Chapter 4, Problem 4.2EP

For the circuit shown in Figure 4.1, $V_{DD}=3.3\text{V}$ and $R_D=10\text{k}\Omega$ . The transistor parameters are $V_{TN}=0.4\text{V}$ , ${k^{\prime }}_n=100\mu {\text{A/V}}^{\text{2}}$ , $W/L=50$ , and $\lambda =0.025{\text{V}}^{-1}$ . Assume the transistor is biased such that $I_{DQ}=0.25\text{mA}$ . (a) Verify that the transistor is biased in the saturation region. (b) Determine the small-signal parameters $g_m$ and $r_o$ . (c) Determine the small-signal voltage gain. (Ans. (a) $V_{GSQ}=0.716\text{V}$ and $V_{DSQ}=0.8\text{V}$ so that $V_{DS}>V_{DS}\left(\text{sat}\right)$ ; (b) $g_m=1.58\text{mA/V,}{r}_o=160\text{k}\Omega$ ; (c) −14.9)

To determine

The region of operation of NMOS.

Answer to Problem 4.2EP

The transistor operates in saturation region.

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{DD}=3.3\text{V},R_D=10\text{k}\\ V_{TN}=0.4\text{V},k_n{}^{\text{'}}=100\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ \frac{W}{L}=50,\lambda =0.025{\text{V}}^{\text{-1}},I_{DQ}=0.25\text{mA}\end{array}$

Calculation:

Assuming the transistor operates in saturation region.

The expression of drain current is:

  $\begin{array}{l}{I}_D=\frac{k_n\text{'}}{2}\left(\frac{W}{L}\right){\left(V_{GS}-V_{TN}\right)}^2\\ 0.25\times {10}^{-3}=\frac{100\times {10}^{-6}}{2}\times 50\times {\left(V_{GS}-V_{TN}\right)}^2\\ 0.25\times {10}^{-3}=25\times {10}^{-4}\times {\left(V_{GS}-V_{TN}\right)}^2\\ {\left(V_{GS}-V_{TN}\right)}^2=\frac{25\times {10}^{-5}}{25\times {10}^{-4}}\\ {\left(V_{GS}-V_{TN}\right)}^2=\frac{1}{10}\\ V_{GS}-V_{TN}=\sqrt{\frac{1}{10}}\\ V_{GS}-V_{TN}=0.316\text{V}\\ V_{GS}=0.316+V_{TN}\\ =0.316+0.4\\ =0.716\text{V}\\ V_{DS}\left(\text{sat}\right)=V_{GS}-V_{TN}\\ =0.316\text{V}\end{array}$

From the circuit:

  $\begin{array}{l}{V}_{DS}=V_{DD}-I_D{R}_D\\ =3.3-\left(\text{0}\text{.25}\text{m}\right)\left(\text{10}\text{k}\right)\\ =3.3-2.5\\ =0.8\text{V}\end{array}$

From the above result:

  $\begin{array}{l}{V}_{DS}>V_{DS}\left(\text{sat}\right)\\ 0.8\text{V}>0.316\text{V}\end{array}$

Hence, the NMOS operates in saturation region.

To determine

The values of small signal parameters $g_m,r_o$ .

Answer to Problem 4.2EP

  $\begin{array}{l}{g}_m=1.58\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\\ r_o=160\text{kΩ}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{DD}=3.3\text{V},R_D=10\text{k}\\ V_{TN}=0.4\text{V},k_n{}^{\text{'}}=100\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ \frac{W}{L}=50,\lambda =0.025{\text{V}}^{\text{-1}},I_{DQ}=0.25\text{mA}\end{array}$

Calculation:

The value of transconductance is determined as follows:

  $\begin{array}{l}{I}_D=\frac{k_n\text{'}}{2}\left(\frac{W}{L}\right){\left(V_{GS}-V_{TN}\right)}^2\\ \frac{\partial I_D}{\partial V_{GS}}=\frac{k_n\text{'}}{2}\left(\frac{W}{L}\right)\times 2\left(V_{GS}-V_{TN}\right)\\ g_m=k_n\text{'}\left(\frac{W}{L}\right)\left(V_{GS}-V_{TN}\right)\\ g_m=100\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\times 50\times 0.316\text{V}\\ g_m=1580\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\\ g_m=1.58\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\end{array}$

The small-signal output resistance is determined as follows:

  $\begin{array}{l}{I}_D=\frac{k_n\text{'}}{2}\left(\frac{W}{L}\right){\left(V_{GS}-V_{TN}\right)}^2\left(1+\lambda V_{DS}\right)\\ {\frac{\partial I_D}{\partial V_{DS}}|}_{V_{GS}=\text{constant}}=\frac{\lambda k_n\text{'}}{2}\left(\frac{W}{L}\right){\left(V_{GS}-V_{TN}\right)}^2\\ \frac{1}{r_o}=\frac{\lambda k_n\text{'}}{2}\left(\frac{W}{L}\right){\left(V_{GS}-V_{TN}\right)}^2\\ \frac{1}{r_o}=\lambda I_D\left(I_D=\frac{k_n\text{'}}{2}\left(\frac{W}{L}\right){\left(V_{GS}-V_{TN}\right)}^2\right)\\ r_o=\frac{1}{\lambda I_D}\\ r_o=\frac{1}{\left(0.025{\text{V}}^{\text{-1}}\right)\left(0.25\text{mA}\right)}\\ r_o=\frac{1}{6.25\times {10}^{-3}}\text{k}\Omega \\ r_o=\frac{1000}{6.25}\text{k}\Omega \\ r_o=160\text{k}\Omega \end{array}$

To determine

The small signal voltage gain.

Answer to Problem 4.2EP

  $A_V=-14.87$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{DD}=3.3\text{V},R_D=10\text{k}\\ V_{TN}=0.4\text{V},k_n{}^{\text{'}}=100\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ \frac{W}{L}=50,\lambda =0.025{\text{V}}^{\text{-1}},I_{DQ}=0.25\text{mA}\end{array}$

Calculation:

It is a linear circuit and superposition is applied. The DC voltage source is short circuited.

It is a common source amplifier. The small signal equivalent model of the circuit is as follows:

  

The $r_o$ and $R_D$ are in parallel their equivalent is:

  $\begin{array}{l}{R}_{eq}=\left(r_o\parallel R_D\right)\\ R_{eq}=\frac{r_o{R}_D}{r_o+R_D}\\ R_{eq}=\frac{160\times 10}{170}\\ R_{eq}=9.412\text{k}\end{array}$

The expression of output voltage is:

  $V_o=-g_m{V}_{gs}{R}_{eq}\mathrm{...}\left(1\right)$

The expression of voltage gain is:

  $A_V=\frac{V_o}{V_i}$

From the circuit:

  $V_i=V_{gs}$

From equation (1):

  $\begin{array}{l}{V}_o=-g_m{V}_{gs}{R}_{eq}\\ \frac{V_o}{V_{gs}}=-g_m{R}_{eq}\end{array}$

The small signal voltage gain is:

  $\begin{array}{l}{A}_V=\frac{V_o}{V_{gs}}\\ =-g_m{R}_{eq}\\ =-\left(1.58\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\right)\left(9.412\text{k}\right)\\ =-14.87\end{array}$

The minus sign indicates that output signal has 180⁰phase shift with respect to input signal.