Robert Ammerman wrote: > Storing power in CAPs may not be as bad an idea as you think. > > The magic formula is: > > 1 farard = 1 amp-second Oops... didn't somebody say recently on this lest to carry along your units? :) 1 F = 1 A * 1 s / 1 V > You need 3 amps for 0.2 second or 0.6 amp-seconds or 0.6 Farad There's one crucial part missing here: the allowed voltage drop. Let's say you need 3A for 200ms, and the voltage may drop a maximum of 3V in this process, the cap needed is 3A*200ms/3V=200mF. However, if the maximum allowed voltage drop is only 300mV, you get 3A*200ms/300mV=2F. The thing to note here is that the voltage to be used is not the total voltage, it is the voltage drop during discharge. Gerhard -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist