A bridge rectifier turns AC into DC, but the output is still a long way from the smooth DC a circuit needs: it falls to zero 100 times every second. The fix is a single capacitor across the output. This post works out the numbers for a real 12 V supply — how big the ripple is, what the average DC voltage becomes, and how to choose the capacitor so the ripple stays small.
Figure 1: The bridge rectifier with a capacitor filter (a), and what the capacitor does to the output waveform (b). Without the capacitor the voltage falls to zero 100 times a second; with it, the output stays near the peak. (Schematic, not to scale.)
What the bridge does on its own
Take a transformer secondary of 12 V RMS at 50 Hz. The peak of each half-cycle is 12 × √2 = 16.97 V. In a bridge, two diodes conduct at a time, and each drops about 0.7 V, so the peak that reaches the load is 16.97 − 1.4 = 15.57 V. The bridge flips the negative half-cycles upright, so the output is a series of humps 10 ms apart — but between humps it still falls all the way to zero. That is pulsating DC, not the steady DC a radio or a microcontroller wants.
The capacitor filter — charge up, leak down
Connect a capacitor across the output, as in Figure 1(a), and the picture changes. Each time the rectified voltage rises to its peak, the capacitor charges up to 15.57 V almost instantly. Then, as the rectified voltage falls away, the capacitor takes over: it supplies the load current itself, and its voltage falls only slowly as it discharges. Before it has fallen far, the next hump arrives and tops it up again. The result, in Figure 1(b), is a voltage that wobbles gently around a high average instead of collapsing to zero.
The wobble is the ripple voltage. It depends on three things: how much current the load draws, how big the capacitor is, and how often it gets topped up. For a full-wave rectifier on 50 Hz mains the top-ups come 100 times a second, and the ripple is approximately
Vripple = I / (f × C)
where I is the load current, f is the ripple frequency (100 Hz) and C is the capacitance. The approximation is good when the ripple is small compared with the peak voltage — which is exactly the case we want.
Worked example — 0.5 A load, 2200 µF capacitor
Let the load draw I = 0.5 A and use C = 2200 µF:
Vripple = 0.5 / (100 × 2200 × 10−6) = 0.5 / 0.22 = 2.27 V peak-to-peak
The output therefore swings between 15.57 V and 15.57 − 2.27 = 13.30 V. Its average (DC) value is roughly the peak minus half the ripple:
Vdc ≈ 15.57 − 2.27 / 2 = 14.43 V
The load resistance that draws 0.5 A at 14.43 V is RL = 14.43 / 0.5 = 28.9 Ω (about 29 Ω), and the power delivered is 14.43 × 0.5 = 7.2 W. As a sanity check, the discharge time constant is RL × C = 28.9 × 2200 µF = 63.6 ms — much longer than the 10 ms between top-ups, which is why the capacitor holds the voltage up so well.
What if the capacitor is too small?
Repeat the calculation with C = 470 µF and the same 0.5 A load:
Vripple = 0.5 / (100 × 470 × 10−6) = 10.64 V · Vdc ≈ 15.57 − 10.64 / 2 = 10.25 V
The output now swings from 15.57 V down to 4.93 V — a ripple of almost 11 V, and the average DC has fallen by more than 4 V. A regulator fed from this supply would drop out every half-cycle. The capacitor is doing the same job in both cases; it is simply too small for the load.
Design — choose the capacitor for the ripple you can accept
Turn the formula around to size the capacitor. Suppose the load draws 1 A and the ripple must stay below 1 V:
C ≥ I / (f × Vripple) = 1 / (100 × 1) = 0.01 F = 10,000 µF
So a 10,000 µF capacitor (or larger) is needed. That is a big electrolytic — ripple current and physical size are why high-current supplies often use a switching regulator instead of a bigger capacitor. Note the pattern: doubling the load current doubles the ripple, and doubling the capacitance halves it.
Two ratings to check before you buy the parts
First, the diodes. Each diode in the bridge sees a peak inverse voltage equal to the secondary peak, 16.97 V, so a common 1N4001 (rated 50 V) has plenty of margin. Each diode conducts on alternate half-cycles, so its average current is about half the load current — 0.25 A here — and a 1 A diode is comfortable.
Second, the capacitor's voltage rating. It charges to the full peak of 15.57 V, so its rating must be higher than that — use a 25 V capacitor, not a 16 V one. (A capacitor run close to its rated voltage ages fast, and mains voltage can drift a few percent above nominal.)
Key points to remember
- A 12 V RMS secondary peaks at 16.97 V; after the two conducting diodes of a bridge, 15.57 V reaches the load — but as pulsating DC that falls to zero 100 times a second.
- A capacitor across the output charges to the peak and discharges slowly into the load, so the output wobbles only slightly: ripple Vripple = I / (f × C), with f = 100 Hz for full-wave on 50 Hz mains.
- Worked example: 0.5 A and 2200 µF give 2.27 V peak-to-peak ripple and Vdc ≈ 14.43 V; the same load with 470 µF gives 10.64 V of ripple and only 10.25 V DC.
- To design for a ripple limit, use C ≥ I / (f × Vripple): 1 A with ripple under 1 V needs at least 10,000 µF.
- Check the ratings: diode PIV is the secondary peak (16.97 V — a 50 V diode is fine), each diode carries about half the load current, and the capacitor's voltage rating must exceed the 15.57 V peak — use 25 V.