Newbie (I hope not to dumb) question....

[Alphawolf]

I'm just getting into this all and know just enough electronics to be dangerous! I have a 5600 mcd ultra-white led that needs 3.2-4.o volts. My question.....Is there any "room" for over-volting these things? Say to 4.5 volts? Or do I need to add a resistor to keep it safe? Thanks in advance. (where I'm going with this is to place the LED in my mini-mag AA light with 3 1.5 volt N cells for power.)

 

[lambda]

Check out http://www.otherpower.com/otherpower_lighting_flashlites.html this guy has pics and instructions for what you want to do. And to answer your question, it's not the volts you need to measure, its the milliamps. Batteries drop voltage when current is drawn, so its varable. Read this guy's site, he explains a lot.

 

[remuen]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Alphawolf:
I'm just getting into this all and know just enough electronics to be dangerous! I have a 5600 mcd ultra-white led that needs 3.2-4.o volts. My question.....Is there any "room" for over-volting these things? Say to 4.5 volts? Or do I need to add a resistor to keep it safe? Thanks in advance. (where I'm going with this is to place the LED in my mini-mag AA light with 3 1.5 volt N cells for power.)<HR></BLOCKQUOTE>

As lambda wrote: It depends on the current not on the voltage - but the voltage says what current flows. In the practical use you have always to measure the current through the LED to avoid any damages. Up to a current of 40 milliamps you will have (almost) no problems with the white 5600mcd LED but at a higher current the LED gets warm and the life time will be reduced - in the worst case it will die after a few seconds!

In your Minimag mod with 3 N cells you don't need any resistor because the internal resistance of the N cells is high so the voltage and therefore the current drops down to a level where the LED will not be damaged. I didn't measure the current but estimate it's somewhere around 30 milliamps with new batteries. This easy mod works pretty well but the light isn't very bright and the longer you are using it the more it will dim.

With other batteries eg. 3 AA or 3 DD cells you always need a resistor in series with the LED. That's because the internal resistance of these batteries is lower means the voltage an a certain load is higher means they can supply much a higher current which would damage the LED.

Another point: The higher the current the lower is the efficiency of the LED. This means in a practical usage: Two LED powered with 20 milliamps give more light than one LED with 40 milliamps. Somewhere in the CPF you will find a post explaining and proofing this with detailled measurment results. Use the search function to find it (eg. with the search term 'efficiency').

 

[remuen]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by remuen:

In your Minimag mod with 3 N cells you don't need any resistor because the internal resistance of the N cells is high so the voltage and therefore the current drops down to a level where the LED will not be damaged. I didn't measure the current but estimate it's somewhere around 30 milliamps with new batteries.
<HR></BLOCKQUOTE>

I just measured the current for you in this Minimag mod.

With 3 N cells (not new) and 1 LED the current is 17 millamps, with 3 LED parallel switched 28 milliamps (means only 9 millamps through each single LED). But the light with 3 LED is reasonable brighter than with only one LED. Unfortunatly I don't have a Lux meter to measure the light.

The voltage drops from 4.4 volts (that is the voltage on the 3 N cells without any load) to 3.24 volts with 3 LED's and 3.47 volts with one LED. This voltage drop is caused by the internal resistance of these N cells.

So I think the maximum current for 1 LED with new batteries will be somewhere between 20 and max 25 milliamps.