0
2.6kviews
For a n-MOS draw (a) basic small signal Model (b) small signal model considering channel length modulation (c) considering body effect

Subject: CMOS VLSI Design

Topic: CMOS analog building blocks

Difficulty: Medium

1 Answer
1
108views

1) n-MOS

(a) Basic small signal model:


enter image description here

(b) considering channel length modulation :

enter image description here

${r_0\ =\frac {\partial V_{TH}}{\partial I_D}\\ \quad \ \ =\frac{1}{\frac {\partial V_{TH}}{\partial I_D}}\\ \quad \ \ =\frac{1}{\frac{1}{2}\frac{W}{L}\mu_nC_{ox}(V_{GS}-V_T)^2}}\\ { \therefore\ r_0\ =\frac{1}{\lambda I_D}}$


(c) Body Effect :

enter image description here

${g_{m b}=\frac {\partial I_D}{V_{BS}}\ = \frac {W}{L}\mu_n C_{ox}(V_{GS}-V_{Th})(-\frac{\partial V_{Th}}{\partial V_{BS}})\\ We\ \ also\ \ have\ ,\frac{\partial V_{Th}}{\partial V_{BS}}\ =\ -\frac{\partial V_{Th}}{\partial V_{SB}}\ =\ -\frac{\gamma}{2}(2\phi_F+V_{SB})^\frac{-1}{2}\\ Thus\ ,\ g_{mb}\ =g_m\frac{\gamma}{2 \sqrt {2\phi_F+V_{SB}}}\\ \quad \quad \quad \quad \quad \quad \quad=\eta g_m\\ \quad \quad \quad \quad \quad \therefore \eta \ =\frac{g_{mb}}{g_m}}$

Please log in to add an answer.