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Chapter 4Controlled Rectifier

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Overview

Introduction to phase angle control.

Single phase half wave full controlled

rectifier

Single phase full wave full controlled rectifierThree phase three pulse full controlled

rectifier

Three phase six pulse full controlled rectifier

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Learning Outcome

After completing this chapter, student will be able to :

1. Draw the types of phase controlled rectifier.

2. Calculate the Vo(dc)

, Vo(rms)

for the controlled rectifier.

3. Explain the operation of controlled rectifier.

4. Sketch the output waveform for phase controlledrectifier.

5. Calculate the performance parameter of controlled

rectifier.

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4

RECTIFIERS

Converts ac to dc voltage

To produce an output that is purely dc

Classified into half-wave and full-wave

Load that can be either purely resistive or

resistive-inductive Either controlled or uncontrolled

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9

Diodes

On and off states controlled by the power circuit

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Performance parameter

Ideal rectifier

Efficiency, rectification ratio = 100%

AC output voltage, Vo(ac)= 0Ripple factor, RF = 0

Harmonic factor, HF = THD = 0

Power factor, PF = 1

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11

HALF-WAVE RECTIFIERS

To conduct current in one direction and block

current in the other direction

Convert ac voltage into dc voltage

Connected to filter and regulator in a basic

power supply

Filtereliminates fluctuations in the rectifiedvoltage

Regulatormaintains a constant dc voltage

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Diode rectifier with R load

12

Single Phase Half WaveRectifier R load

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Single Phase Half Wave RectifierR

load

Peak reverse blocking voltage, PIV

The average output voltage, Vo(dc).

The average current voltage, Io(dc).

The rms output voltage, Vo(rms)

.

The rms current voltage, Io(rms).

mVPIV

2/

0mmo(dc) V0.318dttsinV

T1V T

m

2/12/

0

2

mo(rms) V5.0dtt)sin(VT

1V

T

R

V0.318I mo(dc)

R

V5.0I mo(rms)

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14


When the input voltage

is positive, the diode is

forward-biased andcurrent is conducted

When the input voltage

is negative, the diode isreverse-biased and

current is not conducted

Vave= Vp/Vp(out) = Vp(in) - 0.7 VPIV = Vp(in)

D1

R

Vs

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15


Half-wave rectifier

with transformer-

coupled input

voltageAllows voltage to

be stepped up or

stepped down

AC source is

electrically isolated

from the rectifier

Vsec= nVpriVp(out)= Vp(sec) - 0.7 VPIV = Vp(sec)

Vs

D1

R

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16

UNCONTROLLED HALF-WAVE

RECTIFIER

2sin210

2,

mmrmso VtdtVV

mmavgo

VttdVVV

0

sin2

1

R

VI rmsrms

Output waveform

VS

Vm

Vo

Vd

2 t

2

2

t

tR

VRIP rmsrms

22

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17


RECTIFIER Example

A circuit is connected as shown infigure to a 240V 50Hz supply.

Neglecting the diode voltage drop,determine the current waveform &

the mean load current for a load of apure resistor of 10

Vs

240V10

AI

I

RI

R

V

I

VVV

o

o

o

oo

rmsmo

8.10

10

2402

2402

2

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Single Phase Half Wave RectifierR-L

load Diode rectifier with R-L load

Vo(dc)is lower than case no L-load

LRs vvv

dt

di(t)

Li(t)R)sin(Vm t 18

R

L

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Single Phase Half Wave RectifierR-L

load




The rms output voltage, Vo(rms).


mVPIV

0

mmo(dc) )]cos(-[1

2Vt)d(tsin

2VV

19

2

)2sin(

2

VV mo(rms)

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)R.JEYAGOPI 19


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20

Single Phase Half WaveRectifier R L loadDiode rectifier with R-L load

Until certain time (VR(hence VL= Vs-VRis positive), the

current builds up and inductor stored energy increases.

At maximum of VR, Vs=VRhence, VL=0V.

Beyond this point, VLbecomes negative (means releasing stored

energy), and current begins to decrease.

After T= , the input, Vs becomes negative but current still positive

and diode is still conducts due to inductor stored energy. The load

current is present at certain period, but never for the entire period,

regardless of the inductor size.

This will results on reducing the average output voltage due to the

negative segment. The larger the Inductance, the larger negative

segment

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Single Phase Half Wave RectifierR-L load

Diode rectifier with R-L load From the response waveform, it is observed that the current does not

go to zero when the applied voltage changes it polarity. The current in the inductor forces the diode to keep conducting even

when it is reverse biased.

Such an operation in a practical circuit should be avoided at all costs. Note that the diode does not quit conduction at wt = but keep

conducting until wt = . It stops conduction only when the current in the diode goes to zero. The angle at which the diode stops conduction is called the

extinction angle.

The extinction angle depends upon the time constant of the RL circuit.

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22


RECTIFIER RESISTIVE-INDUCTIVE LOAD

Almost all dc load contains some

inductance

Current flow will begin directly,

the supply voltage goes positive,

but the presence of the

inductance will delay the current

change. The current is still

flowing at the end of the half

cycle, the diode remains on & the

load sees the negative supply

voltage until the current drops to

zero

D1

Vs

R

L

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23


RECTIFIERVs Vs

Vo

VR

VL

Vd

i i

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24


RECTIFIER

The KVL equation that describes

the current in the current for the

forward-biased ideal diode is:

dt

tdiLtRitVm sin

where tititi nf if(t) = force response is the steady- state sinusoidal current that

would exist in the circuit if the diodes are not present

in(t) = natural response is the transient that occurs when the load is

energized. It is the solution to the differential equation for the circuit

without the source or diode

t

t

eitZ

Vti

eZVti

tiL

R

dt

tdi

dt

tdi

tiL

R

dt

tdiLtRi

tZ

Vti

Z

V

Z

Vti

om

mn

m

f

mmf

sin

sin

0

0

sin

0

0)(

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25


RECTIFIER

ktL

R

ti

tdi

dtL

R

ti

tdi

tiL

R

dt

tdidt

tdi

L

Lti

L

R

dt

tdiLtRi

0

0

to

t

o

o

LR

LR

eiti

ei

ti

tL

R

i

ti

itL

Rti

ik

kL

Ri

tit

ktL

Rti

ln

0lnln

ln

00ln

0,0

ln

0

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26

UNCONTROLLED HALF-WAVE RECTIFIER

sinsin

sin0

0sin0

00,0

sin

0

Z

V

Z

Vi

iZ

V

eiZ

V

it

eitZ

Vti

tititi

eiti

R

L

mmo

om

om

om

nf

o

t

t

2

0

sinsin

sin

t

eZVt

ZVti

eitZ

Vti

t

t

mm

om

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27


A circuit shown in figure isconnected to a 240V 50Hzsupply neglecting the diodevoltage drop, determine themean load current for a load ofan inductance of 0.1H in series

with a 10resistor

D1

Vs

R

L

35.72sin29.10

35.7229.10

35.7297.32

041.339

041.3392240

35.7297.32

42.3110

1.050210

tti

I

I

V

ZVI

Z

jZ

jZ

LjRZ

f

f

f

m

mf

R.JEYAGOPI 27


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28


81.935.72sin29.10

81.9

35.72sin29.10

35.72sin29.100

35.720sin29.10035.72sin29.10

,0

0

tti

tititi

i

i

i

eiieitti

tititi

teiti

nf

o

o

o

o

o

nf

on

t

t

sR

L01.0

10

1.0

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29

HALF-WAVE RECTIFIER WITH FREEWHEELING

DIODE The freewheeling diode is

connected across R-L load togive a release path for thecurrent across diode D1

When the source is positive,D1 will be on and DFWwill beoff

When the source is negative,D1 will be off and DFWwill beon

Load voltage: same as sourcevoltage when source is positive,zero when source is negative

Load current: zero when the

circuit is first energized,reaches steady-state after a fewperiod depending on the timeconstant

D1

Vs

R

L

DFW

...6,4,2

020 cos

1

2sin

2 n

mmmtn

n

Vt

VVtv

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30

HALF-WAVE RECTIFIER WITH

FREEWHEELING DIODE

Thewaveform

for output

voltage,output

current,

andcurrent

across

diodes

Vo

iD1

iDFW

io

t

t

t

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Single Phase Half Wave RectifierR-C Load

31

In some applications in

which a constant output is

desirable, a series inductor

is replaced by a parallel

capacitor. The purpose of capacitor is

to reduce the variation in

the output voltage, making

it more like dc.

The resistance mayrepresent an external load,

while the capacitor is a filter

of rectifier circuit.

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32

Half-wave, R-C load

+

vs

_

+vo

_

iD

2 3 4

Vm

Vmax

vs

vo

Vmin

/2

iD

3 /2

a

D Vo

sinOFFisdiodewhen

ONisdiodehenw)sin(/

m

RCtm

o

VV

eV

tVv

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33

Ripple

fRC

VV

VVV

fRC

V

RCVV

RCe

eVeVVV

eVeVv

tVV

VVVVVVV

tV

mm

omo

mmo

RC

RCm

RCmmo

RCm

RCmo

m

mmmmo

22

:asgivenisvoltageloadaverageThe

2

211

:thatNote

1

:asedapproximatisvoltagerippleThe

)2(

:is2atevaluatedtageoutput volThe2.thensuch thatlargeisCand2,andIf

sin)2sin(

:isripplethediagram,toReffering.2atoccurstageoutput volMin.istageoutput volMax

2

22

2222

minmax

max

D

D

D

D

D

a

aa

aa

a

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34

Example

A half-wave rectifier has a 120V rms

source at 60Hz. The load is =500 Ohm,

C=100uF. Assume aand are

calculated as 48 and 93 degrees

respectively. Determine (a) Expression

for output voltage (b) peak-to peak ripple

VufRC

V

RCVV

VVVVVV

VVV

e

t

eV

ttVtv

VradV

radradVV

mmo

mmmmo

o

t

RCt

m

m

o

m

oo

m

7.5610050060

7.1692

:ionApproximatUsing

43sin)2sin(

:Using

:(b)Ripple

(OFF)5.169

(ON))sin(7.169

(OFF)sin

(ON))sin(7.169)sin()(

:tageOutput vol(a);5.169)62.1sin(7.169sin

843.048;62.193;7.1692120

minma x

)85.18/(62.1

/

D

D

D

aa

a

2 3 4

Vm

Vmax

vs

vo

Vmin

/2

iD

3 /2

a

DVo

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Half Wave Rectifier Disadvantages

1. High ripple factor,

2. Low rectification efficiency,

3. Low transformer utilization factor, and,4. DC saturation of transformer secondary

winding.

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36

FULL-WAVE RECTIFIERS

Allows unidirectional (one-way)current through the load during the

entire 360 input cycle

Two types of full-wave rectifier:

(i) center-tapped (transformer)

rectifier, and(ii) bridge rectifier

Vave

= 2Vp/

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37

UNCONTROLLED FULL-WAVE RECTIFIERSCENTER

TAPPED

Center-tapped rectifier

uses two diodes connected

to the secondary of a center-

tapped transformer

When the input voltage ispositive, D

1is forward-biased

and current is conducted

through load

When input voltage is

negative, D2is forward-biased and current is

conducted through load

Vsec

= nVpriVp(out)= (Vp(sec) - 1.4 V) / 2

PIV = 2Vp(out) + 0.7 V

R

L

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UNCONTROLLED FULL WAVE RECTIFIERS CENTER


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UNCONTROLLED FULL-WAVE RECTIFIERSCENTER

TAPPEDVS

Vo

VD1

VD2

is

io

iD1

iD2

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39

UNCONTROLLED FULL-WAVE RECTIFIERS - BRIDGE

Bridge rectifier

uses four diodes

When the input

voltage is positive,

D1and D2isforward-biased

When the input

voltage is negative,

D3and D

4is

forward-biased

Vp(out)

= Vp(sec)

- 1.4 VPIV = Vp(out) + 0.7 V

D1

D2

D3

D4R

L

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40

UNCONTROLLED FULL-WAVE RECTIFIERS - BRIDGE

VS

Vo

VD1,VD2

VD3,VD4

is

io

iD1,iD2

iD3,iD4

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41

UNCONTROLLED FULL-WAVE

RECTIFIERS

2

sin1

2

sin1

sin

sin

0

2

,

0

mmrmso

mo

mo

m

m

o

VdttVV

VV

ttdVV

tV

tVtV

RESISTIVE LOAD R-L LOAD

For 0t

For t2

n

nn

oo

mn

m

n

noo

Z

VI

R

VI

nn

VV

VV

tnVVtV

1

1

1

12

2

cos

0

,..4,2

a

R.JEYAGOPI 41

O O


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42

UNCONTROLLED FULL-WAVE

RECTIFIERS

Example:

Referring to the circuit shown, a 240V 50Hz

supply is fed to the primary winding of a

step-down transformer which has a 20:1

turn ratio. The output voltage from the

secondary winding is connected to a fullbridge rectifier circuit.

a) Calculate the peak and average values of the

output voltage if the diode is silicon

b) What is the PIV rating for each diode?

c) If a capacitor filter was added to the circuit,

calculate the value of capacitor to beinstalled to produce a load current of 10mA

and a ripple of less than 2%.

Vp(out) = Vp(sec) - 1.4 V

PIV = Vp(out) + 0.7 V

D1

D2

D3

D4R

L

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UNCONTROLLED FULL WAVE RECTIFIERS


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43

UNCONTROLLED FULL-WAVE RECTIFIERS

mFC

mAVR

CfR

VV

VVPIV

VV

V

VVV

VVN

N

V

VV

L

L

mo

po

po

aveo

speako

pp

s

s

peakp

12.1702.099150

97.16

99110

91.9

02.0

27.167.0

91.92

57.154.1

97.1641.33920

1

41.3392240

)(

)(

)(

)(

D

a)

b)

c)

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44

Three Phase Rectifier

D1&D6 D1&D2 D2&D3 D3&D4 D4&D5 D5&D6

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Three-phase rectifiers


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45

Three phase rectifiersD1

vo =vp vn

+

vo

_

vpn

vnn

io

D3

D2

D6

+ vcn -

n+ vbn -

+ van -

D5

D4

20 4

Vm

Vm

van vbn vcn

vn

vp

vo =vp - vn

3

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Three Phase Rectifiers

Three phase rectifier are commonly used in

industries to produce dc voltage and current

for large loads.

Single-phase supply circuits are adequate

below a few kilowatts. Three phase rectifier circuit

1. Half wave or three pulse rectifier

2. Full wave or six pulse rectifier3. Twelve pulse circuits

R.JEYAGOPI 47

Th Ph H lf W R ifi R


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Three Phase Half Wave RectifierR

load

The three pulse or half wave rectifier for R

load

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Three Phase Half Wave Rectifier

- R load

Waveform

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Three Phase Half Wave RectifierR

load






mV3PIV

50

mo(rms) V8407.0V

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)

m

6/5

6/mo(dc) V827.0dsinV

3/21V

**note: Vmis maximum per phase (line to

neutral) voltageR.JEYAGOPI 50

Si l Ph F ll W R tifi R


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Single Phase Full Wave RectifierR

load

The six pulse or full wave rectifier for R load

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52

Waveform

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53

(D1 & D4), (D3 & D6), and (D5 & D2) cannot conduct at the same time.

The output voltage across the load is one of the line-to-line voltages ofthe source.

There are six combinations of line-to-line voltages (three phases taken

two at a time). Considering one period of the source to be 3600

, atransition of the highest line-to-line voltage must take place every 3600/ 6= 600

Because of the six transitions that occur for each period of the sourcevoltage, the circuit is called a six- pulse rectifier.

The diodes conducts in pairs (6,1), (1,2), (2,3), (3,4), (4,5), (5,6),(6,1),.

Single Phase Full WaveRectifier R load

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54

At any instant barring the change-over periods when

current flow gets transferred from diode to another,

only one of the following pairs conducts at any time.

Single Phase Full WaveRectifier R load

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Full Controlled Rectifier

Single phase half wave rectifier

Single phase full wave rectifier

Three phase three pulse rectifierThree phase six pulse rectifier

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Silicon Controlled Rectifier: SCR

56R.JEYAGOPI 56

i l h lf ll d


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Single Phase Half Wave Controlled

RectifierR load

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RectifierR load






mVPIV

a

a

)cos(12

Vt)d(tsinV

2

1V mmo(dc)

58

2/1

m

2/1

2

mo(rms) 2

2sin1

2

V

t)d(t)sin(V2

1

V

a

a

a

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)

R.JEYAGOPI

CONTROLLED HALF-WAVE


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59


RECTIFIER

a

a

cos12

sin21

m

o

mo

VV

ttdVV

SCR

Vs RGatecontrol

PURE RESISTIVE LOAD

a

a

a

2

2sin1

2

sin2

1 2

mrms

mrms

VV

tdtVV

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60

CONTROLLED HALF-WAVE RECTIFIER

t

t

t

t

Vs Vs

Vo VSCR

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CONTROLLED HALF WAVE


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62


RECTIFIER

0

sinsin

coscos2

sin2

1

a

a

a

a

t

etVti

VV

ttdVV

m

mo

mo

For a t

Otherwise

R.JEYAGOPI

CONTROLLED HALF WAVE


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63


RECTIFIER

t

t

eitZ

Vti

eiti

tZ

Vti

Z

V

Z

Vi

tititi

om

on

mf

mmf

nf

sin

sin

0

a

a

a

a

a

a

a

aa

aa

eZ

Vi

e

Z

V

i

ZVei

eiZ

Vi

it

mo

m

o

mo

om

.sin

sin

sin

sin

0,

R.JEYAGOPI

CONTROLLED HALF WAVE RECTIFIER


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64


a

a

a

a

a

a

t

t

t

t

eZ

Vt

Z

Vti

eeZ

Vt

Z

Vti

tititi

tZ

V

ti

eeZ

Vti

eiti

mm

mm

nf

m

f

mn

on

sinsin

.sinsin

sin

.sin

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65


Example:

Determine the mean load voltage &current if the load is 10in serieswith an inductor of 0.1H & thefiring of the thyristor is delayed by90. The ac supply is 240V 50Hz &the thyristor voltage drop is to be

neglected.SCR

VsR

L

Gatecontrol

35.72sin295.10

35.72295.10

35.7297.32

041.339

35.7297.32

1.050210

41.3392402

tti

i

i

Z

jZLjRZ

V

Z

Vti

tititi

f

f

m

mf

nf

R.JEYAGOPI

t


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66

t

t

t

etti

tititi

R

L

eee

i

ei

ei

ei

it

eitti

eiti

nf

o

o

o

o

o

on

14.535.72sin295.10

142.3,01.0

14.512.312.312.3

12.3

35.7290sin295.10

35.7290sin295.100

090,90

35.72sin295.10

142.357.190

90

90

90

90

Must be multiplied with

to be changed into radian

x to change into radian

180

R

L

R.JEYAGOPI


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Rectifier

R-L load + FWD

67R.JEYAGOPI

CONTROLLED FULL-WAVE


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68


RECTIFIER

CENTER-TAPPEDTRANSFORMER

BRIDGE RECTIFIER

R

L

D1

D2

D3

D4R

L

R.JEYAGOPI


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Single Phase Full Wave Controlled

RectifierR load

69

Bridge circuit

R.JEYAGOPI

Si l Ph F ll W C t ll d


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RectifierR load






m2VPIV

a

a

cos1Vt)d(tsinV

1V mmo(dc)

70

2/1

m

2/1

2

mo(rms) 4

2sin

22

1Vt)d(t)sin(V

2

2V

a

a

a

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)

R.JEYAGOPI

CONTROLLED FULL-WAVE RECTIFIER


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71

CONTROLLED FULL-WAVE RECTIFIER

RESISTIVE LOAD

Vo

2 t

a

a

a

a

a

42sin

221

sin1

cos1

sin1

,

2

,

mrmso

mrmso

mo

mo

VV

tdtVV

VV

ttdVV

+

R.JEYAGOPI


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RectifierR-L load

72R.JEYAGOPI



73/99

RectifierR-L load

73R.JEYAGOPI

L


74/99

74

CONTROLLED FULL-

WAVE RECTIFIER

R

VI

III

VV

ttdVV

tnVVV

R

L

n

nrms

m

m

n

nno

0

0

...6,4,2

2

2

0

0

0

1

0

1

2

cos2

sin1

cos

tan,

a

aa

a

a

R-L LOAD (CONTINUOUSCURRENT)

......6,4,2

11sin

11sin2

1

1cos

1

1cos2

22

n

nn

nnVb

n

n

n

nVa

baV

mn

mn

nnn

aa

aa

R.JEYAGOPI



75/99

75

CONTROLLED FULL WAVE

RECTIFIER

a

a

a

a

a

t

etZ

Vti

VVtdtVV

m

mmmdco

sinsin

coscoscossin1

)(

a t

R-L LOAD (DISCONTINUOUS CURRENT)

For

R.JEYAGOPI 75


76/99

76

Example 2

The full wave controlled bridge rectifier has an ac

input of 120Vrms at 60Hz and a 20load

resistor. The delay angle is 40o. Determine

a) average and rms current in the loadb) power absorbed by load and source (volt-

ampere)

c) power factor

R.JEYAGOPI


77/99

77

Solution 2

rad698.040o a

V8.8240cos)7.169(2

Va) oo(dc)

V6.984

2sin

22

1VV

2/1

mo(rms)

a

a

A9.420

98.6Io(rms)

A14.42082.8Io(dc)

VA14.483IVS

W480.2RIPb)

o(rms)o(rms)

2

o(rms)

994.0S

PPFc)

V7.1690.707

VV s(rms)m

Given

R.JEYAGOPI


78/99

78

RECTIFIERS

Ripple voltage, VrVariation in the

capacitor voltage due

to the charging and

dischargingUndesirable the

smaller the better

Ripple factor, r

Indication of the

effectiveness of the

filter and defined as

1

2

dc

rms

V

Vr

R.JEYAGOPI


79/99

79

RECTIFIERS

half-wave rectifier full-wave rectifier

VC VC

Vr(pp)

R.JEYAGOPI


80/99

Controlled three-phase


81/99

81

Controlled three phase

+vo_

vpn

vnn

ioD3

D2

D6

+ vcn -

n

+ vbn -

+ van -

D5

D4

Vmvan vbn vcna

vo

a

a

a

cos3

)sin(3

1

:ascomputedbecanvoltageAverage

32

3

LLLLoV

tdtVV

R.JEYAGOPI

Three Phase Half Wave


82/99

82

Three Phase Half WaveControlled Rectifier R load

R.JEYAGOPI

Three Phase Half Wave Controlled


83/99


RectifierR load






mV3PIV

aa

6/m

mo(dc)6

cos12

3Vt)d(tsinV

2

3V

83

2/1

m

2/1

6/

2

mo(rms) 2

3

sin

8

1

424

5V3t)d(t)sin(V

2

3V

a

a

a

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)

** Vm is the peak phase voltageR.JEYAGOPI

Th Ph H lf W C ll d


84/99


RectifierR-L load

84R.JEYAGOPI


85/99

Three Phase Half Wave

Controlled RectifierR-Lload

85

Larger firing angle

Largest firing angle

R.JEYAGOPI



86/99


RectifierR-L load






mV3PIV

a

a

a

6/5

6/

mmo(dc) cos

2

V33t)d(tsinV

2

3V

86

2/1

m

2/16/5

6/

2

mo(rms) cos2

8

3

6

1V3t)d(t)sin(V

2

3V

a

a

a

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)


Th Ph F ll W C t ll d


87/99

Three Phase Full Wave Controlled

RectifierR-L load

87R.JEYAGOPI

Three phase full wave controlled rectifier : R


88/99

load

88R.JEYAGOPI

Three Phase Full Wave


89/99

Controlled RectifierR-L

load

89

Large firing

angle Small firingangle

R.JEYAGOPI



90/99


RectifierR load






mV3PIV

a

aa

2/

6/

mmo(dc) cos

V33t)d(

6tsinV3

3V

90

2/1

m

2/12/

6/

22

mo(rms) cos2

4

33

2

1V3t)d(

6

tsin3V3

V

a

a

a

R

VI

o(dc)

o(dc)

R

VI

o(rms)

o(rms)


C t ll d th h R l d


91/99

Controlled three-phase R load

Average line-to-line half-wave output voltageis given as

rms line-to-line half-wave output voltage

91

a

a 6

cos1

2

3sin

2

3

6

m

m

VtdtV

21

6

2 23

sin81

42453sin

23

a

a

a

mm VdttV

R.JEYAGOPI 91

Controlled three phase RL load


92/99

Controlled three-phase RL load

Average line-to-line half-wave output voltageis given as

rms line-to-line half-wave output voltage

92R.JEYAGOPI 92

Controlled three phase R load


93/99

Controlled three-phase R load

Average line-to-line full-wave output voltagecould be given as

rms line-to-line full-wave output voltage

93

a

a

a

cos

33

6sin3

3 2

6

m

m

V

tdtV

R.JEYAGOPI 93



94/99


Example:

A three phase half-wave rectifier is operated from a three phase

star-connected 208V 60Hz supply and load resistance 10. If it is

required to obtain an average output voltage 50% maximum output

voltage, calculate:

a) the delay angle,

b) the rms and average output currents

94

R.JEYAGOPI 94



95/99


95

AV

R

VI

AV

R

VI

VV

VV

VVV

VV

VVV

VVV

VV

rmso

rmso

dco

dco

rmso

mrmso

mdco

dco

mm

dco

Sm

S

919.9

10

19.99

566.710

66.75

19.991137.0383.169103.5089.0208.0383.169

3

180

)84.63(2sin8

1

4

180

84.63

24

5

383.16923sin8

1

424

5

3

84.63

84.9330

067.01933.0183.1693

266.75

6cos

6cos1

2

366.755.0

3.151

866.12

83.16930

6cos1

2

3

6cos1

2

3

83.1692

1.1203

208

,

,

,

,

3

,

,

,

,

,

a

a

a

a

a

a

a

R.JEYAGOPI 95

RECTIFIERS -


96/99

96

IMPORTANT INFORMATION & EQUATIONS

Average output voltage which is the DC

voltage component of three-phase rectifier is

much higher compared to single phase

Considers only one of the six segments of

output voltage

Rectification ratio=rmso

aveo

ac

dc

PP

PP

,

,

R.JEYAGOPI

V

:loadRwithrectifiereduncontrollwave-Half


97/99

97

RIP

R

VI

R

VI

2

VV

V

VV

2rm s

rm srm s

o

mrm s

maveo

e)(sinZ

V)-t(sin

Z

Vi(t)

e)(sinZV(t)i

)-t(sinZ

V

Z

V

Z

0V(t)i

:loadL-Rwithrectifiereduncontrollwave-Half

t-

mm

t

-mn

mmmf

R.JEYAGOPI

:loadRwithrectifiercontrolledwave-Half


98/99

98

]e)-(sin-)-t[sin(Z

Vi(t)

]cos-[cos

2

VV

:loadL-Rwithrectifiercontrolledwave-Half

t)-(

m

mo

a

a

a

a

a

a

2

2sin-1

2

VV

]cos1[

2

VV

mrms

mo

R.JEYAGOPI

2V

:loadRwithrectifiereduncontrollwave-Full


99/99

RIP

2II

2VV

2

rms

mrms

mo

radian)inis(4

2sin

2-

2

1

R

VI

R

V

I

)cos1(VV

:loadRwithrectifiercontrolledwave-Full

mrms

o

o

m

o

a

a

a

a

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