# hw2_su10_sol - ECE 442 Summer 2010 HW#2 Solutions ...

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Unformatted text preview: ECE 442 Summer 2010 HW#2 Solutions  3.1. The diode can be reverse-biased and thus no current would ow, or forwardbiased where current would ow. a) Reverse-biased: I = 0 A, V = 1.5 V b) Forward-biased: I = 1.5 A, V = 0 V 3.6. ABXY 0000 0101 1001 1111 X=A·B Y=A+B X and Y are the same for A=B X and Y are opposite for A=B 3.10. D D I = (10 26 0)+20 = 0.225 mA 20 V = (10 20)+20 × 6 = 4.5 V 1 (diode is o) I = 0A V = −2 V 3.23. Since the diodes are in series, and are identical, the voltage across each diode is V /3. O VO /3 2/3 I = Is e nVT = 10−14 e 0.025 = 3.81 mA ∆V = V2 − 2 = −22.8 mV 3.26. With the given pairs of I-V values, we can calculate n: 2 i = IS eV /nVT ⇒ 10 × 10−3 = IS e0.7/(n×0.025) 100 × 10−3 = IS e0.8/(n×0.025) (1), and (2) Dividing (2) with (1) gives: 10 = e 0.1/(n×0.025) ⇒ n = 1.737 V = V2 − V1 = nVT × ln( i2 ) = 80 mV i1 − 1.737 × 0.025 × ln( 0.011 i1 ) = 80 × 10−3 i ⇒ i1 = 1.4 mA R = 80 mV /i1 = 57.1 Ω 3.78. VD = 0.7 V Peak voltage across R is: V R,peak √ = 12 2 − 2VD = 15.57 V 1.4 Θ = sin−1 12√2 = 0.0826 rad Fraction of a cycle in which D and D conduct is: π −2Θ 2π 1 2 × 100 = 47.4 % Note: in the other half cycle, D and D are conducting for the same fraction, so the total conduction interval is 94.8 %. The average voltage across the load is obtained by averaging over one cycle, but since in both half-cycles the voltage across the load has the same form (see Figure 3.27 in the textbook), we can average over one half-cycle to make integration simpler: 3 4 3 vR,avg = 1 2π = 2π vR (φ) dφ = 0 1 π π −Θ √ (12 2sin(φ) − 2VD )dφ Θ √ 1 [−12 2cos(φ) − 1.4φ]π−Θ = 9.445 V Θ π iR,avg = vR,avg = 9.445 mA R 4 ...
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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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hw2_su10_sol - ECE 442 Summer 2010 HW#2 Solutions ...

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