with $L = 0.36 \mu m$ and $W/L = 10$; it was fabricated in a $0.18- \mu m$ CMOS process for which $\mu_n C_ox = 387 \mu A/ V^2$ and $V^`_A = 5 V/ \mu m$. Find the values of $g_m$ and $A_0$ obtained at $I_D = 10 \mu A, 100 \mu A$ and $1 mA$.

Since,

$W/L = 10$ and $L = 0.36 \mu m$

Therefore, $W = 3.6 \mu m$

For CS Amplifier :-

$g_m = \sqrt {2 \mu_n C_{ox} (W/L) I_D}$

$A_o = \dfrac {V^`_A \sqrt {2 \mu_n C_{ox} (WL)}}{\sqrt I_D}$

(i) For $I_D = 10 \mu A,$

$g_m = \sqrt {(2)(387 \times 10^{-6}) (10) (10 \times 10^{-6})} = 0.28 mA/V$

$A_o = \dfrac {(5 \times 10^6) \sqrt {(2)(387 \times 10^{-6}) (3.6 \times 0.36)}}{\sqrt {10 \times 10^{-6}}} = 50 V/V$

(ii) For $I_D = 100 \mu A,$

$g_m = \sqrt {(2)(387 \times 10^{-6}) (10) (100 \times 10^{-6})} = 0.88 mA/V$

$A_o = \dfrac {(5 \times 10^6) \sqrt {(2)(387 \times 10^{-6}) (3.6 \times 0.36)}}{\sqrt {100 \times 10^{-6}}} = 15.8 V/V$

(iii) For $I_D = 1 mA,$

$g_m = \sqrt {(2)(387 \times 10^{-6}) (10) (1 \times 10^{-3})} = 2.78 mA/V$

$A_o = \dfrac {(5 \times 10^6) \sqrt {(2)(387 \times 10^{-6}) (3.6 \times 0.36)}}{\sqrt {1 \times 10^{-3}}} = 5 V/V$