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Subject: Re: V regulator input cap size?
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> <firstname.lastname@example.org>
Date: Tue, 29 Oct 2002 00:06:10 GMT
NNTP-Posting-Date: Mon, 28 Oct 2002 19:06:10 EST
John Fields wrote:
> On Sat, 26 Oct 2002 23:28:43 GMT, email@example.com 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
> John Fields
> Professional circuit designer
I was scratching my head about your response - until
I realized that I was not clear. When I said there
is a better way, I was not referring to your circuit.
I was referring to Dave's question about diodes:
"What about using fewer rectifiers? " etc It seemed
that he did not want to pursue your idea of using
a cap & was still thinking of the 7812 + 7824
combination. I was proposing a better way, where
the better way referred to using diodes to lower the voltage
and feeding two regulators.
I did not mean for it to come across that I was
proposing a better circuit than yours, and apologize
for creating that impression.
As to the zener - I don't see it being worthwhile.
You are quite right that if you disconnect the load
from the regulator the thing will be exposed to >35
volts. But the OP intends to build a single circuit
and power it from a permanently installed xformer.
The probability is very low that he will build the
regulator separable from the load. As to current
lower than expected - it would have to drop below
12 ma to expose the 7812 to >35 volts. Again,
extremely low probability. Finally, why a zener?
A 1.5K resistor to ground after the voltage dropping
resistor will keep the voltage under 33 at the
input to the 7812 with no load. At full load,
the 220 resistor would drop about 21 volts, still
leaving the 7812 with a bit over 16. If there was
any chance of load disconnect, I'd use the second
resistor instead of the zener.
The nice thing about your circuit with the cap in place
of the first resistor is that it avoids all of the above
paragraph - no need for the zener or a second resistor.
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