_{FE}{V}_T}{I_{C3}}$

Substitute $120$ for $h_{FE}$ , $0.026$ for $V_T$ and $2.03\text{mA}$ for $I_{C3}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 3}=\frac{\left(120\right)\left(0.026\text{V}\right)}{2.03\text{mA}}\\ =1.54\text{k}\Omega \end{array}$

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

  $g_{m3}=\frac{I_{C3}}{V_T}$

Substitute $2.03\text{mA}$ for $I_{C3}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m3}=\frac{2.03\text{mA}}{0.026\text{V}}\\ =78.08\text{mA}/\text{V}\end{array}$

Conclusion:

Therefore, the trans-conductance of the first transistor $Q_1$ is $32.81\text{mA}/\text{V}$ , transistor $Q_2$ is $19.12\text{mA}/\text{V}$ and for transistor $Q_3$ is $78.08\text{mA}/\text{V}$ .

To determine

The value of the small signal voltage gain $A_{vf}$ .

Answer to Problem 12.37P

The value of small signal voltage gain is 20.7.

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Calculation:

Mark the values and draw the small signal equivalent circuit.

The required diagram is shown in Figure 2

  

The expression for the input voltage $V_i$ is given by,

  $\begin{array}{l}{V}_i=V_{\pi 1}+V_{\epsilon 1}\\ V_{\epsilon 1}=V_i-V_{\pi 1}\end{array}$

Apply KCL at the node 1

  $\frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}=\frac{V_{\epsilon 1}}{R_{]E1}}-\frac{V_{\epsilon 1}-V_o}{R_F}$

Substitute $V_i-V_{\pi 1}$ for $V_{\epsilon 1}$ in the above equation.

  $\begin{array}{l}\frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}=\frac{\left(V_i-V_{\pi 1}\right)}{R_{]E1}}-\frac{\left(V_i-V_{\pi 1}\right)-V_o}{R_F}\\ V_{\pi 1}\left(\frac{1}{r_{\pi 1}}+g_{m1}\right)=\left(V_i-V_{\pi 1}\right)\left(\frac{1}{R_{E1}}+\frac{1}{R_F}\right)-\frac{V_o}{R_F}\end{array}$

Substitute $3.66\text{k}\Omega$ for $r_{\pi 1}$ . $32.81\text{mA}/\text{V}$ for $g_{m1}$ , $0.5\text{k}\Omega$ for $R_{E1}$ and $10\text{k}\Omega$ for $R_F$ in the above equation.

  $\begin{array}{l}{V}_{\pi 1}\left(\frac{1}{3.66\text{k}\Omega }+32.81\text{mA}/\text{V}\right)=\left(V_i-V_{\pi 1}\right)\left(\frac{1}{0.5\text{k}\Omega }+\frac{1}{10k}\right)-\frac{V_o}{10\text{k}\Omega }\\ 35.18V_{\pi 1}=2.10V_i-0.10V_o\end{array}$ ........... (1)

Apply KCL at node 2.

  $V_{\pi 2}=-\left(g_{m1}{V}_{\pi 1}\right)\left(\frac{R_{C1}{r}_{\pi 2}}{R_{C1}+r_{\pi 2}}\right)$

Substitute $6.28\text{k}\Omega$ for $r_{\pi 2}$ . $32.81\text{mA}/\text{V}$ for $g_{m1}$ and $8.8\text{k}\Omega$ for $R_{C1}$ in the above equation.

  $\begin{array}{c}{V}_{\pi 2}=-32.81\text{mA}/\text{V}\left(V_{\pi 1}\right)\left(\frac{\left(8.8\text{k}\Omega \right)\left(6.28\text{k}\Omega \right)}{\left(8.8\text{k}\Omega \right)+\left(6.28\text{k}\Omega \right)}\right)\\ =-120.2V_{\pi 1}\end{array}$

Apply KCL at the node 3

  $g_{m2}{V}_{\pi 2}+\frac{V_{\pi 3}+V_o}{R_{C2}}+\frac{V_{\pi 3}}{r_{\pi 3}}=0$

Substitute $1.54\text{k}\Omega$ for $r_{\pi 3}$ . $19.12\text{mA}/\text{V}$ for $g_{m2}$ and $13\text{k}\Omega$ for $R_{C2}$ in the above equation.

  $\begin{array}{l}\left(19.12\text{mA}/\text{V}\right)V_{\pi 2}+V_{\pi 3}\left(\frac{1}{13\text{k}\Omega }+\frac{1}{1.54\text{k}\Omega }\right)+\frac{V_o}{13\text{k}\Omega }=0\\ \left(19.12\text{mA}/\text{V}\right)V_{\pi 2}+0.7263V_{\pi 3}+0.07692V_o=0\end{array}$

Substitute $-120.2V_{\pi 1}$ for $V_{\pi 2}$ in the above equation.

  $\begin{array}{l}\left(19.12\text{mA}/\text{V}\right)\left(-120.2V_{\pi 1}\right)+0.7263V_{\pi 3}+0.07692V_o=0\\ V_{\pi 3}=\left(316.42\right)V_{\pi 1}-\left(0.1059\right)V_o\end{array}$ ........... (2)

Apply KCL at the node 4

  $\frac{V_{\pi 3}}{r_{\pi 3}}+g_{m3}{V}_{\pi 3}=\frac{V_o}{R_{E3}}+\frac{V_o-V_{\epsilon 1}}{R_F}$

Substitute $V_i-V_{\pi 1}$ for $V_{\epsilon 1}$ in the above equation.

  $\frac{V_{\pi 3}}{r_{\pi 3}}+g_{m2}{V}_{\pi 3}=\frac{V_o}{R_{E3}}+\frac{V_o-\left(V_i-V_{\pi 1}\right)}{R_F}$

Substitute $1.4\text{k}\Omega$ for $R_{E3}$ , $1.54\text{k}\Omega$ for $r_{\pi 3}$ . $78.08\text{mA}/\text{V}$ for $g_{m3}$ and $10\text{k}\Omega$ for $R_F$ in the above equation.

  $\begin{array}{l}\frac{V_{\pi 3}}{1.54\text{k}\Omega }+\left(78.08\text{mA}/\text{V}\right)V_{\pi 3}=\frac{V_o}{1.4\text{k}\Omega }+\frac{V_o-\left(V_i-V_{\pi 1}\right)}{10\text{k}\Omega }\\ 78.73V_{\pi 3}=\left(0.8143\right)V_o-0.1V_i+0.1V_{\pi 1}\end{array}$ ........... (3)

Substitute $\left(316.42\right)V_{\pi 1}-\left(0.1059\right)V_o$ in $V_{\pi 3}$ in equation (2)

  $\begin{array}{l}78.73\left(\left(316.42\right)V_{\pi 1}-\left(0.1059\right)V_o\right)=\left(0.8143\right)V_o-0.1V_i+0.1V_{\pi 1}\\ V_{\pi 1}\left(2.49\times {10}^5\right)V_{\pi 1}=9.152V_o-0.1V_i\\ V_{\pi 1}=\left(3.674\times {10}^{-5}\right)V_o-\left(4.104\times {10}^{-7}\right)V_i\end{array}$

Substitute $\left(3.674\times {10}^{-5}\right)V_o-\left(4.104\times {10}^{-7}\right)V_i$ for $V_{\pi 1}$ in equation (1).

  $\begin{array}{l}35.18\left(\left(3.674\times {10}^{-5}\right)V_o-\left(4.104\times {10}^{-7}\right)V_i\right)=2.10V_i-0.10V_o\\ \frac{V_o}{V_i}=20.7\end{array}$

Conclusion:

Therefore, the value of small signal voltage gain is 20.7.

To determine

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

Answer to Problem 12.37P

The value of the input resistance is $62.4\text{k}\Omega$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Calculation:

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

  $I_{R_{TH}}=\frac{V_i}{R_{TH}}$

Substitute $63.2\text{k}\Omega$ for $R_{TH}$ in the above equation.

  $\begin{array}{c}{I}_{R_{TH}}=\frac{V_i}{63.2\text{k}\Omega }\\ =1.582\times {10}^{-5}{V}_i\text{A}\end{array}$

The value of $V_{\pi 1}$ from equation (1) is given by,

  $\begin{array}{l}35.18V_{\pi 1}=2.10V_i-0.10V_o\\ V_{\pi 1}=0.0596V_i-\left(2.843\times {10}^{-3}\right)V_o\end{array}$

Substitute $20.7V_i$ for $V_o$ in the above equation.

  $\begin{array}{c}{V}_{\pi 1}=0.0596V_i-\left(2.843\times {10}^{-3}\right)20.7V_i\\ =7\times {10}^4{V}_i\end{array}$

The expression for the current $I_{r_{\pi 1}}$ is given by,

  $I_{r_{\pi 1}}=\frac{V_{\pi 1}}{r_{\pi 1}}$

Substitute $7\times {10}^4{V}_i$ for $V_{\pi 1}$ and $3.66\text{k}\Omega$ for $r_{\pi 1}$ in the above equation.

  $\begin{array}{c}{I}_{r_{\pi 1}}=\frac{7\times {10}^4{V}_i}{3.66\text{k}\Omega }\\ =1.913\times {10}^{-7}{V}_i\end{array}$

The expression for the input current $I_i$ is given by,

  $I_i=I_{R_{TH}}+I_{r\pi 1}$

Substitute $1.582\times {10}^{-5}{V}_i\text{A}$ for $I_{R_{TH}}$ and $1.913\times {10}^{-7}{V}_i$ for $I_{r_{\pi 1}}$ in the above equation.

  $\begin{array}{l}{I}_i=1.582\times {10}^{-5}{V}_i\text{A}+1.913\times {10}^{-7}{V}_i\\ \frac{V_i}{I_i}=62.4\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the input resistance is $62.4\text{k}\Omega$ .

To determine

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

Answer to Problem 12.37P

The value of the output resistance is $1.39\Omega$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Calculation:

Consider the equation (1).

  $\begin{array}{l}78.73V_{\pi 3}+I_x=\left(0.8143\right)V_o-0.1\left(0\right)+0.1V_{\pi 1}\\ 78.73V_{\pi 3}+I_x=\left(0.8143\right)V_o+0.1V_{\pi 1}\end{array}$ ........... (4)

Substitute $0$ for $V_i$ in equation (3).

  $\begin{array}{l}35.18V_{\pi 1}=2.10\left(0\right)-0.10V_o\\ V_{\pi 1}=-\left(0.002843\right)V_o\end{array}$

The equation for $V_{\pi 2}$ is given by,

  $V_{\pi 2}=-120.2V_{\pi 1}$

Substitute $-\left(0.002843\right)V_o$ for $V_{\pi 1}$ in the above equation.

  $\begin{array}{c}{V}_{\pi 2}=-120.2\left(-\left(0.002843\right)V_o\right)\\ =0.3417V_o\end{array}$

Consider the equation given,

  $\left(19.12\text{mA}/\text{V}\right)V_{\pi 2}+0.7263V_{\pi 3}+0.07692V_o=0$

Substitute $0.3417V_o$ for $V_{\pi 2}$ in the above equation.

  $\begin{array}{l}\left(19.12\text{mA}/\text{V}\right)\left(0.3417V_o\right)+0.7263V_{\pi 3}+0.07692V_o=0\\ V_{\pi 3}=-9.1V_o\end{array}$

Substitute $-9.1V_o$ for $V_{\pi 3}$ and $-\left(0.002843\right)V_o$ for $V_{\pi 1}$ in equation (4).

  $\begin{array}{l}78.73\left(-9.1V_o\right)+I_x=\left(0.8143\right)V_o+0.1\left(-\left(0.002843\right)V_o\right)\\ \frac{V_o}{I_x}=1.39\times {10}^{-3}\text{k}\Omega \\ \frac{V_o}{I_x}=1.39\Omega \end{array}$

Conclusion:

Therefore, the value of the output resistance is $1.39\Omega$ .