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Subject: Re: Best way to power array of LEDs?
References: <0001HW.B9C26E7400A7161B165FEAC0@news.covad.net> <2Qan9.3129$cS4.firstname.lastname@example.org> <3DC843FE.4A25FFF3@fanwap.com>
Date: Tue, 05 Nov 2002 23:26:11 GMT
NNTP-Posting-Date: Tue, 05 Nov 2002 18:26:11 EST
Organization: Road Runner
Nucharin W. Jansen wrote:
> What's happen when voltage drop across LEDs are not same ?
> I have problem with white LEDs in series.
> I can't control the current.
> I have to adjust in real work one by one in each branch.
> Ban wrote:
>>"Jerry G." schrieb im Newsbeitrag
>>>Each LED must have its own series resistor. Most of these LED's work
>>>at about 30 to 40 ma with about 1.6 VDC across them. You should be able
>>>to use a 320 ohm 1/2 watt resistor in series with each LED if these are
>>>in the spec range that I used as an example.
>>>LED's cannot be wired in parallel to each other being fed from one
>>>resistor. They can not also be wired in series with each other.
>>Nope, of cause they can be wired in series if the available voltage allows
>>Check the datasheet for the typical forward voltage. Red leds have around
>>1.7V, yellow/green ones are 1.9-2.3V, blue/white ones even higher 3.2-3.8V.
>>If you use just a resistor in series with each chain, you should calculate
>>it for the max. supply voltage which I would suggest is 14.0V.
>>Now if you put 6 red leds in series with 20mA current rating, you get
>>6*1.7V=10.2V; 14V-10.2V= 3.8V; 3.8V/20mA=190 ohms; take 180 or 200 ohms. The
>>resistor can be a normal 1/4W 0207 type.
>>electronic hardware designer
The best way to power a bunch of LED's is to put them in series. You do
not need a regulator, but if you want to keep the brightness from going
down when battery voltage starts to go down, you can use a 78xx
regulator, and use it as a current regulator. There are many examples of
this online. Just do a search. Also, by putting them in series, you get
all the LED's lit up with only the current load of ONE! So, it is still
only 20ma for ALL the LED's together, making it much more efficient for