For CMOS Inverter with following parameters

$V_{TO},n =0.6 V \mu_nC_{OX}= 60 µA/V^2 (W/L)_n=8$ $V_{TO},n = -0.7 V \mu_nC_{OX}= 20 µA/V^2 (W/L)_n=12$

Calculate noise margins and the switching threshold of the inverter. The power supply voltage $V_{DD} =3.3 V$

Subject :- VLSI Design

Topic :- MOSFET Inverters

Difficulty :- Medium

Given,

$V_{TO,n}=0.6V$

$V_{TO,p}=0.7V$

$V_{DD}=3.3 V$

$_{\mu n}C_{ox}=60 \mu A/V^2$

$_{\mu p}C_{ox}=20 \mu A/V^2$

$(W_n/L_n)=8$

$(W_p/L_p)=12$

To calculate: Noise Margins

Formula: Noise Margins

$\hspace{1.4cm}N_{HL}=V_{IL}-V_{OL}$

$\hspace{1.4cm}N_{HH}=V_{OH}-V_{IH}$

$V_{IL}=\frac{2V_{out}+V_{top}-V_{DD}+K_R\,V_{TON}}{1+K_R}$

$V_{IH}=\frac{V_{top}+V_{DD}+K_R\,(2V_{out}+V_{TON})}{1+K_R}$

but, $\hspace{1cm} _{\mu n}C_{ox}=60 \mu A/V^2$,

$_{\mu p}C_{ox}=20 \mu A/V^2$

$K_R=\frac{ _{\mu n}C_{ox}}{_{\mu p}C_{ox}}\frac{(W_n/L_n)}{(W_p/L_p)}$

$=\frac{60}{20}*\frac{8}{12}=2$

$V_{IL}=\frac{2 V_{out}-0.7-33+(2*0.6)}{1+2}$

$V_{IL}=0.6V_{out}-0.93 \hspace{2cm}$ ------ (1)

$\frac{K_R^2}{2}(V_{IL}-V_{ON})^2$

$=\frac{K_R^2}{2}[2(V_{in}-V_{DD}-V_{TON})(V_{out}-V_{DD})-(V_{out}-V_{DD})^2]$

$\therefore 4(0.6V_{out}-0.93-0.6)^2$

$=[2(0.67\,V_{out}-2.6)(V_{out}-3.3)-(V_{out}-3.3)^2]$

$=[2(0.67\,V_{out}^2-2.6\,V_{out}-2.21\,V_{out}+8.58)-$

$(V_{out}^2-6.6\,V_{out}+10.89)]$

$\therefore 4(0.6V_{out}-1.53)^2$

$=[2(0.67\,V_{out}^2-4.811V_{out}+8.58)-(V_{out}^2-6.6\,V_{out}+10.89)]$

$\therefore 4(0.36V_{out}^2-1.836V_{out}+2.34)$

$=1.44V_{out}^2-7.34V_{out}+9.36=0.34V_{out}^2-3.02V_{out}+6.27$

$1.1\,V_{out}^2-10.36\,V_{out}+3.09=0$

$\therefore V_{out}=9$

$V_{IL}=0.6\,V_{out}-1.33$
$\therefore V_{IL}=4.07\,V$

$V_{IH}=\frac{V_{top}+V_{DD}+K_R\,(2V_{out}+V_{TON})}{1+K_R}$

$=\frac{3.3-0.7+2(2\,V_{out}+0.6)}{3}$

$=\frac{3.8+4\,V_{out}}{3}$

$\therefore V_{IH}=1.26+1.33\, V_{out}$

$V_{out}=0.3\,V$

$\therefore V_{IH}=1.26+1.33*0.3$
$\therefore V_{IH}=1.6\,V$

$N_{HL}=V_{IL}-V_{OL}=4.07-0=4.07\,V$

$N_{HH}=V_{ON}-V_{IH}=3.3-1.6=1.7\,V$