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Subject: Re: V regulator input cap size?
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Date: Sat, 26 Oct 2002 23:28:43 GMT
NNTP-Posting-Date: Sat, 26 Oct 2002 19:28:43 EDT
> What about using fewer rectifiers? Wouldn't using one or two diodes rather
> than a full bridge to rectify the 28vac result in a lower dc voltage? Ripple
> will be lower freq, and necessarily, require a larger filter cap, but
> wouldn't this be a cheap way to reduce input to the regulator(s)?
Dave, I think it was mentioned before in the thread, but it bears
repeating: adding more series diodes on the input side of
the supply lowers the DC voltage. With a single diode, you
would figure dc output as about:
(ac input - 1 diode voltage drop) * 1.4
If you use a bridge, it's the same as above, except you use
2 diode drops. A diode drops about .6 to .7 volts.
But there is yet a better way. Use the standard bridge circuit
(or whatever, doesn't matter). You have a steady state current
draw of about 51 ma, whether the relay is on or off. Call it
50 ma, and drop some voltage in a series resistor. Your
supply produces about (28 - 1.2) * 1.4 or 37.52 volts - call
it 38.A 220 ohm resistor will drop 11 volts at 50 ma, and 16.72
volts at 76 ma. That leaves the 7812 exposed to 27 (38-11) volts
at 50 ma (1.35 watts) and 21.28 (38-16.72) volts at 76 ma (1.62
watts). Heat sink the 7812. The resistor will need to dissipate
1.27 watts at 76 ma, so use a 5 watt resistor.
You avoid the 7824 that way - the 220 ohm 5 watt resistor is
cheaper & easier to use, and the heatsink can be smaller.
> Note that my return address is corrupted in an attempt to reduce spam. If you
> choose to e-mail me, please correct my address as described below.
> Dave Carpenter
> Sound Logic
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