Bridge Rectifier Calculator
This bridge rectifier calculator turns the AC voltage from a transformer secondary into the DC figures you actually get out of a full-wave bridge and smoothing capacitor. Enter the RMS voltage of your secondary winding, the mains frequency, the load current your circuit draws, the smoothing capacitor value and the forward voltage drop of the diodes, and the calculator returns the peak voltage the capacitor charges to, the peak DC after the two conducting diodes take their share, the peak to peak ripple riding on top, and the average DC voltage your load really sees. A full-wave bridge conducts on both halves of the mains cycle, so it ripples at twice the line frequency, which is why the ripple formula uses two times the frequency. The defaults describe a common 12 V RMS, 50 Hz supply with a 1 A load and a 4700 microfarad reservoir capacitor using ordinary silicon diodes, the sort of thing behind a small linear bench supply. Change the diode drop to about 0.3 V per diode if you are using a Schottky bridge, and raise the capacitance if the ripple is too high for your regulator to swallow. Use it when sizing a reservoir capacitor, checking headroom for a linear regulator, or sanity checking a power supply design before you order parts.
Two diodes conduct at once in a bridge, so the total drop is twice this value. Use about 0.3 V for a Schottky bridge.
Estimate for a full-wave bridge with a single reservoir capacitor. Real ripple depends on capacitor equivalent series resistance and load behaviour, and the average sits between the peak and the trough of the ripple. Estimate only.
How it works
The peak voltage is the RMS secondary voltage times the square root of two, because a rectifier charges the capacitor to the crest of the sine wave rather than its RMS value. Two diodes conduct on each half cycle in a bridge, so the peak DC is that peak minus two forward voltage drops. The reservoir capacitor discharges into the load between peaks, and for a full-wave rectifier the peak to peak ripple is the load current divided by two times the mains frequency times the capacitance. The average DC your load sees sits roughly half the ripple below the peak DC, which is the figure headroom calculations for a linear regulator should use.
Worked example
Take a 12 V RMS secondary at 50 Hz feeding a 1 A load through a 4700 microfarad capacitor, with 0.7 V silicon diodes. The peak is 12 times 1.414, which is 16.97 V. Subtract two diode drops, 1.4 V, to get a peak DC of 15.57 V. The ripple is 1 A divided by (2 times 50 times 0.0047 F), which is 2.13 V peak to peak. The average DC is the peak DC minus half the ripple, about 14.51 V, so a 12 V regulator needs its dropout checked against the 15.57 V minus the ripple trough.
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