How this calculator works
A bridge reverses the negative half-cycles so load current always has the same polarity. Averaging the rectified sine gives 2/pi of its peak; RMS remains peak divided by sqrt(2). Two diode forward resistances are included in the current path when specified. Constant diode threshold drops, source impedance and capacitor charging pulses are not included.
Formula
Vdc = 2 Vpk/pi; Ipk = Vpk/(RL+2Rf); Idc = 2 Ipk/pi; Irms = Ipk/sqrt(2); ripple = sqrt[(Irms/Idc)²-1].
Worked example
A 10 V peak sine gives 6.3662 V average after ideal full-wave rectification. With an 8 ohm load and 1 ohm per conducting diode, peak current is 1 A and RMS current is 0.707107 A.
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Frequently asked questions
Why is the DC voltage below the peak?
This is the average of an unsmoothed full-wave waveform. A reservoir capacitor changes both the output voltage and diode current waveform.
Is the ripple factor a percentage?
The output is a dimensionless RMS ripple-to-DC ratio. Multiply by 100 for percent.
Are diode forward resistance and forward voltage the same?
No. This model uses a resistive diode approximation. A fixed diode voltage drop requires waveform conduction-angle analysis.