Subject: CMOS VLSI Design

Topic: CMOS analog building blocks

Difficulty: Difficult

$I_D\ of\ M_4$

$I_{Ref}=I_{D_1}$

$I_{D_2}=I_{Ref}\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}\ (M_1\ \&\ M_2\ \to current\ mirror)$

Now, $I_{D_2}=I_{D_3}$

$\therefore I_{D_3}=I_{Ref}\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}$

And $I_{out}=I_{D_4}=I_{D_3}\frac{(\frac{W}{L})_4}{(\frac{W}{L})_3}\ (M_3\ \&\ M_4 \ forms\ \ current \ \ mirror)$

$\therefore I_{out}=I_{Ref}\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}\frac{(\frac{W}{L})_4}{(\frac{W}{L})_3}$