Re: ATM A little electronic help on my Foucault tester

[Steve JB Bouton <sjbbouton@juno.com>]

Danger Will Robinson!  Danger!

You must at least include a small minimum resistance to limit MAXIMUM
current, or you risk popping the LED- especially if the power supply is
capable of delivering a high current.  The only exception may be if the
rated drop of the diode is greater than the voltage of the power source. 
And even then, this should only be done with small batteries.  I've been
experimenting with different color light and night vision, and the blue
LED I've been using has a voltage rating greater than the 2 AA batteries
powering it.  No problem without a resistor (but I still put a 10 Ohm
resistor in series).  Do this with a 9V battery, and you could pop it.

And to address the worry about "wasting power", well that is trivial.  A
pair of AA batteries powering a 2V / 20mA-rated LED requires a resistor
that "wastes" 20 mW.  20 thousanths of a watt!  It will take a lot of use
before you will even notice a drop in brightness, and this at maximum
current.

On Tue, 04 Dec 2001 15:40:51 -0600 Joe Mayenschein <mayen1@mwt.net>
writes:
> 
> The easiest way is to forget all this fixed value resistor garb and 
> get yourself
> a Pot and be done with it. then you can use every last bit of energy 
> the battery
> has to offer.
> 
> Joe
> 
> Leftfieldstar@aol.com wrote:
> 
> > Mike,
> >
> >  That's a nice handy method for figuring the resister values. Do 
> you have a
> > reference for other convenient data computation like that?
> >
> > <<Take the battery voltage, subtract the LED voltage, and divide 
> by the
> > nominal
> > current to find the minimum resistor you need in series.>>
> >
> > Dominic DiLeo
> 
> 

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