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

Topic :- Single Stage Amplifiers

Difficulty :- High

cmos(48) • 96 views
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modified 6 days ago by gravatar for dukare030296hemant dukare030296hemant20 written 6 weeks ago by gravatar for awari.swati831 awari.swati83190
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• $V_{out} \gt V_{in} - V_{tn}$ ⇒ 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, $k=\frac{\beta_{driven}}{\beta{load}}$

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written 6 days ago by gravatar for dukare030296hemant dukare030296hemant20
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