Subject: CMOS VLSI Design

Topic: CMOS analog building blocks

Difficulty: High

Solution:

we have,

$V_x = V_y....................(1)$

For, $M_1,\quad$ $V_{G{S_1}}=V_{D{S_1}}=V_x$

but, $I_{D_1}=\frac{1}{2}\mu_n C_{ox} (\frac{W}{L})_1(V_{G{S_1}}-V_{Th})^2$

$\therefore V_{G{S_1}}=\sqrt {\frac{2I_{D_1}}{[\mu_n C_{ox} (\frac{W}{L})_1]}}+V_{Th}$

$V_{G{S_1}}=\sqrt {\frac{2I_{Ref}}{[\mu_n C_{ox} (\frac{W}{L})_1]}}+V_{Th}$

$\therefore\ V_x=V_y=\sqrt {\frac{2I_{Ref}}{[\mu_n C_{ox} (\frac{W}{L})_1]}}+V_{Th}$

The graph nature will be shown as below :-

To find maximum value of $I_{Ref} :-$

We have,

$V_N=V_{G{S_0}}+V_{G{S_1}}$

$\quad \quad \quad=\sqrt {\frac{2I_{Ref}}{\mu_n C_{ox}}} [\sqrt{(\frac{L}{W})_0}+\sqrt{(\frac{L}{W})_1}]+V_{Th} +V_{Th}$

But, $V_{DD}-V_N=0.5..............(Given)$

$\therefore \ V_{DD}-\sqrt {\frac{2I_{Ref}}{\mu_n C_{ox}}} [\sqrt{(\frac{L}{W})_0}+\sqrt{(\frac{L}{W})_1}]-2V_{Th}=0.5 $

$\therefore \ I_{Ref},max=\frac{\mu_n C_{ox}}{2} \frac{(V_{DD}-0.5-2V_{Th})^2}{ [\sqrt{(\frac{L}{W})_0}+\sqrt{(\frac{L}{W})_1}]^2}$