Re: LEDs - Simple Wireing Questions

From: John Popelish (jpopelish_at_rica.net)
Date: 11/04/04 

Date: Thu, 04 Nov 2004 12:35:34 -0500

ShadowTek wrote:
>
> OK thanks.
> I thought 3 Amps for a AA battery sounded strange but I was confused
> by the markings on the battery.
> They were "LRG AM3 1.5V".
> I guess LRG just means large and AM3 means short curcuit current.

I doubt that the designation LRG AM3 has anything to do with size or
amperage.
Here is a summary of some Duracell specs:
http://www.web-ee.com/Component%20Pages/battery_info.htm
Note that the MN1500 AA cell is rated for 2850 milliampere hours and
its maximum rated load is 43 ohms for a load current of 28
milliamperes. Now, .028*43=1.2 volts, so this implies that a typical
internal resistance must be dropping the rest of 1.5 volts. So the
internal resistance is about .3volts/.028 amperes or 10 ohms. So the
short circuit current should be about 1.5 volts / 10 ohms = .15
amperes. I am quite sure that you will get more than this with a
fresh cell, but this is typical for the mid life point.
Here is another site that summarizes battery characteristics:
http://www.zbattery.com/zbattery/batteryinfo.html

Note the resistance of the alkaline AA cells.

> I was further thrown off by trying to read the current with my
> multimeter from terminal to terminal.
> I guess that I was createing a short curcuit which was trying to empty
> the battery.

Yes. Current meters have very low resistance.
 
> (BTW, my multimeter only reads up to 250mA so do you think reading a
> 3A short curcuit for no more than 2 seconds would have damaged the
> MM?)

If it reads more than zero then you haven't blown the fuse in series
with the current meter function.
 
> The exact specs for the LEDs were 2.8V-Min. 3.2V Typ. 3.8V-Max..

Yes, the required voltage varies with temperature and lot to lot.

> So if intend to run the LED at 2.9V then I should still be able to run
> 1 LED off 2 1.5V cells right?

The two cells with their 20 ohms of total internal resistance will
push some current through the LED, but the exact amount is hard to
predict, except that it will be low.

> .1 divided by .02 = 5 Ohm reistor?
> So 2 AA batteries and a 5 Ohm resistor will run 1 of these LEDs for
> 150 hours?

The resistor is in addition ot the 20 ohms of internal resistance, and
at this low voltage the current will be much lower than 20
milliamperes, so the battery life will be very long.

> Or I could wire several LEDs in parallel so long as a 5 Ohm resistor
> prefixed every led?

Yes. But if the LEDs vary all over the specified voltage range, you
can expect the one with the lowest drop to be drawing much more
current than the one with the highest drop.

This is why I recommended that you use 3 cells and a much higher
resistance in series with each. It makes the current matching much
better and more independent of LED variations and sag in battery
voltage.
 
> Do LEDs suffer any damage from being underpowerd?

Just the opposite. Their light output is roughly proportional to
their current, but their lifetime goes up dramatically as the current
goes down.

> Since we are talking about batteries then the constant drain would be
> a problem.
> I have a 3 white LED headlamp and I have run the batteries in it all
> the way down several times.
> It uses 2 flat 3V lithium batteries.
> It still seems to work ok.
>
> BTW, radio shack sells their white LEDs for 6$ a peice.
> If you are used to payin that much then you should check out the link
> I posted earlier. 

-- 
John Popelish

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