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Unformatted text preview: ECE 442 Summer 2008
HW#3 Solutions
3.1. The diode can be reversebiased and thus no current would ow, or forwardbiased where current would ow.
a) Reversebiased: I = 0 A, V = 1.5 V
b) Forwardbiased: 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 IV 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 halfcycles the voltage across the load has the same form
(see Figure 3.27 in the textbook), we can average over one halfcycle 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 09/07/2011 for the course ECE 442 taught by Professor Schuttaine during the Summer '08 term at University of Illinois at Urbana–Champaign.
 Summer '08
 SCHUTTAINE

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