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
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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
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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
π
2π 0
Output voltage (rms),
π
2
Vm
1
(Vm sin(ωt )dωt ) = = 0.5Vm
Vo , RMS =
2π 0
2
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Half-wave with R-L load
i
+
vR
_
+
vs
_
+
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
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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
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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 :
vL = L
di
dt
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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
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RMS current, Power
The average (DC) current is :
β
1
1 2π
Io =
i (ωt ) dωt =
i (ωt )dωt
2π 0
2π 0
The RMS current is :
β
1 2
1 2π 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 )
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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
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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.
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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
=−
ωRC
Vm sin θ ⋅
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
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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
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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
2π
ωRC
−
= Vm e
The ripple voltage is approximated as :
−
∆Vo ≈ Vm − Vm e
2π
ωRC
−
= Vm 1 − e
−
Using Series expansoin : e
∆Vo = Vm
2π
ωRC
2π
ωRC
=1−
2π
ωRC
V
2π
= m
ωRC
fRC
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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 )
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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
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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
α
θ
Vm = 120 2 = 169.7V ;
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)
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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
fRC 60 × 500 × 100u
ωRC
∆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 ) +
169.7 sin(1.62rad )
500
= (4.26 + 0.34) = 4.50 A
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Controlled half-wave
ig
vs
ia
+
vs
_
+
vo
_
ωt
vo
ωt
v
ig
α
Average voltage :
ωt
Vm
1 π
[1 + cosα ]
Vm sin (ωt )dωt =
Vo =
2π α
2π
RMS voltage
π
Vo, RMS =
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
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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 ) =
Vm
⋅ sin (ωt − θ ) + Ae ωτ
Z
Initial condition : i (α ) = 0,
−α
i(α ) = 0 =
Vm
⋅ sin (α − θ ) + Ae ωτ
Z
−α
A=−
Vm
⋅ sin (α − θ ) e ωτ
Z
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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
Vo =
Vm sin (ωt )dωt = m [cos α − cos β ]
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
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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.
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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
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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.
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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
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3π
4π
24
io
D1
iD1
Full wave rectifier
D3
is
+
vs
_
•
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
+
vo
_
D4
Full Bridge
D2
is
iD1
D1
+
vs1
_
+
vs
_
+
vs2
_
+ vD1 −
− vo
io
+ vD2 −
iD2
Center-tapped
+
D2
For both circuits,
vo =
Vm sin ωt
0 ≤ ωt ≤ π
− Vm sin ωt
π ≤ ωt ≤ 2π
Average (DC) voltage :
Vo =
1π
π
0
Vm sin (ωt )dωt =
2Vm
π
= 0.637Vm
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is
io
D1
iD1
Bridge waveforms
D3
+
vs
_
+
vo
_
Full Bridge
Vm
D4
D2
vs
2π
π
Vm
3π
4π
vo
vD1 vD2
-Vm
vD3 vD4
-Vm
io
iD1 iD2
iD3 iD4
is
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Center-tapped waveforms
is
iD1
D1
+
vs1
_
+
vs
_
+
vs2
_
+
io
+ vD2 −
iD2
Center-tapped
Vm
+ vD1 −
− vo
D2
vs
2π
π
Vm
3π
4π
vo
vD1
-2Vm
vD2
-2Vm
io
iD1
iD2
is
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Full wave bridge, R-L load
iD1
io
+
vR
_
+
vL
_
is
+
vs
_
+
vo
_
vs
π
2π
ωt
iD1 , iD2
iD3 ,iD4
io
vo
is
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Approximation with large L
Using Fourier Series,
∞
vo (ωt ) = Vo +
Vn cos(nωt + π )
n = 2, 4...
where the DC term
Vo =
2Vm
π
and the harmonics terms
2Vm
1
1
−
π n −1 n +1
The DC curent
Vo
Io =
R
The harmonic currents :
V
Vn
In = n =
Z n R + jnωL
Vn =
As n increases, Vn harmonic decreases.
Thus I n decreases rapidly very increasing n.
If ωL is large enough, it is possible to drop all
the harmonic terms, i.e. :
V
2V
i (ωt ) ≈ I o = o = m , for ωL >> R,
R
πR
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R-L load approximation
Approximate current
V
2V
Io = o = m ,
πR
R
(
)
I RMS = I o 2 + I n, RMS 2 = I o
Power delivered to the load :
Po = I RMS 2 R
vs
π
2π
ωt
iD1 , iD2
iD3 ,iD4
io
vo
is
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Examples
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
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T1
io
iD1
Controlled full wave, R load
T3
is
+
vs
_
+
vo
_
T2
T4
Average (DC) voltage :
Vo =
1π
πα
V
Vm sin (ωt )dωt = m [1 + cos α ]
π
RMS Voltage
π
2
1
[Vm sin (ωt )] dωt
=
Vo, RMS
πα
1 α sin (2α )
= Vm
−
+
2 2π
4π
The power absorbed by the R load is :
VRMS 2
Po =
R
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Controlled, R-L load
iD1
io
+
vR
_
is
+
vs
_
+
vL
_
+
vo
_
io
α
π β
π+α
2π
vo
Discontinuous mode
π+α
io
α
π
β
2π
vo
Continuous mode
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Discontinuous mode
Analysis similar to controlled half wave with
R - L load :
i (ωt ) =
[
Vm
⋅ sin(ωt − θ ) − sin(α − θ )e −(ωt −α ) ωτ
Z
for α ≤ ωt ≤ β
Z = R 2 + (ωL) 2
ωL
L
R
R
For discontinous mode, need to ensure :
and θ = tan −1
;τ =
β < (α + π )
Note that β is the extinction angle and
must be solved numerically with condition :
io ( β ) = 0
The boundary between continous and
discontinous current mode is when β in
the output current expression is (π + α ).
For continous operation current at
ωt = (π + α ) must be greater than zero.
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]
Continuous mode
i (π + α ) ≥ 0
sin(π + α − θ ) − sin(π + α − θ )e −(π +α −α ) ωτ ≥ 0
Using Trigonometry identity :
sin(π + α − θ ) = sin(θ − α ),
[
sin(θ − α ) 1 − e −(π
ωτ )
] ≥ 0,
Solving for α
α = tan
−1
ωL
R
Thus for continuous current mode,
−1 ωL
α ≤ tan
R
Average (DC) output voltage is given as :
2V
1 α +π
Vm sin (ωt )dωt = m cos α
Vo =
π
α
π
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Single-phase diode groups
D1
io
+
vs
_
D3
vp
+
vo
_
D4
D2
vn
vo =vp −vn
•
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 D3
reverses (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 vs is (-), D4 carry id. D2 reverses.
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Three-phase rectifiers
D1
+ van
-
io
D3
n
+ vbn
+ vcn
-
D5
-
vpn
D2
D6
vnn
+
vo
_
vo =vp −vn
D4
van
Vm
vbn
vcn
vp
Vm
vn
vo =vp - vn
0
π
2π
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3π
4π
37
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 .
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Three-phase, average voltage
vo
vo
π/3
Vm, L-L
0
π/3
2π/3
Considers only one of the six segments. Obtain
its average over 60 degrees or π 3 radians.
Average voltage :
Vo =
=
=
1
2π 3
π 3π
Vm, L − L sin(ωt )dωt
3
3Vm, L − L
π
3Vm, L − L
π
[cos(ωt )]π2π33
= 0.955Vm, L− L
Note that the output DC voltage component of
a three - phase rectifier is much higher than of a
single - phase.
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003
39
Controlled, three-phase
T1
-
+ van
io
T3
-
+ vbn
T5
n
+ vcn
vpn
+
vo
_
T2
vnn
T6
T4
α
van
vbn
Vm
vcn
vo
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003
40
Output voltage of controlled
three phase rectifier
From the previous Figure, let α be the
delay angle of the SCR.
Average voltage can be computed as :
Vo =
=
•
1
2π 3+α
π 3π
Vm, L − L sin(ωt )dωt
3+α
3Vm, L − L
π
⋅ cos α
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.
Power Electronics and
Drives (Version 3-2003),
Dr. Zainal Salam, 2003
41