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 am a EE turned sheep rancher and needed a little refresher to get a mil surplus DC motor running on power supply....this helped a lot.
Anonymous -June 05, 2006 (Article Rating: )
I need text books or notes on computer Engr.
franklin c Godwin -April 15, 2006
The idea is useful, but almost all linear power supply books would mention this in a certain form in their introduction section, I doubt if many people would find this interesting.
Anonymous -June 06, 2005
It's good help.
Amin -March 27, 2005 (Article Rating: )
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