Bridge Rectifier Ripple and the 2200 µF Filter
A trainee assembles a 12 V, 1 A DC supply for a small motor load. The parts are a 230 V/12 V transformer, a 1 A bridge rectifier, a 2200 µF/25 V electrolytic capacitor and a 12 Ω 50 W load resistor. With no load, the DC output is 16.2 V. With the 12 Ω load connected, the load current is 1.05 A, the DMM reads 12.8 V and a DSO across the load shows 3.8 V peak-to-peak ripple at 100 Hz. The transformer secondary measures 12.0 V rms. The trainee replaces the 2200 µF capacitor with a 1000 µF unit and the ripple rises to 7.4 V peak-to-peak. The trainee concludes that the bridge is faulty because the output is not exactly 12 V and because the capacitor change altered the DC level. A spare 4700 µF/35 V capacitor is available.
1. Calculate the expected ripple with 2200 µF and 1000 µF at 1.05 A. What does the 100 Hz ripple frequency tell you about the rectifier?
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For a full-wave rectifier with a capacitor filter, Vpp ≈ I/(2fC), where f is the 50 Hz mains frequency. With 2200 µF: Vpp = 1.05/(2×50×2200×10^-6) = 4.77 V. With 1000 µF: Vpp = 10.5 V. The measured 3.8 V and 7.4 V are lower than these estimates but the ratio is about 2, as expected; capacitor tolerance, ESR and load variation explain the difference. The 100 Hz ripple means the bridge is supplying the load on both half-cycles, so it is operating as a full-wave rectifier. If one diode were open, the ripple frequency would fall to 50 Hz and the waveform would show one charging pulse per mains cycle. So the 100 Hz reading does not support a faulty bridge.
2. Is the bridge faulty? Explain the no-load 16.2 V and loaded 12.8 V readings.
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No. A 12 V rms transformer has a peak of 12 × 1.414 = 17.0 V. A bridge drops about 1.4 V, leaving about 15.6 V peak. At no load the capacitor charges to the peak, and the transformer may be above 12 V rms off-load, so 16.2 V is plausible. Under load the capacitor discharges between peaks, so the DMM reads an average, not the peak. With 3.8 Vpp ripple, the peak is roughly 12.8 + 1.9 = 14.7 V and the trough is about 10.9 V; that is consistent with a working full-wave bridge and filter. The output of an unregulated transformer-rectifier-filter is not meant to be exactly 12 V. The bridge would be suspect if the ripple frequency were 50 Hz, if one half-cycle were missing on the scope, or if the no-load voltage were far below 15 V.
3. The trainee fits the 4700 µF/35 V capacitor. Calculate the new ripple and check the voltage rating. What else should be checked before calling the supply finished?
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New ripple: Vpp = 1.05/(2×50×4700×10^-6) = 2.23 V. The no-load peak is about 17 V, so a 35 V rating gives more than 2:1 margin; that is acceptable, but confirm the capacitor's ripple-current rating and temperature rating. Check polarity before power-up, because a reversed electrolytic can vent. Also check the bridge peak inverse voltage: each diode sees about 17 V, so a 50 V or 100 V bridge is fine. Check transformer VA rating: 12 V × 1.05 A ≈ 12.6 VA, so a 12 VA transformer is marginal; a 20 VA type would run cooler. Fit a fuse, check output polarity and load regulation, and measure the ripple with the 10X probe to avoid noise. The supply is then an unregulated ~15 V DC source, not a precise 12 V supply.