Chapter 12, Problem 12.53P

For the transistors in the circuit in Figure P 12.53, the parameters are: $h_{FE}=50,V_{BE}(\text{on})=0.7\text{V},$ and $V_A=\infty .$ Using nodal analysis, determine the closed-loop current gain $A_{if}=i_o/i_s$ .

To determine

The value of the closed loop current gain of the circuit.

Answer to Problem 12.53P

The value of the current gain is $5.33$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Calculation:

The small signal model of the above circuit is shown below.

The required diagram is shown in Figure 2

  

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

  $\begin{array}{c}{R}_{TH}=\frac{\left(1.4\text{k}\Omega \right)\left(17.9\text{k}\Omega \right)}{\left(1.4\text{k}\Omega \right)+\left(17.9\text{k}\Omega \right)}\\ =1.298\text{k}\Omega \end{array}$

The Thevenin voltage of the circuit is calculated as,

  $\begin{array}{c}{V}_{TH}=\left(\frac{1.4\text{k}\Omega }{1.4\text{k}\Omega +17.9\text{k}\Omega }\right)10\text{V}\\ =0.725\text{V}\end{array}$

The expression for the base current of the first transistor is given by,

  $I_{B1}=\frac{V_{TH}-V_{BE}\left(\text{on}\right)}{R_{TH}}$

Substitute $1.298\text{k}\Omega$ for $R_{TH}$ , $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ , and $0.725\text{V}$ for $V_{TH}$ in the above equation.

  $\begin{array}{c}{I}_{B1}=\frac{0.725\text{V}-0.7\text{V}}{1.298\text{k}\Omega }\\ =0.0196\text{mA}\end{array}$

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

  $I_{C1}=h_{FE}{I}_{B1}$

Substitute $50$ for $h_{FE}$ and $0.0196\text{mA}$ for $I_{B1}$ in the above equation.

  $\begin{array}{c}{I}_{C1}=50\left(0.0196\text{mA}\right)\\ =0.98\text{mA}\end{array}$

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

  $V_{B2}=10\text{V}-I_{C1}\left(7\text{k}\Omega \right)$

Substitute $0.98\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{V}_{B2}=10\text{V}-\left(0.98\text{mA}\right)\left(7\text{k}\Omega \right)\\ =3.14\text{V}\end{array}$

The expression for the current $I_{E2}$ is given by,

  $I_{E2}=\frac{V_{B2}-V_{BE}\left(\text{on}\right)}{250\Omega +250\Omega }$

Substitute $3.14\text{V}$ for $V_{B2}$ and $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{c}{I}_{E2}=\frac{3.14\text{V}-0.7\text{V}}{250\Omega +250\Omega }\\ =3.25\text{mA}\end{array}$

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

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

Substitute $3.25\text{mA}$ for $I_{E2}$ and $50$ for $h_{FE}$ in the above equation.

  $\begin{array}{c}{I}_{C2}=\left(3.25\text{mA}\right)\left(\frac{50}{1+50}\right)\\ =3.19\text{mA}\end{array}$

The expression for the transconductance of the first transistor is given by,

  $g_{m1}=\frac{I_{C1}}{0.026\text{V}}$

Substitute $0.98\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{g}_{m1}=\frac{0.98\text{mA}}{0.026\text{V}}\\ =37.7\text{mA}/\text{V}\end{array}$

The expression for the small signal resistance is given by,

  $r_{\pi 1}=\frac{h_{FE}}{g_{m1}}$

Substitute $50$ for $h_{FE}$ and $37.7\text{mA}/\text{V}$ for $g_{m1}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 1}=\frac{50}{37.7\text{mA}/\text{V}}\\ =1.33\text{k}\Omega \end{array}$

The expression for the transconductance of the second transistor is given by,

  $g_{m2}=\frac{I_{C2}}{0.026\text{V}}$

Substitute $3.19\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{3.19\text{mA}}{0.026\text{V}}\\ =123\text{mA}/\text{V}\end{array}$

The expression for the small signal resistance is given by,

  $r_{\pi 2}=\frac{h_{FE}}{g_{m2}}$

Substitute $120$ for $h_{FE}$ and $123\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{120}{123\text{mA}/\text{V}}\\ =0.408\text{k}\Omega \end{array}$

Apply KCL at source $i_S$ .

  $i_S=\frac{V_{\pi 1}}{1.4\text{k}\Omega \left|\right|1.79\text{k}\Omega \left|\right|r_{\pi 1}}+\frac{V_{\pi 1}-V_1}{5\text{k}\Omega }$

Substitute $1.33\text{k}\Omega$ for $r_{\pi 1}$ in the above equation.

  $\begin{array}{c}{i}_S=\frac{V_{\pi 1}}{1.4\text{k}\Omega \left|\right|1.79\text{k}\Omega \left|\right|1.33\text{k}\Omega }+\frac{V_{\pi 1}-V_1}{5\text{k}\Omega }\\ =1.722V_{\pi 1}-0.20V_1\end{array}$ …… (1)

Apply KCL at node A.

  $g_{m1}{V}_{\pi 1}+\frac{V_{\pi 2}}{r_{\pi 2}}+\frac{V_{\pi 2}+V_{CE}}{7\text{k}\Omega }=0$

Substitute $37.7\text{mA}/\text{V}$ for $g_{m1}$ and $0.408\text{k}\Omega$ for $r_{\pi 2}$ in the above equation.

  $\begin{array}{l}\left(37.7\text{mA}/\text{V}\right)V_{\pi 1}+\frac{V_{\pi 2}}{0.408\text{k}\Omega }+\frac{V_{\pi 2}+V_{CE}}{7\text{k}\Omega }=0\\ 37.7V_{\pi 1}+2.594V_{\pi 2}+0.1429V_{CE}=0\end{array}$ …… (2)

Apply KCL at node $V_{CE}$ .

  $\frac{V_{\pi 2}}{r_{\pi 2}}+g_{m2}{V}_{\pi 2}+\frac{V_{CE}-V_1}{250\Omega }=0$

Substitute $123\text{mA}/\text{V}$ for $g_{m2}$ and $0.408\text{k}\Omega$ for $r_{\pi 2}$ in the above equation.

  $\begin{array}{l}\frac{V_{\pi 2}}{0.408\text{k}\Omega }+\left(123\text{mA}/\text{V}\right)V_{\pi 2}+\frac{V_{CE}-V_1}{250\Omega }=0\\ 125.5V_{\pi 2}=4V_{CE}-4V_1\end{array}$ ……… (3)

Apply KCL at $V_1$ .

  $\begin{array}{l}\frac{V_{CE}-V_1}{250\Omega }=\frac{V_1}{500\Omega }+\frac{V_1-V_{\pi 1}}{5\text{k}\Omega }\\ V_{CE}=1.55V_1-0.05V_{\pi 1}\end{array}$

Substitute $1.55V_1-0.05V_{\pi 1}$ for $V_{CE}$ in equation (3).

  $\begin{array}{l}37.7V_{\pi 1}+2.594V_{\pi 2}+0.1429\left(1.55V_1-0.05V_{\pi 1}\right)=0\\ 125.5V_{\pi 2}=2.20V_1-0.20V_{\pi 1}\end{array}$ …… (4)

Substitute $1.55V_1-0.05V_{\pi 1}$ for $V_{CE}$ in equation (2).

  $\begin{array}{l}37.7V_{\pi 1}+2.594V_{\pi 2}+0.1429\left(1.55V_1-0.05V_{\pi 1}\right)=0\\ V_1=-170.16V_{\pi 1}+11.71V_{\pi 2}\end{array}$

Substitute $-170.16V_{\pi 1}+11.71V_{\pi 2}$ for $V_1$ in equation (1).

  $\begin{array}{c}{i}_S=1.722V_{\pi 1}-0.20\left(-170.16V_{\pi 1}+11.71V_{\pi 2}\right)\\ =35.75V_{\pi 1}+2.342V_{\pi 2}\end{array}$ ……. (5)

Substitute $-170.16V_{\pi 1}+11.71V_{\pi 2}$ for $V_1$ in equation (4).

  $\begin{array}{l}125.5V_{\pi 2}=2.20\left(-170.16V_{\pi 1}+11.71V_{\pi 2}\right)-0.20V_{\pi 1}\\ V_{\pi 1}=-0.4040V_{\pi 2}\end{array}$

Substitute $-0.4040V_{\pi 2}$ for $V_{\pi 1}$ in equation (5).

  $\begin{array}{c}{i}_S=35.75\left(-0.4040V_{\pi 2}\right)+2.342\left(V_{\pi 2}\right)\\ =-12.10V_{\pi 2}\end{array}$ ….. (6)

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

  $i_O=-g_{m2}{V}_{\pi 2}\left(\frac{2.2\text{k}\Omega }{2.2\text{k}\Omega +2\text{k}\Omega }\right)$

Substitute and $123\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $\begin{array}{l}{i}_O=-\left(123\text{mA}/\text{V}\right)V_{\pi 2}\left(\frac{2.2\text{k}\Omega }{2.2\text{k}\Omega +2\text{k}\Omega }\right)\\ V_{\pi 2}=-0.01552i_O\end{array}$

Substitute $-0.01552i_O$ for $V_{\pi 1}$ in equation (6).

  $\begin{array}{l}{i}_S=-12.10\left(-0.01552i_O\right)\\ \frac{i_S}{i_O}=5.33\end{array}$

Conclusion:

Therefore, the value of the current gain is $5.33$ .