CSAmpEx - Copyright 2008. W. Marshall Leach, Jr.,...

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© Copyright 2008. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. Common-Source Amplifier Example K prime 0.002 W1 L1 V TO 1.75 λ 0.016 V plus 24 V minus 24 R 1 510 6 . R 2 110 6 . R D 10 10 3 . R S 310 3 . R 3 50 R L 20 10 3 . R s 3 . R p xy , () . DC Bias Solution 1
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V GG V plus R 2 . V minus R 1 . R 1 R 2 V GG 16 = V SS V minus R SS R S V 1 V GG V SS V TO V 1 6.25 = KK prime W L . I D 1 2K . R S 2 . 12 K . V 1 . R S . 1 2 . I D 1.655 10 3 = V D V plus I D R D . V D 7.454 = V S V minus I D R S . V S 19.036 = V DS V D V S V DS 26.491 = V GS V GG V S V GS 3.036 = V GS V TO 1.286 = Because V DS > V GS V TO , 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 I D 13 . 10 3 . I D 1 410 3 . 1 0.016 48 I D 13 . 10 3 . . . 3000 2 . 14 1 0 3 . 1 0.016 48 I D 13 . 10 3 . . . 6.25 . 3000 . 1 2 . .0017157743653358533060 This is the exact solution for I D including the Early effect. We will use the approximate solution for the ac analysis below. I D .0017157743653358533060 .0017157743653358533060
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CSAmpEx - Copyright 2008. W. Marshall Leach, Jr.,...

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