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From: "Anthony Q. Bachler"
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
Subject: Re: How to use a zener diode for voltage regulation?
X-Newsreader: Microsoft Outlook Express 5.50.4133.2400
Date: Mon, 09 Sep 2002 09:54:27 GMT
NNTP-Posting-Date: Mon, 09 Sep 2002 02:54:27 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
The LM317 can be used for a wide range of output voltages, making it a good
part to design with. If you have a run of the mill voltage requirement, a
specific output voltage component would be better in the production version.
Its a matter of flexibility versus component count.
The american people did to themselves
what Osama bin Laden could never do,
took away our freedoms in exchange for
a false sense of security.
"The little lost angel" wrote in
> On Tue, 3 Sep 2002 19:24:43 +1000, "Trevor Wilson"
> >**A much better idea (but fractionally more expensive) is to use a
> >current source. Look up the National data on the LM317. You will find
> >you require in there. Set the constant current source for the suggested
> >current rating of your LED. The intensity will not vary, over a wide
> >of voltages.
> Okie, looks like there's no way I can do this with just a diode and
> resistor combo :P
> Why the LM317 and not say the LM1117 since I'm trying to cut down on
> costs/component count, the LM317 requires two external resistor while
> the LM1117 is available in fixed 3.3V output.
> The only question I now have is, is the term "constant current source"
> the crux of the matter? Would a 10 ohm resistor with a LM1117-3.3V and
> the LED work?
> +6V ~ 12V ----LM1117-3.3V----R1---LED---Gnd
> R1=10 ohm
> Or should I stick with a 5V output instead of 3.3V? The reason
> for 3.3V is the dropout at 100mA for LM1117 is 1.1 to 1.3V, so at
> 6.3V, it's barely enough to put out 5V. And I cannot determine if 6.3V
> is the floor voltage.
> Thus I thought it would be best to use a 3.3V regulator since
> the nature of the device is that it would only work if the voltage is
> above 5V.
> Thanks to all who responded!
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