Chapter 4, Problem 4.9EP

Consider the circuit shown in Figure 4.28 with circuit parameters $V_{DD}=5\text{V}$ , $R_S=5\text{k}\Omega$ , $R_1=70.7\text{k}\Omega$ , $R_2=9.3\text{k}\Omega$ , and $R_{Si}=500\Omega$ . The transistor parameters are: $V_{TP}=-0.8\text{V}$ , $K_p=0.4{\text{mA/V}}^{\text{2}}$ ,and $\lambda =0$ . Calculate the small−signal voltage gain $A_{\upsilon }={\upsilon }_o/{\upsilon }_i$ and the output resistance $R_o$ seen looking back into the circuit. (Ans. $A_{\upsilon }=0.817$ , $R_o=0.915\text{k}\Omega$ )

To determine

The values of $A_v,R_o$ .

Answer to Problem 4.9EP

  $\begin{array}{l}{A}_v=0.77\\ R_o=0.914\text{kΩ}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{DD}=5\text{V},R_S=5\text{kΩ}\\ R_1=70.7\text{kΩ},R_2=9.3\text{kΩ}\\ R_{si}=500\Omega ,\lambda =0\\ V_{TP}=-0.8\text{V},K_P=0.4\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\end{array}$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

  

The value of gate voltage is:

  $\begin{array}{l}{V}_G=\frac{R_2}{R_1+R_2}\times V_{DD}\\ V_G=\frac{9.3}{70.7+9.3}\times 5\\ V_G=\frac{9.3\times 5}{80}\\ V_G=0.58\text{V}\end{array}$

From the circuit:

  $\begin{array}{l}{V}_s=V_{DD}-I_D{R}_S\\ V_s=5-5\times {10}^3{I}_D\end{array}$

The value of $V_{SG}$ is:

  $\begin{array}{l}{V}_{SG}=V_S-V_G\\ V_{SG}=5-5\times {10}^3{I}_D-0.58\\ V_{SG}=4.42-5\times {10}^3{I}_D\end{array}$

Assuming the transistor operates in saturation region:

  $\begin{array}{l}{I}_D=K_P{\left(V_{SG}+V_{TP}\right)}^2\\ I_D=0.4\times {10}^{-3}{\left(4.42-5\times {10}^3{I}_D-0.8\right)}^2\\ I_D=0.4\times {10}^{-3}{\left(3.62-5\times {10}^3{I}_D\right)}^2\\ {10}^4{I}_D{}^2-15.48I_D+5.24\times {10}^{-3}=0\\ I_D=1.048\text{mA},0.5\text{mA}\end{array}$

The value of $V_{SG}$ is:

  $\begin{array}{l}{V}_{SG}=4.42-5\times {10}^3{I}_D\\ I_D=1.048\text{mA,}\\ V_{SG}=4.42-5\times {10}^3\times 1.048\times {10}^{-3}\\ V_{SG}=-0.82\\ V_{SG}<\left|V_{TP}\right|\end{array}$

Hence, the transistor would be in cutoff mode for $I_D=1.048\text{mA}$ .

Plugging $I_D=0.5\text{mA,}$

  $\begin{array}{l}{V}_{SG}=4.42-5\times {10}^3\times 0.5\times {10}^{-3}\\ V_{SG}=1.92\text{V}\\ V_{SG}>\left|V_{TP}\right|\end{array}$

The value of $V_{SD}$ is:

  $\begin{array}{l}{V}_{SD}=V_S-V_D\\ V_{SD}=5-5\times {10}^3{I}_D-0\\ V_{SD}=5-5\times {10}^3\times 0.5\times {10}^{-3}\\ V_{SD}=2.5\text{V}\end{array}$

  $V_{SD}>V_{SG}+V_{TP}$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

  

The value of $g_m,r_o$ is:

  $\begin{array}{l}{g}_m=2\sqrt{K_P{I}_D}\\ g_m=2\sqrt{0.4\times {10}^{-3}\times 0.5\times {10}^{-3}}\\ g_m=0.8944\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\\ r_o=\frac{1}{\lambda I_D}\\ r_o=\frac{1}{0\times I_D}\\ r_o=\infty \end{array}$

The value of output voltage is:

  $\begin{array}{l}{V}_o=\left(-g_m{V}_{sg}\right)\left(R_s\left|\right|r_o\right)\\ V_o=\left(-g_m{V}_{sg}\right)R_s\mathrm{...}\left(1\right)\end{array}$

Applying Kirchhoff’s voltage law from input to output:

  $\begin{array}{l}{V}_{in}=-V_{sg}+V_o\\ V_{in}=-V_{sg}-\left(g_m{V}_{sg}\right)\left(R_s\right)\\ V_{sg}=-\frac{V_{in}}{1+g_m\left(R_s\right)}\end{array}$

Applying voltage division rule in input:

  $V_{in}=\frac{\left(R_1\left|\right|R_2\right)}{R_{si}+\left(R_1\left|\right|R_2\right)}{V}_i$

Hence, the value of $V_{sg}$ is:

  $V_{sg}=-\frac{\left(R_1\left|\right|R_2\right)}{\left(1+g_m{R}_s\right)\left(R_{si}+\left(R_1\left|\right|R_2\right)\right)}{V}_i$

From equation (1):

  $\begin{array}{l}{V}_o=-g_m\left(R_s\right)\frac{-\left(R_1\left|\right|R_2\right)}{\left(1+g_m{R}_s\right)\left(R_{si}+\left(R_1\left|\right|R_2\right)\right)}{V}_i\\ \frac{V_o}{V_i}=\frac{g_m{R}_s}{\left(1+g_m{R}_s\right)}\left(\frac{R_i}{R_{si}+R_i}\right)\left(R_i=R_1\left|\right|R_2\right)\\ A_v=\frac{0.8944\times 5}{1+0.8944\times 5}\left(\frac{8.22}{0.5+8.22}\right)\\ A_v=0.77\end{array}$

The value of output resistance is:

  

Applying Kirchhoff’s current law at output node:

  $I_x=g_m{V}_{sg}+\frac{V_x}{R_s}+\frac{V_x}{r_o}$

The value of current in input circuit is zero.

  $V_{sg}=V_x$

Plugging the value:

  $\begin{array}{l}{I}_x=g_m{V}_x+\frac{V_x}{R_s}+\frac{V_x}{r_o}\\ I_x=V_x\left(g_m+\frac{1}{R_s}+\frac{1}{r_o}\right)\\ \frac{I_x}{V_x}=g_m+\frac{1}{R_s}+\frac{1}{r_o}\\ \frac{1}{R_o}=g_m+\frac{1}{R_s}+\frac{1}{r_o}\\ R_o=\frac{1}{g_m}\left|\right|R_s\left|\right|r_o\end{array}$

Plugging the values:

  $\begin{array}{l}{R}_o=\frac{1}{g_m}\left|\right|R_s\left|\right|r_o\\ R_o=\frac{1}{0.8944}\left|\right|5\left(r_o=\infty \right)\\ R_o=0.914\text{kΩ}\end{array}$