Subject: Re: V regulator input cap size?
Date: Sat, 26 Oct 2002 20:06:37 -0700
References: <0001HW.B9D991AD000C5917163F7590@news.covad.net> <email@example.com> <0001HW.B9D9CD54001A5D15163F7590@news.covad.net> <0001HW.B9DAE8B7003F6E14163F7590@news.covad.net> <firstname.lastname@example.org> <email@example.com> <0001HW.B9DFC96E0025E158162B2870@news.covad.net> <3DBB253E.A0AE4016@bellatlantic.net>
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On Sat, 26 Oct 2002 16:28:43 -0700, firstname.lastname@example.org wrote
(in message <3DBB253E.A0AE4016@bellatlantic.net>):
> DaveC wrote:
>> 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?
>> 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.
Not so. About 51mA with until the relay kicks in then 76mA. About a 50
> 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.
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