HW12 Solutions 1.(10 points) (a) In the saturation region, VVVVtGSDS20=−≥In the triode region, VVVVtGSDS20=−≤Also check 0tGSVV≥so it’s not operated in the cutoff region (b) In the saturation region, 220)2(5.0)(−=−=GStGSDVVVKi0123456789100510152025Vgs(V)iD (mA)2.(10 points) WhenVVVDSGS5==, the transistor is operated in the saturation region. 20)(2)(tGSDVVKPLWi−=Substitute all the values into the equation and we can solve W/L=1.25. For L=2um, we need W=2.5um. 3.(10 points) The load-line equation is DSDDDDViRV+=, and the plots are: 051015051015Vds(V)iD (mA)Vdd=15VVdd=10VVdd=5V
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