You can buy LEDs with the resistor already integrated. In fact I think with
some you can even buy a PWM switching power supply already integrated so you
can run them at your chosen voltage. Thats what I'd intended to do but not what
actually happened.
The LEDs I ruined were surface mount package which wasn't as easy to solder as
I'd hoped. Looking around I found a bag of through hole LEDs I'd bought but
never used. Dropping one cell from my battery pack gives me 3.6v using
rechargeable batteries which I tested and appeared to be acceptable with my
35ma LEDs so I rewired the thing with those. I finished it last night and
testing shows that it works pretty well for a reading light (its intended
purpose) although it could stand a little work aiming the LEDs better.
I think you missed the point of my diode, I've got a single pole single throw 3
position switch. On low it activates one string of lights, on high it activates
both. The diode sits on the jumper from the high side so that it doesn't turn
on when low is called for.
Also in my experiments LEDs must be used in parallel which matches my
understanding of how they work. Because they are one way valves the electrical
pixies won't pass through like they would a normal valve. Certainly when I
strung them up in series they wouldn't do anything...
-Curt
From: fmiser <fmi...@gmail.com>
To: mercedes@okiebenz.com
Cc: Curt Raymond <curtlud...@yahoo.com>
Sent: Sunday, January 10, 2016 7:35 PM
Subject: Re: [MBZ] Using LEDs - was OT: Politics are angrier polls
> Curt wrote:
>
> Tonight while I was not watching television I learned how
> to wire in a diode. I'm rebuilding the light I use over my
> bed at camp converting it from fluorescent to LED.
> I'd wired the LEDs in 2 strings with a 3 position switch, I
> wanted one position to turn on one string and the other
> position to turn on both. I harvested a diode from an old
> computer power supply to make that work and I think it
> would work except I discovered that what I THOUGHT were 6v
> LEDs turned out to be 3.6v. You push 6v (ie 4 D cell
> batteries in series) into a 3.6v LED and it cooks off
> pretty quick, whoops! I'm curious how bright the LEDs will
> be at 3v, they were SUPER bright at 6.
LEDs don't work like that. It is the _current_ (amps) that
matter. The 3.6 V is not supply voltage - it is the voltage
drop across the diode when conducting.
And if you are supplying DC, you don't need a diode with it.
And it's usually not a good idea to wire LEDs in parallel.
So - do you know the current rating of those LEDS? 20 ma is
typical for "old fashioned" general purpose LEDS. Newer,
brighter LEDs can be designed for a whole lot more.
The you subtract the LED voltage from the supply voltage (6
minus 3.6 equals 2.4) and then using ohm's law figure out
what size resistor you need to drop 2.4 V at the design
current. So if those are 200 ma LEDs, and R=E/I, then
2.4/0.2 = 12 ohm. Then we need to be sure you won't fry the
resistor. Watts = current X voltage, so 2.4 X 0.2 = 0.48
Watt. A half watt resistor would be big enough - just. But
it's a good idea to always overrate wattage a bit. I would
use a 1 W resistor.
With a 6 V supply, for each of the 3.6 V LEDs (if they are
rated for 200 ma), you would need a single one watt 12 ohm
resistor in series with it.
If that's not clear enough, let me know and I'll answer
questions or go into more detail. *smiles*
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