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 -----
From: "Mircea Chiriciuc" <sascha@ROL.RO>
To: <PICLIST@MITVMA.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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