From: "Sir Charles W. Shults III"
References: <email@example.com> <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org> <email@example.com>
Subject: Re: current source for array of infrared leds
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Date: Tue, 03 Dec 2002 17:14:05 GMT
NNTP-Posting-Date: Tue, 03 Dec 2002 12:14:05 EST
Organization: RoadRunner - Central Florida
Add to that the fact that running a string of series LEDs will automatically
increase the efficiency because the resistive drop is much smaller- if you are
within a LED drop or so of the supply voltage there really isn't much lost as
heat. Working this from the voltage side is bass-ackwards at the least.
For 12 volt supplies, assuming that they are regulated and you LEDs have a
1.5 volt drop (which is typical for many new models) and you run 50 mA (also
pretty typical for bright units) then you only see a tiny part of the power as
For instance, you might put 7 LEDs in series for a total of 10.5 volts,
leaving you with some comfortable headroom. Now, with 50 mA being drawn, your
resistor is only 30 ohms, not too low or too high. The LEDs are consuming 0.525
watts, and the resistor is only radiating 0.075 watts, a skimpy 75 milliwatts,
as heat. Overall, (not counting LED light efficiency), you are delivering 87.5%
of the power to the LEDs.
This is comparable to many power supplies' efficiencies. Doing it with
nothing more than a resistor has a great deal going for it- efficient, small,
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