Chapter 12, Problem 12.36P

(a)

To determine

The quiescent collector currents and the dc voltage at the output.

Answer to Problem 12.36P

The value of $I_{C1}$ is $0.5\text{mA}$ and $I_{C2}$ is $0.5\text{mA}$ , the dc voltage at the output is $0$ and the value of $I_{C3}$ is $2\text{mA}$ .

Explanation of Solution

Given:

The give circuit is shown in Figure 1

  

Calculation:

The value of the collector current of the first transistor is calculated as,

  $\begin{array}{c}{I}_{C1}=\frac{1\text{mA}}{2}\\ =0.5\text{mA}\end{array}$

The value of the collector current of the second transistor is calculated as,

  $\begin{array}{c}{I}_{C2}=\frac{1\text{mA}}{2}\\ =0.5\text{mA}\end{array}$

The value of the collector voltage of the second transistor is given by,

  $V_{C2}=12\text{V}-I_{C2}\left(22.6\text{k}\Omega \right)$

Substitute $0.5\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{V}_{C2}=12\text{V}-\left(0.5\text{mA}\right)\left(22.6\text{k}\Omega \right)\\ =0.7\text{V}\end{array}$

The expression for the output voltage is calculated as,

  $v_O=V_{C2}-V_{BE}{\left(\text{on}\right)}_{Q_3}$

Substitute $0.7\text{V}$ for $V_{C2}$ and $V_{BE}{\left(\text{on}\right)}_{Q_3}$ for $V_{BE}{\left(\text{on}\right)}_{Q_3}$ in the above equation.

  $\begin{array}{c}{v}_O=0.7\text{V}-0.7\text{V}\\ =0\end{array}$

The collector current of the third transistor is equal to its emitter current and is given by,

  $I_{C3}=2\text{mA}$

Conclusion:

Therefore, the value of $I_{C1}$ is $0.5\text{mA}$ and $I_{C2}$ is $0.5\text{mA}$ , the dc voltage at the output is $0$ and the value of $I_{C3}$ is $2\text{mA}$ .

(b)

To determine

The value of the small signal voltage gain.

Answer to Problem 12.36P

The value of small signal voltage gain is $5.68$ .

Explanation of Solution

Given:

The give circuit is shown in Figure 1

  

Calculation:

The expression to determine the value of the small signal input resistance of the first transistor is given by,

  $r_{\pi 1}=\frac{h_{FE}{V}_T}{I_{C2}}$

Substitute $100$ for $h_{FE}$ , $0.026\text{V}$ for $V_T$ and $0.5\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 1}=\frac{\left(0.26\text{V}\right)\left(100\right)}{0.5\text{mA}}\\ =5.2\text{k}\Omega \end{array}$

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

  $g_{m1}=\frac{I_{C1}}{V_T}$

Substitute $0.026\text{V}$ for $V_T$ and $0.5\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{g}_{m1}=\frac{0.5\text{mA}}{0.026\text{V}}\\ =19.23\text{mA}/\text{V}\end{array}$

The expression to determine the value of the small signal input resistance of the second transistor is given by,

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

Substitute $100$ for $h_{FE}$ , $0.026\text{V}$ for $V_T$ and $0.5\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{\left(0.26\text{V}\right)\left(100\right)}{0.5\text{mA}}\\ =5.2\text{k}\Omega \end{array}$

The expression to determine the trans-conductance of the second transistor is given by,

  $g_{m2}=\frac{I_{C2}}{V_T}$

Substitute $0.026\text{V}$ for $V_T$ and $0.5\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{0.5\text{mA}}{0.026\text{V}}\\ =19.23\text{mA}/\text{V}\end{array}$

The expression to determine the value of the small signal input resistance of the third transistor is given by,

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

Substitute $100$ for $h_{FE}$ , $0.026\text{V}$ for $V_T$ and $2\text{mA}$ for $I_{C3}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 3}=\frac{\left(0.26\text{V}\right)\left(100\right)}{2\text{mA}}\\ =1.3\text{k}\Omega \end{array}$

The expression to determine the trans-conductance of the third transistor is given by,

  $g_{m3}=\frac{I_{C3}}{V_T}$

Substitute $0.026\text{V}$ for $V_T$ and $2\text{mA}$ for $I_{C3}$ in the above equation.

  $\begin{array}{c}{g}_{m3}=\frac{2\text{mA}}{0.026\text{V}}\\ =76.92\text{mA}/\text{V}\end{array}$

Mark the values and draw the small signal equivalent circuit.

The required diagram is shown in Figure 2

  

Apply KCL at the node 1

  $\begin{array}{l}\frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}+g_{m1}{V}_2+\frac{V_{\pi 2}}{r_{\pi 1}}=0\\ V_{\pi 2}=-V_{\pi 1}\end{array}$

The expression to determine the input voltage is given by,

  $V_i=\frac{V_{\pi 1}}{r_{\pi 1}}\left(R_S+r_{\pi 1}\right)-V_{\pi 2}{+}_{b2}$

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

  $\begin{array}{c}{V}_i=\frac{V_{\pi 1}}{r_{\pi 1}}\left(R_S+r_{\pi 1}\right)+V_{\pi 1}+V_{b2}\\ =V_{\pi 1}\left(2+\frac{R_S}{r_{\pi 1}}\right)+V_{b2}\end{array}$

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

  $\begin{array}{c}{V}_i=\frac{V_{\pi 1}}{r_{\pi 1}}\left(R_S+r_{\pi 1}\right)+V_{\pi 1}+V_{b2}\\ =-V_{\pi 2}\left(2+\frac{R_S}{r_{\pi 1}}\right)+V_{b2}\end{array}$

Substitute $1\text{k}\Omega$ for $R_S$ and $5.2\text{k}\Omega$ for $r_{\pi 1}$ in the above equation.

  $\begin{array}{l}{V}_i=\frac{V_{\pi 1}}{r_{\pi 1}}\left(1\text{k}\Omega +r_{\pi 1}\right)+V_{\pi 1}+V_{b2}\\ V_i=-V_{\pi 2}\left(2+\frac{1\text{k}\Omega }{5.2\text{k}\Omega }\right)+V_{b2}\\ V_i=-V_{\pi 2}\left(2.192\right)+V_{b2}\\ V_{b2}=V_i+V_{\pi 2}\left(2.192\right)\end{array}$

Apply KCL at the node 2

  $\begin{array}{l}\frac{V_{o2}}{R_C}+g_{m1}{V}_{\pi 2}+g_{m1}{V}_{\pi 2}+\frac{V_{o2}-V_o}{r_{\pi 3}}=0\\ V_{o2}\left(\frac{1}{R_C}+\frac{1}{r_{\pi 3}}\right)+g_{m1}{V}_{\pi 2}-\frac{V_o}{r_{\pi 3}}=0\end{array}$

Substitute $19.23\text{mA}/\text{V}$ for $g_{m1}$ , $22.6\text{k}\Omega$ for $R_C$ and $1.3\text{k}\Omega$ for $r_{\pi 3}$ in the above equation.

  $\begin{array}{l}{V}_{o2}\left(\frac{1}{22.6\text{k}\Omega }+\frac{1}{1.3\text{k}\Omega }\right)+\left(19.23\text{mA}/\text{V}\right)V_{\pi 2}-\frac{V_o}{1.3\text{k}\Omega }=0\\ V_{o2}=0.9455V_o-23.64V_{\pi 2}\end{array}$

The expression to determine the value of $V_{\pi 3}$ is given by,

  $V_{\pi 3}=V_{o2}-V_o$

Apply KCL at the node 3

  $\begin{array}{l}\frac{V_{\pi 3}}{r_{\pi 3}}+g_{m3}{V}_{\pi 3}=\frac{V_o}{R_L}+\frac{V_o-V_{b2}}{R_2}\\ V_{\pi 3}\left(\frac{1}{r_{\pi 3}}+g_{m3}\right)=V_o\left(\frac{1}{R_L}+\frac{1}{R_2}\right)-\frac{V_{b2}}{R_2}\end{array}$

Substitute $V_{o2}-V_o$ for $V_{\pi 3}$ and $V_i+V_{\pi 2}\left(2.192\right)$ for $V_{b2}$ in the above equation.

  $\begin{array}{l}\left(V_{o2}-V_o\right)\left(\frac{1}{r_{\pi 3}}+g_{m3}\right)=V_o\left(\frac{1}{R_L}+\frac{1}{R_2}\right)-\frac{\left(V_i+V_{\pi 2}\left(2.192\right)\right)}{R_2}\\ V_{o2}\left(\frac{1}{r_{\pi 3}}+g_{m3}\right)-V_o\left(\frac{1}{r_{\pi 3}}+g_{m3}+\frac{1}{R_L}+\frac{1}{R_2}\right)=-\frac{1}{R_2}\left(V_i+V_{\pi 2}\left(2.192\right)\right)\end{array}$

Substitute $4\text{k}\Omega$ for $R_L$ , $76.92\text{mA}/\text{V}$ for $g_{m3}$ , $50\text{k}\Omega$ for $R_2$ and $1.3\text{k}\Omega$ for $r_{\pi 3}$ in the above equation.

  $\begin{array}{l}\left[\begin{array}{l}{V}_{o2}\left(\frac{1}{1.3\text{k}\Omega }+76.92\text{mA}/\text{V}\right)-V_o\left(\frac{1}{1.3\text{k}\Omega }+76.92\text{mA}/\text{V}+\frac{1}{4\text{k}\Omega }+\frac{1}{50\text{k}\Omega }\right)\\ =-\frac{1}{50\text{k}\Omega }\left(V_i+V_{\pi 2}\left(2.192\right)\right)\end{array}\right]\\ V_{o2}\left(77.69\times {10}^{-3}\right)=\left\{-\left(0.02\times {10}^{-3}\right)\left[V_i+2.192V_{\pi 2}\right]\right\}+V_o\left(77.96\times {10}^{-3}\right)\end{array}$ …… (1)

Apply KCL at the node 4

  $\frac{V_{b2}-V_o}{R_2}+\frac{V_{b2}}{R_1}+\frac{V_{\pi 2}}{R_{\pi 1}}=0$

Substitute $V_i+2.192V_{\pi 2}$ for $V_{b2}$ , $5.2\text{k}\Omega$ for $r_{\pi 1}$ , $50\text{k}\Omega$ for $R_2$ and $10\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{l}\frac{V_i+2.192V_{\pi 2}-V_o}{50\text{k}\Omega }+\frac{V_i+2.192V_{\pi 2}}{10\text{k}\Omega }+\frac{V_{\pi 2}}{5.2\text{k}\Omega }=0\\ \left(0.12\times {10}^{-3}\right)V_i+\left(0.4553\times {10}^{-3}\right)V_{\pi 2}-\left(0.02\times \right)V_o=0\end{array}$

Substitute $0.9455V_o-23.64V_{\pi 2}$ for $V_{o2}$ in equation (1).

  $\left[\left(\begin{array}{l}0.9455V_o\\ -23.64V_{\pi 2}\end{array}\right)\right]\left(77.69\times {10}^{-3}\right)=\left\{-\left(0.02\times {10}^{-3}\right)\left[\begin{array}{l}{V}_i\\ +2.192V_{\pi 2}\end{array}\right]\right\}+V_o\left(77.96\times {10}^{-3}\right)$

Substitute $\left(0.04393\times {10}^{-3}\right)V_o-\left(0.2636\times {10}^{-3}\right)V_i$ for $V_{\pi 2}$ in the above equation.

  $\begin{array}{l}\left(0.12\times {10}^{-3}\right)V_i+\left(0.4553\times {10}^{-3}\right)\left(\left(0.04393\times {10}^{-3}\right)V_o-\left(0.2636\times {10}^{-3}\right)V_i\right)-\left(0.02\times \right)V_o=0\\ V_o\left(4.505+80.677\right)-\left(484.101+0.02\right)V_i=0\\ \frac{V_o}{V_i}=5.68\end{array}$

The expression for the small signal voltage gain is given by,

  $A_f=\frac{V_o}{V_i}$

Substitute $5.68$ for $\frac{V_o}{V_i}$ in the above equation.

  $A_f=5.68$

Conclusion:

Therefore, the value of small signal voltage gain is $5.68$ .