3 Phase Rectifier with Resistive Load
D1
D2
D3
a
V1
+
V2
R1
10
b
n
+
vo(t)
V3
c
D4
D5
D6
Voltage Waveforms:
Line to Neutral Voltages:
V(a)
100V
V(b)
V(c)
80V
60V
40V
20V
0V
-20V
-40V
-60V
-80V
-100V
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
D1,D2,D3 select the highest positive voltage for Output + while D4,D5,D6, select the
lowest negative voltage for Output - :
V(+)
100V
V(-)
80V
60V
40V
20V
0V
-20V
-40V
-60V
-80V
-100V
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
The output voltage is the difference between these two: Please compare the output
voltage with eth line to neutral voltage Van and the line to line voltage Vab, The
ouput voltage rides along the peaks of the line to line voltages. The output voltage has
six ripple pulses per cycle. This is often called a six pulse rectifier.
V(+,-)
180V
V(a,b)
V(a)
150V
120V
90V
60V
30V
0V
-30V
-60V
-90V
-120V
-150V
-180V
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
Current Waveforms:
The output current follows io(t) follows the voltage across the resistor:
20.0A
19.0A
18.0A
17.0A
16.0A
15.0A
14.0A
13.0A
12.0A
11.0A
10.0A
9.0A
8.0A
7.0A
6.0A
5.0A
4.0A
3.0A
2.0A
1.0A
0.0A
0ms
I(R1)
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
When Van is the highest phase voltage D1 conducts and the load current flows out of
phase a supplies the load current. When Van is the lowest phase voltage D4 conducts
and the load current flows back into Phase a. It is important to note taht only one top
diode conducts at a time and only one bottom diode conducts at a time. The following
plot shows the currents in D1 and D4 overlayed onto the Van waveform.
V(a)
100V
I(D1)
I(D4)
18A
80V
16A
60V
14A
40V
12A
20V
10A
0V
8A
-20V
6A
-40V
4A
-60V
2A
-80V
-100V
0ms
0A
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
-2A
100ms
And the AC current ia(t) looks like:
-I(V1)
18A
15A
12A
9A
6A
3A
0A
-3A
-6A
-9A
-12A
-15A
-18A
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
For reference notice how the three ac currents Ia, Ib and Ic line up to provide a
continuous outward and return path for the load current:
-I(V1)
18A
-I(V2)
-I(V3)
15A
12A
9A
6A
3A
0A
-3A
-6A
-9A
-12A
-15A
-18A
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
Analysis of 3 phase rectifier with resistive load:
Notation:
Let Vm = Peak line to neutral voltage
π
Useful Integration formula:
∫ π cos
6
−
2
(ωt )dωt =
6
π
6
+
3
4
3 × Vm
1. Peak Output Voltage = peak of the line of line voltage =
2. Average Value of the output voltage may be got by averaging over a single
output pulse and using the fact that the output voltage follows a line to line voltage
π
∫π
6
−
waveform for each pulse. Vo =
3 × Vm cos(ωt )dωt
6
=
2π
6
3 3
π
Vm = 1.654Vm
3. Similarly we can calculate the rms output voltage by integrating over a single
pulse:
π
∫π(
6
Vo ( rms ) =
−
)
2
3 × Vm cos(ωt ) dωt
6
= Vm
2π
6
3 9 3
+
= 1.655Vm
2 4π
4. Since the output current for a resistive load is just vo(t)/R we can now calculate the
rectification ratio (efficiency)
Vo I o
V2 R
Re ctification Ratio =
= 2o
=
Vo ( rms ) I o ( rms ) Vo ( rms ) R

3 3


Vm

 π


2

V 3 + 9 3
 m 2 4π

27




2
=
π2
3 9 3

 +
 2 4π 


= 0.998 or 99.8%
5. In order to calculate the ripple factor we must extract the ac component from Vo
using the relationship: Vo ( ac ) = Vo2( rms ) − Vo2
The Ripple factor =
Vo ( ac )
Vo
=
Vo2( rms ) − Vo2
Vo
=
Vo2( rms )
V
2
o
−1 =
1
− 1 = 0.04 or 4%
0.998
6. In order to calculate the transformer utilisation factor we need to calculate the
rms supply current. The easiest way to do this is to notice that ac input current
waveform is comprised of four pulses of width 2π/6 over a single cycle of width 2π
(remember that positive and negative current contributes equally to the rms).
RMS Ac input current = I s =
2
 3 × Vm cos(ωt ) 
 dω t
4 × ∫ 6π 

− 
R
V
V
3 3
6

= m 1+
= 1.352 m
R
R
2π
2π
π
Transformer Utilisation Factor =
V I
(1.654V m )(1.654Vm / R )
DC output power
= o o =
= 0.954
Vm
Vm
3 xV s I s
Input VA
3( )(1.352 )
R
2
Three Phase Rectifier with highly inductive load.
L1
D1
D2
D3
100mH
a
V1
R1
10
b
V2
+
n
V3
+
vo(t)
c
D4
D5
D6
The ripple frequency in a six pulse rectifier is 6xf where f is the mains frequency. The
1
cut off frequency of an LR filter is
. If this cut off frequency is much lower than
L
2π
R
1
then the ripple will be almost entirely
the ripple frequency i.e. if 6 f >>
L
2π
R
eliminated leaving a smooth DC load current. The output voltage becomes equal to its
average DC value.
Voltage Waveforms:
Plot showing Van, Vab and vo(t)
V(a)
180V
V(a,b)
V(+,-)
150V
120V
90V
60V
30V
0V
-30V
-60V
-90V
-120V
-150V
-180V
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
The smooth DC output voltage results in a pure DC output current so the diode
currents become square pulses and the input ac current is also "squared up".
Current Waveforms with Inductive Load:
Output Current io(t)
20.0A
19.0A
18.0A
17.0A
16.0A
15.0A
14.0A
13.0A
12.0A
11.0A
10.0A
9.0A
8.0A
7.0A
6.0A
5.0A
4.0A
3.0A
2.0A
1.0A
0.0A
0ms
I(R1)
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
Diode Current Waveforms for D1 and D4. Notice how much squarer these are than
the corresponding waveforms for a resistive load:
20.0A
19.0A
18.0A
17.0A
16.0A
15.0A
14.0A
13.0A
12.0A
11.0A
10.0A
9.0A
8.0A
7.0A
6.0A
5.0A
4.0A
3.0A
2.0A
1.0A
0.0A
0ms
I(D1)
10ms
20ms
I(D4)
30ms
40ms
50ms
60ms
70ms
80ms
90ms
AC input current for phase a shown with line to neutral voltage for reference:
V(a)
100V
-I(V1)
18A
15A
80V
12A
60V
9A
40V
6A
20V
3A
0V
0A
-20V
-3A
-6A
-40V
-9A
-60V
-12A
-80V
-100V
0ms
-15A
-18A
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
And all three ac currents are shown meshing together in the next graph:
-I(V1)
18A
-I(V2)
-I(V3)
15A
12A
9A
6A
3A
0A
-3A
-6A
-9A
-12A
-15A
-18A
0ms
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
Analysis of 3 phase rectifier with highly inductive load:
Notation:
Let Vm = Peak line to neutral voltage
1. Peak Output Voltage = average output voltage = the average of the unfiltered
output voltage of the rectifier = 1.654Vm from before
2. Average Value of the output voltage is the same as before (the inductore
attenuates ripple but does not affect average output voltage)
π
∫π
6
Vo =
−
6
3 × Vm cos(ωt )dωt
2π
6
=
3 3
π
Vm = 1.654Vm
3. The rms output voltage is trivial in this case because there is no ripple
Vo ( rms ) = Vo = 1.654Vm
4. Since the output current for a resistive load is just vo(t)/R we can now calculate the
rectification ratio (efficiency)
Vo I o
V2 R
Re ctification Ratio =
= 2o
= 100%
Vo ( rms ) I o ( rms ) Vo ( rms ) R
5. Since the ripple is vanishingly small Vo ( ac ) ≈ 0 ⇒
Vo ( ac )
≈ 0%
Vo
6. In order to calculate the transformer utilisation factor we need to calculate the
rms supply current. The easiest way to do this is to integrate over a half cycle. In each
half cycle there is a square pulse of current of width 2π/3 and amplitude = Io =
1.654Vm/R
π
2
 1.654Vm 
∫−π3  R  dωt
3
RMS Ac input current = I s =
π
=
1.654Vm
R
2 1.35Vm
=
3
R
Transformer Utilisation Factor =
V I
(1.654Vm )(1.654Vm / R )
DC output power
= o o =
= 0.955
Vm
Vm
3 xVs I s
Input VA
3( )(1.35 )
R
2
Conclusion: The three phase bridge rectifier with resistive load has significantly
lower output voltage ripple and significantly better transformer utilisation than an
unfiltered single phase bridge rectifier. The addition of a sufficiently large filtering
inductor can result in negligible output voltage ripple but does not substantially
improve transformer utilisation over the unfiltered three phase rectifier,