Điện tử công suất (AC DC convertor chapter 2 )

Circuit is energised at ω t=0 • Diode becomes forward biased as the source become positive • When diode is ON the output is the same as source voltage.. • The source becomes less than

Chapter 2

AC to DC CONVERSION

(RECTIFIER)

• Single-phase, half wave rectifier

– Uncontrolled: R load, R-L load, R-C load

– Controlled

– Free wheeling diode

• Single-phase, full wave rectifier

– Uncontrolled: R load, R-L load,

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

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

3-• Output can be made fixed or variable

• Applications: DC welder, DC motor drive,

Battery charger,DC power supply, HVDC

Single-phase, half-wave, R-load

m RMS

o

m

m m

avg

o

V

V t

d t V

V

V

V t

d t V

V

V

5

0 2

)

sin(

2

1 ,

(rms), tage

Output vol

318 0 )

sin(

2

average),

or (DC tage

Output vol

2 0

0 1

π

ω

ω π

+

v s _

+

v o _

Half-wave with R-L load

tan

)

(

:

where

) sin(

)

(

: is response forced

diagram,

From

response, natural

response;

forced

:

) ( )

( )

(

: Solution eqn.

al differenti order

First

) ( )

( ) sin(

L R

Z

t Z

V t

i

i i

t i

t i

t

i

t d

t di L R t i t V

v v

v

m f

n f

n f

m

L R

s

ω θ

ω

θ ω

ω

ω ω

ω

ω ω

ω ω

vL

_ i

R-L load

ωτ

ωτ ω

ωτ ω

θ θ

ω ω

θ θ

θ

θ ω

ω ω

ω

τ ω

ω

ω ω

t m

m m

m

t m

n f

t n

e

t Z

V t

i

Z

V Z

V A

Ae Z

V t

i t

i

t

i

R L Ae

t

i

t d

t di L R

=

=

=

= +

=

) sin(

) sin(

)

(

as, given is

current the

Therefore

) sin(

) sin(

) 0

diode the

before

zero

is

current inductor

realising

by solved be

can

) sin(

) ( )

( )

(

Hence

; )

(

: in results

which

0 ) ( )

(

0, source

when is

response Natural

0

R-L waveform

: i.e , decreasing is

current the

because negative

is

:

Note

di L v

= +

0 for

) sin(

) sin(

)

(

load, L

R ith rectfier w the

-summarise

To

and 0

between conducts

diode the

Therefore,

y.

numericall solved

be only

can

0 )

sin(

) sin(

: to reduces

which

0 )

sin(

) sin(

)

(

angle, tion

theextinc as

known is

point

This

OFF.

turns whendiode

is zero reaches

current

point when he

duration)T that

during negative

is source the

(although radians

n longer tha

biased forward

in remains diode

that the

Note

β ω

θ θ

ω ω

β β

θ θ

β

θ θ

β β

β π

ωτ ω

ωτ β

ωτ β

t

e

t Z

V t

i

e

e Z

V i

t m

m

P pf

I V

t d t i

t d t i

I

t d t i t

d t i I

power apparent

the

is

load.

by the absorbed

power the

to equal

which

source,

by the supplied

power real

the is where

: definition from

computed is

Factor

Power

P

: is load

by the absorbed

Power

N CALCULATIO POWER

)

( 2

1 )

( 2

1

: is current RMS

The

)

( 2

1 )

( 2

1

: is current (DC)

average

The

,

2 o

0

2 2

0 2

β π

ω

ω π

ω

ω π

ω

ω π

ω ω

π

Half wave rectifier, R-C Load

ω θ

sin

OFF is

diode when

ON is

diode hen

w )

sin(

/

m

RC t

+

v o _

• Let C initially uncharged Circuit is

energised at ω t=0

• Diode becomes forward biased as the

source become positive

• When diode is ON the output is the same

as source voltage C charges until V m

• After ω t= π /2, C discharges into load (R).

• The source becomes less than the output voltage

• Diode reverse biased; isolating the load

from source.

• The output voltage decays exponentially.

m

RC m

m

RC t

m

RC

t m

m m

V V

RC

RC RC

RC

RC V

V

e RC

V V

t

e RC V

t d

e V

d

t

V t

d

t V

d

=

= +

−

= +

π π

π π

θ

ω

π ω

ω θ

ω θ

ω θ

θ

ω

θ θ

θ ω

ω θ

ω θ

ω ω

ω

ω θ θ

ω θ ω

ω θ ω

sin

and

Therefore wave.

sine the

of peak the

to close very

is

2 2

-tan

: then large,

is circuits,

practical

For

tan tan

1 tan

1

1 sin

cos

1 sin

cos

equal, are

slopes the

,

At

1 sin

) ( sin

and

cos )

(

sin

: are functions the

of slope

The

1 1

/

/ /

Estimation of α

α

θ α

θ α

π

α π

ω

ω θ α π

ω θ α π

for

y numericall solved

be must equation

This

0 )

(sin sin(

or

) sin (

) 2

sin(

, 2

t

At

) 2

(

) 2

(

=

−

= +

+

=

− +

−

− +

−

RC

RC m

m

e

e V

V

Ripple Voltage

fRC

V RC

V V

RC e

e V

e V V

V

e V e

V v

t

V

V

V V

V

V

V V

V

t V

m m

o

RC

RC m

RC m

m o

RC m

RC m

o

m

m m

m m

α π

ω

π α

π θ

α α

π

α π

ω

ω π

ω

π ω

π

ω

π ω

π π

π θ

2

2 1

: expansoin Series

Using

1

: as ed approximat is

voltage ripple

The

) 2

(

: is 2

at evaluated tage

output vol

The

2.

then constant,

is tage output vol

DC

such that large

is C and 2, and

If

sin )

2 sin(

2

at occurs tage

output vol

Min

is tage output vol

Max

2

2 2

2 2

2 2

min max

max

( t ) ( i.e

OFF, is

diode when

sin

) 2

( t )

(2 i.e

ON, is

diode when

) cos(

), ( ng substituti

Then,

OFF is

diode when

sin

ON is diode when

) (

:, of

In terms

) (

) (

: as expressed be

can capacitor

in the current

The

/

/

α π ω

θ

θ

θ π ω

α π

ω ω

ω

ω

θ

ω ω

ω

ω ω

ω

ω

ω θ ω

ω θ ω

RC t

m o

o c

o c

e R

V

t CV

t

i

t v

e V

t

V t

v

t d

t dv

C t

i

t

t d

t dv C

t

i

Peak Diode Current

R

V CV

i

R

V R

V i

CV CV

I

i i

i

i

m m

peak

D

m

m R

m m

peak

c

C R

D

s

α α

ω

α α

π α

π

α π

α ω

α π

ω

α π

sin cos

: is current peak

diode

The

sin )

(2

sin )

(2

.

: obtained be

can )

(2

at current Resistor

cos )

(2 cos

Hence .

) (2

at occurs current

diode peak

The

: that

+

= +

) sin(

7

169

(OFF)

sin

(ON)

) sin(

7 169 )

sin(

) (

: tage Output vol

(a)

; 5 169 )

62 1 sin(

7 169 sin

843 0 48

; 62 1 93

; 7 169 2

120

/ RC t

m

m o

m

o o m

t

e V

t t

V t

v

V rad

V

rad rad

V V

ωθω

ω

θ

ω

ω ω

θ α

R

V CV

i

e t

e R

V

t

CV t

i

V u

fRC

V RC

V

V

V V

V V

V

V

V V

V

m m

peak

D

t

RC t

m

m c

m m

o

m m

m m

o

o

50 4 ) 34 0

7 169 )

843 0 cos(

7 169 ) 100 )(

60 2

(

sin cos

: current

diode

Peak

(d)

(OFF)

A 339

.

0

(ON)

A )

) sin(

(ON)

)

cos(

: current Capacitor

(c)

7 56 100

500 60

7 169 2

: ion Approximat

Using

43 sin

) 2

sin(

:

Using

: (b)Ripple

,

)85.18/(

62.1

)/(

minmax

= +

π

ω

ωθω

Controlled half-wave

+

v o _

α ω

ω π

ω

ω π

α π

ω

ω π

π α

π α

π α

2

2 sin 1

2

] 2 cos(

1

[ 4

sin 2

sin 2

1

: voltage Average

2

2 ,

m RMS

o

m m

o

V t

d t V

t d t

V V

V t

d t V

V

ωt v

θ α α

α

θ ω ω

ω ω

=

e Z

V A

Ae Z

V i

i

Ae

t Z

V t

i t i

t

i

m m

t m

n f

sin

sin 0

, 0 :

condition

Initial

sin )

( )

( )

d t i I

V t d t V

V

e Z

V i

e t

Z

V t

o

m

t m

)(

: load

by the absorbed

power

The

2

1

: current

RMS

2

1

: current

Average

cos cos

2

sin 2

1

: voltage

Average

angel conduction

the called is

Angle

sin sin

0

y numericall solved

be must angle

Extinction

otherwise

0

t for

sin sin

g, simplifyin and

for ng Substituti

ω ω π

ω ω π

β α

π ω

ω π

θ β γ

θ β θ

β β

β

β ω α θ

α θ

ω ω

βα

ωτωα

1 A half wave rectifier has a source of 120V RMS at 60Hz R=20 ohm, L=0.04H, and the delay angle is 45 degrees

Determine: (a) the expression for i( ω t) , (b) average

current, (c) the power absorbed by the load.

2 Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply Determine the power factor absorbed by the resistance.

Freewheeling diode (FWD)

• Note that for single-phase, half wave rectifier with R-L load, the load (output) current is

NOT continuos.

• A FWD (sometimes known as commutation

diode) can be placed as shown below to make

Operation of FWD

• Note that both D 1 and D 2 cannot be turned

on at the same time.

• For a positive cycle voltage source,

– D1 is on, D2 is off

– The equivalent circuit is shown in Figure (b)

– The voltage across the R-L load is the same as the source voltage.

• For a negative cycle voltage source,

– D1 is off, D2 is on

– The equivalent circuit is shown in Figure (c)

– The voltage across the R-L load is zero.

– However, the inductor contains energy from

positive cycle The load current still circulates through the R-L path

– But in contrast with the normal half wave

rectifier, the circuit in Figure (c) does not

consist of supply voltage in its loop.

– Hence the “negative part” of vo as shown in the normal half-wave disappear.

• The inclusion of FWD results in continuos load current, as shown below.

• Note also the output voltage has no

Full wave rectifier

+

v s2 _

D2

+ vD1 −

+ vD2 −

• Center-tapped (CT) rectifier requires

center-tap transformer Full Bridge (FB) does not.

• CT: 2 diodes

• FB: 4 diodes Hence, CT experienced only one diode volt-drop per half-cycle

• Conduction losses for CT

is half.

• Diodes ratings for CT is twice than FB

m o

d t V

V

t t

V

t t

V

v

637 0

2 sin

1

: voltage (DC)

Average

2

sin

0 sin

circuits, both

π ω

π ω

π ω

v s2 _

Full wave bridge, R-L load

i o

+

vR

_ +

Approximation with large L

: i.e.

terms, harmonic

the

all drop to

possible is

it enough, large

is

If

increasing

ry rapidly ve decreases

Thus

decreases.

harmonic increases,

As

: currents harmonic

The

curent DC

The

1

1 1

1 2

terms harmonics

the

and

2

term DC

the

where

) cos(

)

(

Series, Fourier

Using

4 , 2

R

L R

V R

V I

t

i

L

n I

V n

L jn R

V Z

V t

v

m o

o

n

n

n n

=

ω π

ω

I I

I I

R

V R

V I

RMS o

o RMS

n o

RMS

m o

o

2

2,

2

: load the

to delivered Power

, 2

current e

Approximat

=

= +

=

=

=

π

Given a bridge rectifier has an AC source Vm=100V at 50Hz, and R-L load with R=100ohm, L=10mH

a) determine the average current in the load

b) determine the first two higher order harmonics of the

load current

c) determine the power absorbed by the load

Controlled full wave, R load

V

t d t

V V

V t

d t V

V

RMS o

m

m RMS

o

m m

o

2

2 ,

: is load R

by the absorbed

2 1

sin 1

Voltage RMS

cos 1

sin 1

: voltage (DC)

α

ω

ω π

α π

ω

ω π

π α

π α

i o

T 1

T 3

i o

+

vR

_ +

Discontinuous mode

zero.

an greater th be

must )

output

the us current mode is when in

discontino boundary between continous and

The

0 )

(

: condition

y with numericall

solved be

must Note that is the extinction angle and

) (

: ensure to

need mode,

us discontino

For

; tan

and

) (

for

) sin(

) sin(

)

(

: load

) (

α π

β

τ

ω θ

ω

β ω

α

θ α θ

i

R

L R

L

L R

Z

t

e

t Z

V t

i

ω α

ω α

α

α θ

α θ

θ α π

θ α π

θ α

π α

π

π α α

ωτ π

ωτ α

α π

cos

2 sin

1

: as given is

tage output vol

(DC) Average

tan

mode, current

continuous for

Thus

tan

for Solving

, 0 1

) sin(

), sin(

) sin(

: identity

ry Trigonomet

Using

0 )

sin(

) sin(

0 )

(

1 1

) (

) (

m m

V

R L

R L e

e i

≥

− +

−

Single-phase diode groups

• In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential Therefore, the

diode with its anode (+) at the highest potential will conduct (carry) id.

• For example, when vs is ( +), D1 conducts id and D3reverses (by taking loop around vs, D1 and D3)

When vs is (-), D3 conducts, D1 reverses.

• In the bottom group, the anodes of the two diodes are at common potential Therefore the diode with its cathode at the lowest potential conducts id.

• For example, when vs (+), D2 carry id D4 reverses When v is (-), D carry i D reverses.

Three-phase waveforms

• Top group: diode with its anode at the

highest potential will conduct The other

two will be reversed.

• Bottom group: diode with the its cathode at the lowest potential will conduct The other two will be reversed.

• For example, if D1 (of the top group)

conducts, vp is connected to van. If D6 (of the

bottom group) conducts, vn connects to vbn All other diodes are off.

• The resulting output waveform is given as:

vo=vp-vn

• For peak of the output voltage is equal to

the peak of the line to line voltage vab .

Three-phase, average voltage

DC output that the

Note

955

0

3

) cos(

: voltage Average

radians.

3

or degrees 60

over average

its Considers only one of the six segments. Obtain

, ,

3

2 3 ,

3 2

3 ,

L L m

L L m

L L m

L L m o

V V

t V

t d t V

ω

ω π

π

π π

π π

Output voltage of controlled

three phase rectifier

α π

ω

ω π

α

α π

α π

cos

3

)

sin(

3 1

: as computed be

can voltage

Average

SCR.

the of

angle delay the previous Figure, let be the

From

,

3 2

L L m o

V

t d t V

V

• EXAMPLE: A three-phase controlled rectifier has

an input voltage of 415V RMS at 50Hz The load R=10 ohm Determine the delay angle required to produce current of 50A