Chapter 10, Problem 10.65P

(a)

To determine

The value of resistor R .

Answer to Problem 10.65P

  $\mathrm{R}=6.12\mathrm{k}\mathrm{\Omega }$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{N}}=0.5\mathrm{V}\\ {\mathrm{V}}_{\mathrm{T}\mathrm{P}}=-0.5\mathrm{V}\\ (1/2){\mathrm{\mu }}_{\mathrm{n}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}=50\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\\ (1/2){\mathrm{\mu }}_{\mathrm{p}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}=20\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\\ {\mathrm{\lambda }}_{\mathrm{n}}={\mathrm{\lambda }}_{\mathrm{p}}=0\\ $ {(\mathrm{W}/\mathrm{L})}_1={(\mathrm{W}/\mathrm{L})}_3={(\mathrm{W}/\mathrm{L})}_4=5/1$$ {(\mathrm{W}/\mathrm{L})}_2=50/1$\end{array}$

Calculation:

The given circuit is,

  

According to the statement ${\mathrm{I}}_{\mathrm{D}1}={\mathrm{I}}_{\mathrm{D}2}=50\mathrm{\mu }\mathrm{A}$ is given, so

  ${\mathrm{K}}_{\mathrm{n}1}=\frac{\mathrm{K}{\text{'}}_{\mathrm{n}}}{2}{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1$

Also,

  $\frac{\mathrm{K}{\text{'}}_{\mathrm{n}}}{2}=50\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2$ and ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1=5$

  ${\mathrm{K}}_{\mathrm{n}1}=\frac{\mathrm{K}{\text{'}}_{\mathrm{n}}}{2}{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1$

  $\begin{array}{l}{\mathrm{K}}_{\mathrm{n}1}=50\mathrm{\mu }\times 5\\ {\mathrm{K}}_{\mathrm{n}1}=250\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\end{array}$

Now calculate R ,

  $\mathrm{R}=\frac{1}{\sqrt{{\mathrm{K}}_{\mathrm{n}1}{\mathrm{I}}_{\mathrm{D}1}}}\left(1-\sqrt{\frac{{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1}{{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_2}}\right)$

Substitute the given values,

  $\begin{array}{l}\mathrm{R}=\frac{1}{\sqrt{250\times {10}^{-6}\times 50\times {10}^{-6}}}\left(1-\sqrt{\frac{5}{50}}\right)\\ \mathrm{R}=\frac{1}{\sqrt{0.25\times 0.05}}\left(1-\sqrt{\frac{5}{50}}\right)\\ \mathrm{R}=8.944\times 0.6838\end{array}$

  $\mathrm{R}=6.12\mathrm{k}\mathrm{\Omega }$

Conclusion:

  $\mathrm{R}=6.12\mathrm{k}\mathrm{\Omega }$

(b)

To determine

The biased voltage difference ( ${\mathrm{V}}^{+}-{\mathrm{V}}^{-}$ ).

Answer to Problem 10.65P

  ${\mathrm{V}}^{+}-{\mathrm{V}}^{-}=1.66\mathrm{V}$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{N}}=0.5\mathrm{V}\\ {\mathrm{V}}_{\mathrm{T}\mathrm{P}}=-0.5\mathrm{V}\\ (1/2){\mathrm{\mu }}_{\mathrm{n}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}=50\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\\ (1/2){\mathrm{\mu }}_{\mathrm{p}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}=20\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\\ {\mathrm{\lambda }}_{\mathrm{n}}={\mathrm{\lambda }}_{\mathrm{p}}=0\\ $ {(\mathrm{W}/\mathrm{L})}_1={(\mathrm{W}/\mathrm{L})}_3={(\mathrm{W}/\mathrm{L})}_4=5/1$$ {(\mathrm{W}/\mathrm{L})}_2=50/1$\end{array}$

Calculation:

The given circuit is,

  

According to the circuit,

  $\begin{array}{l}{\mathrm{V}}^{+}-{\mathrm{V}}^{-}={\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}\left(1\right)\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}={\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}}\left(2\right)\end{array}$

Now the drain current expression is,

  ${\mathrm{I}}_{\mathrm{D}1}={\mathrm{K}}_{\mathrm{p}1}{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$

  $\begin{array}{l}$ 50=20\times 5{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+0.5)}^2$$ 1=2{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+0.5)}^2$\sqrt{\frac{1}{2}}=({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+0.5)\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=1.207\mathrm{V}\end{array}$

From equation (2),

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}={\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}=1.207-0.5\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}=0.707\mathrm{V}\end{array}$

Now the drain current expression is,

  ${\mathrm{I}}_{\mathrm{D}1}={\mathrm{K}}_{\mathrm{n}1}{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$

  $\begin{array}{l}$ 50=50\times 5{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)}^2$$ 1=5{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)}^2$\sqrt{\frac{1}{5}}=({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)\\ 0.447=({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=0.947\mathrm{V}\end{array}$

Now from equation (1),

  $\begin{array}{l}{\mathrm{V}}^{+}-{\mathrm{V}}^{-}={\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}\\ {\mathrm{V}}^{+}-{\mathrm{V}}^{-}=0.71+0.947\\ {\mathrm{V}}^{+}-{\mathrm{V}}^{-}=1.66\mathrm{V}\end{array}$

  ${\mathrm{V}}^{+}-{\mathrm{V}}^{-}=1.66\mathrm{V}$

Conclusion:

  ${\mathrm{V}}^{+}-{\mathrm{V}}^{-}=1.66\mathrm{V}$

(c)

To determine

The ratio ${(\mathrm{W}/\mathrm{L})}_5$ and ${(\mathrm{W}/\mathrm{L})}_6$ .

Answer to Problem 10.65P

  ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5=2.5$

  ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6=7.5$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{N}}=0.5\mathrm{V}\\ {\mathrm{V}}_{\mathrm{T}\mathrm{P}}=-0.5\mathrm{V}\\ (1/2){\mathrm{\mu }}_{\mathrm{n}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}=50\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\\ (1/2){\mathrm{\mu }}_{\mathrm{p}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}=20\mathrm{\mu }\mathrm{A}/{\mathrm{V}}^2\\ {\mathrm{\lambda }}_{\mathrm{n}}={\mathrm{\lambda }}_{\mathrm{p}}=0\\ $ {(\mathrm{W}/\mathrm{L})}_1={(\mathrm{W}/\mathrm{L})}_3={(\mathrm{W}/\mathrm{L})}_4=5/1$$ {(\mathrm{W}/\mathrm{L})}_2=50/1$\end{array}$

Calculation:

The given circuit is,

  

According to the circuit,

  $\begin{array}{l}{\mathrm{V}}^{+}-{\mathrm{V}}^{-}={\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}+{\mathrm{V}}_{\mathrm{G}\mathrm{S}1}\left(1\right)\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}={\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}}\left(2\right)\end{array}$

Now the drain current expression is,

  ${\mathrm{I}}_{\mathrm{D}1}={\mathrm{K}}_{\mathrm{p}1}{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$

  $\begin{array}{l}$ 50=20\times 5{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+0.5)}^2$$ 1=2{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+0.5)}^2$\sqrt{\frac{1}{2}}=({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+0.5)\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=1.207\mathrm{V}\end{array}$

From equation (2),

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}={\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}=1.207-0.5\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}3(\mathrm{s}\mathrm{a}\mathrm{t})}=0.707\mathrm{V}\end{array}$

Now the drain current expression is,

  ${\mathrm{I}}_{\mathrm{D}1}={\mathrm{K}}_{\mathrm{n}1}{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$

  $\begin{array}{l}$ 50=50\times 5{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)}^2$$ 1=5{({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)}^2$\sqrt{\frac{1}{5}}=({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)\\ 0.447=({\mathrm{V}}_{\mathrm{G}\mathrm{S}3}-0.5)\\ {\mathrm{V}}_{\mathrm{G}\mathrm{S}3}=0.947\mathrm{V}\end{array}$

The expression for current ${\mathrm{I}}_{\mathrm{O}1}$ is,

  ${\mathrm{I}}_{\mathrm{O}1}=\left(\frac{1}{2}\right){\mathrm{\mu }}_{\mathrm{n}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5{({\mathrm{V}}_{\mathrm{G}\mathrm{S}1}-{\mathrm{V}}_{\mathrm{T}\mathrm{N}})}^2$

On substituting the given values,

  $\begin{array}{l}25=50{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5{(}^{0.947}\\ 1=2{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5{(}^{0.447}\\ \frac{1}{2\times {(0.447)}^2}={\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5\end{array}$

  ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5=2.5$

The expression for current ${\mathrm{I}}_{\mathrm{O}2}$ is,

  ${\mathrm{I}}_{\mathrm{O}2}=\left(\frac{1}{2}\right){\mathrm{\mu }}_{\mathrm{p}}{\mathrm{C}}_{\mathrm{o}\mathrm{x}}{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$

On substituting the given values,

  $\begin{array}{l}$ 75=20{\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6{(1.207-0.5)}^2$$ 3.75={\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6{(0.707)}^2$\frac{3.75}{0.4998}={\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6\end{array}$

  ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6=7.5$

Conclusion:

  ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_5=2.5$

  ${\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_6=7.5$