Chapter 10, Problem 10.56P

The circuit in Figure P 10.56 is a PMOS version of a two-transistor MOScurrent mirror. Assume transistor parameters of $V_{TP}=-0.4V,{k^{\prime }}_P=60\mu A/V^2$ , and $\lambda =0$ . The transistor width-to-length ratios are

${\left(W/L\right)}_1=25,{\left(W/L\right)}_2=15$ , and ${\left(W/L\right)}_3=5$ . (a) Determine $I_O,I_{REF},V_{SG1}$ , and $V_{SG3}$ . (b) What is the largest value of R such that $M_2$ remainsbiased in the saturation region?

To determine

The value of ${\mathrm{I}}_{\mathrm{O}}$ , ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}$ , ${\mathrm{V}}_{\mathrm{S}\mathrm{G}1}$ and ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}$ .

Answer to Problem 10.56P

  ${\mathrm{I}}_{\mathrm{O}}=0.208\text{mA}$

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=0.3468\text{mA}$

  ${\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=1.08\text{V}$

  ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=1.98\text{V}$

Explanation of Solution

Given:

The circuit parameters are

  $\begin{array}{l}{\mathrm{V}}^{+}=+3\text{V}\\ {\mathrm{V}}^{-}=0\text{V}\end{array}$

The transistor parameters are

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{P}}=-0.4\text{V}\\ \mathrm{K}{\text{'}}_{\mathrm{p}}=60{\text{μA/V}}^{\text{2}}\\ \mathrm{\lambda }=0\end{array}$

  $\begin{array}{}$ {(\mathrm{W}/\mathrm{L})}_1=25$$ {(\mathrm{W}/\mathrm{L})}_2=15$$ {(\mathrm{W}/\mathrm{L})}_3=5$\end{array}$

Calculation:

The given circuit is

  

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

  $\begin{array}{}$ \left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{p}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2=\left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{p}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_3{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(25\right){({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(5\right){({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}-0.4)}^2\left(1\right)$\end{array}$

Now,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{S}\mathrm{G}3}={\mathrm{V}}^{+}-{\mathrm{V}}^{-}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-0\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}\left(2\right)\end{array}$

Put the value of ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}$ from equation (2) toequation (1).

  $\begin{array}{l}$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(25\right){({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(5\right){(3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2$$ \frac{25}{5}{({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2={(3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2$\sqrt{5}({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)=(3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)\\ \sqrt{5}({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)=(2.6-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1})\\ \sqrt{5}\times {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-\sqrt{5}0(.4)=(2.6-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1})\\ 2.23{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=2.6+0.892\\ 3.23{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=3.492\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=\frac{3.492}{3.23}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=1.08\text{V}\left(3\right)\end{array}$

From equation (2) and (3),

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-1.08\end{array}$

  ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=1.98\text{V}$

Now the reference current will be

  $\begin{array}{c}$ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\left(\frac{\mathrm{K}{\text{'}}_{\mathrm{p}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\times \left(25\right)\times {(1.08-0.4)}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\times 25\times {(0.68)}^2$=\left(\frac{60\times {10}^{-6}}{2}\right)\times \left(25\right)\times 0.4624\\ =\left(\frac{60\times {10}^{-6}}{2}\right)\times 11.56\end{array}$

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=0.3468\text{mA}$

Now calculate output current by using aspect ratio.

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{O}}=\frac{{(\mathrm{W}/\mathrm{L})}_2}{{(\mathrm{W}/\mathrm{L})}_1}{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}\\ {\mathrm{I}}_{\mathrm{O}}=\frac{15}{25}(0.3468\text{mA})\end{array}$

  ${\mathrm{I}}_{\mathrm{O}}=0.208\text{mA}$

Conclusion:

  ${\mathrm{I}}_{\mathrm{O}}=0.208\text{mA}$

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=0.3468\text{mA}$

  ${\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=1.08\text{V}$

  ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=1.98\text{V}$

To determine

The largest value of $\mathrm{R}$ when ${\mathrm{M}}_2$ biased in the saturation region.

Answer to Problem 10.56P

  $\mathrm{R}=11.15\mathrm{k}\mathrm{\Omega }$

Explanation of Solution

Given:

The circuit parameters are

  $\begin{array}{l}{\mathrm{V}}^{+}=+3\text{V}\\ {\mathrm{V}}^{-}=0\text{V}\end{array}$

The transistor parameters are

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{T}\mathrm{P}}=-0.4\text{V}\\ \mathrm{K}{\text{'}}_{\mathrm{p}}=60{\text{μA/V}}^{\text{2}}\\ \mathrm{\lambda }=0\end{array}$

  $\begin{array}{}$ {(\mathrm{W}/\mathrm{L})}_1=25$$ {(\mathrm{W}/\mathrm{L})}_2=15$$ {(\mathrm{W}/\mathrm{L})}_3=5$\end{array}$

Calculation:

The given circuit is

  

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

  $\begin{array}{}$ \left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{p}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2=\left(\frac{{\mathrm{K}}^{\text{'}}{}_{\mathrm{p}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_3{({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(25\right){({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(5\right){({\mathrm{V}}_{\mathrm{S}\mathrm{G}3}-0.4)}^2\left(1\right)$\end{array}$

Now,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{S}\mathrm{G}3}={\mathrm{V}}^{+}-{\mathrm{V}}^{-}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-0\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}\left(2\right)\end{array}$

Put the value of ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}$ from equation (2) to (1),

  $\begin{array}{l}$ \left(\frac{60\times {10}^{-6}}{2}\right)\left(25\right){({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2=\left(\frac{60\times {10}^{-6}}{2}\right)\left(5\right){(3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2$$ \frac{25}{5}{({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2={(3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)}^2$\sqrt{5}({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)=(3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)\\ \sqrt{5}({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-0.4)=(2.6-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1})\\ \sqrt{5}\times {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}-\sqrt{5}0(.4)=(2.6-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1})\\ 2.23{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=2.6+0.892\\ 3.23{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=3.492\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=\frac{3.492}{3.23}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}1}=1.08\text{V}\left(3\right)\end{array}$

From equation (2) and(3),

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-{\mathrm{V}}_{\mathrm{S}\mathrm{G}1}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=3-1.08\end{array}$

  ${\mathrm{V}}_{\mathrm{S}\mathrm{G}3}=1.98\text{V}$

Now the reference current will be

  $\begin{array}{c}$ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\left(\frac{\mathrm{K}{\text{'}}_{\mathrm{p}}}{2}\right){\left(\frac{\mathrm{W}}{\mathrm{L}}\right)}_1{({\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}})}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\times \left(25\right)\times {(1.08-0.4)}^2$$ =\left(\frac{60\times {10}^{-6}}{2}\right)\times 25\times {(0.68)}^2$=\left(\frac{60\times {10}^{-6}}{2}\right)\times \left(25\right)\times 0.4624\\ =\left(\frac{60\times {10}^{-6}}{2}\right)\times 11.56\end{array}$

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=0.3468\text{mA}$

Now calculate output current by using aspect ratio.

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{O}}=\frac{{(\mathrm{W}/\mathrm{L})}_2}{{(\mathrm{W}/\mathrm{L})}_1}{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}\\ {\mathrm{I}}_{\mathrm{O}}=\frac{15}{25}(0.3468\text{mA})\end{array}$

  ${\mathrm{I}}_{\mathrm{O}}=0.208\text{mA}$

Calculate the largest value of R when ${\mathrm{M}}_2$ biased in saturation region,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{S}\mathrm{D}2(\mathrm{s}\mathrm{a}\mathrm{t})}={\mathrm{V}}_{\mathrm{S}\mathrm{G}1}+{\mathrm{V}}_{\mathrm{T}\mathrm{P}}\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}2(\mathrm{s}\mathrm{a}\mathrm{t})}=1.08+0.4\\ {\mathrm{V}}_{\mathrm{S}\mathrm{D}2(\mathrm{s}\mathrm{a}\mathrm{t})}=0.68\text{V}\end{array}$

According to Ohm’s law,

  ${\mathrm{V}}_{\mathrm{R}}={\mathrm{I}}_{\mathrm{O}}\mathrm{R}\left(4\right)$

Also,

  $\begin{array}{c}{\mathrm{V}}_{\mathrm{R}}={\mathrm{V}}^{+}-{\mathrm{V}}^{-}-{\mathrm{V}}_{\mathrm{D}\mathrm{S}2(\mathrm{s}\mathrm{a}\mathrm{t})}\\ =3-0-0.68\\ {\mathrm{V}}_{\mathrm{R}}=2.32\text{V}\left(5\right)\end{array}$

From equation (4) and (5),

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{R}}={\mathrm{I}}_{\mathrm{O}}\mathrm{R}\\ \mathrm{R}=\frac{{\mathrm{V}}_{\mathrm{R}}}{{\mathrm{I}}_{\mathrm{O}}}\\ \mathrm{R}=\frac{\text{2}\text{.32V}}{\text{0}\text{.208mA}}\end{array}$

  $\mathrm{R}=11.15\mathrm{k}\mathrm{\Omega }$

Conclusion:

  $\mathrm{R}=11.15\mathrm{k}\mathrm{\Omega }$