A rectifier input capacitor's size is often considered nebulous. Therefore, common practice is to pick a large size, and if the ripple voltage is low enough, all is okay *(see the figure, a)*. If not, it can be increased in size. Other attempts sometimes calculate percent ripple, which I consider a largely useless term because we tend to visualize the waveform as an oscilloscope sees it—a sawtooth waveform with the limit being the minimum capacitor voltage.

When I was in school, I derived this simple relationship for input-capacitor size that has been with me throughout my career. I have used it countless times and would like to share it with the readers:

C = 0.7(I)/ΔE(f)

where C = capacitance in farads, I = dc load current in amperes, ΔE = peak-to-peak ripple voltage, f = ripple frequency (generally 120 Hz for full-wave or 60 Hz for half-wave), and 0.7 is the complement of the rectifier-current duty cycle, which is assumed to be 0.3 *(see the figure, b)*.

The equation is derived from the following basic relationships:

Q = CV (charge = capacitance × voltage) I = Q/T (current = charge/time) f = 1/T (frequency = 1/period)

It assumes that the capacitor delivers the current to the load 70% of the cycle, while the rectifier delivers the current (and charges the capacitor) for the remaining 30%. Plugging in some numbers for a typical case:

I = 1 A ΔE = 1 V f = 120 Hz

results in a calculated C = 5833 µF.