Ripple Twice the Design Value in a Capacitor-Filtered Bridge Rectifier
A trainee at a repair shop builds a 12 V, 200 mA bench supply. The transformer secondary is 12 V rms, the bridge uses four 1N4007 diodes, and the reservoir is a 2200 µF/25 V electrolytic feeding a load that draws a steady 200 mA. The design sheet predicts Vdc = √2(12) − 2(0.7) − Vr/2 = 15.6 − 0.45 ≈ 15.1 V, with ripple Vr = I/(2fC) = 0.2/(2 × 50 × 2200 × 10⁻⁶) ≈ 0.91 V peak-to-peak, which should read about 0.26 V on an AC range. On test the supply delivers 14.9 V DC, close to prediction. However, a true-RMS DMM on its AC range reads 0.52 V across the load, roughly twice what was expected. He replaces the 2200 µF capacitor with a new one and repeats the measurement: 0.52 V again, with 14.9 V DC. He now plans to buy a much larger capacitor. The shop has a 20 MHz oscilloscope and a DMM with a diode-test mode, and the load must keep running at 200 mA for the rest of the day.
1. Convert the AC reading into peak-to-peak ripple and decide, with numbers, whether the capacitor or the rectifier is at fault.
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Step 1 – Convert the meter reading. A true-RMS meter reads the RMS value of the AC component. For a near-triangular ripple, Vrms = Vpp/(2√3), so
Vpp = 2√3 × 0.52 = 3.46 × 0.52 ≈ 1.80 V.
Step 2 – Compare with the two possible modes.
- Full-wave (all four diodes good): Vr = I/(2fC) = 0.2/(100 × 2200 × 10⁻⁶) ≈ 0.91 V.
- Half-wave (one diode open): the capacitor now discharges for a full 20 ms instead of 10 ms, so Vr = I/(fC) = 0.2/(50 × 2200 × 10⁻⁶) ≈ 1.82 V.
The measured 1.80 V matches the half-wave prediction to within 1 %, and is almost exactly double the full-wave value. A degraded capacitor would give some larger, arbitrary ripple, not a clean doubling, and the substitution test already ruled it out.
Step 3 – Why the DC reading does not help. In half-wave only one diode drop is present, so Vdc ≈ 16.27 − 0.91 ≈ 15.4 V, close to the original 15.1 V. The DC output is therefore a poor diagnostic; the ripple amplitude and, above all, its frequency are the give-aways. Conclusion: one arm of the bridge is open, and the supply is running as a half-wave rectifier.
2. Describe the oscilloscope and diode-test checks that would confirm the diagnosis, including the settings you would use and the safety step you would take first.
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Safety first. Switch off and unplug; the 2200 µF capacitor holds charge at about 15 V. Discharge it through a 1 kΩ, 1 W resistor and verify with the DMM that the voltage is below 1 V before touching the bridge.
Scope check (live). Connect the probe across the load, DC-coupled (the DC level is about 15 V, so do not use AC coupling unless you first confirm the input rating). Set 5 V/div vertical and 5 ms/div horizontal.
- Healthy full-wave: a sawtooth repeating every 10 ms (100 Hz), about 0.9 V peak-to-peak riding on 15 V.
- Faulty half-wave: a sawtooth repeating every 20 ms (50 Hz), about 1.8 V peak-to-peak.
Count divisions carefully: the ripple is under 0.2 division tall at 5 V/div, so switch to 0.5 V/div with AC coupling for the ripple, and use a separate DC measurement for the mean level.
Diode test (power off). Each 1N4007 in circuit should read 0.5–0.7 V forward and open-circuit reverse. A reading near 0 V in both directions means a shorted diode; open in both directions means the open diode you suspect. Confirm by lifting one leg of the suspect device, since parallel paths through the transformer winding can mislead you.
3. The supervisor says the supply is fine because it still gives nearly 15 V. Evaluate that claim and state the consequences of leaving the fault in place for a week.
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The claim is wrong on three counts, even though the DC value is close to nominal.
- Ripple quality. Ripple rises from about 0.9 Vpp (6 % of 15 V) to 1.8 Vpp (12 %). A circuit expecting a quiet rail will show audible hum, and any comparator or ADC referenced to the rail will show jitter. The RMS ripple is 0.52 V, so the effective usable DC window shrinks noticeably.
- Component stress. In half-wave operation a single diode pair carries the whole 200 mA, so the average current per conducting diode is about 0.2 A instead of 0.1 A, and each diode's conduction angle is halved. The transformer secondary now carries a DC component, which pushes the core toward saturation, raising primary current and temperature. The 1N4007 surge rating is not the issue; the sustained heating of the transformer is.
- Ripple frequency moves into a worse band. The 100 Hz hum becomes 50 Hz, which is far more audible on a loudspeaker and harder to filter.
Also, the fault is intermittent in character: the surviving diodes are being run harder and one of them may fail next, at which point the output collapses. The correct action is to replace the whole bridge (they cost little and the four diodes are aged together) rather than fit a larger capacitor, which would only partly mask an 1.8 V sawtooth.