Chapter 11, Problem 11.43P

a.

To determine

The value of $v_S,v_{D1}\text{and}{v}_{D2}$ for given parameters.

Answer to Problem 11.43P

  $\begin{array}{l}{v}_S=1.459V\\ v_{D1}=v_{D2}=-4.115V\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $V^{+}=+5V,V^{-}=-5V,R_D=1k\Omega ,R_S=2k\Omega ,V_{TP}=-0.6V,K_p=1.2mA/V^2,\lambda =0$

  $v_1=v_2=0$

Calculation:

Consider the given figure,

Let equate the voltages,

  $\begin{array}{l}5=I_S{R}_S+V_{SG}=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}\\ 5=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}\\ 5=2\left(1.2\right)\left(2\right)\left(V_{SG}{}^2-1.2V_{SG}+0.36\right)+V_{SG}\\ 4.8V_{SG}{}^2-4.76V_{SG}-3.272=0\\ V_{SG}=1.459V=v_s\\ v_S=1.459V\\ I_S=2K_P{\left(V_{SG}+V_{TP}\right)}^2\\ I_S=2\left(1.2\right){\left(1.459-0.6\right)}^2\\ I_S=1.77mA\\ I_{D1}=I_{D2}=0.885mA\\ \text{Now,}\\ v_{D1}=v_{D2}=I_D{R}_D-5\\ =0.885\times 1-5\\ v_{D1}=v_{D2}=-4.115V\end{array}$

Hence,

  $\begin{array}{l}{v}_S=1.459V\\ v_{D1}=v_{D2}=-4.115V\end{array}$

b.

To determine

The value of $v_S,v_{D1}\text{and}{v}_{D2}$ for given parameters.

Answer to Problem 11.43P

  $\begin{array}{l}{v}_S=2.344V\\ v_{D1}=v_{D2}=-4.336V\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $V^{+}=+5V,V^{-}=-5V,R_D=1k\Omega ,R_S=2k\Omega ,V_{TP}=-0.6V,K_p=1.2mA/V^2,\lambda =0$

  $v_1=v_2=1V$

Calculation:

Consider the given figure,

Let equate the voltages,

  $\begin{array}{l}5=I_S{R}_S+V_{SG}=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}+1\\ 5=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}+1\\ 5=2\left(1.2\right)\left(2\right)\left(V_{SG}{}^2-1.2V_{SG}+0.36\right)+V_{SG}+1\\ 4.8V_{SG}{}^2-4.76V_{SG}-2.272=0\\ V_{SG}=1.344V\\ v_s=V_{SG}+1=2.344V\\ v_S=2.344V\\ I_{D1}=I_{D2}=\left(1.2\right){\left(1.344-0.6\right)}^2\\ I_{D1}=I_{D2}=0.664mA\\ \text{now,}\\ v_{D1}=v_{D2}=I_D{R}_D-5\\ =0.664\times 1-5\\ v_{D1}=v_{D2}=-4.336V\end{array}$

Hence,

  $\begin{array}{l}{v}_S=2.344V\\ v_{D1}=v_{D2}=-4.336V\end{array}$

c.

To determine

The value of $v_S,v_{D1}\text{and}{v}_{D2}$ for given parameters.

Answer to Problem 11.43P

  $\begin{array}{l}{v}_S=1.459V\\ v_{D1}=-3.909V\\ v_{D2}=-4.321V\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $V^{+}=+5V,V^{-}=-5V,R_D=1k\Omega ,R_S=2k\Omega ,V_{TP}=-0.6V,K_p=1.2mA/V^2,\lambda =0$

  $v_1=-0.1V,v_2=0.1V$

Calculation:

Consider the given figure,

Let equate the voltages,

  $\begin{array}{l}5=I_S{R}_S+V_{SG}=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}\\ 5=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}\\ 5=2\left(1.2\right)\left(2\right)\left(V_{SG}{}^2-1.2V_{SG}+0.36\right)+V_{SG}\\ 4.8V_{SG}{}^2-4.76V_{SG}-3.272=0\\ V_{SG}=1.459V=v_s\\ v_S=1.459V\\ I_S=2K_P{\left(V_{SG}+V_{TP}\right)}^2\\ I_S=2\left(1.2\right){\left(1.459-0.6\right)}^2\\ I_S=1.77mA\\ I_{D1}=I_{D2}=0.885mA\end{array}$

  $\begin{array}{l}{g}_m=2\sqrt{k_p{I}_{D1}}\\ g_m=2\sqrt{\left(1.2\right)\left(0.885\right)}\\ g_m=2.061mA/V\end{array}$

  $\begin{array}{l}{v}_{D1}=-4.115+g_m{R}_D\frac{v_D}{2}\\ =-4.115+\left(2.061\right)\left(1\right)\left(0.1\right)\\ v_{D1}=-3.909V\\ v_{D2}=-4.115-\left(2.061\right)\left(1\right)\left(0.1\right)\\ v_{D2}=-4.321V\end{array}$

Hence,

  $\begin{array}{l}{v}_S=1.459V\\ v_{D1}=-3.909V\\ v_{D2}=-4.321V\end{array}$

d.

To determine

The value of $v_S,v_{D1}\text{and}{v}_{D2}$ for given parameters.

Answer to Problem 11.43P

  $\begin{array}{l}{v}_S=2.344V\\ v_{D1}=-4.158V\\ v_{D2}=-4.515V\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $V^{+}=+5V,V^{-}=-5V,R_D=1k\Omega ,R_S=2k\Omega ,V_{TP}=-0.6V,K_p=1.2mA/V^2,\lambda =0$

  $v_1=0.9V,v_2=1.1V$

Calculation:

Consider the given figure,

Let equate the voltages,

  $\begin{array}{l}5=I_S{R}_S+V_{SG}=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}+1\\ 5=2K_P{\left(V_{SG}+V_{TP}\right)}^2{R}_S+V_{SG}+1\\ 5=2\left(1.2\right)\left(2\right)\left(V_{SG}{}^2-1.2V_{SG}+0.36\right)+V_{SG}+1\\ 4.8V_{SG}{}^2-4.76V_{SG}-2.272=0\\ V_{SG}=1.344V\\ v_s=V_{SG}+1=2.344V\\ v_S=2.344V\\ I_{D1}=I_{D2}=\left(1.2\right){\left(1.344-0.6\right)}^2\\ I_{D1}=I_{D2}=0.664mA\end{array}$

  $\begin{array}{l}{g}_m=2\sqrt{k_p{I}_{D1}}\\ g_m=2\sqrt{\left(1.2\right)\left(0.664\right)}\\ g_m=1.785mA/V\\ v_{D1}=-4.336+g_m{R}_D\frac{v_D}{2}\\ =-4.336+\left(1.785\right)\left(1\right)\left(0.1\right)\\ v_{D1}=-4.158V\\ v_{D2}=-4.336-\left(1.785\right)\left(1\right)\left(0.1\right)\\ v_{D2}=-4.515V\end{array}$

Hence,

  $\begin{array}{l}{v}_S=2.344V\\ v_{D1}=-4.158V\\ v_{D2}=-4.515V\end{array}$