Chapter 10, Problem 10.51P

(a)

To determine

The reference current ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}$ for the given circuit parameters.

Answer to Problem 10.51P

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=606\text{μA}$

Explanation of Solution

Given:

The circuit parameters are

  $\begin{array}{l}{\mathrm{V}}^{+}=+5\text{V}\\ {\mathrm{V}}^{-}=0\end{array}$

The transistor parameters are

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{N}}=0.7\text{V}\\ \mathrm{K}{\text{'}}_{\mathrm{n}}=60{\text{μA/V}}^{\text{2}}\\ \mathrm{\lambda }=0.015{\text{V}}^{-1}\end{array}$

  $\begin{array}{l}{(}_{\mathrm{W}}=20\\ {(}_{\mathrm{W}}=12\\ {(}_{\mathrm{W}}=3\end{array}$

Calculation:

Consider the circuit shown below.

  

Transistor ${\mathrm{M}}_1$ and ${\mathrm{M}}_3$ are in series then consider $\mathrm{\lambda }=0$ ,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}3}={\mathrm{V}}^{+}-{\mathrm{V}}^{-}\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=5\text{V}\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}\left(1\right)\end{array}$

Also,

  $\begin{array}{}$ \left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2=\left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_3{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(3\right){({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.7)}^2$$ \left(\frac{60}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60}{2}\right)\left(3\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.7)}^2\left(2\right)$\end{array}$

From equation substitute the value of ${\mathrm{V}}_{\mathrm{G}\mathrm{S}3}$ in equation (2),

  $\begin{array}{l}$ \left(\frac{60}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60}{2}\right)\left(3\right)\times {(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2$$ \frac{\left(30\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2}{\left(30\right)\left(3\right)}={(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2$\sqrt{\frac{20}{3}}\times ({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)=(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)\\ \sqrt{\frac{20}{3}}\times {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-\sqrt{\frac{20}{3}}\times 0.7)=(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)\\ \sqrt{\frac{20}{3}}\times {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=\sqrt{\frac{20}{3}}\times 0.7+5-0.7\\ 2.581{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=2.581\times 0.7+4.3\end{array}$

  $\begin{array}{l}3.581{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=6.1067\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=1.705\text{V}\end{array}$

Also,

  ${\mathrm{V}}_{\mathrm{G}\mathrm{S}1}={\mathrm{V}}_{\mathrm{G}\mathrm{S}2}=1.705\text{V}$

The reference current is,

  $\begin{array}{c}$ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\left(\frac{\mathrm{K}{\text{'}}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\left(20\right){(1.705-0.7)}^2${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=606\text{μA}\end{array}$

Conclusion:

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=606\text{μA}$

(b)

To determine

The load current ${\mathrm{I}}_{\mathrm{O}}$ of circuit for given value of ${\mathrm{V}}_{\mathrm{D}\mathrm{S}2}$ .

Answer to Problem 10.51P

  ${\mathrm{I}}_{\mathrm{O}}=362.5\text{μA}$

Explanation of Solution

Given:

The circuit parameters are

  $\begin{array}{l}{\mathrm{V}}^{+}=+5\text{V}\\ {\mathrm{V}}^{-}=0\end{array}$

The transistor parameters are

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{N}}=0.7\text{V}\\ \mathrm{K}{\text{'}}_{\mathrm{n}}=60{\text{μA/V}}^{\text{2}}\\ \mathrm{\lambda }=0.015{\text{V}}^{-1}\end{array}$

  $\begin{array}{l}{(}_{\mathrm{W}}=20\\ {(}_{\mathrm{W}}=12\\ {(}_{\mathrm{W}}=3\end{array}$

Calculation:

Consider the given circuit as shown below.

  

The transistor ${\mathrm{M}}_1$ and ${\mathrm{M}}_3$ are in series then consider $\mathrm{\lambda }=0$ ,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}3}={\mathrm{V}}^{+}-{\mathrm{V}}^{-}\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=5\text{V}\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}\left(1\right)\end{array}$

  $\begin{array}{}$ \left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2=\left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_3{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(20\right)={({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(3\right){({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.7)}^2$$ \left(\frac{60}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60}{2}\right)\left(3\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.7)}^2\left(2\right)$\end{array}$

From equation (1) put the value of ${\mathrm{V}}_{\mathrm{G}\mathrm{S}3}$ in equation (2),

  $\begin{array}{l}$ \left(\frac{60}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60}{2}\right)\left(3\right)\times {(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2$$ \frac{\left(30\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2}{\left(30\right)\left(3\right)}={(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2$\sqrt{\frac{20}{3}}\times ({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)=(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)\\ \sqrt{\frac{20}{3}}\times {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-\sqrt{\frac{20}{3}}\times 0.7)=(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)\\ \sqrt{\frac{20}{3}}\times {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=\sqrt{\frac{20}{3}}\times 0.7+5-0.7\\ 2.581{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=2.581\times 0.7+4.3\end{array}$

  $\begin{array}{l}3.581{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=6.1067\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=1.705\text{V}\end{array}$

Also,

  ${\mathrm{V}}_{\mathrm{G}\mathrm{S}1}={\mathrm{V}}_{\mathrm{G}\mathrm{S}2}=1.705\text{V}$

Now the load current is,

  $\begin{array}{c}$ {\mathrm{I}}_{\mathrm{O}}=\left(\frac{\mathrm{K}{\text{'}}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_2{({\mathrm{V}}_{\mathrm{G}\mathrm{S}2}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\left(10\right){(1.705-0.7)}^2${\mathrm{I}}_{\mathrm{O}}=363.6\text{μA}\end{array}$

The load resistance will be,

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{O}}=\frac{1}{\mathrm{\lambda }{\mathrm{I}}_{\mathrm{O}}}\\ {\mathrm{R}}_{\mathrm{O}}=\frac{1}{(0.015)(363.6\times {10}^{-6})}\\ {\mathrm{R}}_{\mathrm{O}}=183.4\mathrm{k}\mathrm{\Omega }\end{array}$

Now the load current for ${\mathrm{V}}_{\mathrm{D}\mathrm{S}2}=1.5\text{V}$ will be

  $\begin{array}{l}\mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=\frac{\mathrm{\Delta }\mathrm{V}}{{\mathrm{R}}_{\mathrm{O}}}\\ \mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=\frac{{\mathrm{V}}_{\mathrm{D}\mathrm{S}2}-{\mathrm{V}}_{\mathrm{G}\mathrm{S}2}}{{\mathrm{R}}_{\mathrm{O}}}\\ \mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=\frac{1.5-1.705}{183.4\times {10}^3}\\ \mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=-1.1\times {10}^{-6}\text{A}\end{array}$

Now the change in load current,

  $\begin{array}{c}{\mathrm{I}}_{\mathrm{O}}={\mathrm{I}}_{\mathrm{O}}+\mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}\\ =363.6\times {10}^{-6}+(-1.1\times {10}^{-6})\\ {\mathrm{I}}_{\mathrm{O}}=362.5\text{μA}\end{array}$

Conclusion:

  ${\mathrm{I}}_{\mathrm{O}}=362.5\text{μA}$

(c)

To determine

The load current ${\mathrm{I}}_{\mathrm{O}}$ of circuit for given value of ${\mathrm{V}}_{\mathrm{D}\mathrm{S}2}$ .

Answer to Problem 10.51P

  ${\mathrm{I}}_{\mathrm{O}}=370.7\text{μA}$

Explanation of Solution

Given:

The circuit parameters are

  $\begin{array}{l}{\mathrm{V}}^{+}=+5\text{V}\\ {\mathrm{V}}^{-}=0\end{array}$

The transistor parameters are

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{N}}=0.7\text{V}\\ \mathrm{K}{\text{'}}_{\mathrm{n}}=60{\text{μA/V}}^{\text{2}}\\ \mathrm{\lambda }=0.015{\text{V}}^{-1}\end{array}$

  $\begin{array}{l}{(}_{\mathrm{W}}=20\\ {(}_{\mathrm{W}}=12\\ {(}_{\mathrm{W}}=3\end{array}$

Calculation:

Consider the given circuit as shown below.

  

The transistor ${\mathrm{M}}_1$ and ${\mathrm{M}}_3$ are in series then consider $\mathrm{\lambda }=0$ ,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}3}={\mathrm{V}}^{+}-{\mathrm{V}}^{-}\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=5\text{V}\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}\left(1\right)\end{array}$

  $\begin{array}{}$ \left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2=\left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_3{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(20\right)={({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(3\right){({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.7)}^2$$ \left(\frac{60}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60}{2}\right)\left(3\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.7)}^2\left(2\right)$\end{array}$

From equation (1) put the value of ${\mathrm{V}}_{\mathrm{G}\mathrm{S}3}$ in equation (2),

  $\begin{array}{l}$ \left(\frac{60}{2}\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2=\left(\frac{60}{2}\right)\left(3\right)\times {(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2$$ \frac{\left(30\right)\left(20\right)\times {({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2}{\left(30\right)\left(3\right)}={(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)}^2$\sqrt{\frac{20}{3}}\times ({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)=(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)\\ \sqrt{\frac{20}{3}}\times {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-\sqrt{\frac{20}{3}}\times 0.7)=(5-{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-0.7)\\ \sqrt{\frac{20}{3}}\times {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=\sqrt{\frac{20}{3}}\times 0.7+5-0.7\\ 2.581{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=2.581\times 0.7+4.3\end{array}$

  $\begin{array}{l}3.581{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=6.1067\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}1}=1.705\text{V}\end{array}$

Also,

  ${\mathrm{V}}_{\mathrm{G}\mathrm{S}1}={\mathrm{V}}_{\mathrm{G}\mathrm{S}2}=1.705\text{V}$

Now the load current is,

  $\begin{array}{c}$ {\mathrm{I}}_{\mathrm{O}}=\left(\frac{\mathrm{K}{\text{'}}_{\mathrm{n}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_2{({\mathrm{V}}_{\mathrm{G}\mathrm{S}2}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\left(10\right){(1.705-0.7)}^2${\mathrm{I}}_{\mathrm{O}}=363.6\text{μA}\end{array}$

The load resistance will be,

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{O}}=\frac{1}{\mathrm{\lambda }{\mathrm{I}}_{\mathrm{O}}}\\ {\mathrm{R}}_{\mathrm{O}}=\frac{1}{(0.015)(363.6\times {10}^{-6})}\\ {\mathrm{R}}_{\mathrm{O}}=183.4\mathrm{k}\mathrm{\Omega }\end{array}$

Now the load current for ${\mathrm{V}}_{\mathrm{D}\mathrm{S}2}=3\text{V}$ will be

  $\begin{array}{l}\mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=\frac{\mathrm{\Delta }\mathrm{V}}{{\mathrm{R}}_{\mathrm{O}}}\\ \mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=\frac{{\mathrm{V}}_{\mathrm{D}\mathrm{S}2}-{\mathrm{V}}_{\mathrm{G}\mathrm{S}2}}{{\mathrm{R}}_{\mathrm{O}}}\\ \mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=\frac{3-1.705}{183.4\times {10}^3}\\ \mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}=7.06\times {10}^{-6}\text{A}\end{array}$

Now the change in load current,

  $\begin{array}{c}{\mathrm{I}}_{\mathrm{O}}={\mathrm{I}}_{\mathrm{O}}+\mathrm{\Delta }{\mathrm{I}}_{\mathrm{O}}\\ =363.6\times {10}^{-6}+(7.06\times {10}^{-6})\end{array}$

  ${\mathrm{I}}_{\mathrm{O}}=370.7\text{μA}$

Conclusion:

  ${\mathrm{I}}_{\mathrm{O}}=370.7\text{μA}$