DEPARTMENT OF ELECTRICAL ENGINEERING,

INDIAN INSTITUTE OF TECHNOLOGY, DELHI

POWER ELECTRONICS LABORATORY

EXPERIMENT NO. 4
PERFORMANCE ANALYSIS OF THREE PHASE RECTIFIER

OBJECTIVE:

1) PERFORMANCE ANALYSIS OF A THREE PHASE AC TO DC UNCONTROLLED

BRIDGE RECTIFIER
2) TO UNDERSTAND THE IMPACT OF NON-LINEAR LOAD (UNCONTROLLED
RECTIFIER) ON THE UTILITY SUPPLY.

BACKGROUND:

The main source of DC supply is battery. But in absence of that, we can get DC power from AC
supply using a three phase uncontrolled bridge rectifier. Bridge rectifiers cater to virtually all
requirements of a DC power source in the very common situation where nothing other than AC
supply is available. Diode bridges are easy to cool, extremely robust against voltage and current
surges, and quite cheap for the power ratings. The load of a bridge rectifier is not a pure DC, but a
rippled DC i.e. DC and superimposed AC component. To make the diode bridge rectifier a better
candidate for DC power supply, the output is normally passed through an L-C filter.

The impact of non-linear load on utility supply is well known. A resistive load connected to
utility via a three phase rectifier forms a non-linear load. Ideally for resistive loads, the input power
factor should be unity. But because of diode bridge rectifier, which is a non-linear load for the supply,
input power factor doesn’t remain unity. As the power factor is not unity, distortion in input supply
current is seen. This distortion in input supply current is measured in terms of THD i.e. Total
Harmonic Distortion.

Let the instantaneous current flowing through the inductor be ‘iL’. ‘iL’ is composed of both AC
and DC components.

iL = ∆iL + IL (1)

where,

∆iL = Ripple current

IL = Average current

Apply KCL at the junction of R, L and C.

iL = ic + Io (2)

where,

iC = Capacitor current

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 Io = Output Current

Comparing equations (1) and (2)

If Io becomes equal to IL, then ic becomes equal to ∆iL. This means capacitor takes the whole
ripple current and the output current we get is pure DC. But practically, it is not possible to get pure
DC current at output. So, we test with different capacitor values and see how each value affects the
ripple current and the output current.

In the laboratory, output of a rectifier board is tested with four sets of filter values.

Value of Inductance(L) Value of Capacitance(C) Cut-off Frequency(Hz)

Filter 1 5 mh 2200 µF 47.98
Filter 2 8.2 mH 2200 µF 37.47
Filter 3 5 mh 680 µF
Filter 4 8.2 mH 680 µF

IMPORTANT PRECAUTIONS:

1) Under all conditions, the electrolytic DC capacitor on the board assumes a fairly high voltage,
so that before you change any connections, this must be discharged. For this purpose, a
discharge switch has been provided close to the output of the rectifier board. In each
experiment where you use the rectifier board the following steps must be carried out before
you change any power circuit connections:
a) Turn off the AC supply feeding the rectifier board.
b) Turn the discharge switch ON, and then turn it OFF after a pause.
2) Before energising the AC side again, make sure that the discharge switch is in the ‘OFF’
position.

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CIRCUIT DIAGRAM:

Fig. 1. Circuit diagram of three phase uncontrolled bridge rectifier.

PROCEDURE:

1) Make the connections as per the circuit diagram.

2) Select the connecting cables according to the current ratings of the various equipments.
3) Initially keep the auto-transformer at zero-position and the rheostat value should be set such
that current in the ammeter doesn’t cross 2A.
4) For 1 set of filter values, apply input voltage of 50V to the rectifier.
5) Now vary the rheostat so that the load current goes from 0 to 2 A. And for different values of
load current, measure average load voltage and load ripple voltage (peak-peak) from DSO
and measure input power factor and THD of input source current from power analyser.
6) Repeat the above step with input voltage of 100 V.
7) Plot a graph: Load Current vs % of Ripple Voltage
8) Now, connect L and C of the second filter set.
9) Repeat the steps 4,5,6 and 7.

% of Ripple Voltage = Load ripple voltage (Peak-peak)/Average load voltage

For the FFT analysis:

1. Set Autotransformer to desired AC voltage.

2. Connect the differential probe to channel-1 (x100 setting).
3. PRESS Channel-1 Button→Coupling→set AC.
4. Move the channel-1 cursor to ground line (move the waveform to top portion of screen) .
5. Press “MATH” button→ set operator FFT.
6. Press “SPAN” button and adjust 1 KHz, centre frequency to zero (now the FFT spectrum in
light pink colour will appear on the screen) .
7. Press”more FFT” button → set vertical units to V RMS.
8. To adjust the magnitude of the FFT waveform use “PUSH for FINE” button (set to 100 mV).
9. To adjust the offset of the FFT waveform use “PUSH to ZERO” button (set to 300 mV).
10. Press cursor button and bring the X1 cursor to middle of the scope screen, then press X2
cursor and adjusting it on the screen. With this X1 and X2 one can able to identify different
harmonic frequency components on the scope screen.
11. To move the” X1, X2”cursors rotate the “cursors knob.”

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Formula used: Average load voltage = (3×Vml) /Π; where Vml = Peak of input line voltage

OBSERVATION TABLE:

For L = 5 mH, C = 2200 µF ; Cut-off Frequency = 47.98 Hz

AC source voltage = 50 V, Average load voltage =

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current

AC source voltage = 100 V, Average load voltage =

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current

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 For L = 8.2 mH, C = 2200 µF ; Cut-off Frequency = 37.47 Hz

AC source voltage = 50 V, Average load voltage =

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current

AC source voltage = 100 V, Average load voltage =

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current

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Ripples in Output
Voltage

Ripples in
Inductor Current

Capacitor Current

INFERENCE:
After conducting the experiment, it has been seen that the load voltage of a bridge rectifier is
not a pure DC, but a rippled DC i.e. DC and superimposed AC component. To minimise these ripples,
the output is passed through an L-C filter. After using the filter, it has been observed that the % of
ripple voltage has reduced. A resistive load with diode bridge rectifier forming a non-linear load and
hence considerable amount of waveform distortion was seen in the utility supply.

It is also seen that with higher values of capacitor, ripples in the output current reduces as
more and more ripples in the source current flows through the capacitor.

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SAMPLE READINGS:

For L = 5 mH, C = 2200 µF ; Cut-off Frequency = 47.98 Hz

AC source voltage = 50 V, Average load voltage = 67.8 V

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current
1) 0.4 912.5 mV 0.66

2) 0.6 1.275 0.93

3) 0.8 1.475 1.08

4) 1 1.6375 1.19

5) 1.2 1.8375 1.34

6) 1.3 1.912 1.39

AC source voltage = 100 V, Average load voltage = 137

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current
1) 0.6 987.5 mV 0.72
2) 0.8 1.175 0.86
3) 1.2 1.3375 0.98
4) 1.4 1.4375 1.05
5) 1.6 1.625 1.19
6) 1.8 1.75 1.28

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 Fig. 2. Ripple voltage variation with load current/power.

For L = 8.2 mH, C = 2200 µF, Cut-off Frequency = 37.47 Hz

AC source voltage = 50 V, Average load voltage = 67.8 V

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current
1) 0.4 787.5 mV 1.16

2) 0.5 937.5 mV 1.38

3) 0.6 1 1.47
4) 0.7 1.1 1.62
5) 0.8 1.225 1.801
6) 0.9 1.325 1.95

AC source voltage = 100 V, Average load voltage = 137

S.No Load Current Load Ripple Voltage % of Ripple Input Power THD of
(in Amperes) (Peak-Peak) Voltage Factor Input Source
(in Volts) Current
1) 0.6 1.025 0.804

2) 0.8 1.1375 0.83

3) 1 1.3625 0.99
4) 1.2 1.425 1.04

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5) 1.4 1.5375 1.12
6) 1.6 1.6 1.17

Fig. 2. Ripple voltage variation with load current/power.

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Experiment: Measurement of Harmonic components in the AC to DC rectifier output
waveform

1. Turn on the Keysight DSOX3014A.

2. Connect the differential probe on channel1.
3. Setup DSO to the default (for removing previous data or settings from DSO memory),
then do auto setup, you should get AC waveform like as shown in fig. 1, go to
measure and measure the frequency and peak to peak voltage (as shown in the right
corner of fig. 1).

Fig. 1

4. Go to mode/coupling located on the front panel or press probe key and select
coupling-AC.
5. Press Math Function and select operator FFT; you will get a pink trace as shown in
fig. 2.
6. Set centre frequency to 0 Hz and span to 2 Khz.

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 FFT
Magnitude
adjustment

FFT frame
offset
adjustment

Cursor
adjustment

Harmonic
magnitude
measurements

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 Fig. 2

7. Go to option: More FFT and then set the vertical units to Vrms (Not dB-scale), then
math function will show you rms value of all the frequency components then this
will look like as shown in fig. 3 .

Fig. 3

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8. Set the offset and scale using keys available on front panel. Here FFT resolution is
important factor which can be changed using horizontal scale key .

Fig. 4

9. Turn on the cursors and select the source to Math and record the readings on Cursor
X1, X2 and Y1, Y2. In this way one can analyze DC component in AC
output(harmonic magnitudes).

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