Chapter 12, Problem 12.40P

(a)

To determine

Value of current I_DQ1 and I_CQ2 .

Answer to Problem 12.40P

Value of current is,

  $\begin{array}{l}{I}_{DQ1}=0.4056\text{mA}\\ I_{CQ2}=0.83\text{mA}\end{array}$

Explanation of Solution

Given:

The given values are:

  $\begin{array}{l}{V}^{+}=5\text{V}\\ V_{GG}=2.5\text{V}\\ R_{D1}=5\text{kΩ}\\ R_{E2}=1.6\text{kΩ}\\ R_L=1.2\text{kΩ}\\ K_n=1.5\left(\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\right)\\ V_{TN}=0.5\text{V}\\ \lambda =0\\ h_{FE}=120\\ V_{EB}\left(\text{on}\right)=0.7\text{V}\\ V_A=\infty \end{array}$

The given circuit is shown below.

  

Calculation:

Modified circuit diagram is shown below.

  

Neglecting base current.

Equation of current I_D1 is,

  $\begin{array}{l}{I}_{D1}=\frac{5-V_D}{R_{D1}}\\ I_{D1}=\frac{5-V_D}{5}\left(1\right)\end{array}$

Equation of voltage V_D is,

  $V_D=5-5I_{D1}\left(2\right)$

Equation of current I_C2 is,

  $\begin{array}{l}{I}_{C2}=\frac{5-\left(V_D+0.7\right)}{R_{E2}}\\ I_{C2}=\frac{4.3-V_D}{1.6}\left(3\right)\end{array}$

Value of voltage V_GS is,

  $\begin{array}{l}\text{Using}\text{equation}\left(2\right)\\ I_{C2}=\frac{4.3-\left(5-5I_{D1}\right)}{1.6}\\ I_{C2}=-0.4375+3.125I_{D1}\left(4\right)\\ I_{D1}+I_{C2}=\frac{V_{GG}-V_{GS}}{R_L}\\ \text{Using}\text{equation}\left(4\right)\\ I_{D1}-0.4375+3.125I_{D1}=\frac{2.5-V_{GS}}{1.2}\\ 4.95I_{D1}=3.025-V_{GS}\\ V_{GS}=3.025-4.95K_n{\left(V_{GS}-0.5\right)}^2\\ V_{GS}=3.025-4.95\times 1.5{\left(V_{GS}-0.5\right)}^2\\ 7.425V_{GS}{}^2-6.425V_{GS}-1.16875=0\\ V_{GS}=\frac{-\left(-6.425\right)\pm \sqrt{{\left(-6.425\right)}^2-4\times 7.425\times -1.16875}}{2\times 7.425}\\ V_{GS}=1.02\text{V}\end{array}$

Value of current I_DQ1 is,

  $\begin{array}{l}{I}_{DQ1}=K_n{\left(V_{GS}-0.5\right)}^2\\ I_{DQ1}=1.5{\left(1.02-0.5\right)}^2\\ I_{DQ1}=0.4056\text{mA}\end{array}$

Value of current I_CQ2 is,

  $\begin{array}{l}{I}_{CQ2}=-0.4375+3.125I_{D1}\\ I_{CQ2}=-0.4375+3.125\times 0.4056\\ I_{CQ2}=0.83\text{mA}\end{array}$

(b)

To determine

Value of small signal voltage gain.

Answer to Problem 12.40P

Value of small signal voltage gain is 0.88.

Explanation of Solution

Given:

Given values are:

  $\begin{array}{l}{V}^{+}=5\text{V}\\ V_{GG}=2.5\text{V}\\ R_{D1}=5\text{kΩ}\\ R_{E2}=1.6\text{kΩ}\\ R_L=1.2\text{kΩ}\\ K_n=1.5\left(\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\right)\\ V_{TN}=0.5\text{V}\\ \lambda =0\\ h_{FE}=120\\ V_{EB}\left(\text{on}\right)=0.7\text{V}\\ V_A=\infty \\ I_{DQ1}=0.4056\text{mA}\\ I_{CQ2}=0.83\text{mA}\\ \end{array}$

Calculation:

Small signal circuit is shown below.

  

Value of trans-conductance of transistor 1 is,

  $\begin{array}{l}{g}_{m1}=2\sqrt{K_n{I}_{D1}}\\ g_{m1}=2\sqrt{1.5\times 0.4056}\\ g_{m1}=1.56\left(\text{mA/V}\right)\end{array}$

Value of trans-conductance of transistor 2,

  $\begin{array}{l}{g}_{m2}=\frac{I_{C2}}{V_T}\\ g_{m2}=\frac{0.83}{0.026}\\ g_{m2}=31.92\left(\text{mA/V}\right)\end{array}$

Value of small signal resistance is,

  $\begin{array}{l}{r}_{\pi 2}=\frac{\beta V_T}{I_{C2}}\\ r_{\pi 2}=\frac{120\times 0.026}{0.83}\\ r_{\pi 2}=3.759\text{kΩ}\end{array}$

Applying KCL at node (2),

  $\begin{array}{l}\frac{V_A+V_{\pi }}{R_{E2}}+\frac{V_{\pi }}{r_{\pi }}+g_{m2}{V}_{\pi }=0\\ \frac{V_A}{1.6}=-V_{\pi }\left(\frac{1}{1.6}+\frac{1}{3.759}+31.92\right)\\ V_A=-52.49V_{\pi }\left(1\right)\end{array}$

Equation of output voltage is,

  $V_o=\left(g_{m1}{V}_{gs}+g_{m2}{V}_{\pi }\right)R_L\left(2\right)$

Applying KCL at node (1),

  $\begin{array}{l}\frac{V_A}{R_{D1}}+g_{m1}{V}_{gs}=\frac{V_{\pi }}{r_{\pi }}\\ \text{Using}\text{equation}\left(1\right)\\ \frac{-52.49V_{\pi }}{5}+1.56V_{gs}=\frac{V_{\pi }}{3.759}\\ V_{gs}=6.944V_{\pi }\left(3\right)\\ V_{\pi }=0.144V_{gs}\\ \text{Using}\text{equation}\left(V_{gs}=V_i-V_o\right)\\ V_{\pi }=0.144\left(V_i-V_o\right)\left(4\right)\end{array}$

Value of small signal voltage gain is,

  $\begin{array}{l}\text{Using}\text{equation}\left(\text{3}\right)\text{and}\left(\text{4}\right)\text{in}\text{equation}\left(2\right).\\ V_o=\left(1.56\times 6.944\times 0.144\left(V_i-V_o\right)+31.92\times 0.144\left(V_i-V_o\right)\right)\times 1.2\\ V_o=1.87\left(V_i-V_o\right)+5.51\left(V_i-V_o\right)\\ V_o+1.87V_o+5.51V_o=1.87V_i+5.51V_i\\ \frac{V_o}{V_i}=0.88\end{array}$

(c)

To determine

Value of small signal output resistance.

Answer to Problem 12.40P

Value of small signal output resistance is $143\Omega$ .

Explanation of Solution

Given:

Given values are,

  $\begin{array}{l}{V}^{+}=5\text{V}\\ V_{GG}=2.5\text{V}\\ R_{D1}=5\text{kΩ}\\ R_{E2}=1.6\text{kΩ}\\ R_L=1.2\text{kΩ}\\ K_n=1.5\left(\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\right)\\ V_{TN}=0.5\text{V}\\ \lambda =0\\ h_{FE}=120\\ V_{EB}\left(\text{on}\right)=0.7\text{V}\\ V_A=\infty \\ I_{DQ1}=0.4056\text{mA}\\ I_{CQ2}=0.83\text{mA}\\ \end{array}$

Calculation:

Small signal circuit for output resistance calculation ( V_i =0) is,

  

Applying KCL at node 3 for output resistance,

  $\begin{array}{l}{I}_x+g_{m2}{V}_{\pi }+g_{m1}{V}_{gs}=\frac{V_x}{R_L}\\ \text{Using}\text{equation}\\ \left(\begin{array}{l}{V}_x=-V_{gs}\\ V_{\pi }=0.144V_{gs}=-0.144V_x\end{array}\right)\\ I_x=-31.92\times -0.144V_x-\left(1.56\times -V_x\right)+\frac{V_x}{1.2}\\ R_o=\frac{V_x}{I_x}=143\Omega \end{array}$