Chapter 4, Problem 4.8EP

The circuit and transistor parameters for the source−follower amplifier shown in Figure 4.29 are $R_S=2\text{k}\Omega$ , $V_{TP}=-1.2\text{V}$ , ${k^{\prime }}_p=40\mu {\text{A/V}}^{\text{2}}$ , and $\lambda =0$ . (a) Design the transistor width−to−length ratio such that $I_{DQ}=1.5\text{mA}$ . (b) Find the small−signal voltage gain. (c) Using the results of part (a), determine the value of $R_L$ that will result in a 10 percent reduction in voltage gain. (Ans. (a) $\left(W/L\right)=117$ ,(b) $A_{\upsilon }=0.882$ , (c) $R_L=2.12\text{k}\Omega$

Figure 4.29 Figure for Exercise Ex 4.8

To determine

The value of width to length ratio of MOSFET.

Answer to Problem 4.8EP

  $\frac{W}{L}=117$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{R}_s=2\text{kΩ,}{V}_{TP}=-1.2\text{V}\\ k_P\text{'}=40\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.,\lambda =0\\ I_{DQ}=1.5\text{mA}\end{array}$

Calculation:

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

The modified figure is:

  

From the circuit:

  $\begin{array}{l}{V}_s=5-I_D{R}_s\\ V_s=5-1.5\times 2\\ V_s=2V\\ V_{SG}=V_s-V_G\\ V_{SG}=2-0\\ V_{SG}=2\text{V}\\ V_{SD}=V_S-V_D\\ V_{SD}=2-\left(-5\right)\\ V_{SD}=7V\\ V_{SG}>\left|V_{TP}\right|\\ V_{SD}>V_{SG}-\left|V_{TP}\right|\end{array}$

Hence, transistor operates in saturation region.

The expression of drain current is:

  $\begin{array}{l}{I}_D=K_p{\left(V_{SG}-\left|V_{TP}\right|\right)}^2\\ 1.5\times {10}^{-3}=\frac{1}{2}{k}_p\text{'}\left(\frac{W}{L}\right){\left(2-1.2\right)}^2\\ 1.5\times {10}^{-3}=\frac{1}{2}\times 40\times {10}^{-6}\times \left(\frac{W}{L}\right)\times 0.64\\ \frac{W}{L}=\frac{3000}{40\times 0.64}\\ \frac{W}{L}=117\end{array}$

To determine

The value of small signal voltage gain.

Answer to Problem 4.8EP

  $A_v=0.882$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{R}_s=2\text{kΩ,}{V}_{TP}=-1.2\text{V}\\ k_P\text{'}=40\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.,\lambda =0\\ I_{DQ}=1.5\text{mA}\end{array}$

Calculation:

The value of $g_m,r_o$ is:

  $\begin{array}{l}{g}_m=2\sqrt{K_p{I}_D}\\ g_m=2\sqrt{\frac{1}{2}{k}_P\text{'}\left(\frac{W}{L}\right)I_D}\\ g_m=\sqrt{2\times 40\times {10}^{-6}\times 117\times 1.5\times {10}^{-3}}\\ g_m=3.75\raisebox{1ex}{$\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 coupling capacitor acts like short circuit for AC value calculation. The DC voltage source is short circuited.

The modified figure is:

  

The value of output voltage is:

  $V_o=\left(-g_m{V}_{sg}\right)\left(R_s\left|\right|R_L\right)\mathrm{...}\left(1\right)$

Applying Kirchhoff’s voltage law from input to output:

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

From equation (1):

  $\begin{array}{l}{V}_o=\frac{g_m\left(R_s\left|\right|R_L\right)}{1+g_m\left(R_s\left|\right|R_L\right)}\times V_i\\ \frac{V_o}{V_i}=\frac{g_m\left(R_s\left|\right|R_L\right)}{1+g_m\left(R_s\left|\right|R_L\right)}\\ A_v=\frac{g_m\left(R_s\left|\right|R_L\right)}{1+g_m\left(R_s\left|\right|R_L\right)}\end{array}$

Plugging the values:

  $\begin{array}{l}{A}_v=\frac{3.75\left(2\right)}{1+3.75\left(2\right)}\left(R_L=\infty \right)\\ A_v=0.882\end{array}$

To determine

The value of load resistance $R_L$ .

Answer to Problem 4.8EP

  $R_L=2.12\text{kΩ}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{R}_s=2\text{kΩ,}{V}_{TP}=-1.2\text{V}\\ k_P\text{'}=40\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.,\lambda =0\\ I_{DQ}=1.5\text{mA}\end{array}$

Calculation:

The new value of voltage gain is:

  $\begin{array}{l}{A}_v{}^{\text{'}}=A_v-A_v\left(10\%\right)\\ A_v{}^{\text{'}}=0.882-0.1\times 0.882\\ A_v{}^{\text{'}}=0.794\end{array}$

The value of load resistance is determined as follows:

  $\begin{array}{l}{A}_v{}^{\text{'}}=\frac{g_m\left(R_s\left|\right|R_L\right)}{1+g_m\left(R_s\left|\right|R_L\right)}\\ g_m\left(R_s\left|\right|R_L\right)=A_v{}^{\text{'}}+A_v{}^{\text{'}}{g}_m\left(R_s\left|\right|R_L\right)\\ g_m\left(R_s\left|\right|R_L\right)-A_v{}^{\text{'}}{g}_m\left(R_s\left|\right|R_L\right)=A_v{}^{\text{'}}\\ g_m\left(R_s\left|\right|R_L\right)\left(1-A_v{}^{\text{'}}\right)=A_v{}^{\text{'}}\\ g_m\left(R_s\left|\right|R_L\right)=\frac{A_v{}^{\text{'}}}{1-A_v{}^{\text{'}}}\\ g_m\left(R_s\left|\right|R_L\right)=\frac{0.794}{1-0.794}\\ 3.75\left(2\left|\right|R_L\right)=3.854\\ 2\left|\right|R_L=1.028\\ \frac{2R_L}{2+R_L}=1.028\\ 2R_L=2.056+1.028R_L\\ R_L=\frac{2.056}{0.972}\\ R_L=2.12\text{kΩ}\end{array}$