Chapter 4, Problem 4.3EP

The parameters of the circuit shown in Figure 4.14 are $V_{DD}=5\text{V}$ , $R_1=520\text{k}\Omega$ , $R_2=320\text{k}\Omega$ , $R_D=10\text{k}\Omega$ , and $R_{Si}=0$ . Assume transistor parameters of $V_{TN}=0.8\text{V}$ , $K_n=0.20{\text{mA/V}}^{\text{2}}$ , and $\lambda =0$ . (a) Determine the small−signal transistor parameters $g_m$ and $r_o$ . (b) Find the small−signal voltage gain. (c) Calculate the input and output resistances $R_i$ and $R_o$ (see Figure 4.15). (Ans. (a) $g_m=0.442\text{mA/V,}{r}_o=\infty$ ; (b) $A_{\upsilon }=-4.42$ (c) $R_i=198\text{k}\Omega ,R_o=R_D=10\text{k}\Omega$ )

To determine

The small-signal transistor parameters $g_m$ and $r_o$ .

Answer to Problem 4.3EP

  $\begin{array}{l}{g}_m=0.442\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\\ r_o=\infty \end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{DD}=5\text{V},R_1=520\text{kΩ},R_2=320\text{kΩ}\\ R_D=10\text{kΩ},R_{Si}=0\\ V_{TN}=0.8\text{V},K_n=0.20\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ \lambda =0\end{array}$

Calculation:

The coupling capacitor $C_{c1}$ behaves like open circuit for DC value calculation.

The value of $V_{GSQ}$ is:

  $\begin{array}{l}{V}_{GSQ}=\left(\frac{R_2}{R_1+R_2}\right)V_{DD}\\ V_{GSQ}=\left(\frac{320}{320+520}\right)\times 5\\ V_{GSQ}=\left(\frac{320}{840}\right)\times 5\\ V_{GSQ}=1.905\text{V}\end{array}$

The value of drain current is:

  $\begin{array}{l}{I}_{DQ}=K_n{\left(V_{GS}-V_{TN}\right)}^2\\ I_{DQ}=\left(0.2\right){\left(1.905-0.8\right)}^2\\ I_{DQ}=0.244\text{mA}\end{array}$

The value of $V_{DSQ}$ is:

  $\begin{array}{l}{V}_{DSQ}=V_{DD}-I_{DQ}{R}_D\\ V_{DSQ}=5-\left(0.244\text{mA}\right)\left(\text{10}\text{k}\right)\\ V_{DSQ}=2.56\text{V}\end{array}$

Since,

  $\begin{array}{l}{V}_{DSQ}>V_{DSQ}\left(\text{sat}\right)\\ V_{DSQ}>\left(V_{GS}-V_{TN}\right)\\ 2.56\text{V}>1.1\text{V}\end{array}$

The transistor operates in saturation region.

The value of transconductance $g_m$ is:

  $\begin{array}{l}{g}_m=2\sqrt{K_n{I}_{DQ}}\\ g_m=2\sqrt{\left(0.2\right)\left(0.244\right)}\\ g_m=0.442\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\end{array}$

The value of small signal output resistance is:

  $\begin{array}{l}{r}_o=\frac{1}{\lambda I_{DQ}}\\ r_o=\frac{1}{0\times 0.244}\\ r_o=\infty \end{array}$

To determine

The value of small signal voltage gain.

Answer to Problem 4.3EP

  $A_V=-4.42$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{DD}=5\text{V},R_1=520\text{kΩ},R_2=320\text{kΩ}\\ R_D=10\text{kΩ},R_{Si}=0\\ V_{TN}=0.8\text{V},K_n=0.20\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ \lambda =0\end{array}$

Calculation:

The coupling capacitor $C_{c1}$ behaves like short circuit for small signal value calculation.The transistor behaves as linear amplifier circuit, so superposition theorem is applied. The DC voltage $V_{DD}$ is short circuited and the modified figure is:

  

The output voltage is:

  $V_o=V_{ds}=-\left(R_D\left|\right|r_o\right)g_m{V}_{gs}\mathrm{...}\left(1\right)$

The value of $V_{gs}$ is:

Applying voltage division rule:

  $V_{gs}=\left(\frac{R_1\left|\right|R_2}{\left(R_1\left|\right|R_2\right)+R_{Si}}\right)V_i$

Putting the value of $V_{gs}$ in equation 1:

  $\begin{array}{l}{V}_o=-\left(R_D\left|\right|r_o\right)g_m\left(\frac{R_1\left|\right|R_2}{\left(R_1\left|\right|R_2\right)+R_{Si}}\right)V_i\\ \frac{V_o}{V_i}=-\left(R_D\left|\right|r_o\right)g_m\left(\frac{R_1\left|\right|R_2}{\left(R_1\left|\right|R_2\right)+R_{Si}}\right)\\ \frac{V_o}{V_i}=-\left(R_D\right)g_m\left(\frac{R_1\left|\right|R_2}{\left(R_1\left|\right|R_2\right)}\right)\left(\begin{array}{l}{r}_o=\infty \\ R_{Si}=0\end{array}\right)\\ \frac{V_o}{V_i}=-g_m{R}_D\end{array}$

The value of small signal voltage gain is:

  $\begin{array}{l}{A}_V=\frac{V_o}{V_i}\\ =-g_m{R}_D\\ =-0.442\left(\text{mA}\right)\times 10\left(\text{kΩ}\right)\\ =-4.42\end{array}$

To determine

The input and output resistance.

Answer to Problem 4.3EP

  $\begin{array}{l}{R}_i=198\text{}\text{kΩ}\\ R_o=10\text{kΩ}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below,

  

  $\begin{array}{l}{V}_{DD}=5\text{V},R_1=520\text{kΩ},R_2=320\text{kΩ}\\ R_D=10\text{kΩ},R_{Si}=0\\ V_{TN}=0.8\text{V},K_n=0.20\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ \lambda =0\end{array}$

Calculation:

The small signal equivalent circuit is:

  

The input resistance is determined as follows:

  $\begin{array}{l}{R}_i=R_1\left|\right|R_2\\ R_i=\frac{R_1{R}_2}{R_1+R_2}\\ R_i=\frac{520\times 320}{520+320}\\ R_i=198\text{}\text{kΩ}\end{array}$

The output resistance is determined as follows:

  $\begin{array}{l}{R}_o=R_D\left|\right|r_o\\ R_o=R_D\left(r_o=\infty \right)\\ R_o=10\text{kΩ}\end{array}$