Question: Compare resistive load inverter saturated load inverter and CMOS inverter on the basis of Noise margins, power dissipation, area and delay
0

Subject :- VLSI Design

Topic :- MOSFET Inverters

Difficulty :- High

cmos(48) • 1.5k views
ADD COMMENTlink
modified 14 months ago by gravatar for Sanket Shingote Sanket Shingote250 written 16 months ago by gravatar for awari.swati831 awari.swati831250
0

enter image description here

• $V_{out} \gt V_{in} - V_{tn}$ ⇒ driver transistor in saturation – When Vin is small
• Load transistor permanently in saturation Load transistor permanently in saturation $– V_{dsp} = V_{gsp} ∴ -V_{dsp} \lt V_{gsp}, \,\,\,\, - V_{tp}\,\,\, or \,\,0 \lt - V_{tp}$ ⇒ Saturated region
• When $V_{in}$ is Small

$I_{ds,driver}=\frac{\beta_{driver}}{2}(V_{in}-V_{tn})^2$

Load in Saturation:
$I_{ds,load}=\frac{-\beta_{load}}{2}(V_{out}-V_{DD}-V_{tp})^2$

Equating the currents:
$V_{out}=V_{DD}+V_{tp}+\sqrt{k} (V_{in}-V_{tn})$
where, $k=\frac{\beta_{driven}}{\beta{load}}$

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

ADD COMMENTlink
written 15 months ago by gravatar for dukare030296hemant dukare030296hemant90
Please log in to add an answer.