Chapter 11, Problem 11.89P

a.

To determine

The value of the resistor R₁ and R₂ in a way that the dc value of the output voltage is 0.

Answer to Problem 11.89P

The value of the resistances are:

  $\begin{array}{l}{R}_1=27.2\text{kΩ}\\ R_2=5\text{kΩ}\end{array}$

Explanation of Solution

Given:

The circuit is given as:

  

The circuit parameters:

  $\begin{array}{l}{I}_1=0.25\text{mA}\\ I_2=1\text{mA}\\ V_{TN}=0.8\text{V}\\ {\text{V}}_{TP}=-0.8\text{V}\\ {\lambda }_n={\lambda }_p=0.02\text{V-1}\\ k_n=0.5{\text{mA/V}}^2\\ k_p=1{\text{mA/V}}^2\end{array}$

The drain current of $M_2$ :

  $\begin{array}{c}{I}_{D2}=I_2\\ =1\text{mA}\end{array}$

Considering the expression for drain current of $M_2$ to evaluate the value of $R_2$ :

  $\begin{array}{c}{v}_0=V^{+}-I_2{R}_2\\ R_2=\frac{V^{+}-v_0}{I_2}\\ R_2=\frac{5-0}{1\times {10}^{-3}}\\ =5000\\ =5\text{kΩ}\end{array}$

Hence, the value of the resistor $R_2$ is $5\text{kΩ}$

Evaluating the value of source gate voltage of $M_2$ :

  $\begin{array}{c}{I}_{D2}=K_p{\left(V_{SG2}+V_{TP}\right)}^2\\ V_{SG2}=\sqrt{\frac{I_{D2}}{K_p}}-V_{TP}\\ V_{SG2}=\sqrt{\frac{1\times {10}^{-3}}{1\times {10}^{-3}}}-\left(-0.8\right)\\ =1+0.8\\ =1.8\text{V}\end{array}$

Considering the expression for the drain current of $M_1$ to evaluate the value of $R_1$ :

  $\begin{array}{c}-V_{SG2}=V^{+}-I_1{R}_1\\ R_1=\frac{V^{+}+V_{SG2}}{I_1}\\ R_1=\frac{5+1.8}{0.25\times {10}^{-3}}\\ =27.2\times {10}^3\\ =27.2\text{kΩ}\end{array}$

Hence, the value of the resistor $R_1$ is $27.2\text{kΩ}$ .

b.

To determine

To sketch: The small signal equivalent circuit and then find the small signal transistor parameters.

Explanation of Solution

Given:

The circuit is given as:

  

The circuit parameters:

  $\begin{array}{l}{I}_1=0.25\text{mA}\\ I_2=1\text{mA}\\ V_{TN}=0.8\text{V}\\ {\text{V}}_{TP}=-0.8\text{V}\\ {\lambda }_n={\lambda }_p=0.02\text{V-1}\\ k_n=0.5{\text{mA/V}}^2\\ k_p=1{\text{mA/V}}^2\end{array}$

Evaluating the value of trans-conductance $M_1$ , $g_{m1}$ :

  $g_{m1}=2\sqrt{K_n{I}_1}$

Substituting $\text{0}{\text{.5mA/V}}^{\text{2}}\text{for}{\text{K}}_{\text{n}}\text{and}\text{0}\text{.25mA}{\text{forI}}_{\text{1}}$ .

  $\begin{array}{c}{g}_{m1}=2\sqrt{\left(0.5\times {10}^{-3}\right)\left(0.25\times {10}^{-3}\right)}\\ =\mathrm{0.7.71}\times {10}^{-3}\\ =0.7071\text{mA/V}\end{array}$

Evaluating the value of trans-conductance of $M_2$ , $g_{m2}$ :

  $\begin{array}{c}{g}_{m2}=2\sqrt{K_p{I}_2}\\ g_{m2}=2\sqrt{\left(1\times {10}^{-3}\right)\left(1\times {10}^{-3}\right)}\\ =2\times {10}^{-3}\\ =2\text{mA/V}\end{array}$

Evaluating the value of the resistor $r_{o1}$ :

  $\begin{array}{c}{r}_{o1}=\frac{1}{{\lambda }_n{I}_1}\\ r_{o1}=\frac{1}{\left(0.02\right)\left(0.25\times {10}^{-3}\right)}\\ =200\times {10}^3\\ =200\text{kΩ}\end{array}$

Evaluating the value of the resistor $r_{o2}$ :

  $\begin{array}{c}{r}_{o2}=\frac{1}{{\lambda }_p{I}_2}\\ r_{o2}=\frac{1}{\left(0.02\right)\left(1\times {10}^{-3}\right)}\\ =50\times {10}^3\\ =50\text{kΩ}\end{array}$

c.

To determine

The small signal voltage gain.

Answer to Problem 11.89P

The small signal gain of the circuit is -15.25V/V.

Explanation of Solution

Given:

The circuit is given as:

  

The circuit parameters:

  $\begin{array}{l}{I}_1=0.25\text{mA}\\ I_2=1\text{mA}\\ V_{TN}=0.8\text{V}\\ {\text{V}}_{TP}=-0.8\text{V}\\ {\lambda }_n={\lambda }_p=0.02\text{V-1}\\ k_n=0.5{\text{mA/V}}^2\\ k_p=1{\text{mA/V}}^2\end{array}$

Drawing the small signal equivalent model of the circuit:

  

Applying nodal analysis at the input node:

  $\begin{array}{c}\frac{V_{O1}}{r_{o1}}+\frac{V_{O1}}{R_1}+g_{m1}{V}_{in}=0\\ V_{O1}\left(\frac{1}{r_{o1}}+\frac{1}{R_1}\right)=-g_{m1}{V}_{in}\\ V_{O1}\left(\frac{1}{200\times {10}^3}+\frac{1}{27.2\times {10}^3}\right)=-\left(0.7071\times {10}^3\right)V_{in}\\ V_{O1}=-16.93V_{in}\end{array}$

Referring to the above diagram:

  $V_{sG2}=V_O-V_{O1}$

Applying the nodal analysis at the output node:

  $\begin{array}{c}\frac{V_O}{r_{o2}}+\frac{V_O}{R_2}+g_{m2}{V}_{sg2}=0\\ V_O\left(\frac{1}{r_{o2}}+\frac{1}{R_2}\right)+g_{m2}\left(V_O-V_{O1}\right)=0\\ V_O\left(\frac{1}{r_{o2}}+\frac{1}{R_2}+g_{m2}\right)=g_{m2}{V}_{O1}\\ V_O\left(\frac{1}{50\times {10}^3}+\frac{1}{5\times {10}^3}+\left(2\times {10}^{-3}\right)\right)=\left(2\times {10}^{-3}\right)\left(-16.93V_{in}\right)\\ V_O\left(\frac{1}{50}+\frac{1}{5}+2\right)=-33.86V_{in}\\ V_O\left(2.22\right)=-33.86V_{in}\\ \frac{V_O}{V_{in}}=-15.25\text{V/V}\end{array}$

Hence, the small signal gain of the circuit is -15.25V/V.

d.

To determine

The output resistance.

Answer to Problem 11.89P

The value of the output resistance is $0.45\text{kΩ}$ .

Explanation of Solution

Given:

The circuit is given as:

  

The circuit parameters:

  $\begin{array}{l}{I}_1=0.25\text{mA}\\ I_2=1\text{mA}\\ V_{TN}=0.8\text{V}\\ {\text{V}}_{TP}=-0.8\text{V}\\ {\lambda }_n={\lambda }_p=0.02\text{V-1}\\ k_n=0.5{\text{mA/V}}^2\\ k_p=1{\text{mA/V}}^2\end{array}$

Evaluating the output resistance $R_o$ :

  $R_o=r_{02}𑨀R_2𑨀\frac{1}{g_{m2}}$

Substituting $\text{50kΩ}\text{for}{\text{r}}_{\text{o2}}\text{,}\text{5kΩ}\text{for}{\text{R}}_{\text{2}}\text{and}\text{2mA/V}\text{for}{\text{g}}_{\text{m2}}$

  $\begin{array}{c}{R}_o=\left(50\times {10}^3\right)𑨀\left(5\times {10}^3\right)𑨀\left(\frac{1}{2\times {10}^{-3}}\right)\\ =\left(50\times {10}^3\right)𑨀\left(5\times {10}^3\right)𑨀\left(500\right)\\ =\left(50\times {10}^3\right)𑨀\left[\frac{\left(5\times {10}^3\right)\left(500\right)}{\left(5\times {10}^3\right)+\left(500\right)}\right]\\ =\left(50\times {10}^3\right)𑨀\left(454.54\right)\\ =\frac{\left(50\times {10}^3\right)\left(454.54\right)}{\left(50\times {10}^3\right)+\left(454.54\right)}\\ =450.45\Omega \\ =0.45\text{kΩ}\end{array}$

Hence, the value of the output resistance is $0.45\text{kΩ}$ .