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NNTP-Posting-Date: Sun, 27 Oct 2002 17:19:24 -0600
From: jfields@texas.net (John Fields)
Newsgroups: sci.electronics.components,sci.electronics.design,alt.engineering.electrical
Subject: Re: V regulator input cap size?
Date: Sun, 27 Oct 2002 22:56:58 GMT
Organization: Austin Instruments, Inc.
Message-ID: <3dbc6d10.24635892@news.texas.net>
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>
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On Sat, 26 Oct 2002 23:28:43 GMT, ehsjr@bellatlantic.net wrote:
>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.
---
This is essentially what I suggested earlier, but using a series cap
on the AC input of the bridge instead of a resistor. Using a
capacitor will allow the desired excess voltage to be dropped across
the capacitor losslessly. Your circuit also requires a Zener diode
across the output of the bridge should the load ever become
disconnected or the load current drop to a lower value than
expected.
---
John Fields
Professional circuit designer
http://www.austininstruments.com
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