Diode rectifiers

- A rectifier is a circuit that converts an ac signal into a unidirectional signal.- A rectifier is a type of ac-dc converter.

Rectifiers are classified according to type of input supply:

Single phase rectifiersThree phase rectifiers

Single phase rectifier

Single phase half-wave rectifiersSingle phase full-wave rectifiers

Single phase half-wave rectifiers classified according to load:

1-Resistive load2 RL load

a- Without free wheeling diodeb- With free wheeling diode

3- EMF load

Performance parameters of a rectifier

Average output (load) voltage, Vdc

Average output (load) current, Idc

Output dc power Pdc=Vdc Idc

rms value of the output voltage, Vrms

rms value of the output current, Irms

Output ac power Pac=Vrms Irms

ac

dc

PP

=ηEfficiency (or rectification ratio) of a rectifier:

The output voltage can be considered as composed of 2 components: (1) the dc value and (2) the ac component or ripple.

- The ripple voltage 22dcrmsac VVV −=

- The form factor is a measure of the shape of output voltage:dc

rms

VVFF =

- The ripple factor is a measure of the ripple content:

dc

ac

VV

RF = or 11 22

−=−⎟⎟⎠

⎞⎜⎜⎝

⎛= FF

VVRF

dc

rms

ss

dc

IVP

TUF =- The transformer utilization factor is defined as

Vs and Is are the rms voltage and current of the transformer secondary respectively

- Crest factor: a measure of peak input current Is(peak) w.r.t. its rms value Is,

s

peaks

II

CF )(=

Half-wave rectifier supplying a resistive load

Figure 1

π

ππ

ωωπ

π

m

m

mdc

V

V

tdtVV

=

+−=

= ∫

))0cos(cos(2

)sin(21

0

RVI m

dc π=

2)sin(

21

0

2 mmrms

VtdtVV == ∫ ωωπ

π

RVI m

rms 2=

%53.40

2*

2

*

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc ππη

57.12

2 ====π

πm

m

dc

rmsV

V

VVFF

211.1157.11 22 =−=−== FFVVRF

dc

ac

Peak reverse (inverse) voltage (PIV) of diode D1 is Vm

Is(peak)=Vm/R and Is=0.5Vm/R. The CF of the input current is CF=Is(peak)/Is=1/0.5=2

Figure 2

Half-wave rectifier supplying an RL load

Without free-wheeling diode

RVI dcdc /=

β

β

Due to inductive load, the conduction period of diode D1extends beyond π (until the current equals zero at ωt = π+σ).

The average voltage across the inductor is zero.

)]cos(1[2

]cos[2

)(sin2 0

0σπ

πω

πωϖ

πσπ

σπ+−=−== +

+

∫ mmmdc

VtVttdVV

) (sin

:periodon During

solutionstransientstatesteadyesincorporatdtdILIRtVm +⇒+=ω

RLast

ZVI m

SSωφφω 1tan )sin( −=−=

tLR

ts eAI−

= .&

φ=∴==

+φ−ω=+==−

sinZ

VA 0t at 0I :conditions initial from

e.A )tsin(Z

VIIII

mS

tLR

mtsssoS

φσφπβτ ≅+≅⇒<<= :: , ) ( , eiloadsinductivehighlyLRFor

Disadvantages:

1. Discontinuous current

2. High ripple content (FF>1)

3. Presence of DC component in supply circuit.

With free-wheeling diode

Figure 3

- It can be noted from equation (1) that the average voltage (and load current) can be increased by making σ=0, which is possible by adding a freewheeling diode Dm as shown in Figure 3 with dashed lines.

-Diode “Dm” prevents appearing of negative voltage across the load.

-At ωt = π the current from D1 is transferred to Dm ; this process is called commutation of diodes .

- The load current i0 is discontinuous with resistive loads and continuous with highly inductive loads. R

VIVV mdc

mdc ,

ππ==

For highly inductive loads, the current doesn’t reach zero, instead, the current lies between two border values after a few cycles of transients.

]1

1)[sin( 2

maxLR

LR

m

e

eZ

VIωπ

ωπ

φ−

−

−

+=

LR

eII ωπ

maxmin .−

=&

tLR

mm eAt

ZVi

dtdILIRtV

−+−=⇒+=

→→→

. )sin( sin

,.........54,32,0 :periodon During

φωω

πππππ

EMF load

Figure 4

If output is connected to a battery, rectifier operates as battery charger.

EVm

For vs>E, diode D1conducts. The angle α at which diode starts conduction can be found as follows:

=αsinmV

E1sin −=α⇒

Diode D1 is turned off when vs<E at απβ −=

REtV

REv

i mso

−=

−=

ωsinThe charging current io can be found from: For α<ωt<β

Example: The battery voltage in figure 4 is E=12V and its capacity is 100Wh.The average charging current should be Idc=5A. The primary input voltage is Vp=120 V, 60 Hz, and the transformer has a turn ratio of n=2:1. Calculate:

(a) Conduction angle δ of the diode, (b) Current limiting resistance R,(c) Power rating PR of R(d) Charging time ho in hours, (e) Rectifier efficiency η,(f) PIV of the diode.

Solution

VVV

Vn

VVVVVE

sm

psp

85.846022

602

120 120 12

=×==

=====

o

o

o

74.16313.887.17187.17113.8180

13.885.84

12sin 1

=−=−=

=−=

== −

αβδ

β

α(a)

),2cos2(2

1

)(sin

21

EEVR

I

tdR

EtVI

mdc

mdc

πααπ

ωω

πβ

α

−+=

−= ∫

For β=π-α

(b)

Which gives

Ω=×−××+×××

=

−+=

26.4)121419.012213.8cos85.842(52

1

)2cos2(2

1

ππ

πααπ

o

mdc

R

EEVI

R

(c)

2.84.67

4.67

cos42sin2

)2(22

1

)()sin(21

2

22

2

22

2

22

==

=

⎥⎦

⎤⎢⎣

⎡−+−⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

−= ∫

rms

rms

mmm

rms

mrms

I

AI

EVVEVR

I

tdR

EtVI

αααππ

ωωπ

βα

The power rating of R is,

WPR 4.28626.42.8 2 =×=

(d) The power delivered Pdc to the battery is

hP

h

PhWEIP

dco

dco

dcdc

667.160

100100100

60512

===

==×==

(e) The rectifier efficiency is

%32.174.28660

60=

+=

+=

=

Rdc

dc

PPP

powerinputtotalbatterythetodeliveredpower

η

η

(f) The peak inverse voltage PIV of the diode is

VEVPIV m 85.961285.84 =+=+=

Single-phase full-wave rectifiers

-A full-wave rectifier circuit with a center-tapped transformer is shown in figure 5.

-Each half of the transformer with its associated diode acts as a half-wave rectifier and the output of a full-wave rectifier is shown in figure 6. To get the same output as bridge rectifier, the source voltage should be doubled.

The average output voltage is: ∫ ===π

πω

π o mm

mdc VVdttVV 6366.02sin1

Figure 5

tVV mS ωsin2=

Figure 6

During the positive half cycle of the input voltage, the power is supplied to the load through diodes D1 and D2.

During the negative cycle, free-wheeling diodes D3 and D4 conduct.

..)( lno

o

VV

m

oo

VXI

00

1 2

1

Bridge rectifier:

Half-wave with free-wheeling diode

Full-wave bridge

Figure 7

Figure 8

mmm

mdc VVVtdtVV 6366.02))0cos(cos(2

)sin(1

0

==+−== ∫ ππ

πωω

π

π

RV

RVI mdc

dc6366.0

==

mm

mrms VVtdtVV 707.02

)sin(1

0

2 === ∫ ωωπ

π

RVI m

rms707.0

=

%81707.0*707.0

6366.0*6366.0

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dcη

11.16366.0707.0

===m

m

dc

rms

VV

VVFF

482.0111.11 22 =−=−== FFVVRF

dc

ac

Is(peak)=Vm/R and Is=0.707Vm/R.

The CF of the input current is CF=Is(peak)/Is=1/0.707= 2

Single-phase Full-wave rectifier with RLE load

Figure 10

Figure 9

If vs=Vm sin ωt is the input voltage, the load current io can be found from:

tVERidtdiL mo

o ωsin2=++ 0≥oifor

Which has a solution of the form

REeAt

ZVi tLRm

o −+−= − )/(1)sin(2 θω

Where22 )( LRZ ω+=

RLωθ 1tan −=

Vs is the rms value of the input voltage

Case 1 continuous load current

)(

1 sin2 LR

mo e

ZV

REIA ω

π

θ ⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

A1 in equation 3 can be determined from the condition: at ωt=π, io=Io.

, Substitution of A1 in equation 3 yields:

REe

ZV

REIt

ZVi

tL

Rm

om

o −⎟⎟⎠

⎞⎜⎜⎝

⎛−++−=

− )]([sin2)sin(2 ωπ

ωθθω

At steady-state: io(ωt=0)=io(ωt=π). ⇒ io(ωt=π)=Io.

RE

e

eZVI R

R

mo −

−

+=

−

−

)L (

)L (

1

1sin2

ωπ

ωπ

θ

REe

et

ZVi tLR

LRm

o −⎥⎦⎤

⎢⎣⎡

−+−= −

−)/(

)/)(/( sin1

2)sin(2 θθω ωπ

After substituting Io in equation 4 and simplification:

Case 2 discontinuous load current

The load current flows only during the period α ≤ ωt ≤ β.

Define x=E/Vm As (EMF) is constant, diodes start to conduct at ωt=α

)(sinsin 11 xVE

m

−− ==α

At ωt=α, io(ωt)=0 and equation 3 gives

)(

1 )sin(2 LR

m eZV

REA ω

α

θα ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

Which, after substituting in equation 1, yields the load current

REe

ZV

REt

ZVi

tL

Rss

o −⎟⎟⎠

⎞⎜⎜⎝

⎛−−+−=

− )]([)sin(2)sin(2 ωα

ωθαθω

To solve for io, assume initially the current is continuous and proceed with the solution. If the assumption is not correct, the load current is considered equal to zero (discontinuous current equation), and β is found accordingly.

Fourier analysis of single-phase full-wave rectifier output:

∑∞

=++=

,..3,2,1)sincos()(

nnndco tnbtnaVtV ωω

Even function β ⇒ bn=0 & repetitive every π ⇒ n=2,4,6,8,….

∑∞

=+=

,....6,4,2cos)(

nndco tnaVtV ω

πωω

πωω

π

ππm

mmdcVtdtVtdtVV 2.sin1.sin

21

0

2

0=== ∫∫

∞=+−

−== ∫ ,.....,8,6,4,2n, )1)(1(

4.cos.sin1 2

0 nnVdttntVa m

mn πωω

π

π

..........4cos1542cos

342)( −−−= tVtVVtV mmm

o ωπ

ωππ

The second harmonic (with frequency 2f is dominant)

3 Phase half-wave rectifier

Disadvantages:

1. DC current component in supply

2. Low source power-factor

(poor utilization:1/3 cycle diode operation).

3. Peak inverse voltage of diodes = line voltage

Load

mLoadsource

sourcermsLoad

RVII

II acbacbaLoad IIIIIIII

33

332222

_

==∴

=== ++⇒++=→→→→

To get source current Ia or Ib or Ic

[ ] mmmmdc VVtVtdtVV 827.03

sin3sin3)(cos3/2

2 3/

0

3/0 ==== ∫

π π ππ

ωπ

ωωπ

RV

RVI mdc

dc827.0

==

RVI m

rms84068.0

=mmrms

mmrms

VVV

ttVtdtVV

84068.03

2sin21

323

42sin

23)(cos

3/22

21

21

3/

0

21

3/

0

22

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦⎤

⎢⎣⎡ +=⎥⎦

⎤⎢⎣⎡= ∫

πππ

ωωπ

ωωπ

ππ

%77.9684068.0*84068.0

827.0*827.0

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dcη

0165.1827.0

84068.0===

m

m

dc

rms

VV

VVFF

1824.010165.11 22 =−=−== FFVVRF

dc

ac

6647.03P

PT.U.F.)( source

dc ===SS

dcdc

IVIVfactornutilizatiorTransforme

Three phase Bridged Rectifier

The diode behaves as a window, permitting the load to “see” the available line voltage.

Advantages:

1. Bi-directional source current;

(no source dc component)

1. Higher diodes utilization

2. Higher source power-factor

The average output voltage is found from:

mm

mdc

VV

tdtVV

654.133

cos36/2

2 60

==

= ∫

π

ωωπ

π

Where Vm the peak phase voltage.

The rms output voltage is:

mmmrms VVtdtVV 6554.14

3923cos3

6/22 2

12

16

0

22 =⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎥⎦

⎤⎢⎣⎡= ∫ π

ωωπ

π

RVI mm /3=If the load is purely resistive, the peak current through a diode is:

The rms value of the diode current is:

mmmd IItdtII 5518.06

2sin21

61cos

24 2

12

1

60

22 =⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ +=⎥⎦

⎤⎢⎣⎡= ∫

πππ

ωωπ

π

The rms value of the transformer secondary current:

mmms IItdtII 7804.06

2sin21

62cos

28 2

12

1

60

22 =⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ +=⎥⎦

⎤⎢⎣⎡= ∫

πππ

ωωπ

π

Example 2: A three phase bridge rectifier has a purely resistive load of Rohms. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor,(d) the TUF, (e) the peak inverse voltage of each diode, and (f) the peak current through a diode. The rectifier delivers Idc=60A at output voltage Vdc=280.7Vand the source frequency is 60Hz.

∫ === 6

0654.133cos3

6/22 π

πωω

π mmmdc VVtdtVV

mmmrms VVtdtVV 6554.14

3923cos3

6/22 2

12

16

0

22 =⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎥⎦

⎤⎢⎣⎡= ∫ π

ωωπ

π

RVI m

rms6554.1

=

%83.996554.1*6554.1

654.1*654.1

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dcη

0008.1654.1

6554.1===

m

m

dc

rms

VV

VVFF

04.010008.11 22 =−=−== FFVVRF

dc

ac

ss

dc

IVPTUF

3=

RVVIV m

mss ××××= 37804.0707.033

9542.07804.0707.033

654.1 2

=×××

=TUF

The peak line-to-neutral voltage is

VVm 7.169654.1

7.280==

VVPIV m 7.2937.16933 =×==

The average current through each diode is

mmd

md

III

tdtII

3183.06

sin2

cos24 6

0

==

= ∫π

π

ωωπ

π

The average current through each diode is

AIAI

m

d

83.623183.0/20203/60

====