hw5_su10_sol - ECE 442 Summer 2010 Homework # 5 Solution...

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Unformatted text preview: ECE 442 Summer 2010 Homework # 5 Solution 5.13 (a) Transistor is operated in the forward active mode Using the Ebers-Moll model, from eq. (5.28) in the textbook we have: iB = IS βF × eVBE /VT − 1 + IS βR × eVBC /VT − 1 ⇒ Note that VBC is negative in the forward active mode −15 eVBE /0.025 − 1 + ⇒ 10 × 10−6 = 10 100 −15 10 e−1/0.025 − 1 ≈ 0 100 10−15 0.11 e−1/0.025 − 1 ⇒ ⇒ VBE = 0.691 [V ] IC = IS × eVBE /VT = 10−15 e0.691/0.025 ⇒ C = 1 [mA] C IB = Iβ = 0.01 [mA] IE = IC + IB = 1.01 [mA] (b) Device oprates in the reverse active mode from eq. (5.28) I I IB = βS × eVBE /VT − 1 + βS × eVBC /VT − 1 F R −17 = −10 + 0.00909 = 9.1 [mA] from eq. (5.28) I IC = Is × eVBE /VT − 1 − αS × eVBC /VT − 1 R = −10−15 − 0.01009 = −10.1 [mA] IE = IC + IB = −1.01 [mA] 1 (rst term is very small) 5.26 (i) For nite β IB = 1 .5 10 = 0.15 [mA] IE = (VE − (−9))/103 = 0.68 [mA] VE = −1.5 − 0.7 = −2.2 [V ] IE IB =β+1⇒ α= β β +1 0.68 0.15 = β + 1 ⇒ β = 3.53 ⇒ α = 0.779 VC = 9 − 10 × (αIE ) ⇒ VC = 3.7 [V ] (ii) For β = ∞ β = ∞ ⇒ IB = 0 ⇒ VB = 0 VE = VB − VE = −0.7 [V ] IE = 0.83 [mA] ⇒ VC = 9 − 10 × (αIE ) ⇒ VC = 0.7 [V ] 2 5.128 Ic = 0.2 [mA] T rπ = β × VC = 100 × I 25 0 .2 = 12.5 [k Ω] = Rin R0 = RC r0 ≈ RC = 24 [k Ω] gm = β rπ = 8 [mho] AV0 ≈ −gm × RC = −8 × 10−3 × 24 × 103 = −192 [V /V ] RL 5k 10k AV = AV0 × RinRin sig × RL +R0 = −192 × 12.12.+10k × 10k+24k = −31.4 [V /V ] +R 5k 5.143 3 (a) IE = 9−0.7 1+100/(β +1) ⇒ β = 40: for IE = 2.41 [mA]; VE = 1 × 2.41 = 2.41 [V ] VB = 2.14 + 0.7 = 3.11 [V ] β = 200: for IE = 5.54 [mA]; VE = 5.54 [V ] VB = 6.24 [V ] (b) Ri = 100k (β + 1) × [re + (1k 1k )] = 100k (β + 1) × [re + 0.5k ] for β = 40: IE = 2.41 [mA]; re = 10.37 [Ω] Ri = 17.3 [k Ω] for β = 200: IE = 5.54 [mA]; re = 4.51 [Ω] Ri = 50.3 [k Ω] (c) vb vS v0 vS = for v0 vb v0 vS = 0.621 [V /V ] for β = 200: v0 vS = 0.827 [V /V ] = Ri RS +Ri (1 1) (1 1)+re β = 40: × × 4 4.79 For common-source amplier we have: gm = 2 [mA/V ], r0 = 50 [k Ω], RD = 10 [k Ω] RG = 10 [M Ω], Rsig = 0.5 [M Ω], RL = 20 [k Ω] from Eq. (4.82): GV = − RGRG sig × gm (r0 RD RL ) ⇒ GV = −11.2 [V /V ] +R 4.82 From Eq. (4.90): GV = RG RG +Rsig × gm (RD RL ) 1+gm RS ⇒ GV = −16 [V /V ] reduced by the factor of 4: GV2 GV1 = = 1 1+gm RS ⇒ 1 4 1 1+2RS ⇒ RS = 1.2 [k Ω] 5 ...
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This note was uploaded on 02/08/2012 for the course ECE 342 taught by Professor Nareshshanbhag during the Spring '11 term at University of Illinois, Urbana Champaign.

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