Solutions Manual for Power Electronics Circuits Devices and Applications 4th Edition by Rashid IBSN 9780133125900 Full download: http://downloadlink.org/p/solutions-manual-for-power-electronicscircuits-devices-and-applications-4th-edition-by-rashid-ibsn9780133125900/ Chapter 2 – Diodes Circ uits Prob 2.1 −6
6
di_dt := 80⋅10
trr := 5⋅ 1 0 (a)
Eq. (2.10)
QRR := 0.5⋅ di_dt⋅ trr 2 (b)
6
QRR⋅ 10 = 1 × 10
3
μC
Eq. (2-11)
IRR :=
2⋅ Q RR⋅ di_dt
IRR = 400
A
Prob 2.2 −6
trr := 5⋅ 10
Ifall_rate := 800⋅ 10
s
6
A
SF := 0.5
s
trr
ta := 1 + SF
ta = 3.333 × 10
tb := SF⋅ ta
tb = 1.667 × 10
−6 −6
IRR := Ifall_rate⋅ ta
3
IRR = 2.667 × 10
Using Eq. (2-7), (a)
QRR :=
1
(
QRR :=
)(
)
⋅ Ifall_rate⋅ ta ⋅ ta + tb
2
1
(
)
⋅I ⋅ t + t 2 RR a b
Using Eq. (2-6),
QRR⋅ 106 = 6.667 × 103 6
μC
QRR⋅ 10 = 6.667 × 10
3
μC
(b)
IRR := Ifall_rate⋅ ta
3
IRR = 2.667 × 10
Prob 2.3 −6
trr := 5⋅ 1 0
SF := 0.5
trr ta := 1 + SF
ta = 3.333 × 10
tb := SF⋅ ta
tb = 1.667 × 10
m :=
1
−6 −6
⋅ ta⋅ trr
m
8.333 =
2
12
10 ×
−
QRR (x) := m⋅ x Plot of the charge storage verus di/dt
Charge storage
0.008
0.006 QRR ( x) 0.004
0.002
0
8
2 .10
8
4 .10
6 .10 x di/dt
8
8
8 .10
9
1 .10
IRR (x) := ta⋅ x Plot of the charge storage verus di/dt 4000
Charge storage
3000
IRR ( x)
2000
1000
0 8 1 .10
8
2 .10
8
3 .10
4 .10
8
5 .10
8
8
6 .10 x di/dt
8
7 .10
8
8 .10
−3
Prob 2.5
VT := 25.8⋅ 10 VD1 := 1.2 VD2 := 1.6
ID2 := 1500
ID1 := 100
Using Eq. (2-3), (a)
η :=
x :=
(b)
VD2 −VD1
VD1
x = 8.124
η ⋅ VT
Using Eq. (2-3), IS :=
⎛ ID1⎞
VT⋅ η ⋅ ln ⎜
⎟ = 1.2
⎝ IS ⎠
η = 5.725
⎛ ID2⎞ VT⋅ ln⎜ ⎟ I D1 ⎝ ⎠
ID1 x
e
IS = 0.03
8
9 .10
9
1 .10
Prob 2-7 VD1 := 2200 IS1 := 20⋅ 10
VD2 := 2200 −3
IS2 := 35⋅ 10
R1 := 100⋅ 10
3
−3
(a) VD1
IR1 := R1 Usi ng Eq. (2-13), (b)
IR1 = 0.022
IR2 := IS1 + IR1 − IS2
R2 :=
VD2
IR2 = 7 × 10
−3
R2 = 3.143 × 105
IR2
Prob 2.11 IT := 300
VD := 2.8
IT I1 := 2
I1 = 150
I2 := I1
I2 = 150
VD1 := 1.4
R1 :=
R2 :=
VD2 := 2.3
VD −VD1 I1 VD −VD2
IT I1 := 2
I2
R1 = 9.333 × 10
−3
−3
R2 = 3.333 × 10
Prob 2-13 R1 := 50⋅ 10
3
R2 := 50⋅ 10
−3
3
VS := 10⋅ 10
3
−3
Is1 := 20⋅ 10
Is2 := 30⋅ 10
Usi ng Eq. (2-14),
VS IS1 − IS2 + R1 VD2 := 1 1 + R1 R2
3
VD2 = 4.625 × 10
VD1 = 5.375 × 103
VD1 := VS − VD2 Prob 2-15 Ip := 500
T :=
f := 500
1 f
t1 := 100⋅
−6
10
3
T⋅ 10 = 2 t
Ip ⌠ 1 IAVG := ⋅ ⎮ sin ( 2⋅ π ⋅ f ⋅ t) dt T ⌡0
IAVG = 3.895
t
IRMS := Ip Ipeak := Ip
1 1 ⌠ 2 ⋅ ⎮ sin ( 2⋅ π ⋅ f ⋅ t) dt T ⌡0
IRMS = 20.08 Ipeak = 500
Prob 2-16 IRMS := 120 T :=
−6
f := 500
t1 := 100⋅ 10
1 f
3
T⋅ 10 = 2 IRMS
Ip :=
Ip = 2.988 × 103
t
1 1 ⌠ 2 ⋅ ⎮ sin ( 2⋅ π ⋅ f ⋅ t) dt T ⌡0 t
IRMS := Ip
1 1 ⌠ 2 ⋅ ⎮ ( sin ( 2⋅ π ⋅ f ⋅ t) ) dt T ⌡0
IRMS = 120
t
Ip ⌠ 1 IAVG := ⋅ ⎮ sin ( 2⋅ π ⋅ f ⋅ t) dt T ⌡0
IAVG = 23.276
Prob 2-17 IAVG := 100 T :=
Ip :=
−6
f := 500
t1 := 100⋅10
1 f
3
T⋅ 10 = 2 IAVG
⎛ ⌡ t1 ⎜⎜ 1 ⌠ 0 T ⋅⎮ Ip
⎞ sin ( 2⋅ π ⋅ f ⋅ t) dt
4
⎟ ⎟⎠
t
⌠1 IAVG := ⋅ ⎮ sin ( 2⋅ π ⋅ f ⋅ t) dt T ⌡0 t1 1 ⌠ 2 IRMS := Ip ⋅ ⎮ ( sin ( 2⋅ π ⋅ f ⋅ t) ) dt T ⌡
⎝
Ip = 1.284 × 10
0
IAVG = 100
IRMS = 515.55
Prob 2-18 t1 := 100⋅ 10 − 6
t2 := 200⋅ 10
−6
t3 := 400⋅ 10
−6
t4 := 800⋅ 10
−6
−3
t5 := 1⋅ 10
f := 250
Ia := 150
Ib := 100
(a) I AVG := Ia⋅ f ⋅ t3 + Ib⋅ f ⋅ t5 − t4 + 2⋅ Ip − Ia ⋅ f ⋅
(
(
)
(
)
( t2 − t1) π
)
IAVG = 22.387 Ir1 = 16.771
Ir1 := Ip − Ia ⋅ (b)
Ip := 300
( t2 − t1) f⋅ 2 Ir2 := Ia⋅ f ⋅ t3
Ir2 = 47.434
(
)
Ir3 := Ib⋅ f ⋅ t5 − t4 Irms :=
2
Ir3 = 22.361
2
2
Ir1 + Ir2 + Ir3
Irms = 55.057
Prob 2-19 t1 := 100⋅ 10 − 6
t2 := 200⋅ 10
−6
t3 := 400⋅ 10
−6
t4 := 800⋅ 10
−6
−3
t5 := 1⋅ 10
f := 250
Ia := 150
Ib := 100
(a) I AVG := Ia⋅ f ⋅ t3 + Ib⋅ f ⋅ t5 − t4 + 2⋅ Ip − Ia ⋅ f ⋅
(
(b)
(
)
Ir1 := Ip − Ia ⋅
f⋅
)
2
π
IAVG = 20
Ir1 = 0 Ir2 = 47.434
(
)
Ir3 := Ib⋅ f ⋅ t5 − t4 2
)
( t2 − t1)
( t2 − t1)
Ir2 := Ia⋅ f ⋅ t3
Irms :=
(
Ip := 150
2
Ir3 = 22.361 2
Ir1 + Ir2 + Ir3
Irms = 52.44
Prob 2-20 t1 := 100⋅ 10 − 6
t2 := 200⋅ 10
−6
t3 := 400⋅ 10
−6
t4 := 800⋅ 10
−6
−3
t5 := 1⋅ 10
f := 250
Ia := 150
Ib := 100
Ip := 150
Irms := 180 Ir2 = 47.434
Ir2 := Ia⋅ f ⋅ t3
(
Ir3 = 22.361
)
Ir3 := Ib⋅ f ⋅ t5 − t4 Ir1 :=
2
2
Ir1
(a) Ip := f⋅
(
Irms :=
Ir1 = 172.192 Ip = 1.69 × 10
+ Ia
( t2 − t1)
3
2
)
Ir1 := Ip − Ia ⋅
(b)
2
Irms − Ir2 − Ir3
f⋅
( t2 − t1) 2
Ir1 = 172.192
Ir12 + Ir22 + Ir32
(
Irms = 180
)
(
)
IAVG := Ia⋅ f ⋅ t3 + Ib⋅ f ⋅ t5 − t4 + 2⋅ Ip − Ia ⋅ f ⋅
( t2 − t1) π IAVG = 44.512
Prob 2-21 t1 := 100⋅ 10 − 6
t2 := 200⋅ 10
−6
t3 := 400⋅ 10
−6
t4 := 800⋅ 10
−6
−3
t5 := 1⋅ 10
f := 250
Ia := 150
Ib := 100
Ip := 150
IAVG := 180 (a)
Iav1 := Ia⋅ f ⋅ t3
Iav1 = 15
(
)
Iav2 := Ib⋅ f ⋅ t5 − t4
Iav2 = 5
Iav3 := IAVG − Iav1 − Iav2
Ip :=
⎡ ⎢ 2f ⋅ ⎣
Iav3 = 160
Iav3 + Ia t2 − t1 ⎤
(
π
)⎥
Ip = 1.02 × 10
⎦
(
)
(
)
IAVG := Ia⋅ f ⋅ t3 + Ib⋅ f ⋅ t5 − t4 + 2⋅ Ip − Ia ⋅ f ⋅
4
( t2 − t1) π IAVG = 180
(b)
(
)
Ir1 := Ip − Ia ⋅ f ⋅
( t2 − t1) 2
Ir1 = 1.124 × 10
Ir2 := Ia⋅ f ⋅ t3
Ir2 = 47.434
(
)
Ir3 := Ib⋅ f ⋅ t5 − t4 Irms :=
2
3
2
Ir1 + Ir2 + Ir3
Ir3 = 22.361 2
Irms = 1.125 × 10
3
Prob 2-22 VS := 220
R := 4.7
−6
C := 10⋅ 1 5 τ = 4.7 × 10 0 −
t := 2⋅ 1
τ := R⋅ C Usi ng Eq. (2-20), (a) (b)
VS Ip := R
Ip = 46.809
VO := VS W := 0.5⋅ C⋅ VO
2
W = 0.242
Usi ng Eq. (2-21),
(c)
− t⎞ ⎛ ⎜ τ ⎟ V c := V S⋅ ⎝ 1 − e ⎠
Prob 2-24 R := 4.7
VS := 110 τ :=
R L
Vc = 9.165 −3
L := 6.5⋅ 10
τ = 723.077
Usi ng Eq. (2-25),
(a) (b)
VS ID := R
ID = 23.404
IO := ID W := 0.5⋅ L⋅ IO
2
W = 1.78
Usi ng Eq. (2-27),
(c)
di :=
VS L
di = 1.692 × 104
−6
0
Prob 2-25
−3
R := 4.7
VS := 220
L := 6.5⋅ 10
R L Usi ng Eq. (2-25), τ :=
τ = 723.077
VS ID := R
(a)
ID = 46.809
IO := ID
(b)
W := 0.5⋅ L⋅ IO
2
W = 7.121
Usi ng Eq. (2-27),
di :=
(c)
VS
di = 3.385 × 10
L
4
Prob 2-29 VS := 110 Usi ng Eq. (2-32),
(a)
(b)
C Ip := VS⋅ L t1 := π ⋅ L⋅ C
−6
C := 10⋅ 1 0
L := 50⋅ 1
−
0 Ip = 49.193 t1 = 7.025 ×
Usi ng Eq. (2-35),
(c)
V C := 2⋅ VS
6
VC = 220
−5
10
Example 2.31 L
4 10 − := ⋅
(a)
Vs := 220
−6
3
C := 0.05⋅ 10
R := 160
R 2⋅ L
α :=
4
α = 2 × 10
Usi ng Eq. (2-41),
1
ω o :=
L⋅ C 2
ω r := A2 :=
4
ω o = 7.071 × 10
ωo − α
2
4
ω r = 6.782 × 10
Vs A2 = 0.811
ω r⋅ L
π t1 := ωr
(b)
vc (t) := e
t1⋅ 106 = 46.32
− α⋅t
μs
( )
⋅ A2⋅ sin ω r⋅ t
Probl 2-32 L
2 10 − := ⋅
(a)
−6
3
C := 0.5⋅ 10
R := 160
α :=
4
α = 4 × 10
4
ω o = 3.162 × 10
L⋅ C
s1 := −α + s2 := −α −
at t = 0
R 2⋅ L
1
ω o :=
at t = 0
Vs := 220
2
α − ωo 2
2
α − ωo
i := 0 di := 0
α > ωo 4
s1 = −1.551 × 10 2
4
s2 = −6.449 × 10 0 ≡ A1 + A2 ≡ A 1⋅ s1 + A2⋅ s2 V VLs
A1 :=
Vs
(
A1 = 2.245
)
L⋅ s1 − s2
A2 = −2.245
A2 := −A1
(b)
(
⋅ A1⋅ e
vc (t) := e
−e
s1⋅ t
− α⋅t
s2⋅ t −6
Probl 2-33 L
) Vs := 220
3
2 10 −
C := 0.05⋅ 10 := ⋅ (a) R := 16 α := R 2⋅ L
1 ω o :=
(b)
2
ωo − α
2
4
ω r = 9.992 × 10
Vs A2 = 1.101
ω r⋅ L
t1 :=
π
t1⋅ 106 = 31.441
ωr
vc (t) := e Prob 2-34
5
ω o = 1 × 10
L⋅ C
ω r := A2 :=
3
α = 4 × 10
− α⋅t
μs
( )
⋅ A2⋅ sin ω r⋅ t
VS := 110
L := 1⋅ 1
VS ⋅t IO := L 1 W := 0.5⋅ L⋅ IO
−
0
3
t1 := 100⋅
IO = 11 2
W = 7.121
−6
10
Example 2.36 −6
Vs := 220 a :=
Lm := 450⋅10
N2 := 100
N1 := 10
t1 := 50⋅
−6
10
N2
a = 10 N1 Usi ng Eq. (2-52), 3
vD = 2.42 × 10
vD := V s⋅ (1 + a)
(a)
Usi ng Eq. (2-55),
(b)
(c)
Io = 24.444
Vs
Io := ⋅t Lm 1 Io Io_peak := a
Io_peak = 2.444
Usi ng Eq. (2-58),
(d)
(e)
t2 := W :=
a⋅ Lm⋅ Io
6
t2⋅ 10 = 500 μs
Vs 1 2
2
⋅ Lm⋅ Io
W = 0.134
J
Example 2.37 Vs := 220 a :=
−6
Lm := 250⋅10
N1 := 10
N2 a=1
N1
Usi ng Eq. (2-52),
(a)
vD := V s⋅ (1 + a)
Usi ng Eq. (2-55),
(b)
Vs Io := Lm
⋅ t1
vD = 440 Io = 44
N2 := 10
t1 := 50⋅
−
10
Io Io_peak := a
(c)
Io_peak = 44
Using Eq. (2-58), (d)
a⋅ Lm⋅ Io
t2 :=
Vs 1 2 ⋅ Lm⋅ Io 2
W :=
(e)
6
t2⋅ 10 = 50
μs
W = 0.242
J
Example 2.38 Vs := 220 a := (a)
Lm := 250⋅ 10
t1 := 50⋅ 10
a = 100
N1
vD := V s⋅ (1 + a)
(c) (d)
a⋅ Lm⋅ Io
(e)
N2 := 1000
N1 := 10
N2
Vs ⋅t Io := Lm 1 Io Io_peak := a
(b)
−6
t2 := W :=
4
vD = 2.222 × 10 Io = 44 Io_peak = 0.44
Vs 1 2
⋅ Lm⋅ Io
6
2
μs
3
t2⋅ 10 = 5 × 10 W = 0.242
J
−6
Solutions Manual for Power Electronics Circuits Devices and Applications 4th Edition by Rashid IBSN 9780133125900 Full download: http://downloadlink.org/p/solutions-manual-for-power-electronicscircuits-devices-and-applications-4th-edition-by-rashid-ibsn9780133125900/
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