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

This solution manual was provided by Maria Geven at Assam University for Electrical Circuital Analysis course. It includes: Load, Driver, Saturation, Nonsaturation, Logic, NMOS, Transistor, Series, Parallel, Composite

Typology: Exercises

2011/2012

Uploaded on 07/06/2012

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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