Solution Manual Electric Circuit Analysis and Design ch 16, Exercises of Electronic Circuits Analysis

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Electronic Circuit Analysis and Design, 2" edition Solutions Manual Chapter 16 Exercise Solutions Et6.3 tsa P Sin - Yoo > in = > = 150 nA E161 4. Driver in consaturation: so = B(F) (e-02-o87 -(& YZ), - Vso =" i= 05 = Fisilats — oy(0.18) - (0.15) = = 87.5[1.2373] m Ap = 44.0 kR b. From Equation (16-10): (SF) ovuea, -n97" + (Vin 0.8) =5 =0 3,920(¥y. ~ 0.8)? + (Vi, — 0.3) —S3 20 rls 1 + 43.9215) 23.52) Vie — 0.8 = 1.0 >¥y, = 18 V Vn 08 = VY =i0V E162 ak wy =Os> my =4V iL) «=ovy = 4 V, driver in nonsaturation (FE) etm (FD) Beam Foe 2S — vy — 1)? = (16)[2(4 ~ te — 16 — By +09 = B(6u9 — vf) 9g ~ 56m +16 = 4 oan — Yo] 56 4/ (56)? — 4(9)(18) 2(3) n= ve = 0.30 ¥ bP =ip- Von 38 2 in = FS — 0.30 ~ 1)" = 479 WA = (0.479)(5) = Psismw Ww w « 180 = as0( 2) > (%) = 0.536 SIE pe-rank 150 = BF) [a(a.2 — 0.8(0.2) - 0.277] w wy 150 = 23.1- (5), > (7), = 6.49 E164 a. Load ig saruration: driver in nonsaturation: (FLT) carr “(BLE h=-torn A —[-L.5))? = (6) [2(3 - 0.T}x0 — v8) 4.5 = 6(8.6%y — 17) Gg — $1.6r +45 = 0 51.6 & /(S1.8)? — 4(6)(¢.5) ae 26) vg = 0.0881 ¥ b. Load Vor ™ Von +a, Equation (16.26(b)) m5 1S > oy HOS V Prom Equation (16 — 28(b)} : K, [e605 ¥a0}= Poe : JSten 0.7} = -(-1.8) = 1.5 ‘Load: ep SL STV Poe = 3.5 V Driver: Py = LST Bop = 0.87 V & ips Faas? = 78.75 uA P =Ip-Vpp = (78.75)(8) =e P = 394 pw