Chapter 11, Problem 11.46P

a.

To determine

The value of $I_S,I_{D1,}{I}_{D2,}\text{and}{v}_{o2}$ .

Answer to Problem 11.46P

  $\begin{array}{l}{I}_S=0.141mA\\ I_{D1}=I_{D2}=70.5\mu A\\ v_{o2}=3.24V\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $\begin{array}{l}{V}^{+}=+5V,V^{-}=-5V,R_D=25k\Omega ,R_S=20k\Omega ,V_{TP}=-0.6V,K_{n1}=K_{n2}=50mA/V^2,\\ {\lambda }_1={\lambda }_2=0.02V^{-1},V_{TN1}=V_{TN2}=0\end{array}$

  $v_1=v_2=0$

Calculation:

Consider the given figure,

Let calculate gate-to source voltage,

  $\begin{array}{l}{V}_{GS}=\sqrt{\frac{I_D}{k_n}}+V_{TN}\\ V_{GS}=\sqrt{\frac{I_s}{2k_n}}+V_{TN}\left[\therefore I_D=I_s/2\right]\\ \text{apply}\text{KVL}\\ V_{GS}+I_s{R}_s+V^{-}=0\\ V_{GS}+I_s{R}_s-5=0\\ I_s{R}_s=5-V_{GS}\\ I_s=\frac{5-V_{GS}}{R_s}\\ \text{now,}\\ V_{GS}=\sqrt{\frac{\frac{5-V_{GS}}{R_s}}{2k_n}}+V_{TN}\\ V_{GS}-V_{TN}=\sqrt{\frac{5-V_{GS}}{2k_n{R}_s}}\\ V_{GS}-1=\sqrt{\frac{5-V_{GS}}{2\left(25\times {10}^3\times 50\times {10}^{-6}\right)}}\\ {\left(V_{GS}-1\right)}^2={\left(\sqrt{\frac{5-V_{GS}}{2.5}}\right)}^2\\ V_{GS}{}^2-2V_{GS}+1=\frac{5-V_{GS}}{2.5}\\ 2.5V_{GS}{}^2-4V_{GS}-2.5=0\\ \text{Apply}\text{quadratic}\text{formula,}\\ V_{GS}=2.175V\end{array}$

Now calculate source current,

  $\begin{array}{l}{I}_S=\frac{5-2.175}{20\times {10}^3}\\ I_S=0.141mA\end{array}$

  $\begin{array}{l}{I}_{D1}=I_{D2}=\frac{I_S}{2}\\ I_{D1}=I_{D2}=0.141/2mA\\ I_{D1}=I_{D2}=70.5\mu A\end{array}$

Now find output voltage,

  $\begin{array}{l}{v}_{o2}=V^{+}-I_{D2}{R}_D\\ v_{o2}=5-\left(70.5\mu A\right)\left(25k\Omega \right)\\ v_{o2}=3.24V\end{array}$

Hence,

  $\begin{array}{l}{I}_S=0.141mA\\ I_{D1}=I_{D2}=70.5\mu A\\ v_{o2}=3.24V\end{array}$

b.

To determine

The differential-mode voltage gain and common-mode voltage gain along with CMRR_dB

Answer to Problem 11.46P

  $\begin{array}{l}{A}_d=1.487\\ A_{cm}=-0.88\\ CMR{R}_{dB}=4.56dB\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $\begin{array}{l}{V}^{+}=+5V,V^{-}=-5V,R_D=25k\Omega ,R_S=20k\Omega ,V_{TP}=-0.6V,K_{n1}=K_{n2}=50mA/V^2,\\ {\lambda }_1={\lambda }_2=0.02V^{-1},V_{TN1}=V_{TN2}=0\end{array}$

Calculation:

Consider the small − signal equivalent circuit.

  

Apply KCL at node $V_S$ we get,

  $\begin{array}{l}{g}_{m1}{V}_{GS1}+g_{m2}{V}_{GS2}=\frac{V_S}{R_S}\\ g_{m1}\left(v_1-V_S\right)+g_{m2}\left(v_2-V_S\right)=\frac{V_S}{R_S}\left[\therefore V_{GS1}=\left(v_1-V_S\right),V_{GS2}=\left(v_2-V_S\right)\right]\\ g_{m1}{v}_1-g_{m1}{V}_S+g_{m2}{v}_2-g_{m2}{V}_S=\frac{V_S}{R_S}\\ g_m{v}_1-g_m{V}_S+g_m{v}_2-g_m{V}_S=\frac{V_S}{R_S}\left[\because g_{m1}=g_{m2}=g_m\right]\\ g_m{R}_S\left(v_1+v_2\right)-2g_m{R}_S{V}_S=V_S\\ V_S=\frac{g_m{R}_S\left(v_1+v_2\right)}{2g_m{R}_S+1}\\ V_S=\frac{\left(v_1+v_2\right)}{2+1/g_m{R}_S}\end{array}$

  $\begin{array}{l}{v}_{o2}=-g_m{R}_D\left(v_2-V_S\right)\\ v_{o2}=-g_m{R}_D\left(v_2-\frac{v_1+v_2}{2+1/g_m{R}_S}\right)\\ v_{o2}=-g_m{R}_D\left(\frac{v_2\left(g_m{R}_S+1\right)-v_1\left(g_m{R}_S\right)}{1+2g_m{R}_S}\right)\\ v_{o2}=-g_m{R}_D\left(\frac{v_{cm}-\frac{v_d}{2}\left(g_m{R}_S+1\right)-\left(v_{cm}+\frac{v_d}{2}\right)\left(g_m{R}_S\right)}{1+2g_m{R}_S}\right)\left[v_2=v_{cm}-\frac{v_d}{2},v_1=v_{cm}+\frac{v_d}{2}\right]\\ v_{o2}=\left(\frac{g_m{R}_D}{2}\right)v_d-\left(\frac{g_m{R}_D}{1+2g_m{R}_S}\right)v_{cm}\end{array}$

Now,

  $\begin{array}{l}{g}_m=2\sqrt{k_n{I}_{DQ}}\\ g_m=2\sqrt{50\times {10}^{-6}\left(70.5\times {10}^{-6}\right)}\\ g_m=0.119mA/V\\ \text{now},\\ v_{o2}=\frac{0.119\times {10}^{-3}\times 25\times {10}^3}{2}\left(v_d\right)-\frac{0.119\times {10}^{-3}\times 25\times {10}^3}{1+2\times 0.119\times {10}^{-3}\times 25\times {10}^3}\left(v_{cm}\right)\\ v_{o2}=1.487v_d-0.88v_{cm}\\ A_d=\frac{v_{o2}}{v_d}=1.487\\ A_{cm}=\frac{v_{o2}}{v_{cm}}=-0.88\\ CMR{R}_{dB}=20\log \left|\frac{A_d}{A_{cm}}\right|\\ CMR{R}_{dB}=20\log \left|\frac{1.487}{0.88}\right|\\ CMR{R}_{dB}=4.56dB\end{array}$

Hence,

  $\begin{array}{l}{A}_d=1.487\\ A_{cm}=-0.88\\ CMR{R}_{dB}=4.56dB\end{array}$