Re: ATM batteries for LED

[Tom Stokes <tstokes@pacbell.net>]

Does the resistance of the batteries come into play anywhere?
Would two new AA cells be as bright as two new D cells?

Tom

----- Original Message -----
From: "Dave McCarter" <dmccarter@sympatico.ca>
To: "Bob May" <bobmay@nethere.com>; <atm@shore.net>
Sent: Tuesday, June 18, 2002 8:02 PM
Subject: Re: ATM batteries for LED


>
> > The desire for max life of the battery is more the reason for the 9V
> battery
>
> That is a false economy. Lets look at the math.
>
> Suppose the battery is 9.3 volts new. The LED requires 2.3 volts and draws
> 10mA.
> There is a 9.3-2.3 = 7 volt drop across the resistance.
> The LED produces 2.3v x 0.01A = 23mW of light and heat.
>
> The resistance produces 7v x 0.01A = 70mW of heat, which unless the mirror
> testing room is cold, is total waste. The circuit is only 23/93 or about
25
> percent efficient. It is true that as the battery voltage drops if the
> resistance can be lowered to keep LED current constant the resistance will
> dissipate less and less power and total efficiency will improve. But most
> battery discharge curves are pretty steep near the end of service, so this
> is overall not very important. By the way, a nine volt battery is made up
of
> cells with quite low current capacity, so life length is short.
>
> What if the battery choice is a pair of AA cells with 4 to 5 times the
> current capacity? The new potential is 1.65 x 2 = 3.3v, so the drop across
> the resistance will be 1 volt, and the heat produced will be 1v x 0.01A =
> 10mW. Again, if the resistance is variable the LED will continue to
operate
> until the batteries get down to just over 2.3 volts, or 1.15 volts each.
Now
> that is a dead AA battery, and through it's life 23/33 or 70% of the
energy
> was delivered to the LED. And the AA cells will last many times longer.
>
> Dave
>
>