In Reply to: "Critical capacitance" posted by Lew on December 30, 2014 at 11:42:19:
You size the first capacitor to give the amount of ripple that you can tolerate for a given load current. The formula for the charge Q on a capacitor is:
Q=V*C where Q is in coulombs, V volts and C farads.
Expressing this eqn. in differential form we have:
dQ/dt=V*dC/dt+C*dV/dt Assuming C is constant this gives dQ/dt (current I)=c*dV/dt
For a bridge rectifier that doubles the ripple frequency to 120Hz dt=.00833 (1/120). Solving for the ripple voltage dV we get:
dV=.00833 I/C where I is in amps and C is in farads!
Using a larger capacitor to increase your minimum voltage can be counter productive. Doing so increases the charging current and the internal resistance of your transformer and diodes may not let you get all that current. Also it sounds bad. The bigger the input cap the smaller the conduction angle. The current is drawn from the PS transformer in very short, sharp pulses which generate lots of very high frequency harmonics in the ripple current. If not careful this will radiate as RF and contaminate the audio circuits.
"It is better to remain silent and thought a fool, then speak and remove all doubt." A. Lincoln
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Follow Ups
- RE: "Critical capacitance" - JKT 12/30/1414:29:08 12/30/14 (9)
- Thanks for that informative response - Lew 14:38:34 12/30/14 (8)
- RE: The other half...inductors - Russ57 15:20:24 12/30/14 (7)
- My original question - Lew 16:40:22 12/30/14 (5)
- RE: My original question - jazbo8 10:51:07 01/1/15 (2)
- RE: My original question - JKT 19:17:32 12/30/14 (1)
- RE: My original question - Lew 08:45:49 01/1/15 (0)
- RE: The other half...inductors - JKT 16:04:30 12/30/14 (0)