Solutions manual for power electronics circuits devices and applications 4th edition by rashid ibsn

Page 1

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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