power electronics- chapter 3

BEE4223 Power Electronics & Drives Systems

Chapter 3: AC TO DC CONVERSION (RECTIFIER)

LEARNING OBJECTIVES

• Upon completion of the chapter the student should be able to: – State the operation and

characteristics of diode rectifier. – Discuss the performance

parameters and use different technique for analyzing and design of diode rectifier circuits.

– Simulate different arrangement of diode rectifiers by using PSpice.

Overview

• Single-phase, half wave rectifier – Uncontrolled– R load– R-L load– R-C load– Controlled– Free wheeling diode

• Single-phase, full wave rectifier

– R load– R-L load, – Controlled R, R-L

Load– continuous and

discontinuous current mode

•Three-phase rectifier – uncontrolled – controlled

Rectifiers

• DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches.

• Basic block diagram

Rectifiers

• Input can be single or multi-phase (e.g. 3-phase).

• Output can be made fixed or variable

• Applications: – DC welder, DC motor drive, Battery charger, DC power

supply, HVDC

Root-Mean-Squares (RMS)

2

0

.2

1td

2(.)

Root Mean Squares of f

2)( fStep 1:

2

0

2)(2

1tdfStep 2:

2

0

2)(2

1tdfStep 3:

Concept of RMS

tv

v2 Average of v2

Square root of the average of v2 Average

of v=0

Ideal Rectifier:Single-Phase, Half-Wave R-Load

• Considering the diode is ideal, the voltage at R-load during forward biased is the positive cycle of voltage source, while for negative biased, the voltage is zero. m

mo

mDCaveo

ss

VV

V

tdtVVVV

voltageoutputDC

tVtvSourceVoltageGiven

318.0

)sin(2

1

,""

),sin()(,

0

R

V

R

V

R

VI

currentoutputDC

mmoo

1

,""

R

V

R

VI

VtdtVVwhere

R

VRIP

mrmso

rmso

msrmso

rmsrms

2

2)()]sin([

2

1,

resistor,by absorbedpower Average

,,

0

2,

22

rmsrms

dcdc

ac

dc

IV

IV

P

P

Efficiency

,

• We observe that:

– DC voltage is fixed at 0.318 or 31.8% of the peak value

– RMS voltage is reducedRMS voltage is reduced from 0.707 (normal sinusoidal RMS) to 0.5 or 50% of peak value.

– Half wave is not practicalHalf wave is not practical because of high distortion supply current. The supply current contains DC component that may saturate the input transformer

Ideal Rectifier:Single-Phase, Half-Wave R-Load

Example 1

• Consider the half-wave rectifier circuit with a resistive load of 25 and a 60 Hz ac source of 110Vrms.– Calculate the average values of Vo and Io. Justify the

significant value of Vo and Io.– Calculate the rms values of Vo and Io.– Calculate the average power delivered to the load.

Example 1 (Cont)

• Solution

• (i) The average values of Vo and Io are given by

AR

VI

and

VV

V

oo

mo

98.125

52.49

,

52.49)110(2

In this case, for the particular circuit, possible dc output voltage obtained from the circuit is 49.52V and dc output current is 1.98A. That means, for any dc application within this value, this circuit can be used.

Example 1 (Cont)

• (ii) The rms value of the of Vo and Io

AR

VI

and

VV

V

rmso

rmso

mrms

11.325

78.77

,

78.772

)110(2

2

,,

WRIP

OR

WR

VP

rmso

rmsoo

24225)11.3(

24225

)78.77(

22

2,

2

(iii) average power delivered to the load over one cycle

Example 2

• For the half-wave rectifier, the source is a sinusoid of 120Vrms at a frequency of 60Hz. The load resistor is 5. Determine

(i) the average load current,

(ii) the average power absorbed by the load, and

(iii) the power factor of the circuit.

Example 2 (Cont)

Solution

• (i) The average load current

AR

VI

and

VV

so

m

8.10)5(

7.169

,

7.169)120(2

W

R

VP

and

VV

V

rms

mrms

14405

9.84

9.842

)120(2

2

22

(ii) The average power absorbed by the load

Example 2 (Cont)

707.0

)5(2)120(2

)120(

1440

2,,,

RV

V

P

IV

P

S

Ppf

mrmsm

rmsmrmsm

(iii) Power Factor

Note:Note: The power factorpower factor at the input of the rectifier circuit is poorpoor even for resistive load and decreases as triggering angle for controlled rectifier is delayed.

Half-wave with R-L load

• Industrial load typically contain inductance as well as resistance.

• By adding an inductor in series with the load resistance causes an increase in the conduction period of the load current, hence resulting the half-wave rectifier circuit working under an inductive load.

dt

diLRiVm

• That means, the load current load current flows not only during Vflows not only during Vs s > 0> 0, but also for a portion of Vs < 0. This is due to


dt

diLVL

Mohd Rusllim Mohamed

• Until certain time (<), VVss>V>VRR (hence VL= Vs-VR is positive), the current builds up and the current builds up and

inductor stored energy increasesinductor stored energy increases.

• At maximum of Vmaximum of VRR, V, Vss=V=VR R hence, Vhence, VLL =0V =0V.• Beyond this point, VL becomes negative (means releasing stored energy), and current begins to decrease.

•After T=T=, the input, V, the input, Vs s becomes negative but becomes negative but

current still positive and diode is still conducts current still positive and diode is still conducts due to inductor stored energydue 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 results on reducing the average output voltageoutput voltage due to the negative segment. The larger the Inductance, the larger negative segment


• The point when the current reaches zero, is when the diode turns off, given by

0)sin()sin(

e

R

L

R

LLRZand

tfor

tforeZ

Vwt

Z

Vwti

wheret

mm

,tan,)(

20

0)sin()0sin()(

,

122

(0).zero isinductor by absorbed

power average theSince;

)()()(2

1

)()(2

1

,

2

2

0

2

0

RIP

OR

tdtitv

tdtpP

loadbyabsorbedpoweraverageThe

rms

)()(2

1

,,

)()(2

1)()(

2

1

,

0

0

22

0

2

tdtiI

currentaverageand

tdtitdtiI

currentrms

o

rms

Example 3

• For half-wave rectifier with R-L load, R=100, L=0.1H, =377rad/s, and Vs=100V. Determine

• An expression for the current in this circuit

• The point where diode turns off

• The average current

• The rms current

• The power absorbed by the R-L load, and

• The power factor

Example 3 (cont)

SolutionSolutionFor parameter given

radR

L

radR

L

LRZ

o

377.0100

1.0)377(

361.07.20100

)1.0)(377(tantan

9.106)]1.0)(377[()100()(

11

2222

tfor

Aettit

0

331.0)361.0sin(936.0)( 377.0

(i) Current Equation

(ii) (diode stop)

0)361.0sin()361.0sin( 377.0

e

Using numerical root finding, is found to be 3.50 rads or 201o

Example 3 (cont)

iii) Average current

A

tdet

tdtiI

t

o

308.0

)(]331.0)361.0sin(936.0[2

1

)()(2

1

2377.0

50.3

0

0

2

A

tdetIt

rms

474.0

)(]331.0)361.0sin(936.0[2

1 2377.0

50.3

0

iv) rms current

v) Power absorbed by resistor

W

RIP rms

4.22

)100(]474.0[ 2

2

vi) Power factor

67.0

474.02

100

4.22,,

rmsmrmsm IV

P

S

Ppf

Half-wave with R-C load

• In some applications in which a constant output is desirable, a series inductor is replaced by a parallel capacitor.


• The purpose of capacitor is capacitor is to reduce the variation in the to reduce the variation in the output voltageoutput voltage, making it more like dcdc.

• The resistance may represent an external load, while the capacitor is a filtercapacitor is a filter of rectifier circuit.


• Assume the capacitor is uncharged, and as source positively increased, diode is forward biased


• Capacitor is charged to VCapacitor is charged to Vmm as input voltage reaches its positive peak at t = t = /2/2.

• As diode is ondiode is on, the output output voltagevoltage is the same as same as source voltagesource voltage, and capacitor chargescapacitor charges.

• As source decreases after As source decreases after t = t = /2/2, the capacitor discharges the capacitor discharges into load resistorinto load resistor. As diode is reversed biaseddiode is reversed biased, the load is load is isolated from sourceisolated from source, and the output voltageoutput voltage (capacitive charge) decaying exponentially with time constant RCdecaying exponentially with time constant RC.


• The angle t = is the point when diode turns off.

• The diode will stay off until the capacitor and input voltages become

equal again.

sin,

2

22sin)(

/)(

s

RCt

m

o

VVwhere

teV

ttVtV

• The effectiveness of capacitor filtereffectiveness of capacitor filter is determined bydetermined by the variation in output voltage, or expressed as maximum and minimum output voltage, which is peak-to-peak ripple peak-to-peak ripple voltagevoltage.

Half-wave with R-C load (Ripple Voltage)

• The ripple:

)sin1(

sinminmax

m

mm

o

V

VV

VVV

fRC

V

RCVV m

mo 2

• if VVm and /2, then ripple can be approximated as

• The output voltage ripple The output voltage ripple is reduced by increasing the filter capacitor, C. by increasing the filter capacitor, C. Anyhow, this results in a larger peak diode current. larger peak diode current.

RCV

R

VCVpeakI

m

mmD

sincos

sincos,

Example 4

• The half-wave rectifier has 120Vrms source at 60Hz, R=500, C=100F and delay when diode turns on is given 48o. Determine– The expression of output voltage– Ripple voltage– Peak diode current– Sketch and label the output waveform– Value of C as ripple voltage is 1% of Vm, and hence

find new under this condition.

Example 4 (cont)


VV

radAngle

radAngle

radRC

VV

m

o

m

5.16962.1sin)7.169(sin

843.048

62.1)85.18(tan

85.18)101)(500)(60(2

7.1692120

1

6

(i) Output Voltage

(ii) Ripple Voltage

25.169

22sin7.169)(

85.18/)62.1( te

tttV

to

V

VV mo

43

)]843.0sin(1[7.169

)sin1(

(iii) Peak diode current

A

I peakD

50.4

500

)843.0sin()843.0cos()10(3777.169, 4

Example 4 (cont)

(v) Capacitor value

o

m

o

m

m

o

m

mo

fRC

V

V

hence

FV

V

VfR

VC

VVFor

9.81

103333)(500)(60(

11sin

11sin

1sin

,

3333)01.0)(500)(60(

01.0

61

1

1

(iv) Waveform must be properly labeledmust be properly labeled according to data

RL Source Load

• To supply a dc source from an ac source

• The diode will remain off as long as the voltage of ac source is less than dc voltage.

• Diode starts to conduct at t=. Given by,

m

dc

dcm

V

V

OR

VV

1sin

sin

RL Source Load

dcrms

dcRac

o

rms

IVRI

PPPPower

tdtiI

currentaverageand

tdtiI

currentrms

2

22

,

)()(2

1

,,

)()(2

1

,

Example 5

• The RL half-wave rectifier has 120Vrms source at 60Hz, R=2, L=20mH, Vdc =100V with extinction angle given by 193o. Determine– The expression of current in the circuit– Power absorbed by resistor– Power absorbed by dc source– Power supplied by ac source– Power factor– Draw the waveform

Example 5 (cont)



ii) Power absorbed by resistor

Example 5 (cont)

iii) Power absorbed by dc source

iv) Power supplied

v) Power factor

v) Waveform

- Refer notes

Freewheeling Diode (FWD)

• Note that, previously discussed uncontrolled half-wave RL load rectifier allows load current to present at certain period (current decreasing by time since opposing negative cycle of input), hence reducing the average output voltage due to the negative segment.

• In other word, for single-phase, half wave rectifier with R-L load, the load (output) current is NOT CONTINUOUS.

• A FWD (sometimes known as commutation diode) can be placed in parallel to RL load to make the load (output) current continuous.


Freewheeling Diode (FWD)• Note that both Dboth D11and Dand D2 2 cannot be cannot be

turned on at the same timeturned on at the same time.

• For a positive cyclepositive cycle voltage source,– D1 is onon, D2 is offoff– The voltage The voltage across the R-L load is R-L load is

the same the same as the source voltage. source voltage.

– D1 is offoff, D2 is onon

– The voltagevoltage across the R-L load is zeroR-L load is zero.

– However, the inductor the inductor contains energy from contains energy from positive cyclepositive cycle. The load current still circulatesstill circulates through the R-L path.

• For a negative cyclenegative cycle voltage source,


Freewheeling Diode (FWD)

– But in contrast with the normal half wave rectifier, negative cycle of FWD does not consist of supply FWD does not consist of supply voltage in its loop.voltage in its loop.

– Hence the “negative part” of V“negative part” of Voo as as shown in the normal half-wave shown in the normal half-wave disappeardisappear.

RIPR

VI

VV rmso

oo

mo

2;;

• negative cyclenegative cycle voltage source (cont),

• IIrmsrms is determined from Fourier is determined from Fourier

component of currentcomponent of current

...6,4,2;

1

2;

2

nLjnRZ

n

VV

Z

VI

on

mn

n

nn

1

2,

2,k

korms rmsIIIHence

2)()]sin([

2

1

0

2, 1

msrms

VtdtVV

n

- same as uncontrolled RLoad Rectifier

Example 6

• Uncontrolled R-L load rectifier, has a problem of discontinuous load current. Suggest a solution to the problem by justifying your answer through its principles of operation and waveform.

SolutionSolutionOperation of FWD and its waveform (refer notes)Operation of FWD and its waveform (refer notes)

Example 7

• Determine the average load voltage and current, and determine the power absorbed by the resistor in the FWD circuit, where R=2 and L=25mH, Vm=100V; 60Hz.

AR

VI

VV

V

oo

mo

9.152

8.31

8.31100

SolutionSolution•The average load voltage and current,

Example 7 (cont)

The ac voltage amplitudes,The ac voltage amplitudes,

.....6,4,2

).......(425.921

200

).......(10256022

).......(1

)100(2

2

3

2

n

iiinjnZ

VI

iijnZ

in

V

n

nn

n

n

Fourier ImpedanceFourier Impedance

Note:Note: angle note included in calculation

VV

VV

VV

VV

82.116

)100(2

24.414

)100(2

2.2112

)100(2

502

100

26

24

22

1

58.56425.962

75.37425.942

96.18425.922

63.9425.912

2425.902

6

4

2

1

0

jZ

jZ

jZ

jZ

jZ

Example 7 (cont)

• rms current

A

IIIk

korms rms

34.16

2

11.0

2

12.1

2

19.59.15

222

2

1

2,

2

n Vn (V) Zn () IInn (A) (A)

0 31.8 2 15.915.9

1 50 9.63 5.195.19

2 21.2 18.96 1.121.12

4 4.24 37.75 0.110.11

6 1.82 56.58 0.030.03

Resulting Fourier Terms are as follows:Resulting Fourier Terms are as follows:• Power Absorbed

W

RIP rmso

534

)2(34.16 2

2

The Controlled Half-wave Rectifier

• PreviouslyPreviously discussed are classified as uncontrolled uncontrolled rectifiers.rectifiers.

• Once the source and load parameters are established, the dc leveldc level of the output and power transferred to the load are fixed quantitiesfixed quantities.

• A way to control the output is to use SCRcontrol the output is to use SCR instead of diode. Two condition must be met before SCR can conduct:– The SCR must be forward biased (VSCR>0)

– Current must be applied to the gate of SCR

Controlled, Half-wave R load

]cos1[2

)sin(2

1

,""

s

mDCaveo

V

tdtVVVV

voltageoutputDCAverage

2

)2sin(1

2

)()]sin([2

1,

resistor,by absorbedpower Average

0

2,

22

m

mrmso

rms

V

tdtVVwhere

R

VRIP rms

• A gate signal is applied at t = , where is the delay/firing angle.

R

V

R

VI srmso

rmso 2,

,

Example 8

• Design a circuit to produce an average voltage of 40V across 100 load resistor from a 120Vrms 60 Hz ac source. Determine the power absorbed by the resistor and the power factor.

Briefly describe what happen if the circuit is replaced by diode to produce the same average output.

Example 8 (Cont)

• SolutionSolution

rad

VV

o

so

07.12.61

]cos1[2

212040

]cos1[2

In such that to achieved 40V average voltage, the delay angle must be

• If an uncontrolled diode is used, the average voltage would be

• That means, some reducing average resistor to the design must be made. A series resistor or A series resistor or inductor could be added to an inductor could be added to an uncontrolled rectifieruncontrolled rectifier, while controlled rectifier has advantage of not altering the load or introducing the losses

V

VV m

rmso

6.752

)07.1(2sin07.11

2

2120

2

)2sin(1

2,

WR

VP rms 1.57

100

6.75 22

63.0

1006.75

)120(

1.57

pf

VV

V so 54

)120(2

Controlled, Half-wave R-L load

R

L

R

LLRZand

otherwise

tforewtZ

Vwti

t

m

,tan,)(

0

)sin()[sin()(

122

• The analysis of the circuit is very much similar to that of uncontrolled rectifier.

)()(2

1

,,

)()(2

1)()(

2

1

,

22

2

tdtiI

currentaverageand

tdtitdtiI

currentrms

o

rms

Controlled, Half-wave R-L load

;

,

]cos[cos2

)sin(

2

1

,

2 RIP

loadbyabsorbedpoweraverageThe

VtdtV

V

voltageoutputaverageThe

rms

mm

o

Example 9

• For controlled RL rectifier, the source is 120Vrms at 60Hz, R=20, L=0.04H, delay angle is 45o and extinction angle is 217o. Determine

i. An expression for i(t)

ii. Average current and voltage

iii. Power absorbed by load

iv. Power factor

Example 9 (cont)



Example 9 (cont)

ii) Average current and voltage

iii) rms current

iv) Power absorbed by resistor

v) Power factor

power electronics- chapter 3

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