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Microelectronics Homework and solutions
Typology: Exercises
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vo = gm
ro|| (^) sC^1 L
vin = (^) 1+gsrmrooCL Av (s) = (^) vvino = (^) 1+gmsrrooCL for λ => 0 , ro => ∞ ∴ Av (s) = (^) sCgmL (Ideal Integrator)
(11.19) Av = −gmro ∴ Cin = (1 + Av ) CF = (1 + gmro) CF (11.28)
2 πfT = (^) Cgmgs = μnCox(^ WL )(VGS −Vth) (^23) (W L)Cox^ =^32 μ Ln 2 (VGS −^ Vth)
(4.10) VCC = 0. 2 V ∴ VCC = IC RC + 0. 2 VB = VT ln I ICS = VCC − I βC (10k) ∴ VB = 0.2 + IC (1k) − IC (100) = 0.2 + 900IC Let VB = 0. 7 ∴ I C(0) = 0.^7900 −^0.^2 V (^) b(1) = 0. 734 V I(1) C = 5. 94 × 10 −^4 A
V (^) b(2) = 0. 736 V I(2) C = 5. 96 × 10 −^4 A V (^) b(3) = 0. 736 V I(3) C = 5. 96 × 10 −^4 A ∴ VCC = 0. 795 V for R=>∞ IC = 0. 533 mA, VB = 0. 733 V = VCC
(4.26)
(a) IC = IS exp
VT
= 5 × 10 −^17 exp (^80026 mVmV^ )^ ' 1. 15 mA Vx = VCC − RC IC = 2. 5 − 1 × 1. 15 Vx = 1. 35 V Transistor is in Forward Active Region
(b) IC = IS exp
VT
IC = 5 × 10 −^17 exp (^80026 ) [1 + V 5 X^ ]^ =>Eqn 1 Also we know: Vx = VCC − RC IC => IC = VCC R^ −C Vx=> Eqn 2 Equations 1 and 2 => VCC R^ −C Vx= 5 × 10 −^17 exp (^80026 ) [1 + V 5 x^ ] => Vx + 5 × 10 −^17 exp (^80026 ) [1 + V 5 x^ ]^ = 2. 5
