Điện tử công suất (AC DC convertor chapter 2 )

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,
– Controlled
– Continuous and discontinuous current mode
• Three-phase rectifier
– uncontrolled
– controlled

Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

1


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

AC input

DC output

• Input can be single or multi-phase (e.g. 3phase).
• Output can be made fixed or variable
• Applications: DC welder, DC motor drive,
Battery charger,DC power supply, HVDC
Power Electronics and
Drives (Version 3-2003),
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2


Single-phase, half-wave, R-load
+
vs
_

+
vo
_

vs

π
vo

ωt

2π

io

Output voltage (DC or average),
π

V
Vo = Vavg = 1 Vm sin(ωt )dωt = m = 0.318Vm
2π 0
π
Output voltage (rms),
π

2

Vm
1
(Vm sin(ωt )dωt ) = = 0.5Vm
Vo , RMS =
2π 0
2
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

3


Half-wave with R-L load

i

+
vs
_

+
vR
_

+

+
vL
_

_

vo

KVL : vs = v R + v L

di (ωt )
dωt
First order differential eqn. Solution :

Vm sin(ωt ) = i (ωt ) R + L
i (ωt ) = i f (ωt ) + in (ωt )

i f : forced response; in natural response,

From diagram, forced response is :
i f (ωt ) =

Vm
⋅ sin(ωt − θ )
Z

where :
Z = R 2 + (ωL) 2

θ = tan −1

ωL
R
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

4


R-L load
Natural response is when source = 0,
di (ωt )
i (ωt ) R + L
=0
dωt
which results in :
in (ωt ) = Ae −ωt ωτ ; τ = L R
Hence

Vm
i (ωt ) = i f (ωt ) + in (ωt ) =
⋅ sin(ωt − θ ) + Ae −ωt ωτ
Z
A can be solved by realising inductor current
is zero before the diode starts conducting, i.e :
Vm
⋅ sin(0 − θ ) + Ae −0 ωτ
Z
V
V
A = m ⋅ sin(−θ ) = m ⋅ sin(θ )
Z
Z

i ( 0) =

Therefore the current is given as,

[

Vm
i (ωt ) =
⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ
Z
Power Electronics and
Drives (Version 3-2003),
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]

5


R-L waveform
vs,
io
β

vo

vR

vL

0

2π

π

3π

4π

ωt

Note :
v L is negative because the current is decreasing, i.e :
di
vL = L

dt
Power Electronics and
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Dr. Zainal Salam, 2003

6


Extinction angle
Note that the diode remains in forward biased
longer than π radians (although the source is
negative during that duration)The point when
current reaches zero is whendiode turns OFF.
This point is known as theextinction angle, β .

[

]

Vm
⋅ sin( β − θ ) + sin(θ )e − β ωτ = 0
Z
which reduces to :

i(β ) =

sin( β − θ ) + sin(θ )e − β ωτ = 0
β can only be solved numerically.
Therefore, the diode conducts between 0 and β
To summarise the rectfier with R - L load,

[

Vm
⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ
Z
i (ωt ) = for 0 ≤ ωt ≤ β
0

]

otherwise
Power Electronics and
Drives (Version 3-2003),
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7


RMS current, Power
The average (DC) current is :
β
1 2π
1
Io =
i (ωt ) dωt =
i (ωt )dωt
2π 0
2π 0

The RMS current is :
β
1 2π 2
1 2
I RMS =
i (ωt ) dωt =
i (ωt )dωt
2π 0
2π 0

POWER CALCULATION
Power absorbed by the load is :
Po = ( I RMS )2 ⋅ R
Power Factor is computed from definition :
P
S
where P is the real power supplied by the source,
which equal to the power absorbed by the load.
pf =

S is the apparent power supplied by the
source, i.e
S = (Vs, RMS ).( I RMS )
pf =

P
(Vs,RMS ).(I RMS )
Power Electronics and
Drives (Version 3-2003),
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8


Half wave rectifier, R-C Load
+
vs
_

iD

+
vo
_

vs

Vm

π /2

2π 3π /2

π

3π

4π

vo

Vmax
Vmin

∆Vo

iD
α

θ

Vm sin(ωt )
when diode is ON
vo =
V e −(ωt −θ ) / ωRC
when diode is OFF
θ

vθ = Vm sin θ
Power Electronics and
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9


Operation
• Let C initially uncharged. Circuit is
energised at ωt=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 Vm
• After ωt=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.

Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

10


Estimation of θ
The slope of the functions are :

d (Vm sin ωt )
= Vm cos ωt
d (ωt )
and
d V sin θ ⋅ e −(ωt −θ ) / ωRC

(m

d (ωt )

)

1
⋅ e −(ωt −θ ) / ωRC
ωRC
At ωt = θ , the slopes are equal,
= Vm sin θ ⋅ −

Vm cosθ = Vm sin θ ⋅ −

1
⋅ e −(θ −θ ) / ωRC
ωRC

Vm cosθ
1
=−
Vm sin θ ⋅
ωRC
1
1
=
tan θ − ωRC

θ = tan −1 (− ωRC ) = − tan −1 (ωRC ) + π
For practical circuits, ωRC is large, then :
π
π
θ = -tan(∞ ) + π = − + π =

2
2
θ is very close to the peak of the sine wave. Therefore
and Vm sin θ = Vm
Power Electronics and
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11


Estimation of α
At ωt = 2π + α ,
Vm sin(2π + α ) = (Vm sin θ )e −( 2π +α −θ ) ωRC
or
sin(α − (sin θ )e −( 2π +α −θ ) ωRC = 0
This equation must be solved numerically for α

Power Electronics and
Drives (Version 3-2003),
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12


Ripple Voltage
Max output voltage is Vmax .

Min output voltage occurs at ωt = 2π + α

∆Vo = Vmax − Vmin
= Vm − Vm sin(2π + α ) = Vm − Vm sin α

If Vθ = Vm and θ = π 2, and C is large such that
DC output voltage is constant, then α ≈ π 2.

The output voltage evaluated at ωt = 2π + α is :
vo (2π + α ) = Vm e

−

2π +π 2−π 2
ωRC

= Vm e

2π
ωRC

−

The ripple voltage is approximated as :
∆Vo ≈ Vm − Vm e

−

2π
ωRC

= Vm 1 − e

Using Series expansoin : e
∆Vo = Vm

−

2π
ωRC

V
2π
= m
ωRC
fRC
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

−

2π
ωRC

=1−

2π
ωRC

13

Capacitor Current
The current in the capacitor can be expressed as :
dvo (t )
d (t )
In terms of ωt , :

ic (t ) = C

ic (ωt ) = ωC

dvo (ωt )
d (ωt )

But
vo (ωt ) =

Vm sin(ωt )

Vm sin θ ⋅ e −(ωt −θ ) / ωRC

when diode is ON
when diode is OFF

Then, substituting vo (ωt ),

ωCVm cos(ωt )
when diode is ON,
i.e (2π + α ) ≤ ωt ≤ (2π + θ )
ic (ωt ) =

Vm sin θ −(ωt −θ ) / ωRC
−
⋅e
R
when diode is OFF,
i.e (θ ) ≤ ωt ≤ ( 2π + α )
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

14


Peak Diode Current
Note that :
is = iD = iR + iC
The peak diode current occurs at (2π + α ). Hence.
I c, peak = ωCVm cos(2π + α ) = ωCVm cos α
Resistor current at (2π + α ) can be obtained :
.
V sin (2π + α ) Vm sin α
iR (2π + α ) = m
=
R
R
The diode peak current is :
V sin α
iD, peak = ωCVm cos α + m
R

Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003

15


Example
A half-wave rectifier has a 120V rms source at 60Hz. The
load is =500 Ohm, C=100uF. Assume α and θ are calculated
as 48 and 93 degrees respectively. Determine (a) Expression
for output voltage (b) peak-to peak ripple (c) capacitor
current (d) peak diode current.
vs
Vm

π /2

2π 3π /2

π

3π

4π

vo

Vmax
Vmin

∆Vo

iD

Vm = 120 2 = 169.7V ;

α

θ

θ = 93o = 1.62rad ;
α = 48o = 0.843rad
Vm sin θ = 169.7 sin(1.62rad ) = 169.5V ;
(a) Output voltage :
Vm sin(ωt ) = 169.7 sin(ωt )
vo (ωt ) =
V sin θ ⋅ e −(ωt −θ ) / ωRC

(ON)
(OFF)

m

=

169.7 sin(ωt )
169.5e −(ωt −1.62 ) /(18.85)

(ON)

(OFF)

Power Electronics and
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16


Example (cont’)
(b)Ripple :
Using : ∆Vo = Vmax − Vmin
∆Vo = Vm − Vm sin( 2π + α ) = Vm − Vm sin α = 43V
Using Approximation :
V
2π
169.7
= m =
= 56.7V
ωRC
fRC 60 × 500 × 100u

∆Vo = Vm

(c) Capacitor current :

ωCVm cos(ωt )
ic (ωt ) = Vm sin(θ ) −(ωt −θ ) /(ωRC )
−

R

⋅e

6.4 cos(ωt ) A
=
− 0.339 ⋅ e −(ωt −1.62 ) /(18.85)

(ON)
(OFF)
(ON)

A

(OFF)

(d) Peak diode current :
V sin α
iD, peak = ωCVm cos α + m
R
= (2 × π × 60)(100u )169.7 cos(0.843rad ) +
= (4.26 + 0.34) = 4.50 A

Power Electronics and
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169.7 sin(1.62rad )
500

17


Controlled half-wave
ig

vs

ia
+
vs
_

+
vo
_

ωt

vo

ωt
v
ig

α

Average voltage :

ωt

Vm
1 π
[1 + cosα ]
Vo =
Vm sin (ωt )dωt =
2π α
2π
RMS voltage
Vo, RMS =

π

2

1
[Vm sin (ωt )] dωt
2π α

Vm2 π
Vm
α sin (2α )
=
[1 − cos(2ω t ] dωt =
1− +
4π α
2
π
2π
Power Electronics and

Drives (Version 3-2003),
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18


Controlled h/w, R-L load
i

+
vR
_

+
vs
_

+
vL
_

+
vo
_

vs

π

ωt

2π

vo
io
β

α

−ωt

i (ωt ) = i f (ωt ) + in (ωt ) =
Initial condition : i (α ) = 0,
i(α ) = 0 =

Vm
⋅ sin (ωt − θ ) + Ae ωτ
Z
−α

Vm
⋅ sin (α − θ ) + Ae ωτ
Z
−α

A=−

Vm
⋅ sin (α − θ ) e ωτ
Z

Power Electronics and
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19


Controlled R-L load
Substituting for A and simplifying,
−(α −ωt )

Vm
⋅ sin (ωt − θ ) − sin (α − θ )e ωτ
Z

i (ωt ) =

0

for α ≤ ωt ≤ β

otherwise

Extinction angle β must be solved numerically
V
i (β ) = 0 = m
Z

−(α − β )

sin (β − θ ) − sin (β − θ )e ωτ

Angle γ = (β − θ ) is called the conduction angel.
Average voltage :
β

V
1
Vm sin (ωt )dωt = m [cos α − cos β ]
Vo =
2π α
2π
Average current :
β

1
Io =
i (ωt )dω
2π α
RMS current :
1 β 2
I RMS =
i (ωt )dω
2π α
The power absorbed by the load :
Po = I RMS 2 ⋅ R
Power Electronics and
Drives (Version 3-2003),
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20


Examples
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(ωt), (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.

Power Electronics and
Drives (Version 3-2003),
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21


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
it continuos
io

+

vR
_

+
vs
_

+
vL
_

+
vo
_

io

io

vo = 0
+
vs
_

vo = vs

+
vo

+

vo

io

_

_

D1 is on, D2 is off

D2 is on, D1 is off

Power Electronics and
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22


Operation of FWD
• Note that both D1 and D2 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.
Power Electronics and
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23


FWD- Continuous load current
• The inclusion of FWD results in continuos
load current, as shown below.
• Note also the output voltage has no
negative part.

output

vo
io
ωt

iD1

Diode
current

iD2
0

π

2π

Power Electronics and
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3π

4π

24


is

D3

io

D1

iD1

Full wave rectifier

+
vs
_

D4

•
•

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

D2

is

iD1

D1
+ vD1 −

+
vs1
_

+
vs
_

+
vs2
_

− vo

+

io

+ vD2 −
iD2

Center-tapped

D2

For both circuits,
Vm sin ωt

− Vm sin ωt

0 ≤ ωt ≤ π

π ≤ ωt ≤ 2π

Average (DC) voltage :
Vo =

Center-tapped
(CT) rectifier
requires
center-tap
transformer.
Full Bridge

(FB) does not.

+
vo
_

Full Bridge

vo =

•

1π

π

0

Vm sin (ωt )dωt =

2Vm

π

= 0.637Vm

Power Electronics and
Drives (Version 3-2003),
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25