Bridge Rectifier That Ripples at 50 Hz
A student builds a power supply using a 12 V RMS transformer, a four-diode bridge made from 1N4007 diodes, a 1000 µF/25 V electrolytic capacitor, and a 100 Ω 5 W load resistor. With no load, the output is about 15.5 V DC. With the 100 Ω load connected, a digital multimeter reads 14.8 V DC and 3.2 V AC. The student has borrowed an oscilloscope and sees a sawtooth ripple of about 3.2 V peak-to-peak at 50 Hz, not the 100 Hz expected for a full-wave rectifier. Two of the four diodes feel noticeably warmer than the other two. The capacitor is not hot and shows no bulge. The student wants to replace the capacitor because the DC output is 'too low and too ripple-rich'. The transformer secondary resistance and diode forward drops are not negligible. Mains frequency is 50 Hz.
1. Calculate the expected full-wave DC output and ripple for this supply and compare them with the 14.8 V DC and 3.2 V peak-to-peak readings. Show the formulas used.
Read the worked answer
The peak secondary voltage is:
Vm = 12 V RMS × √2 = 16.97 V.
In a bridge, two diodes conduct in series, so the capacitor charges to approximately:
Vpeak ≈ 16.97 − 2 × 0.7 = 15.57 V.
The load current is approximately:
IL = 14.8 V / 100 Ω = 0.148 A.
For full-wave rectification the ripple frequency is 2f = 100 Hz, and the peak-to-peak ripple is:
Vr ≈ IL / (2fC) = 0.148 / (100 × 1000 × 10⁻⁶) = 1.48 V peak-to-peak.
The average DC output is about Vpeak − Vr/2 = 15.57 − 0.74 = 14.83 V.
So a healthy full-wave supply should give about 14.8 V DC and 1.5 V peak-to-peak ripple at 100 Hz. The measured DC value agrees, but the measured ripple is about twice as large and at 50 Hz. That points to half-wave operation, not to a design that was wrong from the start.
2. Explain the 50 Hz, 3.2 V peak-to-peak waveform and the warm diodes. Why is this evidence against a capacitor fault?
Read the worked answer
If one diode in the bridge is open, the bridge can no longer conduct on both half-cycles. It behaves like a half-wave rectifier on the remaining half-cycle. The capacitor then has to supply the load for a full 20 ms period instead of 10 ms, so the ripple approximately doubles and its frequency becomes the mains frequency, 50 Hz. The DC output also falls somewhat. That matches the measured 3.2 V peak-to-peak at 50 Hz. The two diodes that conduct on the remaining half-cycle carry the full load current and become warmer than the two that never conduct.
A failed capacitor would give a different signature. An open or dried-out capacitor would leave a 100 Hz full-wave pulsating waveform with ripple almost as large as the peak voltage, and the DC average would fall sharply. A shorted capacitor would usually blow a fuse or overheat the transformer. A capacitor that is merely reduced in value would still show 100 Hz ripple, not 50 Hz. The combination of 50 Hz ripple and two warm diodes is therefore much stronger evidence for an open diode than for a bad capacitor.
3. Describe a safe confirmation and repair procedure, including one check that distinguishes an open diode from an incorrectly reversed diode.
Read the worked answer
First switch off and unplug the supply, then discharge the 1000 µF capacitor through a suitable resistor and verify with a meter that the voltage is near zero. Remove the load or isolate the bridge as needed. Use the diode-test function of a digital multimeter on each 1N4007. A good silicon diode shows about 0.5–0.7 V forward and no conduction reverse. An open diode shows no forward reading. A reversed diode is installed with its band opposite to the PCB marking; it may test good in isolation but the bridge will not rectify correctly, and in some bridge layouts a reversed diode can cause a short and blown fuse.
Replace the faulty diode, preferably all four if they are of unknown origin, and check the capacitor polarity. Reconnect the load and verify with the oscilloscope that the ripple is about 1.5 V peak-to-peak at 100 Hz. The 1N4007 has a 1000 V repetitive peak reverse voltage rating, while the bridge diodes see about 2Vm ≈ 34 V, so the diode type itself is adequate; the fault is the open or wrongly installed part, not an underrated PIV.