Kieren,

Curious what resistor you're mentioning...

You use a 5.1V zener to get 5.1V for your app circuit.
You'll pretty much have a series-circuit "loop" with the
battery, resistor and zener, and then the app circuit is
connected across the zener.  The zener will adjust it's
current to get about 5.1V as the input voltage varies.

To properly design a zener circuit, you need to calculate
the current requirements of your circuit, and then calc
the value of the current-limiting resistor that will keep
down the current/power thru the zener.  Once you've
figured that out, you can calculate the power dissipated
thru the zener and get one with the appropriate rating.

There are some calcs to be done, but it's a simple result
in the end.  Do a web search for "zener circuit" or "zener
design" and you should find a bunch of tools that'll help
you design this part.

Cheers,
-Neil "that'll be Mr. Dude to you :-)" N.


-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Kieren Johnstone
Sent: Friday, July 19, 2002 2:12 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..

-Kieren

----- Original Message -----
From: "Mircea Chiriciuc" <sascha@ROL.RO>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Friday, July 19, 2002 7:42 AM
Subject: Re: [PIC]: Flashing LED Episode II (small attachment)


> It seems that something has passed unnoticed: the MCLR pin is tied to GND
so
> the device will be kept in reset all the time. How does it work then?
> As much as I can understand from the data sheet it should be logic high
for
> the device to function.
>
> Or it's another "patent" idea? ;)
>
> You should also be using a series resistor with the LED to limit the
current
> unless you just want to drain the batteries really quick. A good lecture
on
> this should be the Ohm's Law? The voltage drop on usual LED is approx.
1.8V.
> The Pic pin will handle 25mA max and the LED needs 10-20mA for bright
> results.
>
> So this configuration may make you the surprise of  -GREAT! IT'S WORKING!
> and after two minutes - WHY DID THE CAT DIED? CURIOSITY PERHAPS? (no, too
> much current trough it's fur but didn't realize that until it started
> smelling).
>
> After all why do you need - The Stupendous - PIC 16f877 to make a flashing
> LED?
>
> Do you always need the BIG things to make the SMALL jobs?
>
> Are you perhaps a woman?
>
>
> Wait I have an idea! R=U/I shuld I go for a patent?
>
>
> LPF(jokes),
> Mircea Chiriciuc
> EMCO INVEST
>
> --
> http://www.piclist.com#nomail Going offline? Don't AutoReply us!
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>
>
>

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