Chapter 10, Problem 10.76P

For the circuit shown in Figure P10.76, the transistor parameters are
$V_{TN}=0.5V,V_{TP}=-0.5V,{k^{\prime }}_n=100\mu A/V^2,{k^{\prime }}_P=60\mu A/V^2,{\lambda }_1=0.02V^{-1}$ ,and ${\lambda }_2=0.03V^{-1}$ .The quiescent drain current is $I_{DQ}=200\mu A$ and the quiescent output voltage is $V_O=1.25V$ . (a) Determine ${\left(W/L\right)}_1$ such that the small-signal voltage gain is $A_V=-100$ , (b) determine $V_I$ ,and (c) determine $V_G$ assuming $K_{n1}=K_{P2}$ .

To determine

The ${(W/L)}_1$ ratio.

Answer to Problem 10.76P

  ${\left(\frac{W}{L}\right)}_1=25$

Explanation of Solution

Given:

  $\begin{array}{l}K{\text{'}}_n=100{\text{μA/V}}^{\text{2}}\\ K{\text{'}}_p=60{\text{μA/V}}^{\text{2}}\\ {\lambda }_1=0.02{\text{V}}^{-1}\\ {\lambda }_2=0.03{\text{V}}^{-1}\\ V_{TN}=0.5\text{V}\\ V_{TP}=-0.5\text{V}\\ V_O=1.25\text{V}\\ I_{DQ}=200\text{μA}\end{array}$

  $A_v=-100$

Calculation:

The given circuit is,

  

The small-signal voltage gain is given, the expression for gain is,

  $A_v=-g_{m1}(r_{o1}\parallel r_{o2})$

Substituting the given values,

  $\begin{array}{l}{A}_v=-g_{m1}(r_{o1}\parallel r_{o2})\\ -100=-g_{m1}(r_{o1}\parallel r_{o2})\\ 100=g_{m1}(r_{o1}\parallel r_{o2})\left(1\right)\end{array}$

Substitute $k{\text{'}}_n{\left(\frac{W}{L}\right)}_1(V_I-V_{TN})$ for $g_{m1}$ in equation (1)

  $100=\left[k{\text{'}}_n{\left(\frac{W}{L}\right)}_1(V_I-V_{TN})\right](r_{o1}\parallel r_{o2})\therefore V_I=V_{GS1}$

Now substitute the given values in above expression,

  $100=\left[100\times {10}^{-6}{\left(\frac{W}{L}\right)}_1(V_I-0.5)\right](r_{o1}\parallel r_{o2})\left(2\right)$

It is known that,

  $r_{o1}=\frac{1}{{\lambda }_1{I}_{DQ}}$

  $r_{o2}=\frac{1}{{\lambda }_2{I}_{DQ}}$

On putting the given values,

  $\begin{array}{l}{r}_{o1}=\frac{1}{(0.02)(200\times {10}^{-6})}\\ r_{o1}=250k\Omega \end{array}$

  $\begin{array}{l}{r}_{o2}=\frac{1}{(0.03)(200\times {10}^{-6})}\\ r_{o2}=166.67k\Omega \end{array}$

Now substitute the value of $r_{o1}\text{and}{r}_{o2}$ in equation (2)

  $\begin{array}{l}100=\left[100\times {10}^{-6}{\left(\frac{W}{L}\right)}_1(V_I-0.5)\right](250\times {10}^3\parallel 166.67\times {10}^3)\\ {10}^6={\left(\frac{W}{L}\right)}_1(V_I-0.5)\frac{(250\times {10}^3)\times (166.67\times {10}^3)}{250\times {10}^3+166.67\times {10}^3}\\ {10}^6={\left(\frac{W}{L}\right)}_1(V_I-0.5)(100\times {10}^3)\\ 10={\left(\frac{W}{L}\right)}_1(V_I-0.5)\\ (V_I-0.5)=\frac{10}{{\left(\frac{W}{L}\right)}_1}\left(3\right)\end{array}$

The drain current $I_{DQ}$ for transistor $M_1$ is expressed as,

  $I_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN})}^2(1+{\lambda }_1{V}_{DS1})$

  $=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{G1}-V_{S1}-V_{TN})}^2(1+{\lambda }_1(V_{D1}-V_{S1}))$

Hence $V_{G1}=V_{I,}$ $V_{D1}=V_O$ and $V_{D1}=V_O$ substitute the values,

  $\begin{array}{}$ I_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN})}^2(1+{\lambda }_1{V}_{DS1})$$ 200\times {10}^{-6}=\frac{100\times {10}^{-6}}{2}{\left(\frac{W}{L}\right)}_1{(V_1-0-0.5)}^2(1+0.02(V_O-0))$\end{array}$

  $4={\left(\frac{W}{L}\right)}_1{(V_1-0.5)}^2(1+0.02V_O)$

Substitute $1.25\text{V}$ for $V_O$ ,

  $4={\left(\frac{W}{L}\right)}_1{(V_1-0.5)}^2(1+0.02(1.25))$

  $4={\left(\frac{W}{L}\right)}_1{(V_1-0.5)}^2(1.025)$

  $3.902={\left(\frac{W}{L}\right)}_1{(V_1-0.5)}^2$

Put the value of $(V_1-0.5)$ from equation (3).

  $\begin{array}{l}3.902={\left(\frac{W}{L}\right)}_1{\left(\frac{10}{{\left(\frac{W}{L}\right)}_1}\right)}^2\\ {\left(\frac{W}{L}\right)}_1=\frac{100}{3.902}\\ {\left(\frac{W}{L}\right)}_1=25.63\end{array}$

  ${\left(\frac{W}{L}\right)}_1=25$

Conclusion:

  ${\left(\frac{W}{L}\right)}_1=25$

To determine

The voltage $V_I$ .

Answer to Problem 10.76P

  $V_I=0.9\text{V}$

Explanation of Solution

Given:

  $\begin{array}{l}K{\text{'}}_n=100{\text{μA/V}}^{\text{2}}\\ K{\text{'}}_p=60{\text{μA/V}}^{\text{2}}\\ {\lambda }_1=0.02{\text{V}}^{-1}\\ {\lambda }_2=0.03{\text{V}}^{-1}\\ V_{TN}=0.5\text{V}\\ V_{TP}=-0.5\text{V}\\ V_O=1.25\text{V}\\ I_{DQ}=200\text{μA}\end{array}$

Calculation:

The given circuit is,

  

The small-signal voltage gain is given, the expression for gain is,

  $A_v=-g_{m1}(r_{o1}\parallel r_{o2})$

Substituting the given values,

  $\begin{array}{l}{A}_v=-g_{m1}(r_{o1}\parallel r_{o2})\\ -100=-g_{m1}(r_{o1}\parallel r_{o2})\\ 100=g_{m1}(r_{o1}\parallel r_{o2})\left(1\right)\end{array}$

Substitute $k{\text{'}}_n{\left(\frac{W}{L}\right)}_1(V_I-V_{TN})$ for $g_{m1}$ in equation (1)

  $100=\left[k{\text{'}}_n{\left(\frac{W}{L}\right)}_1(V_I-V_{TN})\right](r_{o1}\parallel r_{o2})\therefore V_I=V_{GS1}$

Now substitute the given values in above expression,

  $100=\left[100\times {10}^{-6}{\left(\frac{W}{L}\right)}_1(V_I-0.5)\right](r_{o1}\parallel r_{o2})\left(2\right)$

It is known that,

  $r_{o1}=\frac{1}{{\lambda }_1{I}_{DQ}}$

  $r_{o2}=\frac{1}{{\lambda }_2{I}_{DQ}}$

On putting the given values,

  $\begin{array}{l}{r}_{o1}=\frac{1}{(0.02)(200\times {10}^{-6})}\\ r_{o1}=250k\Omega \end{array}$

  $\begin{array}{l}{r}_{o2}=\frac{1}{(0.03)(200\times {10}^{-6})}\\ r_{o2}=166.67k\Omega \end{array}$

Now substitute the value of $r_{o1}\text{and}{r}_{o2}$ in equation (2).

  $\begin{array}{l}100=\left[100\times {10}^{-6}{\left(\frac{W}{L}\right)}_1(V_I-0.5)\right](250\times {10}^3\parallel 166.67\times {10}^3)\\ {10}^6={\left(\frac{W}{L}\right)}_1(V_I-0.5)\frac{(250\times {10}^3)\times (166.67\times {10}^3)}{250\times {10}^3+166.67\times {10}^3}\\ {10}^6={\left(\frac{W}{L}\right)}_1(V_I-0.5)(100\times {10}^3)\\ 10={\left(\frac{W}{L}\right)}_1(V_I-0.5)\\ (V_I-0.5)=\frac{10}{{\left(\frac{W}{L}\right)}_1}\left(3\right)\end{array}$

The drain current $I_{DQ}$ for transistor $M_1$ is expressed as,

  $\begin{array}{}$ I_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN})}^2(1+{\lambda }_1{V}_{DS1})$$ =\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{G1}-V_{S1}-V_{TN})}^2(1+{\lambda }_1(V_{D1}-V_{S1}))$\end{array}$

Hence $V_{G1}=V_{I,}$$V_{D1}=V_O$ and $V_{D1}=V_O$ substitute the values,

  $\begin{array}{l}{I}_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(}^{V_{GS1}}(1+{\lambda }_1{V}_{DS1})\\ 200\times {10}^{-6}=\frac{100\times {10}^{-6}}{2}{\left(\frac{W}{L}\right)}_1{(}^{V_1}(1+0.02(V_O-0))\\ 4={\left(\frac{W}{L}\right)}_1{(}^{V_1}(1+0.02V_O)\end{array}$

Substitute $1.25v$ for $V_O$ ,

  $\begin{array}{l}4={\left(\frac{W}{L}\right)}_1{(}^{V_1}(1+0.02(1.25))\\ 4={\left(\frac{W}{L}\right)}_1{(}^{V_1}(1.025)\\ 3.902={\left(\frac{W}{L}\right)}_1{(}^{V_1}\end{array}$

Put the value of $(V_1-0.5)$ from equation (3).

  $\begin{array}{l}3.902={\left(\frac{W}{L}\right)}_1{\left(\frac{10}{{\left(\frac{W}{L}\right)}_1}\right)}^2\\ {\left(\frac{W}{L}\right)}_1=\frac{100}{3.902}\\ {\left(\frac{W}{L}\right)}_1=25.63\end{array}$

  ${\left(\frac{W}{L}\right)}_1=25$

  $(V_I-0.5)=\frac{10}{{\left(\frac{W}{L}\right)}_1}$

Substitute ${\left(\frac{W}{L}\right)}_1=25$ in above expression,

  $\begin{array}{l}(V_I-0.5)=\frac{10}{25}\\ (V_I-0.5)=0.4\\ V_I=0.4+0.5\end{array}$

  $V_I=0.9\text{V}$

Conclusion:

  $V_I=0.9\text{V}$

To determine

The voltage $V_G$ for transistor $M_2$

Answer to Problem 10.76P

  $V_G=1.607\text{V}$

Explanation of Solution

Given:

  $\begin{array}{l}K{\text{'}}_n=100{\text{μA/V}}^{\text{2}}\\ K{\text{'}}_p=60{\text{μA/V}}^{\text{2}}\\ {\lambda }_1=0.02{\text{V}}^{-1}\\ {\lambda }_2=0.03{\text{V}}^{-1}\\ V_{TN}=0.5\text{V}\\ V_{TP}=-0.5\text{V}\\ V_O=1.25\text{V}\\ I_{DQ}=200\text{μA}\end{array}$

  $K_{n1}=K_{p2}$

Calculation:

The given circuit is,

  

The small-signal voltage gain is given, the expression for gain is,

  $A_v=-g_{m1}(r_{o1}\parallel r_{o2})$

Substituting the given values,

  $\begin{array}{l}{A}_v=-g_{m1}(r_{o1}\parallel r_{o2})\\ -100=-g_{m1}(r_{o1}\parallel r_{o2})\\ 100=g_{m1}(r_{o1}\parallel r_{o2})\left(1\right)\end{array}$

Substitute $k{\text{'}}_n{\left(\frac{W}{L}\right)}_1(V_I-V_{TN})$ for $g_{m1}$ in equation (1)

  $100=\left[k{\text{'}}_n{\left(\frac{W}{L}\right)}_1(V_I-V_{TN})\right](r_{o1}\parallel r_{o2})\therefore V_I=V_{GS1}$

Now substitute the given values in above expression,

  $100=\left[100\times {10}^{-6}{\left(\frac{W}{L}\right)}_1(V_I-0.5)\right](r_{o1}\parallel r_{o2})\left(2\right)$

It is known that,

  $r_{o1}=\frac{1}{{\lambda }_1{I}_{DQ}}$

  $r_{o2}=\frac{1}{{\lambda }_2{I}_{DQ}}$

On putting the given values,

  $\begin{array}{l}{r}_{o1}=\frac{1}{(0.02)(200\times {10}^{-6})}\\ r_{o1}=250k\Omega \end{array}$

  $\begin{array}{l}{r}_{o2}=\frac{1}{(0.03)(200\times {10}^{-6})}\\ r_{o2}=166.67k\Omega \end{array}$

Now substitute the value of $r_{o1}\text{and }{r}_{o2}$ in equation (2).

  $\begin{array}{l}100=\left[100\times {10}^{-6}{\left(\frac{W}{L}\right)}_1(V_I-0.5)\right](250\times {10}^3\parallel 166.67\times {10}^3)\\ {10}^6={\left(\frac{W}{L}\right)}_1(V_I-0.5)\frac{(250\times {10}^3)\times (166.67\times {10}^3)}{250\times {10}^3+166.67\times {10}^3}\\ {10}^6={\left(\frac{W}{L}\right)}_1(V_I-0.5)(100\times {10}^3)\\ 10={\left(\frac{W}{L}\right)}_1(V_I-0.5)\\ (V_I-0.5)=\frac{10}{{\left(\frac{W}{L}\right)}_1}\left(3\right)\end{array}$

The drain current $I_{DQ}$ for transistor $M_1$ is expressed as,

  $I_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN})}^2(1+{\lambda }_1{V}_{DS1})$

  $=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{G1}-V_{S1}-V_{TN})}^2(1+{\lambda }_1(V_{D1}-V_{S1}))$

Hence $V_{G1}=V_{I,}$

  $V_{D1}=V_O$ and $V_{D1}=V_O$ substitute the values,

  $\begin{array}{l}{I}_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(}^{V_{GS1}}(1+{\lambda }_1{V}_{DS1})\\ 200\times {10}^{-6}=\frac{100\times {10}^{-6}}{2}{\left(\frac{W}{L}\right)}_1{(}^{V_1}(1+0.02(V_O-0))\\ 4={\left(\frac{W}{L}\right)}_1{(}^{V_1}(1+0.02V_O)\end{array}$

Substitute $1.25\text{V}$ for $V_O$ ,$I_{DQ}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN})}^2(1+{\lambda }_1{V}_{DS1})$

  $200\times {10}^{-6}=\frac{100\times {10}^{-6}}{2}{\left(\frac{W}{L}\right)}_1{(V_1-0-0.5)}^2(1+0.02(V_O-0))$

  $4={\left(\frac{W}{L}\right)}_1{(V_1-0.5)}^2(1+0.02V_O)$

Put the value of $(V_1-0.5)$ from equation (3).

  $\begin{array}{l}3.902={\left(\frac{W}{L}\right)}_1{\left(\frac{10}{{\left(\frac{W}{L}\right)}_1}\right)}^2\\ {\left(\frac{W}{L}\right)}_1=\frac{100}{3.902}\\ {\left(\frac{W}{L}\right)}_1=25.63\end{array}$

  ${\left(\frac{W}{L}\right)}_1=25$

Since $K_{n1}=K_{p2}$ so,

  $\begin{array}{l}\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1=\frac{K{\text{'}}_p}{2}{\left(\frac{W}{L}\right)}_2\\ {\left(\frac{W}{L}\right)}_2=\frac{100}{60}\times 25\\ {\left(\frac{W}{L}\right)}_2=41.67\end{array}$

The drain current $I_{DQ}$ for transistor $M_2$ is expressed as,

  $\begin{array}{l}{I}_{DQ}=\frac{K{\text{'}}_p}{2}{\left(\frac{W}{L}\right)}_2{(}^{V_{SG2}}(1+{\lambda }_2{V}_{SD2})\\ =\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_2{(}^{V_{S2}}(1+{\lambda }_2(V_{S2}-V_{D2}))\end{array}$

On substituting the given values,

  $\begin{array}{l}200\times {10}^{-6}=\frac{60\times {10}^{-6}}{2}\left(41.67\right){(}^{V_{SG2}}(1+0.03(2.5-1.25))\\ V_{SG2}=0.893\text{V}\end{array}$

Now the gate voltage will be,

  $\begin{array}{l}{V}_S-V_G=0.893\\ 2.5-V_G=0.893\end{array}$

  $V_G=1.607\text{V}$

Conclusion:

  $V_G=1.607\text{V}$