fundamentals of microelectronics - previewedition chapter 3 solution manual
TRANSCRIPT
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3.1 (a)
I X =
V X
R1V X 0
V X (V)
I X
Slope = 1/R1
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3.2
I X =
V X
R1V X 0
Plotting I X(t), we have
0
−V 0/R1 I X ( t ) f o r V
B
=
1 V
( S o l i d )
−π/ω 0 π/ωt
−V 0
0
V 0
V X ( t
) ( D o t t e d )
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3.3
I X =
0 V X < V BV X−V BR1
V X > V B
Plotting I X vs. V X for V B = −1 V and V B = 1 V, we get:
−1 1V X (V)
I X
V B = −1 VV B = 1 V
Slope = 1/R1 Slope = 1/R1
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3.4
I X =
0 V X < V BV X−V BR1
V X > V B
Let’s assume V 0 > 1 V. Plotting I X(t) for V B = −1 V, we get
0
(V 0 − V B)/R1
I X ( t ) f o r V B
=
− 1 V
( S o l i d )
−π/ω 0 π/ωt
−V 0
0
V B
V 0
V X ( t
) ( D o t t e d )
Plotting I X(t) for V B = 1 V, we get
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0
(V 0 − V B)/R1
I X ( t ) f o r V B
=
1 V
( S o l i d )
−π/ω 0 π/ωt
−V 0
0
V B
V 0
V X ( t ) ( D o t t e d )
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3.5
I X =
V X−V BR1
V X 0
Plotting I X vs. V X for V B = −1 V and V B = 1 V, we get:
−1V X (V)
−1/R1
1/R1
I X
Slope = 1/R1
Slope = 1/R1
I X for V B = −1 VI X for V B = 1 V
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3.6 First, note that I D1 = 0 always, since D1 is reverse biased by V B (due to the assumption that V B > 0).We can write I X as
I X = (V X − V B)/R1
Plotting this, we get:
V BV X (V)
I X
Slope = 1/R1
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3.7
I X =
V X−V BR1
V X < V BV X−V BR1R2
V X > V B
I R1 = V X − V B
R1
Plotting I X and I R1 for V B = −1 V, we get:
−1V X (V)
I X for V B = −1 VI R1 for V B = −1 V
Slope = 1/R1
Slope = 1/R1 + 1/R2
Plotting I X and I R1 for V B = 1 V, we get:
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3.8
I X =
0 V X <
V B
R1+R2R1
V X
R1+ V X−V B
R2V X >
V B
R1+R2R1
I R1 = V B
R1+R2V X <
V B
R1+R2R1
V X
R1 V X > V B
R1+R2 R1
Plotting I X and I R1 for V B = −1 V, we get:
V BR1+R2
R1
V X (V)V BR1+R2
−V B/R2
I X for V B = −1 VI R1 for V B = −1 V
Slope = 1/R1
Slope = 1/R1 + 1/R2
Plotting I X and I R1 for V B = 1 V, we get:
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V BR1+R2
R1
V X (V)
V BR1+R2
I X for V B = 1 VI R1 for V B = 1 V
Slope = 1/R1
Slope = 1/R1 + 1/R2
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3.9 (a)
V out =
V B V in < V B
V in V in > V B
−5 −4 −3 −2 −1 0 1 2 3 4 5V in (V)
0
1
2
3
4
5
V o u t
( V )
Slope = 1
(b)
V out =
V in − V B V in < V B
0 V in > V B
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−5 −4 −3 −2 −1 0 1 2 3 4 5V in (V)
−7
−6
−5
−4
−3
−2
−1
0
1
2
V o u t
( V )
Slope = 1
(c)
V out = V in − V B
−5 −4 −3 −2 −1 0 1 2 3 4 5V in (V)
−7
−6
−5
−4
−3
−2
−1
0
1
2
3
V
o u t
( V )
Slope = 1
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(d)
V out =
V in V in < V B
V B V in > V B
−5 −4 −3 −2 −1 0 1 2 3 4 5V in (V)
−5
−4
−3
−2
−1
0
1
2
V o u t
( V )
Slope = 1
(e)
V out =
0 V in < V B
V in − V B V in > V B
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−5 −4 −3 −2 −1 0 1 2 3 4 5V in (V)
0
1
2
3
V o u t
( V )
Slope = 1
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3.11 For each part, the dotted line indicates V in(t), while the solid line indicates V out(t). Assume V 0 > V B .
(a)
V out =
V B V in < V B
V in V in > V B
−π/ω π/ωt
−V 0
V B
V 0
V o u t
( t ) ( V )
(b)
V out =
V in − V B V in < V B
0 V in > V B
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−π/ω π/ωt
−V 0 − V B
−V 0
V B
V 0
V o u t
( t ) ( V )
(c)
V out = V in − V B
−π/ω π/ωt
−V 0 − V B
−V 0
V 0 − V B
V B
V 0
V o u t
( t ) ( V )
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(d)
V out =
V in V in < V B
V B V in > V B
−
π/ω π/ω t
−V 0
V B
V 0
V o u t
( t ) ( V )
(e)
V out =
0 V in < V B
V in − V B V in > V B
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−π/ω π/ωt
−V 0
V 0 − V B
V B
V 0
V o u t
( t ) ( V )
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3.12 For each part, the dotted line indicates V in(t), while the solid line indicates V out(t). Assume V 0 > V B .
(a)
V out =
V in − V B V in < V B
0 V in > V B
−π/ω π/ω
t
−V 0 − V B
−V 0
V B
V 0
V o u t
( t ) ( V )
(b)
V out =
V in V in < V B
V B V in > V B
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−π/ω π/ωt
−V 0
V B
V 0
V o u t
( t ) ( V )
(c)
V out =
0 V in < V B
V in − V B V in > V B
−π/ω π/ω
t
−V 0
V 0 − V B
V B
V 0
V o u t
( t ) ( V )
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(d)
V out = V in − V B
−π/ω π/ωt
−V 0 − V B
−V 0
V 0 − V B
V B
V 0
V o u t
( t )
( V )
(e)
V out =V B V in < V B
V in V in > V B
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−π/ω π/ωt
−V 0
V B
V 0
V o u t
( t ) ( V )
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3.16 (a)
I R1 =
I in I in <
V D,on
R1V D,on
R1I in >
V D,on
R1
Slope = 1
V D,on/R1I in
V D,on/R1
I R1
(b)
I R1 =
I in I in < V D,on+V BR1V D,on+V B
R1I in >
V D,on+V B
R1
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Slope = 1
(V D,on + V B) /R1I in
(V D,on + V B) /R1
I R1
(c)
I R1 =
I in I in <
V D,on−V BR1
V D,on−V BR1
I in > V D,on−V B
R1
Slope = 1
(V D,on − V B) /R1
(V D,on − V B) /R1I in
I R1
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(d)
I R1 =
I in I in <
V D,on
R1V D,on
R1I in >
V D,on
R1
Slope = 1
V D,on/R1I in
V D,on/R1
I R1
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3.17 (a)
V out =
I inR1 I in <
V D,onR1
V D,on I in > V D,onR1
−I 0R1
0
V D,on
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
0
V D,on/R1
I 0
I i n
( t ) ( D o t t e d )
(b)
V out =I inR1 I in < V
D,on+V BR1
V D,on + V B I in > V D,on+V B
R1
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−I 0R1
0
V D,on + V B
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
0
(V D,on+V B)/R1I 0
I i n
( t ) ( D o t t e d )
(c)
V out =
I inR1 + V B I in <
V D,on−V BR1
V D,on I in > V D,on−V B
R1
−I 0R1 + V B
0
V D,on
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
0
(V D,on−
V B) /R1
I 0
I i n
( t ) ( D o t t e d )
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(d)
V out =
I inR1 + V B I in <
V D,onR1
V D,on + V B I in > V D,onR1
−I 0R1 + V B
0
V D,on + V B
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
0
V D,on/R1
I 0
I i n
( t ) ( D o t t e d )
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3.20 (a)
V out =
I inR1 I in >
V B−V D,onR1
V B − V D,on I in < V B−V D,on
R1
V B − V D,on
0
I 0R1
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
0
(V B − V D,on) /R1
I 0
I i n
( t ) ( D o t t e d )
(b)
V out =I inR1 + V B I in >
−
V D,on+V BR1
−V D,on I in < −V D,on+V B
R1
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−V D,on
0
I 0R1 + V B
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
0
−(V D,on+V B)/R1
I 0
I i n
( t ) ( D o t t e d )
(c)
V out =
I inR1 + V B I in > −
V D,onR1
V B − V D,on I in < −V D,onR1
V B − V D,on
0
I 0R1 + V B
V o u t
( t ) ( S o l i d )
−π/ω 0 π/ωt
−I 0
−V D,on/R1
0
I 0
I i n
( t ) ( D o t t e d )
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3.23 (a)
V out =
R2R1+R2
V in V in < R1+R2
R2V D,on
V D,on V in > R1+R2
R2V D,on
R1+R2R2
V D,on
V in (V)
V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
(b)
V out =
R2R1+R2
V in V in < R1+R2R1
V D,on
V in − V D,on V in > R1+R2
R1V D,on
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R1+R2R1
V D,on
V in (V)
R2R1V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
Slope = 1
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3.24 (a)
I R1 =
V inR1+R2
V in < R1+R2
R2V D,on
V in−V D,onR1
V in > R1+R2
R2V D,on
I D1 = 0 V in < R1+R2
R2V D,on
V in−V D,onR1
−
V D,onR2 V in >
R1+R2R2 V D,on
R1+R2R2
V D,on
V in (V)
V D,on/R2
Slope = 1/ (R1 + R2)
Slope = 1/R1
Slope = 1/R1
I R1I D1
(b)
I R1 =
V inR1+R2
V in < R1+R2
R1V D,on
V D,onR1
V in > R1+R2
R1V D,on
I D1 =
0 V in <
R1+R2R1
V D,onV in−V D,on
R2−
V D,onR1
V in > R1+R2
R1V D,on
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R1+R2R1
V D,on
V in (V)
V D,on/R1
Slope = 1/ (R1 + R2)
Slope = 1/R2
I R1I D1
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3.25 (a)
V out =
V B +
R2R1+R2
(V in − V B) V in < V B + R1+R2
R1V D,on
V in − V D,on V in > V B + R1+R2
R1V D,on
V B + R1+R2
R1V D,on
V in (V)
V B + R2R1V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
Slope = 1
(b)
V out =
R2R1+R2
V in V in < R1+R2R1 (V D,on + V B)
V in − V D,on − V B V in > R1+R2
R1(V D,on + V B)
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V B + R1+R2
R1(V D,on + V B)
V in (V)
R2
R1 (V D,on + V B)
V o u t
( V )
Slope = R2/ (R1 + R2)
Slope = 1
(c)
V out =
R2R1+R2
(V in − V B) V in > V B + R1+R2
R1V D,on
V in + V D,on − V B V in < V B + R1+R2
R1V D,on
V B + R1+R2
R1V D,on
V in (V)
R2R1V D,on
V o u t
( V )
Slope = 1
Slope = R2/ (R1 + R2)
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(d)
V out =
R2R1+R2
(V in − V B) V in < V B + R1+R2
R1(V D,on − V B)
V in − V D,on V in > V B + R1+R2
R1(V D,on − V B)
V B + R1+R2
R1(V D,on − V B)
V in (V)
R2R1V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
Slope = 1
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3.26 (a)
I R1 =
V in−V BR1+R2
V in < V B + R1+R2
R1V D,on
V D,onR1
V in > V B + R1+R2
R1V D,on
I D1 = 0 V in < V B + R1+R2
R1V D,on
V in−V D,on−V BR2
−
V D,onR1 V in > V B +
R1+R2R1 V D,on
V B + R1+R2
R1V D,on
V in (V)
V D,on/R1
Slope = 1/ (R1 + R2)
Slope = 1/R2
I R1I D1
(b)
I R1 =
V inR1+R2
V in < R1+R2
R1(V D,on + V B)
V D,on+V BR1
V in > R1+R2
R1(V D,on + V B)
I D1 =
0 V in <
R1+R2R1
(V D,on + V B)V in−V D,on−V B
R2−
V D,on+V BR1
V in > R1+R2
R1(V D,on + V B)
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R1+R2R1
(V D,on + V B)
V in (V)
(V D,on + V B) /R1
Slope = 1/ (R1 + R2)
Slope = 1/R2
I R1I D1
(c)
I R1 =
V in−V BR1+R2
V in > V B − R1+R2
R1V D,on
−
V D,onR1
V in < V B − R1+R2
R1V D,on
I D1 =
0 V in > V B −
R1+R2R1
V D,on
−
V in+V D,on+V BR2
−
V D,onR1
V in < V B − R1+R2
R1V D,on
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V B + R1+R2R1 V D,on
V in (V)
−V D,on/R1
I R1I D1
Slope = 1/ (R1 + R2)
Slope = −1/R2
(d)
I R1 =
V in−V BR1+R2
V in < V B + R1+R2
R1(V D,on − V B)
V D,on−V BR1
V in > V B + R1+R2
R1(V D,on − V B)
I D1 =
0 V in < V B +
R1+R2R1
(V D,on − V B)V in−V D,on
R2−
V D,on−V BR1
V in > V B + R1+R2
R1(V D,on − V B)
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V B + R1+R2
R1(V D,on − V B)
V in (V)(V D,on − V B) /R1
I R1I D1
Slope = 1/ (R1 + R2)
Slope = 1/R2
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3.27 (a)
V out =
0 V in < V D,on
R2R1+R2
(V in − V D,on) V in > V D,on
V D,onV in (V)
V o u t
( V )
Slope = R2/ (R1 + R2)
(b)
V out =
−V D,on V in <
−
R1+R2R2 V D,on
R2R1+R2
V in −R1+R2R2
V D,on < V in < R1+R2
R1V D,on
V in − V D,on V in > R1+R2
R1V D,on
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−
R1+R2R2 V D,on
R1+R2R1 V D,on
V in (V)
−V D,on
R2R1V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
Slope = 1
(c)
V out =
R2R1+R2
(V in + V D,on) − V D,on V in < −V D,on
V in V in > −V D,on
−V D,on
V in (V)−V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
Slope = 1
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(d)
V out =
0 V in < V D,on
R2R1+R2
(V in − V D,on) V in > V D,on
V D,onV in (V)
V D,on
V o u t
( V )
Slope = R2/ (R1 + R2)
(e)
V out =
R2R1+R2 (V in + V D,on) V in < −V D,on
0 V in > −V D,on
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−
V D,onV in (V)
V o u t
( V )
Slope = R2/ (R1 + R2)
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3.28 (a)
I R1 =
0 V in < V D,onV in−V D,onR1+R2
V in > V D,on
I D1 = 0 V in < V D,onV in−V D,onR1+R2 V in > V D,on
V D,onV in (V)
Slope = 1/ (R1 + R2)
I R1I D1
(b)
I R1 =
V in+V D,onR1
V in < −R1+R2R2
V D,onV in
R1+R2−
R1+R2R2
V D,on < V in < R1+R2
R1V D,on
V D,onR1
V in > R1+R2
R1V D,on
I D1 =
0 V in < −R1+R2R2
V D,on
0 −R1+R2R2
V D,on < V in < R1+R2
R1V D,on
V in−V D,onR2
−
V D,onR1
V in > R1+R2
R1V D,on
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−
R1+R2R2 V D,on R1+R2R1 V D,on
V in (V)−V D,on/R2
V D,on/R1
Slope = 1/R1
Slope = 1/ (R1 + R2)
Slope = 1
I R1I D1
(c)
I R1 =
V in+V D,onR1+R2
V in < −V D,on
0 V in > −V D,on
I D1 = 0 V in < −V D,on
0 V in > −V D,on
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−
V D,on V in (V)
Slope = 1/ (R1 + R2)
I R1I D1
(d)
I R1 =
0 V in < V D,onV in−V D,onR1+R2
V in > V D,on
I D1 = 0 V in < V D,on
V in−
V D,onR1+R2 V in > V D,on
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V D,onV in (V)
Slope = 1/ (R1 + R2)
I R1I D1
(e)
I R1 =
V in+V D,onR1+R2
V in < −V D,on
0 V in > −V D,on
I D1 =
0 V in < −V D,on
0 V in > −V D,on
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−
V D,on V in (V)
Slope = 1/ (R1 + R2)
I R1I D1
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3.29 (a)
V out =
V in V in < V D,on
V D,on + R2R1+R2
(V in − V D,on) V D,on < V in < V D,on + R1+R2
R1(V D,on + V B)
V in − V D,on − V B V in > V D,on + R1+R2
R1(V D,on + V B)
V D,on V D,on + R1+R2
R1(V D,on + V B)
V in (V)
V D,on
V D,on + R2R1
(V D,on + V B)
V o u t
( V )
Slope = 1
Slope = R2/ (R1 + R2)
Slope = 1
(b)
V out =
V in + V D,on − V B V in < V D,on +
R1+R2R1
(V B − 2V D,on)R2
R1+R2(V in − V D,on) V in > V D,on +
R1+R2R1
(V B − 2V D,on)
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V D,on + R1+R2
R1(V B − 2V D,on)
V in (V)
R2R1
(V B − 2V D,on)
V o u t
( V )
Slope = 1
Slope = R2/ (R1 + R2)
(c)
V out =
V in V in < V D,on + V B
V D,on + V B V in > V D,on + V B
V D,on + V B
V in (V)
V D,on
+ V B
V o u t
( V )
Slope = 1
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(d)
V out =
0 V in < V D,onR2
R1+R2(V in − V D,on) V D,on < V in < V D,on +
R1+R2R2
(V B + V D,on)
V D,on + V B V in > V D,on + R1+R2
R2(V B + V D,on)
V D,on V D,on + R1+R2
R2(V B + V D,on)
V in (V)
V D,on + V B
V o u t
( V )
Slope = R2/ (R1 + R2)
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3.30 (a)
I R1 =
0 V in < V D,onV in−V D,onR1+R2
V D,on < V in < V D,on + R1+R2
R1(V D,on + V B)
V D,on+V BR1
V in > V D,on + R1+R2
R1(V D,on + V B)
I D1 =
0 V in < V D,on + R1+R2
R1 (V D,on + V B)V in−2V D,on−V BR2
− V D,on+V B
R1V in > V D,on +
R1+R2R1
(V D,on + V B)
V D,on V D,on + R1+R2
R1(V D,on + V B)
V in (V)
V D,on + V B
Slope = 1/ (R1 + R2)
Slope = 1/R2
I R1I D1
(b) If V B V D,on
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V D,on
V in (V)
Slope = 1/ (R1 + R2)
I R1I D1
If V B > 2V D,on:
I R1 = I D1 =
V B−2V D,on
R1V in < V D,on +
R1+R2R1
(V B − 2V D,on)V in−V D,onR1+R2
V in > V D,on + R1+R2
R1(V B − 2V D,on)
V D,on + R1+R2
R1(V B − 2V D,on)
V in (V)
V B−2V D,on
R1
Slope = 1/ (R1 + R2)
I R1I D1
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(c)
I R1 =
0 V in < V D,on + V BV in−V D,on−V B
R1V in > V D,on + V B
I D1 =
0 V in < V D,on + V B
0 V D,on + V B < V in 2V D,on + V B
V D,on + V B 2V D,on + V B
V in (V)
Slope = 1/R1
Slope = 1/R2
I R1I D1
(d)
I R1 =
0 V in < V D,onV in−V D,onR1+R2
V D,on < V in < V D,on + R1+R2
R2(V B + V D,on)
V in−2V D,on−V BR1
V in > V D,on + R1+R2
R2(V B + V D,on)
I D1 =
0 V in < V D,onV in−V D,onR1+R2
V D,on < V in < V D,on + R1+R2
R2(V B + V D,on)
V in−2V D,on−V BR1
V in > V D,on + R1+R2
R2(V B + V D,on)
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V D,on V D,on + R1+R2
R2(V B + V D,on)
V in (V)
V B+V D,on
R2 Slope = 1/ (R1 + R2)
Slope = 1/R2
I R1I D1
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3.31 (a)
I D1 = V in − V D,on
R1= 1.6 mA
rd1 = V T
I D1= 16.25 Ω
∆V out = R1rd + R1
∆V in = 98.40 mV
(b)
I D1 = I D2 = V in − 2V D,on
R1= 0.8 mA
rd1 = rd2 = V T
I D1= 32.5 Ω
∆V out = R1 + rd2
R1 + rd1 + rd2∆V in = 96.95 mV
(c)
I D1 = I D2 = V in − 2V D,on
R1= 0.8 mA
rd1 = rd2 = V T
I D1= 32.5 Ω
∆V out = rd2
rd1 + R1 + rd2∆V in = 3.05 mV
(d)
I D2 = V in − V D,on
R1−
V D,on
R2= 1.2 mA
rd2 = V T
I D2= 21.67 Ω
∆V out = R2 rd2
R1 + R2 rd2∆V in = 2.10 mV
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3.32 (a)
∆V out = ∆I inR1 = 100 mV
(b)
I D1 = I D2 = I in = 3 mArd1 = rd2 =
V T
I D1= 8.67 Ω
∆V out = ∆I in (R1 + rd2) = 100.867 mV
(c)
I D1 = I D2 = I in = 3 mA
rd1 = rd2 = V T
I D1= 8.67 Ω
∆V out = ∆I inrd2 = 0.867 mV
(d)
I D2 = I in − V D,on
R2= 2.6 mA
rd2 = V T
I D2= 10 Ω
∆V out = ∆I in (R2 rd2) = 0.995 mV
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3.34
π/ω 2π/ωt
−V p
0.5 V
V D,on + 0.5 V
V p − V D,on
V p
V in(t)
V out(t)
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3.35
π/ω 2π/ωt
−V p
0.5 V
−V D,on + 0.5 V
−V p + V D,on
V p
V in(t)
V out(t)
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3.36
V R ≈ V p − V D,on
RLC 1f in
V p = 3.5 V
RL = 100 Ω
C 1 = 1000 µ Ff in = 60 Hz
V R = 0.45 V
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3.37
V R = I L
C 1f in≤ 300 mV
f in = 60 Hz
I L = 0.5 A
C 1 ≥ I L
(300 mV) f in= 27.78 mF
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3.38 Shorting the input and output grounds of a full-wave rectifier shorts out the diode D4 from Fig. 3.38(b).Redrawing the modified circuit, we have:
+
V in
−D2
D3
RL
+
V out−
D1
On the positive half-cycle, D3 turns on and forms a half-wave rectifier along with RL (and C L, if included). On the negative half-cycle, D2 shorts the input (which could cause a dangerously largecurrent to flow) and the output remains at zero. Thus, the circuit behaves like a half-wave recifier.The plots of V out(t) are shown below.
π/ω 2π/ωt
−V 0
V D,on
V 0 − V D,on
V 0
V in(t) = V 0 sin(ωt)
V out(t) (without a load capacitor)V out(t) (with a load capacitor)
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3.39 Note that the waveforms for V D1 and V D2 are identical, as are the waveforms for V D3 and V D4.
π/ω 2π/ωt
−V 0
−V 0 + V D,on
−V 0 + 2V D,on
−2V D,on
−V D,on
V D,on
2V D,on
V 0 − 2V D,on
V 0
V in(t) = V 0 sin(ωt)
V out(t)
V D1(t), V D2(t)V D3(t), V D4(t)
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3.40 During the positive half-cycle, D2 and D3 will remain reverse-biased, causing V out to be zero asno current will flow through RL. During the negative half-cycle, D1 and D3 will short the input(potentially causing damage to the devices), and once again, no current will flow through RL (eventhough D2 will turn on, there will be no voltage drop across RL). Thus, V out always remains at zero,and the circuit fails to act as a rectifier.
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3.42 Shorting the negative terminals of V in and V out of a full-wave rectifier shorts out the diode D4 fromFig. 3.38(b). Redrawing the modified circuit, we have:
+
V in
−D2
D3
RL
+
V out−
D1
On the positive half-cycle, D3 turns on and forms a half-wave rectifier along with RL (and C L, if included). On the negative half-cycle, D2 shorts the input (which could cause a dangerously largecurrent to flow) and the output remains at zero. Thus, the circuit behaves like a half-wave recifier.The plots of V out(t) are shown below.
π/ω 2π/ωt
−V 0
V 0
V in(t) = V 0 sin(ωt)
V out(t) (without a load capacitor)V out(t) (with a load capacitor)
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3.44 (a) We know that when a capacitor is discharged by a constant current at a certain frequency, theripple voltage is given by I Cf in , where I is the constant current. In this case, we can calculate the
current as approximately V p−5V D,on
R1(since V p − 5V D,on is the voltage drop across R1, assuming
R1 carries a constant current). This gives us the following:
V R ≈ 1
2
V p − 5V D,on
RLC 1f in
V p = 5 V
RL = 1 kΩ
C 1 = 100 µ F
f in = 60 Hz
V R = 166.67 mV
(b) The bias current through the diodes is the same as the bias current through R1, which isV p−5V D,on
R1= 1 mA. Thus, we have:
rd = V T
I D
= 26 Ω
V R,load = 3rd
R1 + 3rdV R = 12.06 mV
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3.45
I D1 =
0 V in < V D,on + V B1V in−V D,on−V B1
R1V in > V D,on + V B1
I D2 = V in+V D,on+V B2
R1V in < −V D,on − V B2
0 V in > −V D,on − V B2
−V 0
0
V 0
V D,on + V B1
−V D,on − V B2
V i n ( t )
−π/ω 0 π/ωt
0
V 0−V B1−V D,on
R1
−V 0+V B1+V D,on
R1
I D 1
( t ) a n d
I D 2
( t )
V in(t)
I D1(t)
I D2(t)
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