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Chapter 3-Diodes Rectifiers
Prob 3-1

Vm := 170

R := 5

f := 60

Using Eq
...
23

π

Vdc := 0
...
22
−3

R := 10

Lc := 0
...
(3-11)
Vdc := 0
...
82

R

Using Eq
...
65

Vo := Vdc − Vx

Vo = 107
...
(3-25)
6
⎛π⎞
Vdc := Vm⋅ ⋅ sin ⎜ ⎟
π
⎝6⎠
Prob 3-4

Vm := 170

f := 60

R := 5

Vdc = 162
...
(3-25)
6

⎛π⎞
⎟
⎝6⎠

Vdc := Vm⋅ ⋅ sin ⎜
π
Idc :=

Vdc = 162
...
47

R

Using Eq
...
84

Vo := Vdc − Vx

Vo = 156
...
5⋅ 10

Prob 3-5

Vs := 280

R := 5

2
3

Vm := 280⋅

f := 60
Vm = 228
...
(3-33)
Vdc := 1
...
18
R := 5

2
3

Vm := 280⋅

−3

f := 60

Lc := 0
...
62

Using Eq
...
18

Vdc := 1
...
64

R

Using Eq
...
61

Vo := Vdc − Vx

Prob 3-7

Vdc := 240

0
...
58

2

Idc :=
Ip :=

Vm = 377

Vm

Vs :=

Diodes

R := 10
Vdc

Vm :=

Vo = 364
...
7

R

Id :=

Idc

Id = 12

2

Chapter 3-Diodes Rectifiers
Page # 3 -2

Ip

IR :=
Transformer

IR = 18
...
58

2
Ip

Is :=

Is = 26
...
(3-1)
Pdc

Using Eq
...
6366⋅ Vm)
(
:=

1
TUF

Prob 3-8
Vdc := 750
Using Eq
...
76 × 10

R

Using Eq
...
107 × 10

Pac = 7
...
8105

Pac
= 1
...
39

1
...
6

2

Ip := Idc

Id :=

3

Ip = 6 × 10

Idc

IR :=

3

Id = 3 × 10

2
Ip

3

IR = 4
...
6

2
3

Is := Ip

Is = 6 × 10

VI := Vs⋅ Is

VI = 1
...
(3-1)

Pdc := Vdc⋅ Idc

Pdc = 4
...
(3-2)

Pac := 3Vs⋅ Is

Pac = 5
...
(3-8)

TUF

6

Pdc

TUF :=

1

6

TUF = 0
...
28

Prob 3-9
Vm := 170

f := 60

R := 10

ω := 2⋅ π ⋅ f

ω = 376
...
04

Guess
−3

L := 5⋅ 10
Given

Using Eq
...
481 = 0
⎝ R ⎠
3
Find ( L ) ⋅ 10 = 158
...
99
RF := 0
...
(3-67)
2

2
⎛ 6⋅ ω⋅ L ⎞
RF⋅ 2⋅ 1 + ⎜
= 0
⎟ −
⎝ R ⎠
35
3
Find ( L ) ⋅ 10 = 7
...
85

= 13
...
73

(

⋅ 2⋅ Vm⋅ cos ( α ) + 2⋅ E⋅ α − π ⋅ E

)

R = 1
...
98
2
PR := Irms ⋅ R
( d)

( e)

PR = 512
...
37

π
π

Idc := 10

Pdc = 200
ho = 1
η ⋅ 100 = 28
...
85

Prob 3-12
E := 12

Cap := 100

Vp
Vs :=
n
( a)

Vp := 120

Vs = 60

α := asin ⎛⎜

⎞
⎟
V
⎝ m⎠
E

α⋅

β := π − α

β⋅

δ := β − α

( b)

( c)
Irms :=

R :=

Vm :=

δ

1
2π ⋅ Idc

180
π

180

Idc := 5

2⋅ Vs

n := 2

f := 60

Vm = 84
...
13

= 171
...
74

π

(

⋅ 2⋅ Vm⋅ cos ( α ) + 2⋅ E⋅ α − π ⋅ E

)

R = 4
...
21
2
PR := Irms ⋅ R
( d)

PR = 287
...
67

Pdc

η ⋅ 100 = 17
...
85

Prob 3-13
−3

L := 4
...
99

Ω

Vs := 120

E := 20

T :=

1

3
T⋅ 10 = 16
...
34
θ = 0
...
02

iL ( t) :=

2⋅

Vs
Z

⎛ R ⎞ ⋅⎛ π ⎞ − ⎛ R ⎞ ⋅t
⎜ ⎟⎜ ⎟ ⎜ ⎟
E
L
ω
L
⋅ sin ( ω⋅ t − θ ) + ( Y) ⋅ e⎝ ⎠ ⎝ ⎠ ⋅ e ⎝ ⎠ −

R
iL ( 0) = 10
...
04

The rms output current can then be found by

Chapter 3-Diodes Rectifiers
Page # 3 -7

Irms := Ir⋅ 2

Irms = 24
...
18 × 10

( )

Imax := i1 t1
tx := t1 −

Imax = 34
...
15

( )

Imin := i1 tx

tx⋅ 1000 = 0
...
93

degrees

Imin = 6
...
6

degrees

R ⎛π⎞
⎤⎤
⎡
⎡
− ⎛⎜ ⎞⎟ ⋅ ⎜ ⎟
⎛ R ⎞ ⋅ ⎛ π − t⎞ ⎢
⎢
⎥⎥
L
⎝ ⎠ ⎝ω⎠
⎜ ⎟⎜
⎟
Vs⋅ 2
E
L ⎠ ⎝ ω ⎠ ⎢1 + e
⎝
⎢
i1 ( t) :=
⋅ sin ( ω⋅ t − θ ) + sin ( θ ) ⋅ e
⋅
− 1⎥ ⎥ −
Z
R ⎛π⎞
⎢
⎢
⎥⎥ R
− ⎛⎜ ⎞⎟ ⋅ ⎜ ⎟
⎢
⎢
⎥⎥
⎝L⎠ ⎝ω⎠
⎣
⎣1 − e
⎦⎦

Chapter 3-Diodes Rectifiers
Page # 3 -8

ta := 0 ,
...
2π

( )

i1 ta

40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0

0

9 18 27 36 45 54 63 72 81 90 99 108 117 126 135 144 153 162 171 180
180
π

Imin = 6
...
6

Imax = 34
...
93

degrees

Prob 3-14: Three-phase Diode Rectifier
Input Phase Voltage

VS := 120

Load Resistance

R := 5

Load Inductance

L := 2
...


Chapter 3-Diodes Rectifiers
Page # 3 -9

1

VS = 120
VAB :=

ω := 2⋅ π ⋅ fs

2
2
Z := ⎡⎣ R + ( ω⋅ L) ⎤⎦

ω = 376
...
09

T :=

1
fs

2

⎛ ω⋅ L ⎞
⎟
⎝ R ⎠

θ := atan ⎜
180
...
67

T = 0
...
71

VAB = 207
...


⎤
⎡
⎡ ⎛ 2
...
⋅ ω ⎠⎥ − E
⋅ ⎢ sin ( ωt − θ ) + ⎢
Z
R ⎛ π ⎞
⎢
⎢
⎥
⎥ R
− ⎛⎜ ⎞⎟ ⋅ ⎜
⎟
⎢
⎢
⎥
⎥
L
1 − e ⎝ ⎠ ⎝ 3
...


R ⎛ π ⎞⎤
⎡
− ⎛⎜ ⎞⎟ ⋅ ⎜
⎟
⎢ ⎛ 2⋅ π
π
⎞
⎛ − θ⎞ ⋅ e ⎝ L ⎠ ⎝ 3⋅ ω ⎠ ⎥
sin
−
θ
−
sin
⎟
⎜
⎟
⎥
2⋅ VAB ⎢ ⎜⎝ 3
3
⎠
⎝
⎠
⎢
⎥−E
⋅
Z
⎢
⎡− ⎛ R ⎞ ⋅ ⎛ π ⎞⎤
⎥ R
⎢
⎜
⎟
⎜
⎟
⎥
⎢
⎥
L
1 − e⎣ ⎝ ⎠ ⎝ 3⋅ ω ⎠⎦
⎣
⎦
I1 = 50
...
38

Chapter 3-Diodes Rectifiers
Page # 3 -10

I1 = 50
...


Calculations for RMS diode Current

1
T
⎛
⎞
⎜ ⌠3
⎟
2 ⎮
2 ⎟
⎜
Ir :=
⋅ ⎮ i ( t ) dt
⎜T ⎮ L
⎟
⎜ ⌡T
⎟
6
⎝
⎠

4
...
11

Calculations for RMS Output Current

Irms :=

3⋅ Ir

Irms = 52
...
99

Vm = 169
...
(3-67)

Ce :=

( b)

f := 60

⎞
⎟
2⋅ RF ⎠

⎛

1

1

⋅ ⎜1 +

4⋅ f ⋅ R ⎝

Vdc := Vm −

6
Ce⋅ 10 = 450
...
5

4⋅ f ⋅ R ⋅ Ce

Prob 3-16
Vs := 120
Vm :=

f := 60

R := 140

2⋅ Vs

ω := 2⋅ π ⋅ f

ω = 376
...
71

( a) Using Eq
...
32

Chapter 3-Diodes Rectifiers
Page # 3 -11

RF :=

5
100

( b)

Vdc := Vm −

Vm

Vdc = 164
...
(3-3)

Using Eq
...
(3-7)

2

( a)

( b)

( c)

η :=

0
...
45

2

0
...
5
FF :=
0
...
23

2
FF − 1

RF⋅ 100 = 121
...
(3-8)

Using Eq
...
318
0
...
5

( d)

TUF :=

( e)

PIV := Vm

(f)

CF := 2

( g)

0
...
707⋅ 0
...
61
PIV = 100

2

Chapter 3-Diodes Rectifiers
Page # 3 -12

PF = 0
...
99

2⋅ Vs

α := asin ( x)

Vm

ω := 2⋅ π ⋅ f

180⋅

α
π

= 16
...
(3-74)

⎛2

k :=

1 − ( x) 2 + ⎜

Ipk :=

Idc

⎝π

−

π⎞

⎟x

100⋅ k = 69
...
78

k

Using Eq
...
64

ω⋅ Ipk

Using Eq
...
91
Irms := kr⋅ Ipk

Irms = 23
...
87
kn := 23
...
25
k = 21
...
54
kn1 := 15
...
58

krn := 31
...
06

100

α := α n +

x = 61
...
84

⎡( krn1 − krn) ⋅ ( k − kn)⎤
⎢
⎥
kn1 − kn
⎣
⎦

kr⋅ Ipk

100

kr = 29
...
69

Chapter 3-Diodes Rectifiers
Page # 3 -15

Problem 3-20
Let t1 and t2 be the charging and discharging time of capacitor
...

t1 + t2 = T/2
...

The peak-to-peak ripple voltage is

t / RC
t / RC
v  v (t  t )  v (t  t )  V  V e 2
 V [1  e 2
]
r o
1
o
2
m m
m
Since, e x  1 - x, vr = Vm (1 - 1 + t2/RC) = Vm t2/RC = Vm/(2fRC)
Thus, the rms value of the output voltage harmonics is

V
v
m
V  r 
ac 2 2 4 2 f R C

Chapter 3- Diode Rectifiers

Page # 3-16

Prob 3-21
Vs := 120
Vm :=

f := 60

2⋅ Vs

Ce :=

R := 10

Vm = 169
...
84

ω = 376
...
808

35⋅ RF

Le :=

a+1

(6⋅ ω)

2

3
Le⋅ 10 = 1
...
(3-37) gives the output voltage as

2
2


v (t )  0
...
 
L
m  35
143

The load impedance, Z  R  j (n L)  R 2  (n L)2 

n

and   tan 1(n L / R)

n

and the load current is

i (t )  I 
L
dc
where

I dc 

0
...
 

Vdc 0
...
9549Vm )2  2 
(0
...

2[ R 2  (6 L)2 ]  35  2[ R 2  (12 L)2 ]  143 
2
ac

Considering only the lowest order harmonic (n = 6) and neglecting others,

I ac 

0
...
02
2
I dc
2  1  (6 L / R)  35 

0
...
022 [1 + (6 x 377 L/200)2] or L = 11
...
 
s
  1
3
5

The rms value of the fundamental current is
I1 = 2Ia/(2)
The rms current is Is = Ia/2
...
9 and HF  (I /I )2 - 1 = 0
...

s 1
(c) For the rectifier input (or secondary) side,
Chapter 3- Diode Rectifiers

Page # 3-19

ao/2 = Ia/2

1
a  0 I cos(n ) d  0
n 
a

I
1
b  0 I sin(n ) d  a (1  cos n )
n 
a
n
n = tan-1 (an/bn) = 0
Cn = (an2 + bn2) and I1 = C1/2 = 2Ia/
and Is = Ia/2
PF = I1/Is = 2/ = 0
...
211
s 1

Problem 3-24
(a)

(b) For the primary (or supply) current: From Eq
...
 

I1 = 4Ia/(2)
The rms current is Is = Ia
...
9 and HF  (I /I )2 - 1 =
s 1
0
...

(c) For the rectifier input (or secondary) current:
ao/2 = Ia/2

1
a  0 I cos(n ) d  0
n 
a

I
1
b  0 I sin(n ) d  a (1  cos n )
n 
a
n
n = tan-1 (an/bn) = 0
Cn = (an2 + bn2) and I1 = C1/2 = 2Ia/
and Is = Ia/2
PF = I1/Is = 2/ = 0
...
211
s 1
Problem 3-25

(a)

Chapter 3- Diode Rectifiers

Page # 3-21

(b) For the primary (or secondary) phase (or line) current:
ao/2 = 0

1  2Ia
1   Ia
a  
cos(n ) d   
cos(n ) d
n 
3

3
2I
n
n
 a cos
sin
n
2
3
2  /6

5 /6
/6

5 /6

1  2I a
1   Ia
b  
sin(n ) d   
sin(n ) d
n 
3

3
2I
n
n
 a sin
sin
n
2
3
2  /6

5 /6
/6

5 /6

2I
n
Cn  an2  bn2  a sin
n
3
n = tan-1 (an/bn) = tan-1(cot n/2)
(c) I1 = C1/2 = 3Ia/(2), 1 = 0 and Is = 2 Ia/3
PF = I1/Is = 33/2 = 0
...
68
s 1

Problem 3-26

(a)

Chapter 3- Diode Rectifiers

Page # 3-22

(b) For the primary line current:
ao/2 = 0

1  Ia
1   Ia
a  
cos(n ) d   
cos(n ) d
n 

3
3
2  /6

5 /6
/6



3 /2

2I a
n
2n 
sin
(1  cos
)
6
3
3 n

1  Ia
1   Ia
b  
sin(n ) d   
sin(n ) d
n 

3
3
2  /6

5 /6
/6



3 /2

2I a
n
2n 
cos
(1  cos
)
6
3
3 n

2I a
2n 
Cn  an2  bn2 
(1  cos
)
3
3 n
n = tan-1 (an/bn) = tan-1(-tan n/6) = -n/6
I1 = C1/2 = 3Ia/(2), 1 = -/6 and Is = 2 Ia/3
(c) For the secondary (or primary) phase current,
ao/2 = 0

1  2I a
1   Ia
a  
cos(n ) d   
cos(n ) d
n 

3 3
3 3
2I a
n
n

cos
sin
2
3
3 n
2  /6

5 /6
/6

5 /6

1  2I a
1   Ia
b  
sin(n ) d   
sin(n ) d
n 

3 3
3 3
2I a
n
n

sin
sin
2
3
3 n
2  /6

5 /6
/6

5 /6

2I
n
Cn  an2  bn2  a sin
3 n
3
n = tan-1 (an/bn) = tan-1(cot n/2) = n/2
I1 = C1/2 = Ia/(2), 1 = 0 and Is = 2 Ia/(33)
PF = I1/Is = 33/2 = 0
...
68
s 1

Problem 3-27
Chapter 3- Diode Rectifiers

Page # 3-23

(a)

(b) For the primary line current:
ao/2 = 0

I


 
a  a   cos(n ) d   cos(n ) d    cos(n ) d 
n 
2I
n
7n 
  a sin
cos
n
2
6
2 /3

2  /3

5 /3

/3

2

I


 
b  a   sin(n ) d   sin(n ) d    sin(n ) d 
n 
I 
n
7n  
 a 1  cos n  2sin
cos

n 
2
6 
2 /3
/3

5 /3

2  /3

2

For n = 1, C1 = (a12 + b12) = 23Ia/
1 = tan-1 (a1/b1) = tan-1(1/3) = /6
I1 = C1/2 = 2 3Ia/, and Is = Ia (2/3)
(c) For the secondary or primary phase current,
ao/2 = 0

Chapter 3- Diode Rectifiers

Page # 3-24

I
 / 3 cos(n ) d   0
a  a  2/ 3/ 3 cos(n ) d  45
/3
n  

I
 / 3 sin(n ) d 
b  a  2/ 3/ 3 sin(n ) d  45

/3
n  
4I
n
n
 a sin
sin
n
2
6
Cn = bn and n = 0
C1 = 2 Ia/, I1 = C1/2 = 2Ia/(2 ), 1 = 0 and Is = Ia/3
PF = I1/Is = 2 3/ = 0
...
803
s 1

Problem 3-28
(a)

(b) For the primary (or secondary) phase (or line) current:
ao/2 = 0

2I 
4I
n
n
a  a  cos(n ) d  a sin
cos
n 
n
3
2
5 /6
/6

Chapter 3- Diode Rectifiers

Page # 3-25

2I 
4I
n
n
b  a  sin(n ) d  a sin
sin
n 
n
3
2
5 /6
/6

4I
n
and n = tan-1 (an/bn) = tan-1(cot n/2)
C  a sin
n n
3
(c) C1 = 23Ia/ and 1 = 0
I1 = C1/2 = 2 3Ia/, and Is = Ia (2/3)
PF = I1/Is = 3/ = 0
...
3108
s 1

Problem 3-29
(a)

(b) For the primary line current,
ao/2 = 0

2I



a  a   cos(n ) d   2cos(n ) d    cos(n ) d 
n
3
8I a
2n 
n
n

cos
sin
cos
3
3
6
3 n
/2

5 /6

7 /6

/6

/2

5 /6

Chapter 3- Diode Rectifiers

Page # 3-26

2I



b  a   sin(n ) d   2sin(n ) d    sin(n ) d 
n
3
8I a
2n 
n
n

sin
sin
cos
3
3
6
3 n
/2

5 /6

7 /6

/6

/2

5 /6

8I a
n
n
C 
sin
cos
n
3
6
3 n
n = tan-1 (an/bn) = tan-1(cot(2n/3))
C1 = 23Ia/ and I1 = C1/2 = 2 3Ia/, 1 = tan-1 (-1/3) = -/6
(c) For the primary (or secondary) phase current,
ao/2 = 0

2I 
4I
n
n
a  a  cos(n ) d  a sin
cos
n 
n
3
2
5 /6
/6

2I 
4I
n
n
b  a  sin(n ) d  a sin
sin
n 
n
3
2
5 /6
/6

4I
n
and n = tan-1 (an/bn) = tan-1(cot n/2)
C  a sin
n n
3
I1 = C1/2 = 2 3Ia/, and Is = Ia (2/3)
PF = I1/Is = 3/ = 0
...
3108
s 1

Chapter 3- Diode Rectifiers

Page # 3-27

Problem 3-30
(a)

(b) For the primary (or secondary) line current,
ao/2 = 0

2I 
4I
n
n
a  a  cos(n ) d  a sin
cos
n 
n
3
2
5 /6
/6

2I 
4I
n
n
b  a  sin(n ) d  a sin
sin
n 
n
3
2
5 /6
/6

4I
n
C  a sin
and n = tan-1 (an/bn) = tan-1(cot n/2)
n n
3
I1 = C1/2 = 2 3Ia/, and Is = Ia (2/3)
(c) For the primary (or secondary) phase current,
ao/2 = 0

2I 


a  a   cos(n ) d   2cos(n ) d    cos(n ) d 
n 3
8I
2n 
n
n
 a cos
sin
cos
3 n
3
3
6
/2

5 /6

7 /6

/6

/2

5 /6

Chapter 3- Diode Rectifiers

Page # 3-28

2I 


b  a   sin(n ) d   2sin(n ) d    sin(n ) d 
n 3
8I
2n 
n
n
 a sin
sin
cos
3 n
3
3
6
/2

5 /6

7 /6

/6

/2

5 /6

8I
n
n
C  a sin
cos
n 3 n
3
6
n = tan-1 (an/bn) = tan-1(cot(2n/3))
C1 = 23Ia/ and I1 = C1/2 = 2 3Ia/, 1 = tan-1 (-1/3) = -/6
Is = 2Ia/3, PF = I1/Is = 3/ = 0
...
3108

Chapter 3- Diode Rectifiers

Page # 3-29

Prob 3-31
q := 12

Vdc := 240

IT := 300

R :=

Vdc

Using Eq
...
764
q
⎛π⎞
Vdc := Vm⋅ ⋅ sin ⎜ ⎟
q
π

⋅ sin ⎜

π
Idc :=

R = 0
...
(3-26)

Irms :=

⎛π

q

Vrms := Vm⋅

2π

⋅⎜

⎝q

Vrms

( b)

( c)

( d)

Vrms = 240
...
016

R

Pdc := Vdc⋅ Idc
η :=

1
⎛ 2⋅ π ⎞ ⎞
⋅ sin ⎜
⎟⎟
2
⎝ q ⎠⎠

+

Pac := Vrms⋅ Irms

Pdc

4

Pdc = 7
...
989

Pac
FF :=

Vrms
FF ⋅ 100 = 100
...
028

Vm
Vs = 171
...
(3-27)
Is :=

Vm
R

VA :=

⋅

q

2

1 ⎛π 1
⎛ 2⋅ π ⎞⎞
⋅ ⎜ + ⋅ sin ⎜
⎟⎟
2⋅ π ⎝ q 2
⎝ q ⎠⎠
Vs⋅ Is

Is = 86
...
92 × 10

Chapter 3- Diode Rectifiers
Page # 6 -30

4

Pac = 7
...
8072

VA

PIV :=

( e)

TUF = 0
...
479

IT
q

Im :=

Id

1
π

Prob 3-32
Using Eq
...
455

f := 60

n := q

Vp = 168
...
(3-69)
V1 :=

( a)
( b)

−2
2
n −1

⎛ n⋅ π ⎞
⎟
⎝ q ⎠

⋅ cos ⎜

VD := Vp⋅ V1

V1 = 0
...
351

f1 := q⋅ f

f1 = 720

Chapter 3- Diode Rectifiers
Page # 6 -31

V



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