Single Phase Rectifiers

8/10/2019 Single Phase Rectifiers

1/22

Dr. Sorin Deleanu Dr. David Carpenter

Power Electronics. Theory and Applications

2-1

2. Single Phase Rectifiers

The learning objectives of this chapter are:

To be able to classify the types of single phase rectifiers

To be able to describe the operation of single-phase rectifiers with different types ofload

To differentiate between the operation of single-phase uncontrolled and controlledrectifiers, following similar load conditions

To compare different types of rectifiers from the prospective of certain performanceparameters

To develop effective calculation methods for analyzing and design of single phase

rectifiers

To be able to simulate single phase rectifiers, perform lab measurements and comparethe results

To be able to formulate conclusions regarding the operation of single-phase rectifiersin normal and abnormal operating conditions

2.1.Single-phase uncontrolled rectifiers

A rectifier is a circuit that converts an AC signal into unidirectional one. In this paragraph

diode rectifiers are considered, and for simplicity ideal diodes are considered. When making

the assumption of ideal diodes, we mean diodes with zero forward voltage drop in

conduction, infinite internal resistance in blocking state (when reverse biased) and

instantaneous transition from on-state to off-state and inversely, depending of biasing

conditions.

2.1.1 Half-Wave Uncontrolled Rectifier

Applications: Low cost, low power supplies for electronics

For the single phase rectifiers, having connected certain loads at the output, the following

objectives are to be achieved:

a. Explain the operation

b.

Calculate the output (load) voltage average value VL,DC

c. Calculate the output (load) voltage effective (RMS) value VL,RMS

d. Calculate the output (load) current average value IL,DC

e.

Calculate the output (load) current effective (RMS) value IL,RMSIn addition to the above-mentioned, most of the characteristics defined in previous chapter

can be calculated or determined upon the necessity. These are:

Efficiency of the rectifier Form factor FF Ripple factor RF

Transformer utilization factor TUF

Peak inverse voltage of the diode PIV

Effective (RMS) value of the diode current

Crest factor of the input current CF Input power factor PF


2/22


3/22

Once in conduction, the ideal diode D will let the power supply voltage to be

applied across the load:

dt

diiRitVvv LLLmSL +=== sin (2.2)

Equation (2.2) has the following solution:

( ) ( )

++

=t

L

R

et

RL

Vti

mL

sinsin222

, with

=

R

L arctan (2.3)

Note: Solving the equation (2.2.) is the key point for further calculations and for this purpose

will use Laplace Transform, when zero initial load current is considered:

( )



2-3

( ) ( ) ( ) ( )

+

=+=

L

LLLL

TsL

sVsIsIsLRsV

1, with

R

LTL = (2.4)

( ) ( ){ }22

+==

s

VtvLsV mL (2.5)

Finally, the operational load current is expressed as:

( )( )

+

+=

L

mL

TsL

s

VsI

1

122

(2.6)

For finding the load current in time domain, we have to express the operational

current from (2.6) as a sum of simple fractions. Once determined, the coefficients A,Band C,

will give the possibility to find out the time expression of the current by applying Laplace

inversions:

( )

+

++

+

=

L

mL

Ts

C

js

B

js

A

L

VsI

1

(2.7)

Bringing to a common denominator (2.6) and (2.7) and imposing to have the same

numerator, a three-equation system is composed:

( ) ( )

( )

=+

=++

=++

1

01

0

2

CBAT

j

BA

T

BAj

CBA

L

L

(2.8)

Solving the above system, we found:

,

=

jT

j

B

L

12

1and

2

2

1

1

+

=

LT

C (2.9)

+

=

jT

j

A

L

12

1

From (2.7) and (2.9), by inverting the Laplace elementary fractions, we obtained:


4/22

( )

+

+

+

+

= LT

t

e

T

tje

jT

j

tje

jT

jL

Vti

LLL

mL

2

2

1

1

12

1

12

1 (2.10)

Applying the definitions of the complex sine and cosine functions:

and ( )j

tje

tje

t2

sin

= , we obtain:( )

2cos

tje

tje

t

+=

( )( )



2-4

+

+

=

sin

sinsin

1

1

2

2

LT

t

et

T

L

Vti

L

mL (2.11)

Finally, by performing the calculations, we obtain:

( ) ( )

++=

tL

R

etRL

V

ti m

L

sinsin222 , with

= R

L

arctan (2.3)

Due to the presence of the inductance, the conduction angle is higher than 1800and is

strongly dependent upon the resistance and inductance of the load.

To determine the value of the conduction angle of the diode, we have to solve the

following equation, where represents the conduction angle in radians:

( ) 0sinsin =

+

t

L

R

e

, (2.12)

which is equivalent to:

0cossin =

+

L

R

eL

R (2.12)

Equation (2.12) shows that the conduction angle is practically determined by the

values of load resistance and reactance. Solving it requires numerical methods, and is very

important when estimating the value of the load average voltage:

b. Calculate the output (load) voltage average value VL,DC

To calculate it, we apply the definition:

( ) ( ) (

cos12

sin2

1

0

.0, === m

mavgDCL

VttdVVV ) (2.13)

The above mentioned relationship can be expressed with respect to the angle , which

represents how far beyond =t , the current goes through the resistive inductive load:


5/22

( ) ( ) (

cos12

sin2

1

0

.0, +=== +

mmavgDCL

VttdVVV ) (2.13)

Note: Meanwhile, it can be evaluated the contribution of the load inductance upon the overall

load voltage, using (2.3) and assuming the substitution += :

( ) ( )

( )

+==

sincos

222

tL

R

eL

Rt

RL

LV

dt

tdiLtv

mLL (2.14)

( ) ( )tdt

L

R

eL

Rt

RL

LVV

mDCXL

+=

+

sincos2

1

0222

, (2.15)

( )

+

+= ++

00

222, sinsin

2

tm

DCXLL

R

et

RL

LVV (2.15)

One step further in (2.15) will lead to:

( ) ( )



2-5

( )

+

+++

=

sinsinsinsin

2 222,

L

R

e

RL

LVV

mDCXL , and after

subtracting the terms:

( ) 0sinsin2 222

, =

+

+=

L

R

e

RL

tVV

mDCXL , according to (2.12)

It is very important to demonstrate that the presence of the inductor modifies the value

of the DC output voltage, despite the fact that the average voltage across the inductive part of

the load is equal to zero.

c.

Calculate the output (load) voltage effective (RMS) value VL,RMS

The definition for the effective value is to be applied, in order to determine the value of the

effective (RMS) value of the load voltage:

( ) ( ) ( )

( ) ( )

( )tdV

tdttV

tdtVV mm

mRMSL

=

== 4

2sin

224

2sin

22sin

2

12

00

2

0

22,

(2.16)


( ) ( ) =

++

= tdt

L

R

et

RL

VI

mDCL

0222

, sinsin2

1


6/22

( ) ( ) ( ) =

++

= tdt

L

R

etdt

RL

VI

mDCL

00222

, sinsin

2

( ) =

++=

0

0222

, sincos2

t

L

R

eR

L

tRL

V

I m

DCL

( )

++

++

+=

L

R

eR

L

RL

VI

mDCL 1sincos1

2222

, (2.17)

e. The output (load) current effective (RMS) value IL,RMS:

( ) ( )=

+

+

=

0

22

222, sinsin

21 tdtL

R

et

RL

VI mRMSL

( ) ( ) ( )

+

+

+

+

+

++

+

+=

sincos

1

sin2sin

2sin

2

24

2sin

2

2

12

22

222

L

RL

R

e

L

RR

LL

R

eR

L

RL

Vm

(2.18)

Note: In order to determine the effective (RMS) value of the load current , the

following preliminary calculation is necessary to be performed:

( ) ( ) ( ) =

+

+ tdt

L

R

ett

L

R

et

0

22 sinsin2sin

2

sin

( ) ( ) ( ) ( ) ( )

+

+

00

2

0

2 sinsin2sin

2

sin tdt

L

R

ettdt

L

R

etdt

Every single term is evaluated separately:

( ) ( ) ( ) ( )

4

2sin

24

22sin

2sin

00

2

++

=

=

+=

tttdt

( )( )



2-6

+

+

=

=

+

22

0

2

0

2sin

2sin

2

2sin

2

2sin

2

R

LL

R

eR

LL

R

eR

Ltd

tL

R

e

t


7/22

( ) ( )( )

+

+

=

sincos

1

sin2sinsin2

2

0L

RL

R

e

L

Rtd

tL

R

et

Finally, the overall integration appears as:

( ) ( ) ( ) ( )

( ) ( )

+

+

+

+

+

+++

=

+

+

sincos

1

sin2sin

2sin

2

2

4

2sin

2sinsin2sin

2

sin

2

22

0

22

L

RL

R

e

L

RR

LL

R

eR

L

tdt

L

R

ett

L

R

et

Example 2-1

For the circuit from Figure 2.1, lets consider: VVS 120= , = 25R , HL 2.0= , .Hzf 60=

Determine:

a. The output (load) voltage average value VL,DC

b. The output (load) voltage effective (RMS) value VL,RMS

c. The output (load) current average value IL,DC

d. The output (load) current effective (RMS) value IL,RMS

e.

The efficiency of the rectifier

f. The form factor FF

g.

The ripple factor RFh. The transformer utilization factor TUF

i. The peak inverse voltage of the diode PIV

j. The effective (RMS) value of the diode current

k. The crest factor of the input current CF

Solution: For a complete illustration, simulated results are presented in Figure 2.2 and

Figure 2.3.

a. First of all, we have to determine the conduction angle, which is beyond radians,

due to the presence of the inductor:

( )radR

fL

R

L25.166.71

25

2.0602arctan

2arctanarctan

0=

=

=

=

So, the total conduction angle is equal to ( )rad25.1+=

( ) ( ) (

cos12

sin2

1

0

.0, === m

mavgDCL

VttdVVV )=35.51V

b. The output (load) voltage effective (RMS) value VL,RMS:

( ) ( ) ( )

( ) ( )

( ) VtdV

tdttV

tdtVV mm

mRMSL 48.684

2sin

224

2sin

22sin

2

12

00

2

0

22, =

=

==



2-7


8/22

c.

The output (load) current average value IL,DC:

( ) ( ) =

++

= tdt

L

R

et

RL

VI

mDCL

0222

, sinsin2

1

( )

++

++

+=

L

R

eR

L

RL

VI mDCL 1sincos1

2 222, =1.194A

d. The output (load) current effective (RMS) value IL,RMS:

( ) ( )=

+

+=

0

22

222, sinsin

21 td

tL

R

et

RL

VI mRMSL

( ) ( ) ( )

AL

RL

R

e

L

RR

LL

R

eR

L

RL

Vm 732.1

sincos

1

sin2sin

2sin

2

24

2sin

2

2

12

22

222=

+

+

+

+

+

++

+

+=

e.

The efficiency of the rectifier:

357.0732.148.68

194.151.35

,,

,,

,

,=

===

AV

AV

IV

IV

P

P

RMSLRMSL

DCLDCL

RMSL

DCL

f. The form factor:

( )928.1

561.35

48.68

.0

,

,

,====

V

V

V

V

V

VFF

avg

RMSL

DCL

RMSL

g.

The ripple factor is equal to:

( )648.11928.111 22

2

,

,

.0

,

,

,===

=== FF

V

V

V

V

V

VRF

DCL

RMSL

avg

ACL

DCL

ACL

h. The transformer utilization factor TUF:

204.0732.1120

194.151.35

,

,,,=

===

AV

VV

IV

IV

IV

PTUF

RMSLS

DCLDCL

SS

DCL

i. The peak inverse voltage of the diode PIV

VPIV 1702120



2-8


9/22

j.

The effective (RMS) value of the diode current:

AII RMSLRMSDIODE 732.1,, ==


7.1732.1

95.2

,

,,max,=====

A

A

I

I

I

I

I

ICF

RMSL

peaks

S

peaks

S

s

In order to calculate the crest factor we have to determine the maximum value of the

input current (which in this case is basically the same like for the output current). This can be

solved by imposing:( )

( ) ( ) 0sincos0 =

=

tt

R

eL

Rt

td

tdiL

. In fact, this represents a

transient equation, which requires numerical methods for solving it. Graphical methods are

less accurate, but still convenient sometimes. So, by plotting the curve )( tiL we can find out

the maximum load current value as being AI peaks 95.2,

2.1.1.2. Half-wave uncontrolled rectifier with resistive load

a. Explain the Operation

When having purely resistive load, then 0=L , which means that: 0= . For such asituation will have the have-wave uncontrolled rectifier with purely resistive load, shown in

figure 2.2. The secondary winding of the transformer is considered as power supply.

The conduction duration equals a half of the cycle of the power supply voltage. During the

first (positive) half of the cycle of the power supply voltage, the diode D is in conduction.

Almost all of the power supply voltage appears across the load (because assuming that D is

ideal, we neglect the on state voltage drop across it). In the second (negative) half of the cycle

of the power supply voltage, the diode is reverse biased, so is blocked and the output voltage

is equal to zero.

b.

Calculate the output (load) voltage average value VL,DC

To calculate it, we apply the definition:

( ) ( ) ( ) ( ) mmm

mLDCL V

V

t

V

tdtVtdtvV 318.0cos2sin2

1

2

1

00

2

0, =====

(2.19)


The definition for the effective value is to be applied, in order to determine the value

of the effective (RMS) value of the load:

( ) ( ) ( ) ( ) mmLRMSL VtdtVtdtvV 5.0sin2

1

2

1

0

22

2

0

2, ===

(2.20)



2-9


10/22


11/22



2-11

Figure2.3Half-Wave Rectifier with resistive load

Example 2-2

For the circuit from Figure 2.1, lets consider: VVS 120= , = 100R , Hzf 60= .

Determine:

a. The output (load) voltage average value VL,DC

b.

The output (load) voltage effective (RMS) value VL,RMS

c. The output (load) current average value IL,DC

d.

The output (load) current effective (RMS) value IL,RMS

e. The efficiency of the rectifier

f. The form factor FF

g. The ripple factor RF

h.

The transformer utilization factor TUFi.

The peak inverse voltage of the diode PIV

j. The effective (RMS) value of the diode current


l. The input power factor PF

m. The harmonic content of the output (load) voltage

Solution: For a complete illustration, simulated results are presented in Figure 2.5 and

Figure 2.6.

a.

The output (load) voltage average value is calculated as:

( ) ( ) ( ) ( ) VVV

tV

tdtVtdtvV mmm

mLDCL 97.53318.0cos2

sin2

1

2

10

0

2

0

, ======

b.

The output (load) voltage effective (RMS) value is calculated as:

( ) ( ) ( ) ( ) VVtdtVtdtvV mmLRMSL 85.845.0sin2

1

2

1

0

22

2

0

2, ====

c. The output (load) current average value is calculated as:

VLvPvS

R

iL

D


12/22

( ) ( ) ( ) ( ) AR

V

R

Vt

R

VtdtItdtiI

mmmmLDCL 54.0318.0cos

2sin

2

1

2

10

0

2

0

, ======

d. The output (load) current effective (RMS) value is calculated as:

( ) ( ) ( ) ( ) AR

VtdtV

Rtdtv

RI

mmLRMSL 8485.05.0sin

2

1

2

1

0

222

0

2, ====

e. The efficiency of the rectifier is equal to:

%45.401001

5.05.0

318.0318.0

100100[%],,

,,

,

,=

===

RVV

R

VV

IV

IV

P

P

mm

mm

RMSLRMSL

DCLDCL

ACL

DCL

f.

The form factor FF is:

57.1318.0

5.0

,

,===

m

m

DCL

RMSL

V

V

V

VFF

Figure2.4 Half-Wave Rectifier with resistive load: Voltage across the Load,

Voltage Across the Diode (anode to cathode) and Load (Diode)

Current(Simulations)



2-12


13/22

g. The ripple factor RF is:

21.1157.11 22

,

,==

==

DCL

RMSL

DC

AC

V

V

V

VRF

h.

The transformer utilization factor TUF

286.05.0

2

318.0318.0

,,=

===

R

VV

R

VV

IV

IV

IV

PTUF

mm

mm

SS

DCLDCL

SS

DC

Note: The last value justifies the fact that we need a transformer with

5.31

TUF

apparent power than the value of the load DC power

i. The peak inverse voltage of the diode PIV, from the waveforms appears as:

mVPIV=

j. The effective (RMS) value of the diode current is:

AR

V

R

VII m

RMSL

RMSLRMSDIODE 8485.05.0,

,, ====

Figure2.5 Half-Wave Rectifier with inductive load: Voltage across the Load,

Voltage Across the Diode (anode to cathode) and Load (Diode)Current(Simulations)



2-13


14/22

Note: The last two calculations represent useful values when wanting to determine

the ratings of the diode used to build the rectifier with. In fact, when choosing a diode

from a manufacturers catalog, we have to impose the following conditions:

)log,, cataRMSDIODERMSDIODE

II andlogcata

PIVPIV

k. The crest factor of the input current is given by:

25.0

,max,====

R

V

R

V

I

I

I

ICF

m

m

S

peaks

S

s

l.

The input power factor can be calculated as following:

707.02

1

5.0

2

5.05.0

cos

,

,1 =====

R

VV

R

VV

IV

R

VV

IV

IVPF

mm

mm

SS

RMSL

RMSL

SS

SS

m. The harmonic content of the output voltage can be determined using Fourier

Series:

(2.23)( ) ( )

=

++=1

, sinsin

n

nnDCLL tnbtnaVtv

The average (DC) component was previously determined: VVV mDCL 97.53318.0, ==

( ) ( ) ( )

=

==== 5,4,3,2:,0

1:,2sinsin

1sin

1

0

2

0 nfor

nforVttdntVttdntva

m

mLn

(2.24)

( ) ( ) ( )

=

==== 8,6,4,2:,

1

2

7,5,3,1:,0

cossin1

cos1

20

2

0nfor

n

V

nfor

ttdntVttdntvb mmLn

(2.25)

Finally, from (2.23), (2.24) and (2.25), well assemble the output voltage as a

harmonic (Fourier) series:

( ) ...8cos63

26cos

35

24cos

15

22cos

3

2sin

2+= t

Vt

Vt

Vt

Vt

VVtv

mmmmmmL

(2.26)

For the data in Example 2.2, we can express (2.26) as:

( ) ...10cos09.18cos71.16cos09.34cos2.72cos36sin85.8497.53 += tttttttvL

Note: For determining the coefficients of the Fourier series, the following calculationsna

have been performed:

For , we have1=n ( ) ( ) ( )24

2sin

2sinsinsin

00

2

0

=

==

tttdtttdnt



2-14


15/22

For will apply the trigonometric identities:,...6,5,4,3,2=n

( )[ ] ( ) tnttnttnttn sinsincoscoscos1cos =+= +

(2.27)

( )[ ]

( ) tnttnttnttn , sinsincoscoscos1cos =+=+

They show to the following integral calculations:

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]

( )



2-15

( ) ( )[ ]

( )[ ]

( ) ( ) Ittdnt

nn

nttdnt

n

tnttdnttdtnttdntttdnt

==+

++

+=

=+

+=+=

00

0000

coscos1

0sin

1

1sincoscos

...01

1sincoscos1coscoscossinsin

Now, if considering: , then( )[ ] ( ) ( ) ( ) =+=

000

2sinsincoscos1cos Ittdntttdnttdtn

because: ( )[ ] ( ) ( )

( )( )( ) ( )

001

0sin

1

1sin01

1sin1cos

0==

=

= Inn

tn

n

tntdtn

Now was clarified that ( ) ,0sinsin1

0

== ttdntVa mn

when ....6,5,4,3,2=n

For determining the coefficients of the Fourier series, the following trigonometricnb

identities are available:

( )[ ] ( ) tnttnttnttn sincoscossinsin1sin =+= +

(2.28)( )[ ] ( ) tnttnttnttn sincoscossinsin1sin =+=+ +

For , we have and after calculating:1=n ( ) ( )[ ] ( ) ( )tdttntdtnttdnt

+=0

cossin

0

1sin

0

cossin

( )[ ] ( ) ( )[ ]

( )

( )

( )

( )

( )

=

+==+

++

+=

+

+=+

....10,7,6,4,2,1

2

9,7,5,3.,1,01

0cos

1

1cos01

1cos1sin

0

nn

nnn

n

n

tntdtn

(2.29)

So, forodd order numbers, ( )[ ] ( ) ( ) ( ) Kttdtnttdnttdtn ===+

000

cossincossin01sin

On the other hand:

( )[ ] ( ) ( )[ ]

( ) ( ) ( ) 02sincoscossin0

01

1cos1sin

000

==+==

= KKttdntttdntn

tntdtn

For even numbers, can ( )[ ] ( ) ( ) ( ) ( )tdtntttdntntdtn

+=+=+ 000sincoscossin1

2

1sin . This


16/22

contributes to express one term under integration with respect to the another like in the next

identity:

( )( )

( )tdtntn

ttdnt

+=00

sincos1

2cossin (2.30)

Furthermore, when integrating the first trigonometric identity from (2.29), then obtain:

( )[ ] ( ) ( )[ ]

( ) ( )



2-16

1

201

1cos1sin

0

=

= nn

tntdtn

(2.31)

From (2.30) and (2.31) applying the identity, we cab express the value of the following

integral as:

( ) ( ) ( ) ( )

( )200

1

2cossincossin2

1

2

1

2

ntdtnttdtnt

nn =

+=

(2.32.)

All of the relationships (2.27) through (2.32), explain the structure of the harmonicdecomposition of the load (output) voltage of the single phase half wave rectifier with

resistive load

2.1.1.3. Half-wave uncontrolled rectifier with resistive inductive load and freewheeling

diode

Figure2.6. Half-Wave Rectifier with resistive-inductive load and freewheeling diode

a. Explain the Operation

For the rectifier from figure 2.7, having the output connection across a resistive-

inductive load, an anti-parallel (freewheeling) diode is connected across the load. The

conduction interval of the diode D is diminished to the half of the cycle.

The freewheeling diode (FWD), actually, is preventing the appearance of the negative

voltage across the load.

When the power supply polarity is changing ( ) =t , the main diode will become

negatively polarized, and due to the magnetic energy accumulated in the inductor, the load

current will find a path through the loop composed by the free wheeling diode and the load.

During the second half of the cycle, when the power supply voltage becomes

negative, across the load will be applied the freewheeling diode voltage drop, which can beneglected. The overall output (load) voltage average value is increased.

FWD

D

VLvPvS

vR

vXL

R

L

iL


17/22

The load current can stay continuous or can be discontinuous depending upon the load

constantR

L .Calculating the load current becomes an important issue in order to be able to

predict all the performance parameters. The power supply voltage is described by (2.1).

In the first half of the cycle, the ideal diode D will let the power supply voltage to

be applied across the load, and equation (2.2) describes the circuit.

In the second half of the cycle, the output (load) voltage dropped to zero:

dt

diiRiv LLLL +== 0 (2.33)

The operation of the circuit is cyclical, so the following conditions must be imposed

for determining the load current:

( )( )

+

+==+

L

R

L

R

mLL

e

e

RL

VIkI

1

11

sin12

222max, (2.34)

( )( )

+

+==+

L

R

L

R

L

R

mLL

e

ee

RL

VIkI

1

1sin

12222

max, (2.35)

It is to remind that

=

R

L arctan . In the above mentioned expressions, krepresents an integer

constant. For conditions like (2.34) and (2.35), the stabilized load current can be expressed

as:

( ) ( ) ( )

( ) ( ) ( )

++

++

=

+

+

+

+

+=

1212,1sin

122,

1

sinsin

222

222222

ktkt

L

R

eL

R

e

RL

Vti

ktke

e

e

RL

Vet

RL

Vti

mL

tL

R

L

R

L

R

mt

L

R

mL

(2.36)

We considered k a multiplication number, which show the fact that the rectifier operates

under steady state conditions, following the first few tens of cycles.

Note: Stabilized (steady state) current means that the rectifier has been operating for asignificant number of cycles.

Note: Solving the equations (2.33.) and (2.34) is the key point for further calculations and for

this purpose will use Laplace Transform, when the initial load current is considered:( )0LI

( ) ( ) ( ) ( ) ( ) ( ) ( )

+

+=+=

L

LLLLLL

TsL

LIsVsILIsIsLRsV

1

00 , with

=

tR

LTL

0

(2.37)

( ) ( ){ }22

+==

s

VtvLsV

mL (2.38)

Finally, the operational load current is expressed as:



2-17


18/22

( )( )

( ) ( )

+

+

+

+=

+

++=

L

L

L

m

L

L

Lm

L

TsL

LI

TsL

s

V

I

TsL

LIs

V

sI1

0

10

1

02222

(2.39)

( ) ) ( )

( )

++

++=

L

LmL

TssL

IsLVsI

1

0

22

22

(2.39)

For the second half of the cycle 2 t .( )



2-18

( ) ( ) ( ) ( )

+

=+=

L

LLLL

TsL

LIsILIsIsLR

10

(2.40)

For finding the load current in time domain, we have to express the operational

current from (2.39) as a sum of simple fractions. Once determined, the coefficientsA,Band

C, will give the possibility to find out the time expression of the current by applying Laplace

inversions:

( )

+

++

+

=

L

L

Ts

C

js

B

js

A

LsI

1

1

(2.41)

Bringing to a common denominator (2.39) and (2.41) and imposing to have the same

numerator, a three-equation system is composed:

( )

( ) ( )

( ) ( )

+=+

=++

=++

220

01

0

Lm

L

L

L

LIVCBAT

j

BAT

BAj

LICBA

(2.42)

Solving the above system, we found:

,

=

jT

j

VB

L

m

12

and ( )2

2

10

+

+=

L

mL

T

VLIC (2.43)

+

=

jT

j

VA

L

m

12

After introducing (2.43) into (2.41), the load current in Laplace form is changed into:

( ) ( )

+

+

+++

+

+

=

LL

mL

L

m

L

mL

Ts

T

VLI

js

Tjj

V

js

Tjj

V

LsI

1

1

10

1

12

1

12

1

2

2

(2.44)

From (2.44) by inverting the Laplace elementary fractions, we obtained:

( ) ( )

+

++

+

+

= LT

t

e

T

VLI

tje

jT

j

Vtje

jT

j

V

Lti

L

mL

L

m

L

mL

2

2 10

12

12

1 (2.45)

Applying the definitions of the complex sine and cosine functions,


19/22

( )2

costj

etj

et

+= and ( )

j

tje

tje

t2

sin

= , after some intermediate calculations will

obtain:

( ) ( )



2-19

LT

t

L

mL

L

m

L

mL e

T

VI

T

tL

VtLTVti

+++

+

=

2

2

2

2 10

11cossin (2.45)

Finally, by performing the calculations, we obtain:

( ) ( ) ( ) tL

R

Lm

L eIt

L

R

et

RL

Vti

+

++

= 0sinsin222

(2.46)

Previously, we described the load as having the power factor angle

=

R

L arctan

In a similar manner, when 2 t , after applying the Laplace transformation,

when describing the operational current we find:

( )( )

( )( ) ( )

tL

R

eIti

Ts

I

sLR

LIsi LL

L

LLL

=

+

=+

=1

(2.47)

When considering the rectifier operating for sufficient time, then we can express the

current for ( ) 12 += kt as:

( )( )

L

R

e

RL

V

k

L

R

e

L

R

e

L

R

e

RL

Vki

mm

kL

+=

+

+

+=+

1

1sin

12

12

1

1sinlim12

222222 (2.48)

When kt 2= , then:

( )

L

R

e

RL

L

R

eVL

R

e

k

L

R

e

L

R

e

L

R

e

RL

Vki

mm

kL

+

=

+

+

+=

1

1sin

12

12

1

1sinlim2

222222(2.49)

b. Calculate the output (load) voltage average value VL,DC

To calculate it, we apply the definition and obtain a similar value like for the half-wave, uncontrolled rectifier with purely resistive load (2.19). This is due to the presence of

the free-wheeling diode which closes the current path for the second half of the cycle when

the power supply voltage comes negative.


The effective value is determined exactly like for the half-wave, uncontrolled rectifier

with purely resistive load (2.20).



20/22

( ) ( ) ( ) ( ) ( ) ( )( ) ( )

++

+

+

+==

0

2

222

2

0

, 122sinsin2

1

2

1tdekItdekIet

RL

VtdtiI

tL

R

L

tL

R

L

tL

R

mLDCL

Considering the values determined in (2.48) and (2.49), we obtain:

+

+

+=

L

R

eL

R

eR

L

RL

VI mDCL

2

2sincos22 222

, (2.50)

e.

The output (load) current effective (RMS) value IL,RMS:

Figure2.7 Half-Wave Rectifier with Resistive inductive load and Free-wheeling

diode: Power Supply Voltage, Load Current, Load Voltage, Voltage

across the Inductive Reactance, Voltage Across the Diode (anode to

cathode)and Voltage across Free-Wheeling Diode(Simulations)



2-20


21/22

Figure2.8 Half-Wave Rectifier with Resistive Capacitive Load: Power Supply

Voltage, Load Voltage, Diode Current, Capacitor Current and

Resistor Current (Simulations)



2-21


22/22

Dr. Sorin Deleanu Dr. David Carpenter2-22

Bibliography:

[1] Agrawal, J. P, Power Electronic Systems. Theory and Design, Prentice Hall, Upper

Saddle River, New Jersey, Columbus, Ohio, 2001

[2] Rashid, M.H. Power Electronics Circuits, Devices and Applications, Pearson

Prentice Hall, Upper Saddle River, NJ 07458, 2003

[3] Deleanu, S. Contribuii Privind Acionarile Electrice de Curent Alternativ in

Traciune. Tezde Doctorat, Universiatea POLITEHNICA, Bucureti, ROMANIA,

Iulie 2001 (In Romanian).

Documents

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