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BCE 3050 — Fall 2002 Page 1 Homework 14.12 Veg = 12022 20k + 62kN (12-0.7-9,07)V tp = NO 907 a Ie = B37 = 12 ~ 3900I, - 82001, = 5.47 V BY 5.1kQ4 (75 + 13.9 BHA | Ie = B37 HA | Vee ® Otc =9.07V | Reg = 20kf/62K0 = 15.12 75(0.025V) _ 4 aoke Forward - active region Is correct. | r, = TPA in = Vy Mea eka =0.938v, | Ry, = IKOII5,1kO = 0,938k2 Ry = Fof8. 2KQ|IOOKN =8. 2kNNOOkR = 7.58kQ | Aya, = pl =-128 Ay =0.938Ay,, =-120 | A, = No -60.9 Rin = 15. 1KQ[3. 49kQ = 2.83 kO | Royr = rof8.2kN = 8.20 Vie = 0.9389, 094s IS =0.739v, | v= aw =6.76 mV Ay =-10V¢¢ = —10(12) = -120. | Gain is identical lo the rule - of - thumb estimate. 14.13 500kQ, = 18———_ -4, 74V 2 = . = Veg Toss im 4.74V | Reg = 500KQ{1.4MO = 368kQ. 4.74 = Vag + 270001 yg = b+ + 27000Ipg ~ Ipg = 104 pA Vog = 18 ~ [pg(75kO+ 27kQ) = 7.99V | Saturation region Is correct. fn = 2(280x10 Yioani0 = 0.228mS 368kQ. = V, 5 = 0.9970, | Ry, = LkQIS68kQ = 0.9970 Vth "Tk + 368h0 o | Run a Ry = Fo f?5kQ[470kQ =75KQ/470K0 = 64.7kO | Avy, = -(0.228m8)\(64.7kQ) ~-14.8 1 75kQ =0.997Avn =-14.7 | A, = 368K0(- 25) ——————— = -1 Ay = 0.997 Avi | Ar = 308K0(-8m) er aOR Riy = 368 kO | Royy = Tof75KQ = 75k0 yy O.912V , ge = 0.9974, | Vas~Vin= O.912V | vy, = 0.2 == = 0.183 V ~ Vou i8 Ay 8 -oa aie = 17 | The rule-of - thumb estimate assumes Vp, Vos- Vin 0.912 We have Vy, = 1041A(75K02) = 7. 80V = 0.433V py The estimate also doesn’ 1 account for the presence of Ry. 
