Message-ID: <3DBC67DB.14E7E059@bellatlantic.net>
From: ehsjr@bellatlantic.net
Reply-To: ehsjr@bellatlantic.net
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Newsgroups: sci.electronics.components,sci.electronics.design,alt.engineering.electrical
Subject: Re: V regulator input cap size?
References: <0001HW.B9D991AD000C5917163F7590@news.covad.net> <3db451b5.4005327@news.texas.net> <0001HW.B9D9CD54001A5D15163F7590@news.covad.net> <0001HW.B9DAE8B7003F6E14163F7590@news.covad.net> <3db5cd86.76507893@news.texas.net> <3db7a7e2.132416061@news.texas.net> <0001HW.B9DFC96E0025E158162B2870@news.covad.net> <3DBB253E.A0AE4016@bellatlantic.net> <0001HW.B9E0A64D003CC7E9162B2870@news.covad.net>
Date: Sun, 27 Oct 2002 22:25:28 GMT
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NNTP-Posting-Date: Sun, 27 Oct 2002 17:25:28 EST
DaveC wrote:
>
> On Sat, 26 Oct 2002 16:28:43 -0700, ehsjr@bellatlantic.net 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?
> >> 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.
>
> Not so. About 51mA with until the relay kicks in then 76mA. About a 50
> percent increase.
Yes, so. You misunderstand the term. Steady state current
refers to the current that flows in the circuit all the time
it is powered on, with insignificant variation, regardless of
input/output changes. Your steady state current draw is 51 ma.
When 76 ma is flowing, 51 ma is flowing. When 76 ma is not
flowing, 51 ma is flowing. 51 ma flows all the time, whether
the relay is on or off.
I hope you didn't miss the rest of the post due to that
misunderstanding.
>
> > 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.
>
> Thanks,
> Dave
> ---
> Dave Carpenter
> Sound Logic
> voicebox@d8n8a8i.com
>
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