written 6.7 years ago by | • modified 6.5 years ago |
Subject :- VLSI Design
Topic :- MOSFET Inverters
Difficulty :- High
written 6.7 years ago by | • modified 6.5 years ago |
Subject :- VLSI Design
Topic :- MOSFET Inverters
Difficulty :- High
written 6.6 years ago by |
• $V_{out} \gt V_{in} - V_{tn}$ ⇒ 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, $k=\frac{\beta_{driven}}{\beta{load}}$