ENEL 585 Experiment 1
The Single-Phase Full-Bridge Diode Rectifier
Purpose:
To become familiar with the single-phase diode bridge rectifier. Operation with three types of passive
filtering (L, C, and LC) is investigated.
Introduction:
In this experiment you will investigate an old method of obtaining a low ripple dc voltage source from an ac
supply. A pure resistor is employed as the load.
The voltage and current waveforms in this experiment are sampled using National Instruments 1 MSps data
acquisition hardware and the Lab Windows CVI software. The waveforms are multiplex sampled, hence the
sampling frequency decreases as more waveforms are displayed. The lab 1 hardware and software,
employed in this lab, were kindly developed by Sean Victor Hum, Ed Evanik, Jason Pannell, Richard
Galambos and Arif Al-Judi. The hardware for Labs 2 onward was kindly developed at the U of Minnesota.
Tutorial:
The use of a four diode full-bridge implies full-wave ie, 120Hz ripple, rectification. The reverse relationship
is not necessarily true: Full-wave rectification may also be accomplished using a half-bridge diode rectifier
fed by a 2-phase ac source (a.k.a. split ac), as may be obtained from a single-phase source using a centretapped transformer. There is another type of half-bridge diode rectifier, consisting of a rectifying diode and
free-wheeling diode, fed directly by a single- phase source. This converter provides half-wave rectification,
ie, with 60Hz ripple. Of course a single diode may be used for half-wave rectification, generally employed
for low-inductance or pure resistive loads. Interestingly the single diode rectifier is not referred to as a
quarter-bridge Rectifier (Topologically, we could refer to the single diode rectifier as a single-phase twowire half-bridge rectifier. What a mouthful huh?).
(a)
(b)
Figure 1: Bridge diode rectifier with possible L and/or C filter component
EXPERIMENT: 1 – The Single-Phase Full-Bridge Diode Rectifier
Shown in Fig. 1 (a) is a schematic of a full-bridge diode rectifier with single-phase ac source, and shown in
Fig. 1 (b) is a schematic of a full-bridge diode rectifier with single-phase ac source. Yup, they are identical.
In ENEL 585 we will use the Fig. 1 (b) representation.
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For completeness of discussion, we can note that there exists an analogy between single-phase and 3-phase
diode bridges. A single-phase full-bridge (2-wire 4 diode) rectifier produces 120Hz ripple. A 3-phase fullbridge (3-wire 6 diode) rectifier produces 360Hz ripple. A single-phase half-bridge (2-wire 1 or 2 diode)
rectifier produces 60Hz ripple. A 3-phase half-bridge (4-wire 3 diode) rectifier produces 180Hz ripple.
To understand the single-phase full-bridge operation, refer to Fig. 1 and consider the case where the line
voltage, aka the source voltage, vs , goes positive. For the moment, assume no filter components are present
and the load is purely resistive. It makes sense that diode, D1 , will tend to turn on (since its anode is going
positive). At the same time, diode, D4, will tend to turn on (since its cathode is going negative). Thus, with
both D1 and D4 on, the output voltage is positive. During the next half cycle, ie, the ac source goes negative,
diodes, D2 and D3 , now turn on simultaneously, and again the output voltage is positive. In this way, the
current that flows into the load is always positive. Referring to Fig. 1 (a), can you see the symmetry of
circuit operation?
When do you use each type of filtering? This is a function of the load requirement. You use as little filtering
as possible, but generally, at lower power levels (less than 1 kW) a single filter capacitor is effective, filterwise and cost-wise. Whereas at higher power levels, a single filter inductor may payoff, since the peak
diode current will be reduced in this case, permitting the use of less expensive diodes. Both Lf and Cf are
used only if a lower ripple voltage is needed, and the peak input current is a concern.
In the lab, we will be calculating input power factor (c.f. Lecture Notes). You may also wish to calculate
some average and rms quantities (c.f. Lecture Notes) and compare to the software calculation.
Pre-Lab (one per lab group collected at the start of the lab):
1. Consider a single-phase diode bridge rectifier having an ac source of 10V amplitude, a resistive load of
22Ω, and no filtering components. Assuming ideal diodes, calculate (approximately if necessary) the
following quantities. Include well-labeled sketches of waveforms for the calculations. Please calculate:
average load voltage,
peak-to-peak (ie, ripple) load voltage,
rms load voltage,
peak diode current,
rms diode current,
rms line (or source) current,
input power factor.
EXPERIMENT: 1 – The Single-Phase Full-Bridge Diode Rectifier
If no filter components (Lf or Cf) are present in the converter, then the load voltage, vo , will simply be a
rectified sine wave, having an amplitude equal to the ac source (less two diode drops). Assuming a resistive
load, R, then having the filter inductor alone present will produce first-order filtering (with time constant
Lf / R) of the rectified sine wave, giving a lower ripple (ie, peak-to-peak) output voltage. One might think
that having the filter capacitor alone present would also produce first-order filtering (with time constant
RCf). This is not the case. To better understand this circuit consider its operation in the time-domain, that is,
as the filter capacitor is charged up, the diode pairs, D1 D4 and D2 D3, need only turn on near the peaks of the
ac source to just "top off" the voltage on the capacitor (remember the load is continually "bleeding" the
capacitor). As a result, the output voltage (ie, the capacitor voltage) is approximately equal to the peak line
voltage and has a more or less saw-tooth ripple component (can you see what the actual shape of the ripple
component is, assuming a resistive load?). It is also possible to build the rectifier with both filter inductor
and filter capacitor present, in which case you roughly obtain 2nd-order filtering of the rectified sine wave
(especially if Lf is big enough).
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2. Repeat the analysis of problem 1, if a filter capacitor of 4,700 µF is added to the converter circuit. Hint: If
you have trouble finding the conduction angle of the diodes, ie, the Δωt that each diode is on for, you may
use the fact that it is about 25° (this value is not exact, but allows you to make calculations).
3. Repeat the analysis of problem 1, if a filter inductor of 0.6H is added to the converter circuit (ie, no filter
capacitor). Hints: To find the output current ripple (and hence the ripple voltage), first realize that the
output of the bridge itself is a rectified sine wave having dc and ac components. Note that the dc component
appears entirely across the resistance in the circuit (this doesn't preclude an ac component across the
resistance as well). Note that the ac component has a fundamental at 120Hz. Assume the peak-to-peak
voltage of the 120Hz fundamental is roughly equal to the peak-to-peak voltage of the ac component itself.
Now - think of the inductor as an impedance: Hint: ZLf ≈ 2πf Lf = 2ωLf , where f=120Hz and ω = 377rad/s
(this is not a typo, the power industry often sets ω = 2π60Hz = 377rad/s as a constant). What is the
impedance of the series LR combination? You should now be able to approximately determine the peak-topeak output ripple current and then use Ohm’s Law to find peak-to-peak ripple load voltage.
4. Repeat the analysis of problem 1, if a filter inductor of 0.6H and a filter capacitor of 4,700µF are used in
the converter circuit. To simplify this analysis, assume the capacitor is ideal (in practice its equivalent
series resistance would be considered), assume the inductor is ideal, and let the load resistance be 22Ω.
There are two possible ways to analyze the problem. One way is to proceed along the lines of problem 3,
noting that the load voltage may be approximated as pure dc. Once the inductor current ripple is obtained,
it may be integrated to obtain the capacitor (ie, load) voltage ripple. Another analysis approach is to
consider the LCR elements as a passive low-pass filter having a 3dB corner frequency of f3dB=1/2π LC
and an attenuation slope of 40dB per decade. If you perform both analyses, you can check your results.
1. A single-phase diode bridge module will be provided to your lab group. This module is supplied by the
120V (rms of course) mains, stepped down by a 120:6.3 transformer (In days of old, 6.3Vrms was a very
common AC Voltage used to heat the filament of a vacuum tube.). The step down transformer provides
isolation, and the desired ac voltage, vs(t), indicated in Fig. 1. For this lab, we can assume that the line
inductance is negligible. There is a power switch on the transformer box. Turn power off whenever making
any circuit changes.
2. Connect a resistive load of 22Ω across the load output terminals of the rectifier module. If unsure about
the wiring have it checked. Apply power.
3. Observe and save the source and load voltage waveforms (it is best to use the print screen key and copy
the waveform into a word document – or you may also try saving the waveform on the U:\ drive). What do
you notice about the output voltage compared to theory where we assumed the converter has idea diodes?
Note the average and rms source and load voltages. This can be done in CVI by expanding on the graphs
and selecting the voltage waveform.
4. Observe the source voltage, with the load and with no load.
5. Observe and save the source and load current waveforms for the resistor load.
6. Now add a large inductor (0.5 H) to the output side of the bridge rectifier. N.B. Never physically
disconnect an inductor while a circuit is operating. This produces a large di/dt and a damaging spark.
7. Observe and save the source, bridge output, and load voltage waveforms. What do you notice about the
bridge voltage compared to theory where we assumed the converter is lossless?
EXPERIMENT: 1 – The Single-Phase Full-Bridge Diode Rectifier
Procedure (a lab book is not required, but neat notes must be taken):
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8. Observe and save the source and load current waveforms. You may also observe and save a diode current
waveform. In what mode is the load inductor operating?
9. Remove the inductor (power should be off) and replace it with a short. Add a filter capacitor of 4700µF.
Watch capacitor polarity as you plug the cap in the circuit!
10. Now that you have a good idea of what you are doing, investigate circuit operation. Note diode on
voltage as a function of time. Remove the load resistance and observe the average output voltage.
11. Finally, investigate rectifier operation with an LC filter.
Analysis:
As you analyze your observations, two points about the measurement of device currents will be of possible
use: (a) the data acquisition system determines device currents by actually measuring the voltage across a
small (0.2Ω) shunt resistance, placed in series with the actual device, and (b) the 4,700µF electrolytic
capacitor used in this experiment has a measured equivalent series resistance (ESR) of about 0.05Ω.
Compare your observations to theory. Consider the following:
1. (step 3 of the procedure) For the bridge rectifier with a pure resistive load and no filtering, how does the
output voltage waveform compare to what you expected with ideal diodes?
3. (step 4 of the procedure) For the bridge rectifier with no load and no filtering, how does the output
voltage waveform compare to what you expected with ideal diodes? Hint: the data acquisition system has
an input impedance of 10kΩ.
4. (step 7 of the procedure) In the case of inductive filtering, why is the average output voltage so much less
than in theory? Hint: examine the experimental voltage waveform across the bridge output.
5. (step 8 of the procedure) How does the inductor current ripple compare to theory. Can you suggest some
explanation for the difference? Hint: Even though the inductor is a high quality inductor, it is designed for
low power operation and has significant leakage as well as resistive and core losses.
6. (step 10 of the procedure) How does the source current waveform compare to theory in the case of pure
capacitive filtering. Can you suggest some explanation for the differences? Hint: Once again, we note that
our 120:6.3 transformer is designed for low power and hence has a high output impedance.
7. Calculate power factor (PF) from experimental data (at least for the LCR case, but time permitting, do the
calculation for all four cases: R, LR, CR, LCR), compare to theory, and discuss. Hint: To calculate PF, you will
need to find Pin , where Pout = η Pin , and η is efficiency, or: Pin =Pout+ Plosses , then find PF=Pin /(Vs(rms) x Is(rms)).
First find Pout= Vout x Iout, then for simplicity consider the power losses are only due to the resistance of the
inductor rL and power losses in the rectifying diodes Pd, neglecting any other losses. Pd can be approximated
by (Iout x 1.4), where the 1.4V is the voltage drop across two diodes in series. Measure the resistance of the
inductor rL using a multimeter (expected value is approximately 10 ohm), and calculate PrL = (Iout)2x rL. Find
Pin then find PF. Alternatively, use the avg diode current and rms inductor current to calculate the losses.
EXPERIMENT: 1 – The Single-Phase Full-Bridge Diode Rectifier
2. (step 4 of the procedure) What happens to the source voltage waveform when the load is removed? Hint:
Our 120:6.3 transformer is designed for low power and hence has high values of leakage as well as high
resistive and core losses.
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