Re: LED question for electronics wizards
- From: Ook <zootal@xxxxxxxxx>
- Date: Mon, 11 Jun 2007 12:15:13 -0700
<ship>
If you want to be more efficient, hook up 2 or 3 LEDS in series. 3 of them
at 3.5 V would be 10.5V, leaving 1.5V across the resistor. The total current
(power) used is the same, but now, instead of heating up a resistor and
lighting 1 LED, we are lighting up 3 LEDs instead of one. I leave finding
the size of the resistor in this case as an exercise for anyone one that
really wants to find out because I gotta go and don't have any more time :(
Here is another thought. Using the math I did, if you had a 12V
battery and 3 LEDS, each 3.5V, then your resistor would be 60 ohm, and
you could use a 1/4 or 1/8 watt resistor. However! If your LEDs turned
out to be 3.0 V and not 3.5, then the current through them would be
50mA, not 25mA. You may be pushing their design limitations and risk
burning them out, and the resistor may get hot depending on the size
your used.
If you want to get fancy, you could build a current regulator and set
it at 25mA. Use this in place of your current limiting resistor. Then
you could use any combination of 1, 2, or 3 LEDs, and the current
would still be properly regulated. An LM317 will do this quite nicely.
The downside of this is that you need to understand a bit about
electronics, or find a friend that does to design it for you. I could
whip one out pretty quickly, but if you don't know a bit about
electronics, you won't know what to do if it's not working, or how to
test that it is indeed working correctly.
.
- References:
- LED question for electronics wizards
- From: Robert Reynolds
- Re: LED question for electronics wizards
- From: Chuck
- Re: LED question for electronics wizards
- From: Robert Reynolds
- Re: LED question for electronics wizards
- From: Tim Wescott
- Re: LED question for electronics wizards
- From: Ook
- LED question for electronics wizards
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