Olin wrote:
> Short answer: This is a really bad idea in this case.

'Really bad'?  Or did you mean 'not the best'?  Assuming
that the correct resistor value is calculated, what
could go wrong?

Cheers,
-Neil.



-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Olin Lathrop
Sent: Friday, July 19, 2002 7:43 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [PIC]: Flashing LED Episode II (small attachment)


> I fixed the MCLR-wrong-polarity thingy, added a resistor, and a zener
diode
> to regulate instead.  But, a question (which my books don't appear to
answer
> properly!) - would I use a 5.1V zener to drop the voltage to 5.1V, or a 2V
> zener to drop the voltage (from 6V) to 4V?; I would assume that the latter
> is true, because a website told me they have a "specified voltage drop" -
> but to quote Mr. Dude, "a 5.1V zener".
> Just to clear that up..

A zener in this context is used as a shunt regulator.  In other words, you
put it accross the PIC and the zener draws enough current to drop the
incoming voltage down to 5.1V or whatever.  This also requires a resistor in
series with the battery for the zener to react against.  Short answer: This
is a really bad idea in this case.  Keep in mind that many people on this
list will offer advice, but the quality of that advice should be suspect
until nobody objects to it for 24 hours or so.

You had the right idea originally by using a voltage regulator.  The LM7805
is not a good fit because it requires too much headroom, but an LDO (Low
Drop Out regulator) will work fine.  You can also use the hack I mentioned
earlier about putting a single diode (such as 1N4001, for example) in series
with the battery.


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