Try this site:
http://led.linear1.org/1led.wiz
If you don't know the diode(LED) forward voltage and current go to
RadioShack and get a new one. The info is on the LED packaging and the
website under Tech specs. for the Power supple voltage, if in series add
the volts if in parallel its the mean(NOT RECOMMENDED to use more than one
cell(battery), unless they are all the same) i.e. voltage, size and amount
charged. or that more powerful battery will try to charge the others.
http://www.radioshack.com/search/ind...rigkw=led&sr=1
In order to change or lower the brightness simply, take the output of the
calculator measured in Ohms, and get a pot or (variable resistor) with the
same value. that way you will get max power and brightness with your chosen
LED.
Hope this Helps
Mark
"Victory" wrote in message
...
I forgot to add - with the 12v, the LED went all the way to it
brightest and dimmest while turning the resistor. It was perfect, but
with the 5V, it peaked near the top of the resistors turn, and then
started to burn out. Not sure why with 5V it would burn out, but with
12V it did not.
On Jan 25, 7:29 pm, "Dave Plowman (News)"
wrote:
In article
,
Victory wrote:
Ok, I tried something with a 10K variable resistor today at a school
today (they were kind to indulge me).
When I hooked up the 2 3V batteries in series to the resistor, it
dimmed immediately when I turned it only slightly. The falloff of
power to the LED went to the dimmest setting right away. On a whim,
we hooked it up to 12V and the dimmer worked all the way through the
entire turn. It was great. I am a little confused why this
happened? The LED comes with 2 3V batteries in series from the $
store and is quite bright. When hooked up to the 10K resistor, it
isn't as bright. So, the next basic question is - what happened?
Should i get a resistor that is less than 10K?
LEDs are current driven. Which in practice means a different value
resistor according to the supply voltage as you need to keep the current
through the LED constant for the same light output.
You can work out this series resistor value easily. The formula is:-
Vs - Vf
R = ------------
I
Where Vs is the supply voltage, Vf the forward voltage drop of the LED
and
I the current in amps. You'd need to know the spec of the actual LED to
be
accurate but making Vs 3V and I 0.020A will do for most white LEDs.
Which gives 150 ohms for 6 volts and 450 ohms for 12
The pot needs to go in series with a fixed resistor of the above value to
set the maximum current - otherwise you could burn out the LED. The pot
should be about 4 x the fixed resistor value to get a reasonably smooth
dimming range.
--
*Funny, I don't remember being absent minded.
Dave Plowman London SW
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