Common-Source Amplifier ExampleKprime0.002 W1 L1 VTO1.75λ0.016χ0Vplus24 Vminus24 R15106.R21106.RD10 103.RS3103.R350 RL20 103.Rs5103.Rpxy,()xy.xyDC Bias Solution1VGGVplusR2.VminusR1.R1R2VGG16=VSSVminusRSSRSV1VGGVSSVTOV16.25=We neglect the Early effect, i.e. set λ = 0 to solve for the drain bias current.KKprimeWL.ID12K.RS2.12K.V1.RS.12.ID1.655 103=VDVplusIDRD.VD7.454=VSVminusIDRS.VS19.036=VDSVDVSVDS26.491=VGSVGGVSVGS3.036=VGSVTO1.286=Because VDS > VGSVTO, the MOSFET is in the active or saturated state.Here is an exact solution for the drain current. Note that MathCad requires numbers for everything except the variable being solved for. The drain-source voltage in the equation is 48 ID13.103.ID14103.1 0.016 48 ID13.103...30002.14103.1 0.016 48 ID13.103...6.25.3000.12..0017157743653358533060This is the exact solution for ID including the Early effect. We will use the approximate solution for the ac analysis below. ID.0017157743653358533060.0017157743653358533060100.3.567=This is the percentage error in neglecting the Early effect in solving for the drain current.2Now for the ac solution.KKprimeWL.1λVDS..K 2.848 103=gm2K.ID.gm3.07 103=rs1gmrs325.758=risrs1χ()ris325.758=No body effect because the body lead is connected to the source lead. This is equivalent to setting χ = 0 in the equations.RtsRpRSR3,Rts49.18=r0λ1VDSIDr05.378 104=ridr01Rtsris.Rtsrid6.195 104=vs1This makes the gain equal to vo. vtgvsRpR1R2,RsRpR1R2,.vtg0.994=RtgRpRsRpR1R2,,Rtg4.97 103=RtdRpRDRL,Rtd6.667 103=3Gmg11χ1risRts.r0r0RprisRts,.idscGmgvtg.idsc2.649 103=voidscRpridRtd,.vo15.945=This is the voltage gain.routRpRDrid,rout8.61 103=rinRpR1R2,rin8.333
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