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'[PIC]: Flashing LED Episode II (small attachment)'
2002\07\18@155156 by Kieren Johnstone

picon face
part 1 1971 bytes content-type:multipart/alternative; (decoded quoted-printable)


------ 2002\07\18@163156 by Karl Seibert
flavicon
face
Kieren,

6 V isn't enough to supply a 7805.  I forget how high of voltage
it needs.  If you want to regulate 6 V down to 5 V, you need
a low-dropout regulator.

I don't know what you mean by an "LED rated 5 V".  Does that mean
it has a built in current limiting resistor?  If it doesn't, you'll
need some type of resistor.

I don't have a PIC datasheet with me, so others will have to comment
on other parts of the circuit.

Karl

Quoting Kieren Johnstone <spam_OUTmisterfugitTakeThisOuTspamHOTMAIL.COM>:

{Quote hidden}

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2002\07\18@170033 by Kieren Johnstone

picon face
The PIC16F877 needs 2-5.5V.  I've never heard of a low-dropout regulator,
and my intro books just talk about voltage regulators like the 7805..  Is
there a simple (like, 3/4 pin) "drop-in" replacement that works like a 7805
(i.e. not too hard to use)?
When I say "LED rated 5V", I mean I went into my Maplin shop, asked for 8
5mm LEDs, the guy asked "what voltage" I said "well, 5" (they don't blow up
from 5v battery supply with no resistor).

Thanks,
Kieren

{Original Message removed}

2002\07\18@170520 by Pic Dude

flavicon
face
Few things...

- Nothing wrong with the way you have oriented the + and - rails, but
convention tends to put the positive rail on top.  It'll just make it easier
for others to understand.  For example, when I first looked at it, I thought
the R & C were reversed.
- ...unless that reverse-convention is what you want to patent...?  :-)  Did
you mean 'copyright' instead?
- Can you really trademark the phrase "The Flashing LED" ?
- Where are you getting the 4-pin version of the 7805? :-)  I know what you
mean, but you should draw it with 3 pins, especially since you even kept the
locations of the PIC pins intact.
- The regulated supply is not wrong, but not necessary.  You can use 3 cells
for 4.5V, a 5.1V zener, etc to make it simpler/cheaper.
- Not familiar with that (new?) version of the 16F877 (~The Stupendous~  PIC
16F877).  What cool features did Microchip add to this version?

All else looks good and non-destructive.  I say go for it, and learn by
actually making your own mistakes, so don't be afraid to plunge in and just
build it.  A racecar driver friend once told me "You're not a real driver
until you've licked a few", so in the PIC world, you're not a real PICster
until you've fried a few.  That of course, makes me a guru.  :-)

Cheers,
-Neil.

 {Original Message removed}

2002\07\18@172914 by Kieren Johnstone

picon face
> Few things...
>
> - Nothing wrong with the way you have oriented the + and - rails, but
> convention tends to put the positive rail on top.  It'll just make it
easier
> for others to understand.  For example, when I first looked at it, I
thought
> the R & C were reversed.

Okies.

> - ...unless that reverse-convention is what you want to patent...?  :-)
Did
> you mean 'copyright' instead?
> - Can you really trademark the phrase "The Flashing LED" ?

I'm patenting the flashing LED idea, its a revolution! :D

> - The regulated supply is not wrong, but not necessary.  You can use 3
cells
> for 4.5V, a 5.1V zener, etc to make it simpler/cheaper.

Oh, good idea - I'll ave another good look at the zener sections of my books
(usually takes 7 times to get through).

> - Not familiar with that (new?) version of the 16F877 (~The Stupendous~
PIC
> 16F877).  What cool features did Microchip add to this version?

It glows in the dark and automatically updates its logic through wireless
internet :)

Ta,
Kieren

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2002\07\18@174354 by Pic Dude

flavicon
face
Good catch on the 2V dropout of the 7805, Karl.
I have a device here called the LM 1117T-5.0 that
I picked up from Digikey (in a TO-220 package,
which is that same as the 7805).  It's pretty
much a drop in replacement for the 7805, with
the added feature that it's low dropout (so you
don't need an additional 2V over the 5V rating
to ensure regulation)

Only difference is that the output tab is not
grounded as with the 7805, but connected to Vout
(IIRC).  So you may need to verify that your
heatsink is not connected to ground elsewhere.

Ahhhh, Maplin.  How I miss London. :-(

Cheers,
-Neil.


{Original Message removed}

2002\07\18@174950 by Pic Dude

flavicon
face
Kieren yelled:
> I'm patenting the flashing LED idea, its a revolution! :D


Well if peanut-butter and jelly sandwiches were patented, then
why not this.

Cheers,
-Neil.

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2002\07\19@024504 by Mircea Chiriciuc

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face
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

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2002\07\19@031240 by Kieren Johnstone

picon face
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 removed}

2002\07\19@033823 by Roman Black

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face
Kieren Johnstone wrote:
>
> 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..


I'm not sure Mr. Dude's suggestion is the best for
your needs. :o)

Using a 5.1v zener with a 6v battery raises many
problems, and you really need a good understanding
of ohms law etc to "tune" it for each circuit.

Since you seem to be starting out with PICs and
with circuit design, I suggest something you can use
on lots of circuit, because the moment your "flashing
led (TM)" is working you will want to build other
stuff with your PIC 16F877.

Maybe use a 12v plugpack (wall wart) and the 7805
that you have already. Another suggestion is to put
a 150 ohm resistor between the +12v and the INPUT
of the 7805. This is very handy when you are first
starting out and will *probably* stop you frying your
expensive 16F877 when something gets connected wrong
(which seems very likely!)  :o)

Is there any reason you're not using a 16F84 or 16F628?
These program in about 20 seconds, compared to about 2
minutes to program a 16F877. You will soon get sick of
THAT wait.
-Roman

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2002\07\19@033842 by Alan B. Pearce

face picon face
>Ahhhh, Maplin.  How I miss London. :-(

You won't miss it at the moment - the tube train staff have been on 24 hr
strike :)))

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2002\07\19@035348 by Mircea Chiriciuc

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part 1 919 bytes content-type:text/plain; (decoded 7bit)

Use this. With a 1V voltage drop across the resistor the current is also
limmited at about  30mA.
The voltage drop parammeter for a zenner diode states for the voltage drop
across the diode when reversed biased. So in the drawing attached the diode
is "keeping" the voltage at a maximum of 5.1V. If the voltage tends to
increase the zenned will permit current to flow trough it so the voltage
drop will remain aprox. constant untill the current limit trough the diode
is passed and the diode will becone a wire. If the diode becomes a wire, the
resistor will get the full current of your batteries I=U/R=6/33=181mA. The
power the resistor will try to excange in heat will be P=U*I=6*0.181=1.09W.
If you use a 1/4W resistor it will provide also protection for your circuit
as it will burn and open the circuit if the current will exceed 50mA.

Good Luck,
Mircea Chiriciuc.



part 2 5828 bytes content-type:image/gif; (decode)


part 3 136 bytes
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2002\07\19@062748 by Kieren Johnstone

picon face
Thanks, this seems most helpful.
Am I to understand that the resistor is there only to blow the circuit up if
there's a short circuit / something drawing too much current?  I mean,
obviously it adds to the global resistance of the circuit, so if I wanted to
use a higher-current device, I'd have to draw it's power off in parallel
from the battery and use a transistor? (that's a question :))  The only
problem in my head with that could be there's a 5V supply connected to the
base of a transistor that's controlling a 6V "load"; would the internals of
the transistor stop current flowing back down the base, or would I need to
add a diode, or..?

Other strange question:
One other thing that's got my head spinning; assuming I've got a pure 5v
power supply.. now, the PIC datasheet says the chip consumes 20uA typically
(at 3V but lets just assume its the same).  Even without using Ohm's Law
it's clear to see that I=V/R is going to be really quite large.  Is my
assumption correct therefore when I assume that the chip is basically just a
resistor?  I mean, using crazy logic I could add a massive resistor to the
circuit, and the PIC would still work OK because it only needs 20uA, but
then my head melts further when I think when the circuit has no resistor,
wouldnt the chip be overloaded with current and burn (5V / 0.00verysmall R =
veryhigh A).

Thanks :/
Kieren

{Original Message removed}

2002\07\19@080206 by Bill & Pookie

picon face
----- Original Message -----
From: "Kieren Johnstone" <misterfugitspamspam_OUTHOTMAIL.COM>
To: <@spam@PICLISTKILLspamspamMITVMA.MIT.EDU>
Sent: Friday, July 19, 2002 3:23 AM
Subject: Re: [PIC]: Flashing LED Episode II (small
attachment)


> Thanks, this seems most helpful.
> Am I to understand that the resistor is there
only to blow the circuit up if
> there's a short circuit / something drawing too
much current?

******

The LED is a "light emitting diode".  So it is a
diode and diodes act as an infinite resistance
till a certain voltage is reached across it, say 1
volt.  Then it becomes an short circuit to any
voltage above that.  So without the "limiting
resister" too much current would flow through the
led and destroy it.  I think I use a 150 ohm
resister in series with a led.  4v/150 ohms = 27
ma of current.

*****

>I mean,
> obviously it adds to the global resistance of
the circuit, so if I wanted to
> use a higher-current device, I'd have to draw
it's power off in parallel
> from the battery and use a transistor? (that's a
question :))  The only
> problem in my head with that could be there's a
5V supply connected to the
> base of a transistor that's controlling a 6V
"load"; would the internals of
> the transistor stop current flowing back down
the base, or would I need to
> add a diode, or..?

**************

You are thinking too deep.  just be sure the power
supply can supply enuff current at the required
voltage.

***********

> Other strange question:
> One other thing that's got my head spinning;
assuming I've got a pure 5v
> power supply.. now, the PIC datasheet says the
chip consumes 20uA typically
> (at 3V but lets just assume its the same).  Even
without using Ohm's Law
> it's clear to see that I=V/R is going to be
really quite large.  Is my
> assumption correct therefore when I assume that
the chip is basically just a
> resistor?  I mean, using crazy logic I could add
a massive resistor to the
> circuit, and the PIC would still work OK because
it only needs 20uA, but
> then my head melts further when I think when the
circuit has no resistor,
> wouldnt the chip be overloaded with current and
burn (5V / 0.00verysmall R =
> veryhigh A).
>
> Thanks :/
> Kieren

****************

The Pic is a "semiconductor" device.  Which means
that it is part conductor, part resistor.  Just
think of it as a variable resistor.  And  what it
is doing will determine it's effective resistance.

Yes, I=V/R is correct,, but R = V/I.  and
5V/0.000020A  or 250 thousand ohms. But when the
led turns on, then the chip would draw 20ma of
current.  That would be 5V/0.020A or 250 ohms.  A
thousand times less.

So you have to insure that your power supply can
supply the  maximum amount of current you may need
while keeping the voltage supply constant at 5V.

Bill

**************

>
> ----- Original Message -----
> From: "Mircea Chiriciuc" <KILLspamsaschaKILLspamspamROL.RO>
> To: <RemoveMEPICLISTTakeThisOuTspamMITVMA.MIT.EDU>
> Sent: Friday, July 19, 2002 8:46 AM
> Subject: Re: [PIC]: Flashing LED Episode II
(small attachment)
>
>
> > Use this. With a 1V voltage drop across the
resistor the current is also
> > limmited at about  30mA.
> > The voltage drop parammeter for a zenner diode
states for the voltage drop
> > across the diode when reversed biased. So in
the drawing attached the
> diode
> > is "keeping" the voltage at a maximum of 5.1V.
If the voltage tends to
> > increase the zenned will permit current to
flow trough it so the voltage
> > drop will remain aprox. constant untill the
current limit trough the diode
> > is passed and the diode will becone a wire. If
the diode becomes a wire,
> the
> > resistor will get the full current of your
batteries I=U/R=6/33=181mA. The
> > power the resistor will try to excange in heat
will be
> P=U*I=6*0.181=1.09W.
> > If you use a 1/4W resistor it will provide
also protection for your
> circuit
> > as it will burn and open the circuit if the
current will exceed 50mA.
> >
> > Good Luck,
> > Mircea Chiriciuc.
> >
> >
>
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2002\07\19@081001 by Olin Lathrop

face picon face
> Got the code simulating right in MPLAB, anyone fancy checking
> the circuit I've got planned? :)  Would be very helpful to me,
> thanks.

I thought you had the RC oscillator wired upside down until I noticed that
you had the - rail at the top and the + rail at the bottom.  Argh!  Don't do
that.  You also managed to confuse yourself with this because you have MCLR
tied to Vss, which will keep the chip permanently in reset.

There are no LEDs "rated at 5V".  Perhaps you mean one with a built in
resistor.  Even so, I would wire up the LED between the + supply and the PIC
pin because the low side drivers are stronger than the high side drivers.


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2002\07\19@082041 by Olin Lathrop

face picon face
> The PIC16F877 needs 2-5.5V.

No, it needs 4 to 5.5 volts.  Only the LF variant can go down to 2.0 at
clock speeds of 4MHz and below.

> I've never heard of a low-dropout regulator,
> and my intro books just talk about voltage regulators like the 7805..  Is
> there a simple (like, 3/4 pin) "drop-in" replacement that works like a
7805
> (i.e. not too hard to use)?

There are many.  I keep a bunch of LP2950 around.

If you're running from a 6V battery, you know the voltage can't get too
high.  In your case you can get away with a diode in series with the
battery.  This will drop the voltage 600mV or so.  This is a hack, but
should work for what you need in this case.


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2002\07\19@084358 by Olin Lathrop

face picon face
> 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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2002\07\19@085318 by Olin Lathrop

face picon face
> One other thing that's got my head spinning; assuming I've got a pure 5v
> power supply.. now, the PIC datasheet says the chip consumes 20uA
typically
> (at 3V but lets just assume its the same).  Even without using Ohm's Law
> it's clear to see that I=V/R is going to be really quite large.  Is my
> assumption correct therefore when I assume that the chip is basically just
a
> resistor?

No!  It draws 20uA in a particular case.  Its current draw can change
dramatically and quickly.  When running, it will require much more than
20uA.


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2002\07\19@102348 by Kieren Johnstone

picon face
Ok, the only "low dropout regulator" maplin has is..

www.maplin.co.uk/products/Module.asp?CartID=0207191520161455610&modul
eno=8063&modulecode=

To quote the link:

     Specifications:
     Output voltage: 5V typical 10.15V max.
     Line regulation: 20mV
     Load regulation: 35mV
     Output impedance: 35mW
     Quiescent current: 10mA max.
     Output noise: 1505V rms
     Ripple rejection: 72dB min.
     Long term stability: 20mV/1000 hrs
     Dropout voltage: 0.5V min.


Does the "droupout voltage 0.5V min." mean I have to supply 5.5V as a
minimum?  Will this work with my 4 AA batteries ? :)

Thanks,
Kieren

{Original Message removed}

2002\07\19@141820 by Pic Dude

flavicon
face
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 removed}

2002\07\19@145428 by Pic Dude

flavicon
face
True, the zener requires some calcs, but I was suggesting
the zener as one of other alternatives since the 7805
seems so overkill.  Personally, for a flashing LED ckt,
no matter how revolutionary, I'd just use 4.5V, which
can be obtained from 3 batteries or a 4.5V wall-wart,
etc.

BTW, I'm starting to get used to the "Mr. Dude" thing. :-)

Cheers,
-Neil.



{Original Message removed}

2002\07\19@151736 by Pic Dude

flavicon
face
Olin wrote:

> There are no LEDs "rated at 5V".  Perhaps you mean one with a built in
> resistor.

Well, for simplification in this circuit, you can assume that there are
LED's rated at 5V, since you can go to Digikey and get a single component
called an LED with a rating of 5V.  Just call me Mr. Technicality. :-)

> Even so, I would wire up the LED between the + supply and the PIC
> pin because the low side drivers are stronger than the high side drivers.

Why do you say the low-side drivers are stronger than the high-side
drivers?  The 16F87x datasheets say that the max current sourced by
an I/O pin = 25ma, and is the same as that sunk by any pin.  Is
there more that matters here in an LED circuit?

Cheers,
-Neil.

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2002\07\19@151944 by Pic Dude

flavicon
face
Olin wrote:

> In your case you can get away with a diode in series with the
> battery.  This will drop the voltage 600mV or so.  This is a hack, but
> should work for what you need in this case.

This I like.  Must remember this for other purposes.

Cheers,
-Neil.

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2002\07\19@173247 by Pic Dude

flavicon
face
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 removed}

2002\07\19@180531 by Dwayne Reid

flavicon
face
At 02:15 PM 7/19/02 -0500, Pic Dude wrote:

>Why do you say the low-side drivers are stronger than the high-side
>drivers?  The 16F87x datasheets say that the max current sourced by
>an I/O pin = 25ma, and is the same as that sunk by any pin.  Is
>there more that matters here in an LED circuit?

N-channel mosfets have a lower on-resistance than a similarly sized
P-channel device.  If you look closely at the pin output curves published
in the data sheet, you will see that the saturation voltage is lower for a
LO than a HI.

It really doesn't matter if all you are doing is driving a single LED at 10
mA.

dwayne

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2002\07\20@100605 by Bill & Pookie

picon face
The old way of using the ic was to sink current
for the led.  And that worked fine, Is there a
need to change now, why not continue?

When people see a led, they assume that it is
hooked to 5v and that a negative output on the ic
is what turns it on.  Why confuse them?

Bill

{Original Message removed}

2002\07\20@100817 by Bill & Pookie

picon face
My pic board used a 9v battery with some cheep
signal diodes (5) in series with the positive
terminal.

It was a 16c54 with a RC oscillator, led and push
button.  Push the button X number of times, then Y
number of times and the led would blink X*Y number
of times.  Had 5 diodes for the 9v supply voltage
drop, 1 decoupling cap across the supply, led and
current limiting resistor, push button and pull up
resistor, as well as the resistor and cap for the
RC oscillator.

Bill

{Original Message removed}

2002\07\20@114644 by Alan B. Pearce

face picon face
.Olin wrote:

>> In your case you can get away with a diode in series with the
>> battery.  This will drop the voltage 600mV or so.  This is a hack, but
>> should work for what you need in this case.

>This I like.  Must remember this for other purposes.

It also has the obvious advantage that it protects the circuit against
reverse polarity batteries - ever the problem with a hobbyist in a hurry :)

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2002\07\20@114700 by Bill & Pookie

picon face
Oh yes.

A flash of light, loud bang, puff of smoke, and
the smell.  How useful the nose is in
troubleshooting.  Pookie can tell by the smell,
what value of resistor crashed and burned.  Which
is useful as the color code on the darn thing
changes a few shades.

And the flashing led.
Didn't Einstein crawl before he walked?  Babbled
before talking?  Made gunpowder before the atomic
bomb?

I think the best way to learn something new is to
start out as simple as possible.  Then after I
accomplish it and get that good feeling, start
adding stuff to it.

Bill

{Original Message removed}

2002\07\20@193833 by David Duffy

flavicon
face
Bill (& Pookie) wrote:
>The old way of using the ic was to sink current
>for the led.  And that worked fine, Is there a
>need to change now, why not continue?

One reason is that the 0V of the board is more likely to be close
to where you want to connect the LED and resistor to especially
if it's ground plane construction. Personally, I tie the LED to 0V
or +5V depending on the circumstances. If it's RA4 (open drain),
the tying the LED (via a resistor or course) to +5V is the way to go.
Almost always, you can flip the polarity in the PIC code anyway.

>When people see a led, they assume that it is
>hooked to 5v and that a negative output on the ic
>is what turns it on.

Do they? I don't. You know what they say about assuming don't you?
This sort of assumption is never good. It's how people make mistakes
and end up wasting time trouble-shooting based on that assumption.

>Why confuse them?

They could always read the schematic couldn't they?  :-)
Regards...

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