Subject: CMOS VLSI Design

Topic: CMOS analog building blocks

Difficulty: High

Solution:

The small signal drain current of $M_1=g_{m_1}V_{in}$

Also $I_{D_1}=I_{D_2} \quad \therefore I_{D_2}=g_{m_1}V_{in}$

$M_2\ \&\ M_3$ forms a basic current mirror

$As,\ I_{D_3}=\frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2} I_{D_2}$

$\therefore I_{D_3}=\frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2}g_{m_1}V_{in}$

$As,\ V_{out}=R_L I_{D_3}$

$\therefore V_{out}=R_L \frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2}g_{m_1}V_{in}$

$\therefore A_v=\frac{V_{out}}{V_{in}}=R_L \frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2}g_{m_1}$