Full-wave rectifier — centre-tapped and bridge, with and without a capacitor filter
Full-wave rectifier experiment: bridge and centre-tapped connections, CRO waveforms, ripple factor and rectification efficiency from your own readings, the effect of a shunt capacitor, PIV, and viva questions with answers.
Aim
To construct full-wave rectifiers in both the centre-tapped and bridge configurations, observe the output waveform on a CRO, and determine the ripple factor, rectification efficiency and percentage regulation with and without a shunt capacitor filter.
Apparatus required
| Apparatus | Specification | Qty |
|---|---|---|
| Step-down transformer | 230 V / 12-0-12 V, 1 A, centre-tapped | 1 |
| Silicon rectifier diodes | 1N4007, 1 A, 1000 V PIV | 4 |
| Electrolytic capacitors | 100 µF and 470 µF, 63 V | 1 each |
| Load resistor / decade box | 220 Ω to 10 kΩ, 5 W | 1 |
| Cathode ray oscilloscope | Dual trace, 20 MHz | 1 |
| Digital multimeter | DC and AC volts, true-RMS preferred | 2 |
| Breadboard and patch cords | — | 1 set |
Theory
A rectifier converts alternating voltage into unidirectional voltage. A half-wave rectifier passes one half of each cycle and wastes the other, so its output pulses at the supply frequency and its DC content is low. A full-wave rectifier uses both halves: the output pulses at twice the supply frequency — 100 Hz from a 50 Hz mains — and delivers twice the average voltage from the same transformer secondary.
There are two ways to do it. The centre-tapped circuit uses two diodes and a transformer whose secondary is split about a common tap; each diode conducts on alternate half cycles, and each must withstand a peak inverse voltage of 2Vm because the non-conducting diode sees the whole secondary. The bridge circuit uses four diodes and no centre tap; two diodes conduct in series on each half cycle, so the PIV per diode is only Vm, but two forward drops — about 1.4 V in silicon — are lost instead of one.
For an ideal full-wave rectifier the average output is V(dc) = 2Vm/π = 0.637Vm and the RMS output is Vm/√2. The ripple factor, defined as the ratio of the RMS value of the AC component to the DC value, follows as γ = √((V(rms)/V(dc))² − 1) = 0.482, and the maximum theoretical rectification efficiency is 81.2 %. Those three numbers are what the experiment sets out to confirm, and the measured values will fall short of them because of diode drops and winding resistance.
A shunt capacitor across the load turns the rectifier into a peak detector: the capacitor charges to near Vm and then discharges into the load between peaks, so the output rises towards Vm and the ripple collapses. For a capacitor-input filter the ripple factor is γ = 1/(4√3 f C R(L)), with f the supply frequency. Every term is a design lever — a bigger capacitor or a lighter load (larger R(L)) gives smoother output — and the formula also explains why regulation gets worse as the load draws more current.
Percentage regulation measures how far the output sags between no load and full load: it is (V(nl) − V(fl))/V(fl) × 100. A rectifier with a capacitor filter regulates poorly because the sag comes from the capacitor having less time to recharge as load current rises, which is why a real supply follows the rectifier with a regulator.
Circuit connections
Check every point below against your board before switching on. There is no diagram here on purpose — a wrong diagram is worse than none, and this is the list a demonstrator actually walks through with you.
- Transformer primary to the 230 V mains through a fuse; secondary ends to the two AC corners of the bridge, or to the two outer ends of the centre-tapped winding for the two-diode version.
- Bridge DC corners to the load: the junction of the two cathodes is positive, the junction of the two anodes is negative. For the centre-tapped version the tap is the negative rail and the joined cathodes are positive.
- Load resistor across the DC output; the filter capacitor, when used, goes in parallel with the load, positive terminal to the positive rail.
- CRO across the load with the probe ground on the negative rail; DMM in DC mode across the load for V(dc) and a second DMM in AC mode across the same points for the ripple.
- Never earth the CRO probe on the transformer secondary while it is also earthed at the load — that shorts part of the winding through the instrument.
Procedure
- 1Test all four diodes with the multimeter's diode range before building: about 0.6 V one way and open-circuit the other. One reversed diode in a bridge is a dead short across the secondary.
- 2Build the bridge circuit without the filter capacitor. Have the connections checked, then switch on.
- 3Record V(dc) on the DC meter and the ripple voltage on the AC meter, and trace the output waveform from the CRO — note that the pulse repetition rate is twice the supply frequency.
- 4Switch off, connect the 100 µF capacitor across the load, switch on and repeat the readings. Repeat once more with 470 µF. The waveform should change from full humps to a sawtooth riding on a DC level.
- 5Vary the load resistance over its range at a fixed capacitance and record V(dc) and the ripple at each setting, so regulation and the load-dependence of ripple can both be plotted.
- 6Take the no-load output voltage with the load disconnected, for the regulation calculation.
- 7Rebuild as a centre-tapped rectifier with two diodes and repeat the measurements, noting that the output voltage is now set by half the secondary.
- 8Compute the ripple factor, efficiency and regulation for each case and compare with the ideal values.
Work out your readings
Type in the numbers off the meters. This fills the tabular column, works the calculation through step by step, plots the characteristic — and tells you when a reading cannot physically be right, which is the part a manual can't do. Everything stays on this device, and it works with the network off.
Nameplate and machine data
Fill in the machine data to see the results, the worked calculation and the curve.
Precautions
- Discharge the filter capacitor through a resistor before touching the circuit. A 470 µF capacitor charged to 17 V holds its charge after the supply is switched off.
- Observe electrolytic capacitor polarity. Reversed electrolytics vent, and they do it loudly.
- Check the diode orientation twice before switching on; a single reversed diode in the bridge short-circuits the transformer secondary.
- Keep the load resistor within its power rating: V(dc)²/R(L) must stay below the wattage marked on it.
- Use an isolation transformer, or keep the CRO ground on the circuit's negative rail only — connecting a mains-earthed CRO ground to a live point earths that point through the instrument.
- Switch off before changing the capacitor or the load, not while the circuit is energised.
Sources of error
Every record asks for these, and every record gets the same three lines copied from the one before. These are the errors this particular experiment actually has.
- Diode forward drops: the bridge loses about 1.4 V and the centre-tapped circuit about 0.7 V, so the measured V(dc) is always below 0.637Vm and the shortfall is largest at low secondary voltages.
- Ordinary averaging multimeters are calibrated for sine waves and misread the ripple of a rectified waveform; a true-RMS meter, or the CRO's own RMS measurement, is needed for an honest ripple figure.
- Transformer winding resistance and leakage reactance, which make the secondary voltage droop as load current rises and inflate the apparent regulation.
- Electrolytic capacitors carry tolerances of −20 % to +80 % and lose capacitance with age, so the measured ripple rarely matches the value computed from the marked capacitance.
- The AC-coupled ripple reading includes mains-borne noise picked up by long leads, which matters once the ripple itself is down in the tens of millivolts.
Viva questions with answers
What is the ripple frequency at the output of a full-wave rectifier fed from a 50 Hz supply?
100 Hz. Both halves of the input cycle produce an output pulse, so the fundamental ripple component is at twice the supply frequency. A half-wave rectifier on the same supply would ripple at 50 Hz, which is one way to tell the two apart on a CRO without tracing the circuit.
Why is the PIV of a centre-tapped full-wave rectifier 2Vm while a bridge is only Vm?
In the centre-tapped circuit the non-conducting diode has the full secondary voltage across it: its cathode sits at +Vm through the conducting diode while its anode is at −Vm, giving 2Vm. In the bridge, the two non-conducting diodes share the reverse voltage with the load, so each blocks only Vm. That is why the bridge is preferred at high voltages.
State the theoretical ripple factor and efficiency of a full-wave rectifier.
The ripple factor is 0.482 and the maximum rectification efficiency is 81.2 %. Both follow from V(dc) = 2Vm/π and V(rms) = Vm/√2 for the unfiltered output. The corresponding half-wave figures are 1.21 and 40.6 %, so full-wave rectification is better on both counts.
How does a shunt capacitor reduce ripple, and what is the penalty?
The capacitor charges to near the peak and supplies the load between peaks, so the output approaches Vm instead of 0.637Vm and the ripple falls as γ = 1/(4√3 f C R(L)). The penalty is that the diodes conduct in short, high current pulses to refill the capacitor, which raises the peak repetitive current rating needed and worsens the transformer's power factor.
Why does the regulation get worse when the filter capacitor is added?
Without a filter the output follows the secondary and sags only by the winding resistance drop. With a capacitor the no-load output rises to nearly Vm, so there is much further to fall; as load current increases the capacitor discharges more between peaks and the average drops sharply. The larger no-load value in the numerator is what makes the percentage regulation figure worse.
What is the advantage of a bridge rectifier over a centre-tapped one apart from PIV?
The bridge uses the whole secondary winding on both half cycles rather than half of it at a time, so for the same DC output it needs a transformer of about half the secondary voltage rating and makes better use of the copper. It also removes the need for a centre tap, which makes off-the-shelf transformers and packaged bridge modules usable.
What would you observe if one diode in a bridge rectifier went open circuit?
The circuit reverts to half-wave operation: the output pulses at 50 Hz instead of 100 Hz, the average voltage roughly halves, and the ripple rises steeply. On the CRO alternate humps are missing. If a diode goes short circuit instead, the transformer secondary is shorted on one half cycle and the fuse should blow.
Why must the CRO ground and the DMM be connected carefully in this experiment?
The CRO's ground lead is bonded to mains earth. Clipping it to any point that is not the circuit's own negative rail earths that point through the instrument, which can short part of the transformer secondary, damage the probe, or give a misleading trace. Measure with the ground on the negative rail, or use an isolation transformer.