Power Electronics Circuits Devices and Applications 4th Edition Rashid Solutions Manual
November 17, 2018 | Author: a260625922 | Category: N/A
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Chapter 2 – Diodes Circuits Prob 2.1
−6
trr := 5⋅ 10 ( a)
di_dt
6
:=
80 80⋅ 10
Eq. (2.10)
2
QRR := 0.5⋅ di_dt⋅ trr ( b)
Q RR ⋅ 10
6
= 1 × 103
μC
Eq. (2-11)
IRR :=
2⋅Q RR ⋅ di_dt
IRR = 400
A
Prob 2.2
−6
trr := 5⋅ 10 ta
:=
t b
s
Ifall_rate
:=
6
800⋅ 10
A
trr 1 + SF
:=
SF⋅ ta
:=
SF
s
−6
ta
=
3.333 × 10
t b
=
1.66 .667 × 10
IRR := Ifall_rate⋅ ta
0 .5
−6 3
IRR = 2.667 × 10
Using Eq. (2-7), ( a)
QRR :=
1 2
QRR :=
⋅ ( Ifall_rate⋅ ta) ⋅ ( ta + b t )
1
⋅I ⋅ t + t 2 RR a b
(
)
QRR ⋅ 10
6
=
3
6.667 × 10
QRR ⋅ 10
6
=
μC 3
6.667 × 10
μC
Using Eq. (2-6), ( b)
IRR := Ifall_rate⋅ ta
3
IRR = 2.667 × 10
Chapter 2-Diodes Circuits Page #2-1
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Prob 2.3
−6
trr := 5⋅ 10 ta
:=
t b
QRR ( x)
:=
0. 5
trr 1 + SF
:=
m :=
SF
SF⋅ ta
1
⋅t ⋅t 2 a rr
m
−6
ta
=
3.333 333 × 10
t b
=
1.6 1.667 × 10
=
8.333 × 10
−6
− 12
:= m⋅ x Plot of the charge storage verus di/dt
0.008
e g 0.006 a r o Q t RR ( x) s e g r a h 0.004 C
0.002
0
2 .10
8
4 .10
8
6 .10 x di/dt
8
8 .10
8
1 .10
9
Chapter 2-Diodes Circuits Page #2-2
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IRR ( x)
:= ta⋅ x Plot of the charge storage verus di/dt 4000
3000 e g a r o t I RR ( x) s 2000 e g r a h C
1000
0 8 1 .10
2 .10
Prob 2.5 VD2
:=
VT
8
3 .10
4 .10
:=
25.8⋅ 10
8
8
5 .10
6 .10 x di/dt
8
8
7 .10
8 .10
8
9 .10
8
8
1 .10
9
−3
1.6
VD1
:=
ID2 := 1500
1.2
ID1
:=
100
Using Eq. (2-3), ( a)
( b)
η :=
x
:=
V D2 − V D1
η =
⎛ ID2 ⎞ V T⋅ ln ⎜ ⎟ ID1 ⎝ ⎠ V
5.725
D1
x
η ⋅ VT
Using Eq. (2-3),
⎛ ID1 ⎞ V T⋅ η ⋅ ln ⎜ ⎟ = 1.2 IS ⎝ ⎠
IS
:=
ID1 x
e
=
8.124
IS = 0.03
Chapter 2-Diodes Circuits Page #2-3
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Prob 2-7 VD1 := 2200 IS1
:=
VD2 := 2200
−3
20⋅ 10
IS2
:=
R 1
:=
3
100⋅ 10
−3
35⋅ 10
( a) IR1
:=
VD1
IR1 = 0.022
R 1 Using Eq. (2-13), ( b)
IR2
R 2
:= :=
IS1 + IR1 − IS2 VD2
IT
:=
I1
300 IT
:=
2
I2
:=
I1
V D1
:=
1.4
R 1
:=
R 2
I1
= 7 × 10− 3
=
3.143 143 × 10
R 2
IR2
Prob 2.11
IR2
VD
:=
:=
VD
:=
I1
=
150
I2
=
150
− VD1 I1
V D − VD2 I2
R 1
2.8
:=
VD2
=
R 2
5
2.3
−3
9.333 333 × 10
=
−3
3.33 .333 × 10
IT 2
Chapter 2-Diodes Circuits Page #2-4
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Prob 2-13
3
R 1
:=
50⋅ 10
Is1
:=
20⋅ 10
R 2
−3
:= 50⋅ 103
Is2
:=
VS
:= 10⋅ 103
−3
30⋅ 10
Using Eq. (2-14),
VS
VD2
:=
IS1 − IS2 + R 1 1
+
R 1
:=
V D1
Prob 2-15 I p
T
:=
1
f
:= 500
1
VD1
:=
t1
3
T⋅ 10
f
3
4.6 4.625 × 10
3
=
5.375 × 10
−6
100⋅ 10
=
2
t
⌠ 1 ⋅ ⎮ sin( 2⋅ π ⋅ f ⋅ t) d t IAVG := T ⌡0 I p
=
R 2
VS − VD2
500
:=
VD2
IAVG = 3.895
t
IRMS
:=
I p
1 1 ⌠ 2 ⋅ ⎮ sin( 2⋅ π ⋅ f ⋅ t) d t T ⌡0
I peak := I p
IRMS = 20.08 I peak = 500
Chapter 2-Diodes Circuits Page #2-5
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Prob 2-16 IRMS
:=
T
I p
:=
120
:=
f
500
1
3
T⋅ 10
f
100⋅ 10
=
IRMS
:=
−6
:=
t1
2
I p
t
⌠ 1 2 ⋅ ⎮ sin( 2⋅ π ⋅ f ⋅ t) d t T ⌡0
3
=
2.9 2.988 × 10
1
t
:=
IRMS
⌠ 1 2 ⋅ ⎮ ( sin( 2⋅ π ⋅ f ⋅ t) ) d t T ⌡0 1
I p
IRMS
120
t
⌠ 1 ⋅ ⎮ sin( 2⋅ π ⋅ f ⋅ t) d t IAVG := T ⌡0 I p
=
IAVG = 23.276
Prob 2-17 IAVG T
I p
:= :=
:=
100
f
:=
500
:=
t1
1
3
T⋅ 10
f
−6
100⋅ 10
=
2
IAVG
⎛ 1 ⌠ t1 ⋅⎮ ⎝ ⌡0 T
⎞ sin( 2⋅ π ⋅ f ⋅ t) d t
=
4
1.284 × 10
⎠
t
⌠ 1 IAVG := ⋅ ⎮ sin( 2⋅ π ⋅ f ⋅ t) d t T ⌡0 I p
I p
IAVG
=
100
t
IRMS
:=
I p
1 1 ⌠ 2 ⋅ ⎮ ( sin( 2⋅ π ⋅ f ⋅ t) ) d t T ⌡0
IRMS = 515.55
Chapter 2-Diodes Circuits Page #2-6
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Prob 2-18 t1
:=
t5 ( a)
−6
100⋅ 10
:=
200⋅ 10
f
:=
250
:= 1⋅10 − 3
t3
Ia
:=
:=
−6
400⋅ 10
150
I b
:=
( I p − Ia) ⋅
Ir1
:=
Ir2
:= Ia⋅
Ir3
f ⋅
I p
:=
300
π IAVG = 22.387 Ir1 = 16.771
2
f ⋅ t3
Ir2 = 47.434
(
)
f ⋅ t5 − t4
:=
2
Ir1
−6
800⋅ 10
( t2 − t1)
( t2 − t1)
:= I b⋅
Irms
:=
t4
100
:= Ia⋅ f ⋅ t3 + I b⋅ f ⋅ ( t5 − t4) + 2⋅ ( I p − Ia) ⋅ f ⋅
IAVG
( b)
−6
t2
Ir3 = 22.361
+ Ir22 + Ir32
Irms = 55.057
Prob 2-19 t1
:=
t5 ( a)
−6
100⋅ 10
−3 := 1⋅10 10 IAVG
( b)
−6
t2
:=
200⋅ 10
f
:=
250
Ia
t3
:=
:=
150
−6
400⋅ 10 I b
:=
:= Ia⋅ f ⋅ t3 + I b⋅ f ⋅ ( t5 − t4) + 2⋅ ( I p − Ia) ⋅ f ⋅
Ir1
:=
Ir2
:= Ia⋅
Ir3 Irms
(
I p − Ia
)⋅
f ⋅
2
−6
800⋅ 10
I p
:=
150
( t2 − t1) IAVG
π
=
20
Ir1
=
0
Ir2 = 47.434
)
:= I b⋅
f ⋅ t5 − t4
:=
2
Ir1
100
:=
( t2 − t1)
f ⋅ t3
(
t4
+ Ir22 + Ir32
Ir3 = 22.361 Irms = 52.44
Chapter 2-Diodes Circuits Page #2-7
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Prob 2-20 t1
−6
:=
t5
100⋅ 10
:=
200⋅ 10
f
:=
250
−3 := 1⋅10 10
Irms
:=
:= Ia⋅
f ⋅ t3
Ir3
:= I b⋅
f ⋅ t5 − t4
:=
( a) I p
:=
Irms
( b)
:=
:=
−6
400⋅ 10
150
I b
:=
t4
100
:=
−6
800⋅ 10
I p
:= 150
Ir2 = 47.434
(
2
Irms
Ir1
:=
IAVG
( t2 − t1)
+ Ia
Ir1 = 172.192 I p
=
3
1.69 × 10
2
( I p − Ia) ⋅ :=
Ir3 = 22.361
)
− Ir22 − Ir32
f ⋅
Ir1
Ia
t3
180
Ir2
Ir1
−6
t2
2
Ir1
f ⋅
( t2 − t1) Ir1 = 172.192
2
+ Ir22 + Ir32
Irms
:= Ia⋅ f ⋅ t3 + I b⋅ f ⋅ ( t5 − t4) + 2⋅ ( I p − Ia) ⋅ f ⋅
=
180
( t2 − t1) π
IAVG = 44.512
Chapter 2-Diodes Circuits Page #2-8
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Prob 2-21 t1
−6
:=
t5
100⋅ 10
:= 1⋅ 10− 3
IAVG ( a)
:=
:=
200⋅ 10
f
:=
250
:= Ia⋅ f ⋅ t3
Iav2
:= I b⋅ f ⋅ ( t5 − t4)
Iav3
:=
IAVG
:=
:=
150
−6
400⋅ 10 I b
t4
+ Ia
⎡ ( )⎤ 2f ⋅ π ⎦ ⎣ t2 − t1
I p
:= Ia⋅ f ⋅ t3 + I b⋅ f ⋅ ( t5 − t4) + 2⋅ ( I p − Ia) ⋅ f ⋅
( b)
( I p − Ia) ⋅
:=
Ir2
:= Ia⋅
Ir3 Irms
−6
800⋅ 10
I p
:=
150
=
Iav1
=
15
Iav2
=
5
Iav3
=
160
4
1.02 × 10
( t2 − t1) π IAVG
Ir1
:=
:= 100
IAVG − Iav1 − Iav2 Iav3
:=
Ia
t3
180
Iav1
I p
−6
t2
f ⋅
2
Ir1
=
3
1.124 × 10
Ir2 = 47.434
)
:= I b⋅
f ⋅ t5 − t4
:=
2
Ir1
180
( t2 − t1)
f ⋅ t3
(
=
+ Ir22 + Ir32
Ir3 = 22.361 Irms
=
3
1.1 1.125 × 10
Chapter 2-Diodes Circuits Page #2-9
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Prob 2-22 VS
:=
220
R
:=
4. 7
C
−6 := 10 1 0⋅ 10
τ := R⋅ C
τ=
−6
t := 2⋅ 10
−5
4.7 × 10
Using Eq. (2-20),
( a)
( b)
I p
VS
:=
:=
VO W
I p = 46.809
R
:=
VS
0.5⋅ C⋅ VO
2
=
W
0.242
Using Eq. (2-21),
( c)
⎛ − t ⎞ τ Vc := VS⋅ ⎝ 1 − e ⎠
Prob 2-24
:=
VS
110
R
Vc = 9.165
:=
4 .7
L
:=
R
τ :=
τ =
L
−3
6.5⋅ 10
723.077
Using Eq. (2-25),
( a)
( b)
ID
:=
IO W
VS
:=
:=
ID = 23.404
R ID
0.5⋅ L⋅ IO
2
W
=
1.78
Using Eq. (2-27),
( c)
di
:=
VS
di
L
=
4
1.692 × 10
Chapter 2-Diodes Circuits Page #2-10
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Prob 2-25
:= 220
VS
R
:=
4.7
L
R
τ :=
:=
τ =
L
−3
6.5⋅ 10
723.077
Using Eq. (2-25),
( a)
ID
( b)
:=
IO W
VS
:=
:=
ID = 46.809
R ID
0.5⋅ L⋅ IO
2
=
W
7.121
Using Eq. (2-27),
VS
di :=
( c)
di
L
=
4
3.385 × 10
Prob 2-29 VS
:=
110
C
:=
−6
10⋅ 10
L
:=
−6
50⋅ 10
Using Eq. (2-32),
( a)
( b)
I p
:=
t1
V S⋅
:= π ⋅
C L L⋅ C
I p = 49.193 t1
=
−5
7.025 × 10
Using Eq. (2-35),
( c)
VC
:= 2⋅ VS
VC
=
220
Chapter 2-Diodes Circuits Page #2-11
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Example 2.31 L
:= 4⋅ 10− 3
( a)
C
:=
R
−6 := 0. 0 .05⋅ 10
α :=
160
Vs
R
:=
220
α = 2 × 104
2⋅ L
Using Eq. (2-41),
1
ω o :=
ωo =
L⋅ C 2
ω r := ω o − α
( b)
t1 vc ( t)
ω r =
Vs
:=
A2
2
4
6.782 782 × 10
A2 = 0.811
ω r ⋅ L :=
4
7.071 × 10
π ω r
6
t1⋅ 10
=
μs
46.32
:= e− α ⋅ t⋅ A2⋅ sin ( ω r ⋅ t)
Probl 2-32 L
:= 2⋅ 10− 3
( a)
R
:=
C
Vs
R
4
ωo =
L⋅ C 2
220
α = 4 × 104
2⋅ L
1
:=
3.162 × 10
2
:= −α + α − ω o
s2 at t = 0 at t = 0
−6
0. 0.5⋅ 10
α :=
160
ω o := s1
:=
2
α > ωo 4
s1 = −1.551 × 10 2
:= −α − α − ω o
4
s2 = −6.449 × 10
i
:=
0
0
di
:=
0
Vs L
≡ ≡
A1 + A 2 A1⋅ s1 + A2⋅ s2
Chapter 2-Diodes Circuits Page #2-12
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A1
:=
A2
( b)
Vs
A1 = 2.245
L⋅ s1 − s2
(
)
:= −A1
vc ( t)
A2 = −2.245
:= e− α ⋅ t⋅ A1⋅
( es1⋅ t − es2⋅ t)
Probl 2-33 L
:= 2⋅ 10− 3
( a)
C
:=
R
−6 := 0. 0 .05⋅ 10
α :=
16
t1 vc ( t)
Prob 2-34
Vs
:=
W
ω r =
4
9.992 992 × 10
π ω r
6
t1⋅ 10
=
31.441
μs
−3
t1
:= e− α ⋅ t⋅ A2⋅ sin ( ω r ⋅ t)
VS
IO
2
A2 = 1.101
ω r ⋅ L :=
220
α = 4 × 103
2⋅ L
L⋅ C 2
( b)
:=
ω o = 1 × 105
ω r := ω o − α :=
R
1
ω o :=
A2
Vs
:=
:=
110
VS L
L
:= 1⋅ 10
⋅ t1
0.5⋅ L⋅ IO
IO 2
=
W
=
:=
−6
100⋅ 10
11 7.121
Chapter 2-Diodes Circuits Page #2-13
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Example 2.36 Vs
:= :=
a
−6
Lm := 450⋅ 10
220 N2
=
a
N1
N 1
:=
N2
10
:=
100
t1
−6
:=
50⋅ 10
t1
:=
10
Using Eq. (2-52),
( a)
vD
:=
Vs⋅ ( 1 + a)
vD
=
3
2.42 × 10
Using Eq. (2-55),
( b)
Io
Vs
:=
⋅t Lm 1
Io = 24.444
Io Io_peak := a
( c)
Io_peak = 2.444
Using Eq. (2-58),
( d )
( e)
a⋅ Lm⋅ Io
t2
:=
t2⋅ 10
W
:= ⋅ Lm⋅ Io2
Vs 1
W
2
=
6
=
500
0.134
μs J
Example 2.37 Vs a
:= :=
220
Lm
N 2 a
N 1
:=
=
−6
250⋅ 10
N1
:=
10
N2
:=
10
−
50⋅ 10
1
Using Eq. (2-52),
( a)
vD
:=
Vs⋅ ( 1 + a)
vD
=
440
Io
=
44
Using Eq. (2-55),
( b)
Io
:=
Vs
⋅t Lm 1
Chapter 2-Diodes Circuits Page #2-14
Full file at https://testbankuniv.eu/Power-Electr https://testbankuniv.eu/Power-Electronics-Circuitsonics-Circuits-Devices-and-Appli Devices-and-Applications-4th-Editio cations-4th-Edition-Rashid-Solutionsn-Rashid-Solutions-Manual Manual
Full file at https://testbankuniv.eu/Power-Electr https://testbankuniv.eu/Power-Electronics-Circuitsonics-Circuits-Devices-and-Appli Devices-and-Applications-4th-Editio cations-4th-Edition-Rashid-Solutionsn-Rashid-Solutions-Manual Manual
Io
Io_peak := a
( c)
Io_peak = 44
Using Eq. (2-58), ( d )
( e)
a⋅L m⋅ Io
t2
:=
W
:= ⋅ Lm⋅ Io2
t2⋅ 10
Vs 1
W
2
=
6
=
μs
50
0.242
J
Example 2.38 Vs
:=
220
a
:=
( a) ( b)
Lm
:=
−6
250⋅ 10
N 2 a
N 1 vD
:=
Io
=
Vs
⋅t Lm 1 Io
( c)
Io_peak := a
( d )
a⋅L m⋅ Io
( e)
N2 := 1000
:=
10
vD
=
2.222 × 10
Io
=
44
t1
:=
−6
50⋅ 10
100
V s⋅ ( 1 + a)
:=
N 1
t2
:=
W
:= ⋅ Lm⋅ Io2
Vs
4
Io_peak = 0.44
1
t2⋅ 10
2
W
=
6
μs
= 5 × 103
0.242
J
Chapter 2-Diodes Circuits Page #2-15
Full file at https://testbankuniv.eu/Power-Electr https://testbankuniv.eu/Power-Electronics-Circuitsonics-Circuits-Devices-and-Appli Devices-and-Applications-4th-Editio cations-4th-Edition-Rashid-Solutionsn-Rashid-Solutions-Manual Manual
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