Chapter 12, Problem 12.42P

To determine

The value of the closed loop small signal voltage gain at the mid band frequency.

Answer to Problem 12.42P

Thevalue of the small signal closed loop voltage gain is $45.4$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

  

Calculation:

The Thevenin resistance of the above circuit is calculated as,

  $\begin{array}{c}{R}_{TH1}=\left(150\text{k}\Omega \right)\left|\right|\left(47\text{k}\Omega \right)\\ =\frac{\left(150\text{k}\Omega \right)\left(47\text{k}\Omega \right)}{\left(150\text{k}\Omega \right)+\left(47\text{k}\Omega \right)}\\ =35.8\text{k}\Omega \end{array}$

The expression to determine the value of the Thevenin voltage is given by,

  $\begin{array}{c}{V}_{TH1}=25\text{V}\left(\frac{47\text{k}\Omega }{150\text{k}\Omega +47\text{k}\Omega }\right)\\ =5.96\text{V}\end{array}$

The expression to determine the value of the current $I_{B1}$ is given by,

  $I_{B1}=\frac{V_{TH1}-V_{BE\left(\text{on}\right)Q_1}}{R_{TH1}+\left(h_{FE}+1\right)\left(4.7\text{k}\Omega +100\Omega \right)}$

Substitute $0.7\text{V}$ for $V_{BE\left(\text{on}\right)Q_1}$ , $5.96\text{V}$ for $V_{TH1}$ , $35.8\text{k}\Omega$ for $R_{TH1}$ and $50$ for $h_{FE}$ in the above equation.

  $\begin{array}{c}{I}_{B1}=\frac{5.96\text{V}-0.7\text{V}}{35.8\text{k}\Omega +\left(50+1\right)\left(4.7\text{k}\Omega +100\Omega \right)}\\ =0.0187\text{mA}\end{array}$

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

  $I_{C1}=h_{FE}{I}_{B1}$

Substitute $0.0187\text{mA}$ for $I_{B1}$ and $50$ for $h_{FE}$ in the above equation.

  $\begin{array}{c}{I}_{C1}=50\left(0.0187\text{mA}\right)\\ =0.935\text{mA}\end{array}$

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

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

Substitute $50$ for $h_{FE}$ , $0.026$ for $V_T$ and $0.935\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 1}=\frac{\left(50\right)\left(0.026\text{V}\right)}{0.935\text{mA}}\\ =1.39\text{k}\Omega \end{array}$

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

  $g_{m1}=\frac{I_{CQ1}}{V_T}$

Substitute $0.935\text{mA}$ for $I_{C1}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m1}=\frac{0.935\text{mA}}{0.026\text{V}}\\ =35.96\text{mA}/\text{V}\end{array}$

The Thevenin resistance of the second transistor is calculated as.

  $\begin{array}{c}{R}_{TH2}=\frac{\left(47\text{k}\Omega \right)\left(33\text{k}\Omega \right)}{\left(47\text{k}\Omega \right)+\left(33\text{k}\Omega \right)}\\ =19.4\text{k}\Omega \end{array}$

The expression to determine the value of the Thevenin voltage of the second transistor is given by,

  $\begin{array}{c}{V}_{TH2}=25\text{V}\left(\frac{33\text{k}\Omega }{33\text{k}\Omega +47\text{k}\Omega }\right)\\ =10.3\text{V}\end{array}$

The expression to determine the value of the current $I_{B1}$ is given by,

  $I_{B2}=\frac{V_{TH2}-V_{BE\left(\text{on}\right)Q_1}}{R_{TH2}+\left(h_{FE}+1\right)\left(4.7\text{k}\Omega \right)}$

Substitute $0.7\text{V}$ for $V_{BE\left(\text{on}\right)Q_1}$ , $10.3\text{V}$ for $V_{TH2}$ , $19.4\text{k}\Omega$ for $R_{TH2}$ and $50$ for $h_{FE}$ in the above equation.

  $\begin{array}{c}{I}_{B2}=\frac{10.3\text{V}-0.7\text{V}}{19.4\text{k}\Omega +\left(50+1\right)\left(4.7\text{k}\Omega \right)}\\ =0.03705\text{mA}\end{array}$

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

  $I_{C2}=h_{FE}{I}_{B2}$

Substitute $0.03705\text{mA}$ for $I_{B2}$ and $50$ for $h_{FE}$ in the above equation.

  $\begin{array}{c}{I}_{C2}=50\left(0.03705\text{mA}\right)\\ =1.85\text{mA}\end{array}$

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

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

Substitute $50$ for $h_{FE}$ , $0.026$ for $V_T$ and $1.85\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{\left(50\right)\left(0.026\text{V}\right)}{1.85\text{mA}}\\ =0.703\text{k}\Omega \end{array}$

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

  $g_{m2}=\frac{I_{C2}}{V_T}$

Substitute $1.85\text{mA}$ for $I_{C2}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{1.85\text{mA}}{0.026\text{V}}\\ =71.15\text{mA}/\text{V}\end{array}$

The diagram for the small signal equivalent circuit is shown in Figure 2

  

Figure 2

The expression to determine the value of the supply voltage is given by,

  $V_S=V_{\pi 1}+V_{\epsilon }$ ……. (1)

Apply KCL at node $\epsilon$ .

  $\frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}=\frac{V_{\epsilon }}{R_1}-\frac{V_{\epsilon }-V_O}{R_2}$

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

  $\begin{array}{l}\frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}=\frac{V_S-V_{\pi 1}}{R_1}-\frac{V_S-V_{\pi 1}-V_O}{R_2}\\ \frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}=\left(V_S-V_{\pi 1}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)-\frac{V_O}{R_2}\end{array}$

Substitute $1.39\text{k}\Omega$ for $r_{\pi 1}$ , $35.96\text{mA}/\text{V}$ for $g_{m1}$ , $100\Omega$ for $R_1$ and $4.7\text{k}\Omega$ for $R_2$ in the above equation.

  $\begin{array}{l}\frac{V_{\pi 1}}{1.39\text{k}\Omega }+\left(35.96\text{mA}/\text{V}\right)V_{\pi 1}=\left(V_S-V_{\pi 1}\right)\left(\frac{1}{100\Omega }+\frac{1}{4.7\text{k}\Omega }\right)-\frac{V_O}{4.7\text{k}\Omega }\\ 46.89\times {10}^{-3}{V}_{\pi 1}=10.213\times {10}^{-3}{V}_S-0.2128\times {10}^{-3}{V}_O\\ V_{\pi 1}=0.2178V_S-\left(4.538\times {10}^{-3}\right)V_O\end{array}$

Apply KCL at node 1

  $g_{m1}{V}_{\pi 1}+\frac{V_{\pi 2}}{10\text{k}\Omega }+\frac{V_{\pi 2}}{R_{TH2}}+\frac{V_{\pi 2}}{r_{\pi 2}}=0$

Substitute $0.703\text{k}\Omega$ for $r_{\pi 2}$ , $35.96\text{mA}/\text{V}$ for $g_{m1}$ and $19.4\text{k}\Omega$ for $R_{TH2}$ in the above equation.

  $\begin{array}{l}\left(35.96\text{mA}/\text{V}\right)V_{\pi 1}+\frac{V_{\pi 2}}{10\text{k}\Omega }+\frac{V_{\pi 2}}{R_{TH2}}+\frac{V_{\pi 2}}{0.703\text{k}\Omega }=0\\ V_{\pi 2}=-22.85\times {10}^{-3}{V}_{\pi 1}\end{array}$

Apply KCL at node 2

  $g_{m2}{V}_{\pi 2}+\frac{V_O}{R_{C2}}+\frac{V_O-V_{\epsilon }}{R_2}=0$

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

  $g_{m2}{V}_{\pi 2}+\frac{V_O}{R_{C2}}+\frac{V_O-\left(V_S-V_{\pi 1}\right)}{R_2}=0$

Substitute $4.7\text{k}\Omega$ for $R_{C2}$ , $71.15\text{mA}/\text{V}$ for $g_{m2}$ and $4.7\text{k}\Omega$ for $R_2$ .in the above equation.

  $\begin{array}{l}\left(71.15\text{mA}/\text{V}\right)V_{\pi 2}+\frac{V_O}{4.7\text{k}\Omega }+\frac{V_O-\left(V_S-V_{\pi 1}\right)}{4.7\text{k}\Omega }=0\\ \left[\begin{array}{l}\left(71.15\text{mA}/\text{V}\right)V_{\pi 2}+V_O\left(0.4255\times {10}^{-3}\right)-\\ \left(0.2128\times {10}^{-3}\right)V_S+\left(0.2128\times {10}^{-3}\right)V_{\pi 1}\end{array}\right]=0\end{array}$

Substitute $-22.85\times {10}^{-3}{V}_{\pi 1}$ for $V_{\pi 2}$ in the above equation.

  $\begin{array}{l}\left[\begin{array}{l}\left(71.15\text{mA}/\text{V}\right)\left(-22.85\times {10}^{-3}{V}_{\pi 1}\right)+V_O\left(0.4255\times {10}^{-3}\right)-\\ \left(0.2128\times {10}^{-3}\right)V_S+\left(0.2128\times {10}^{-3}\right)V_{\pi 1}\end{array}\right]=0\\ -1625.3V_{\pi 1}+V_{O}0.4255-0.2128V_S=0\end{array}$

Substitute $0.2178V_S-\left(4.538\times {10}^{-3}\right)V_O$ for $V_{\pi 1}$ in the above equation.

  $\begin{array}{l}-1625.3\left(0.2178V_S-\left(4.538\times {10}^{-3}\right)V_O\right)+V_{O}0.4255-0.2128V_S=0\\ \frac{V_O}{V_S}=45.4\end{array}$

Conclusion:

Therefore, the value of the small signal closed loop voltage gain is $45.4$ .