Chapter 12, Problem 12.38P

The circuit shown in Figure P12.38 is an ac equivalent circuit of a feedback amplifier. The transistor parameters are $h_{FE}=100$ and $V_A=\infty .$ The quiescent collector currents are $I_{C1}=14.3\text{mA},I_{C2}=4.62\text{mA},$ and $I_{C3}=$

Figure P12.38 $4.47\text{mA}.$ (a) Determine the closed-loop voltage gain $A_{vf}=V_o/V_i.$ Compare this value to the approximate ideal value of $A_{vf}\cong \left(R_F+R_E\right)/R_E$ . (b) Determine the values of $R_{if}$ and $R_{of}$ .

To determine

The value of the closed loop voltage gain.

To compare: The approximate value with the ideal value of the closed loop gain.

Answer to Problem 12.38P

The value of closed loop gain is $23.6$ . The closed loop voltage gain is by formula $\frac{V_o}{V_i}$ is $23.6$ and is less than the approximate ideal value of closed loop voltage gain that is $25$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

  

Calculation:

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

  

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 $14.3\text{mA}$ for $I_{C1}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 1}=\frac{\left(120\right)\left(0.026\text{V}\right)}{14.3\text{mA}}\\ =0.182\text{k}\Omega \end{array}$

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

  $g_{m1}=\frac{I_{CQ1}}{V_T}$

Substitute $14.3\text{mA}$ for $I_{C{Q}_1}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m1}=\frac{14.3\text{mA}}{0.026\text{V}}\\ =550\text{mA}/\text{V}\end{array}$

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 $4.62\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{\left(100\right)\left(0.026\text{V}\right)}{4.62\text{mA}}\\ =0.563\text{k}\Omega \end{array}$

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

  $g_{m2}=\frac{I_{C2}}{V_T}$

Substitute $4.62\text{mA}$ for $I_{C2}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{4.62\text{mA}}{0.026\text{V}}\\ =178\text{mA}/\text{V}\end{array}$

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

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

Substitute $100$ for $h_{FE}$ , $0.026$ for $V_T$ and $4.47\text{mA}$ for $I_{C3}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 3}=\frac{\left(100\right)\left(0.026\text{V}\right)}{4.47\text{mA}}\\ =0.582\text{k}\Omega \end{array}$

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

  $g_{m3}=\frac{I_{C3}}{V_T}$

Substitute $4.47\text{mA}$ for $I_{C3}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{g}_{m3}=\frac{4.47\text{mA}}{0.026\text{V}}\\ =172\text{mA}/\text{V}\end{array}$

The expression for the $V_{\pi 1}$ is given by,

  $V_{\pi 1}=V_i-V_A$

Apply KCL at $V_A$

  $\frac{V_i-V_A}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}=\frac{V_A}{R_E}+\frac{V_A-V_o}{R_F}$

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

  $\left(V_i-V_A\right)\left(\frac{1}{r_{\pi 1}}+g_{m1}\right)=\frac{V_A}{R_E}+\frac{V_A-V_o}{R_F}$

Substitute $0.182\text{k}\Omega$ for $r_{\pi 1}$ and $550\text{mA}/\text{V}$ for $g_{m1}$ in the above equation.

  $\begin{array}{l}\left(V_i-V_A\right)\left(\frac{1}{0.182\text{k}\Omega }+550\text{mA}/\text{V}\right)=\frac{V_A}{50\Omega }+\frac{V_A-V_o}{1.2\text{k}\Omega }\\ V_i\left(555.5\right)+V_o\left(0.8333\right)=V_A\left(576.3\right)\end{array}$ …… (1)

Apply KCL at $V_B$

  $\frac{V_B}{R_{C1}}+g_{m1}{V}_{\pi 1}+\frac{V_B}{r_{\pi 2}}=0$

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

  $\begin{array}{l}\frac{V_B}{R_{C1}}+g_{m1}\left(V_i-V_A\right)+\frac{V_B}{r_{\pi 2}}=0\\ V_B\left(\frac{1}{R_{C1}}+\frac{1}{r_{\pi 2}}\right)g_{m1}\left(V_i-V_A\right)=0\end{array}$

Substitute $0.3\text{k}\Omega$ for $R_{C1}$ , $0.563\text{k}\Omega$ for $r_{\pi 2}$ and $550\text{mA}/\text{V}$ for $g_{m1}$ in the above equation.

  $\begin{array}{l}{V}_B\left(\frac{1}{0.3\text{k}\Omega }+\frac{1}{0.563\text{k}\Omega }\right)\left(550\text{mA}/\text{V}\right)\left(V_i-V_A\right)=0\\ V_B=V_A\left(107.7\right)-V_i\left(107.7\right)\end{array}$

Apply KCL at $V_C$

  $\frac{V_C}{R_{C1}}+g_{m2}{V}_{\pi 2}+\frac{V_C-V_o}{r_{\pi 3}}=0$

Substitute $V_i-V_A$ for $V_{\pi 1}$ and $V_B$ for $V_{\pi 2}$ in the above equation

  $\begin{array}{l}\frac{V_C}{R_{C1}}+g_{m2}{V}_B+\frac{V_C-V_o}{r_{\pi 3}}=0\\ V_C\left(\frac{1}{R_{C2}}+\frac{1}{r_{\pi 3}}\right)+g_{m2}{V}_B-\frac{V_o}{r_{\pi 3}}=0\end{array}$

Substitute $0.65\text{k}\Omega$ for $R_{C2}$ , $0.582\text{k}\Omega$ for $r_{\pi 3}$ and $178\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $\begin{array}{l}{V}_C\left(\frac{1}{0.65\text{k}\Omega }+\frac{1}{0.582\text{k}\Omega }\right)+\left(178\text{mA}/\text{V}\right)V_B-\frac{V_o}{0.582}=0\\ V_C\left(\text{3}\text{.527}\right)+178V_B-V_o\left(1.718\right)=0\end{array}$ …… (2)

Apply KCL at $V_o$ .

  $g_{m3}{V}_{\pi 3}+\frac{V_{\pi 3}}{r_{\pi 3}}=\frac{V_o-V_A}{R_F}$

Substitute $V_C-V_o$ for $V_{\pi 3}$ in the above equation.

  $\begin{array}{l}{g}_{m3}{V}_{\pi 3}+\frac{V_C-V_o}{r_{\pi 3}}=\frac{V_o-V_A}{R_F}\\ \left(V_C-V_o\right)\left(g_{m3}+\frac{1}{r_{\pi 3}}\right)=\frac{V_o-V_A}{1.2}\\ V_C=V_o\left(1.0046\right)-V_A\left(0.004749\right)\end{array}$

Substitute $V_A\left(107.7\right)-V_i\left(107.7\right)$ for $V_B$ and $V_o\left(1.0046\right)-V_A\left(0.004749\right)$ for $V_C$ in equation (2).

  $\begin{array}{l}\left[\begin{array}{l}\left(V_o\left(1.0046\right)-V_A\left(0.004749\right)\right)\left(\text{3}\text{.527}\right)\\ +178\left(V_A\left(107.7\right)-V_i\left(107.7\right)\right)-V_o\left(1.718\right)\end{array}\right]=0\\ V_A=V_i-V_o\left(0.00008106\right)\end{array}$

Substitute $V_i-V_o\left(0.00008106\right)$ for $V_A$ in the equation (1).

  $\begin{array}{l}{V}_i\left(555.5\right)+V_o\left(0.8333\right)=\left(V_i-V_o\left(0.00008106\right)\right)\left(576.3\right)\\ \frac{V_o}{V_i}=23.6\end{array}$

The approximate ideal value of the closed loop voltage gain is given by,

  $\begin{array}{c}{A}_{vf}\cong \frac{\left(R_F+R_E\right)}{R_E}\\ \cong \frac{\left(1.2\text{k}\Omega +50\Omega \right)}{50\Omega }\\ \cong 25\end{array}$

The closed loop voltage gain is by formula $\frac{V_o}{V_i}$ is less than the approximate ideal value of closed loop voltage gain.

Conclusion:

Therefore, the value of closed loop gain is $23.6$ . The closed loop voltage gain is by formula $\frac{V_o}{V_i}$ is $23.6$ and is less than the approximate ideal value of closed loop voltage gain that is $25$ .

To determine

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

Answer to Problem 12.38P

The value $R_{if}$ is $95.1\text{k}\Omega$ and the value of $R_{of}$ is $33.1\text{k}\Omega$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

  

Calculation:

Consider the equation for the voltage $V_A$ is given by,

  $V_A=V_i-V_o\left(0.00008106\right)$

Substitute $23.6$ for $\frac{V_o}{V_i}$ in the above equation.

  $\begin{array}{c}{V}_A=V_i-V_i\left(23.6\right)\left(0.00008106\right)\\ =V_{i}0.99809\end{array}$

The expression for the current $I_i$ from the circuit is given by,

  $I_i=\frac{V_{\pi 1}}{r_{\pi 1}}$

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

  $I_i=\frac{V_i-V_A}{r_{\pi 1}}$

Substitute $0.182\text{k}\Omega$ for $r_{\pi 1}$ and $V_{i}0.99809$ for $V_A$ in the above equation.

  $\begin{array}{l}{I}_i=\frac{V_i-V_{i}0.99809}{0.182\text{k}\Omega }\\ \frac{V_i}{I_i}=95.1\text{k}\Omega \end{array}$

Apply KCL at the output.

  $I_x+g_{m3}{V}_{\pi 3}+\frac{V_{\pi 3}}{r_{\pi 3}}=\frac{V_x-V_A}{R_F}$

Substitute $V_C-V_x$ for $V_{\pi 3}$ in the above equation.

  $I_x+g_{m3}\left(V_C-V_x\right)+\frac{\left(V_C-V_x\right)}{r_{\pi 3}}=\frac{V_x-V_A}{R_F}$ …… (2)

The equation for the $V_C$ is given by,

  $\begin{array}{c}{V}_C=V_o\left(1.0046\right)-V_A\left(0.004749\right)\\ =V_x\left(1.0046\right)-V_A\left(0.004749\right)\end{array}$ …… (3)

Evaluate the equation of the voltage $V_A$ is given by,

  $V_i\left(555.5\right)+V_o\left(0.8333\right)=V_A\left(576.3\right)$

Substitute $0$ for $V_i$ in the above equation.

  $\begin{array}{l}\left(0\right)\left(555.5\right)+V_o\left(0.8333\right)=V_A\left(576.3\right)\\ V_A=V_x\left(0.001447\right)\end{array}$

Substitute $V_x\left(0.001447\right)$ for $V_A$ in equation (3)

  $\begin{array}{l}{V}_C=V_x\left(1.0046\right)-V_x\left(0.001447\right)\left(0.004749\right)\\ V_C=V_x\left(1.0046\right)\end{array}$

Substitute $V_x\left(0.001447\right)$ for $V_A$ , $172\text{mA}/\text{V}$ for $g_{m3}$

  $V_x\left(1.0046\right)$ for $V_C$ , $1.2\text{k}\Omega$ for $R_F$ and $0.582\text{k}\Omega$ for $r_{\pi 3}$ in equation (2).

  $\begin{array}{l}{I}_x+\left(172\text{mA}/\text{V}\right)\left(V_x\left(1.0046\right)-V_x\right)+\frac{\left(V_x\left(1.0046\right)-V_x\right)}{0.582\text{k}\Omega }=\frac{V_x-V_x\left(0.001447\right)}{1.2\text{k}\Omega }\\ \frac{V_x}{I_x}=\frac{1}{0.03024}\\ \\ \frac{V_x}{I_x}=33.1\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value $R_{if}$ is $95.1\text{k}\Omega$ and the value of $R_{of}$ is $33.1\text{k}\Omega$ .