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bob1371 (Industrial) (OP)
11 Apr 06 12:16
I have an illuminated pushbutton switch. It uses a LED which requires 3.5 VDC. I have 2 power source available.
24 VDC and 12 VDC. What size resistor should I use to drop the voltage to this level? The resistance thru the LED is 2.18 Kohm.

thanks,
Bob
VE1BLL (Military)
11 Apr 06 12:32
LEDs don't really have resistance (assuming that it is JUST an LED, without a built-in resistor). They have voltage drops. The voltage drop usually relates to the colour of the LED diode itself (not including ant other tricks).

Take the power supply (12 VDC is reasonable) and subtract the LED drop (at a reasonable current). This leaves the voltage you want across the resistor.

Then use Ohms Law to work out the resistor value based on the desired current and calculated voltage.

Finally, double check the power on the resistor.

Using a lower voltage supply means that your resistor can be dissipating less power as heat. But if you get too close to the LED drop, then the current can get unstable if any voltage changes.

VE1BLL (Military)
11 Apr 06 12:37
Typo: 'any' not 'ant'.
coconutalley (Electrical)
11 Apr 06 12:42
Bob1371,

For 24Vdc, (Vsupply - Vledvf-3.5V)/Iled = resistor
For example, (24 - 2.0-3.5)/.005 = 3.7 kOhms. I guessed
5mA should keep the LED lit and resistor to 1/4W.
3.7 kOhm - 2.18 kohm = is the resistor you need.
Check resistor for heat dissipation.
P = Iled^2 * resistor
e.g. P = .005^2*3700 = 92.5 mWatts
You might still be able to push more current to get brighter
LED.  Same steps for the 12Vdc.
coconutalley (Electrical)
11 Apr 06 12:43
That's assuming your switch has an LED with the series 2.18kohms.
bob1371 (Industrial) (OP)
11 Apr 06 13:10
Thanks for the Info. I should have given a little more detail in first post. Here is a link to the switch Im working with. http://www.bulgin.co.uk/PDFs/CatNo82/CatNo82pages/Page_183_2005_Cat.pdf

it came with no details at all. I dont have internet access at work so I take the first one, hook up to my 24 vdc test box and you can guess the rest Im sure. After finding the company and getting the pdf it says "an appropriate resistor must be series connected by user." that would have been useful info to send with the sensor.

thanks again,
Bob
jimkirk (Electrical)
11 Apr 06 13:58
According to the spec, 3.5 volts is the blue flavor.  LED is rated at 20 mA.

So resistance = (Vsupply - 3.5) / .02 as a minimum value.
For 12 volts that's 245 ohms at .17 watts.
For 24 volts it's 1.025 kOhms at .41 watts.

So using real world values, try 270 ohms 1/4 watt resistor if you've got 12 volts.  For 24 volts, 1 kOhm or 1.2 kOhms.
1/2 watt should work, but a conservative application might want more margin.

These are minimum values.  If it's too bright, increase the resistance.
itsmoked (Electrical)
11 Apr 06 14:11
Cool switches!!! I WANT ONE (or twelve). Where do I get them? How much do they cost? what part actually moves when you activate them?

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com
bob1371 (Industrial) (OP)
11 Apr 06 17:18
I bought them from Allied Electric. Im gonna use them In a motor home Im working on and also have a couple planned for my daily driver (rear window release, fog lamps)
they are about $16 each.

thanks for all the help guys. Not only an answer but also how you came up with the answer so I can do this myself in the future.

Bob
logbook (Electrical)
11 Apr 06 18:08
jimkirk's gone a bit dyslexic there.

He meant 425 ohms at 12V.
itsmoked (Electrical)
11 Apr 06 18:22
So he runs 35mA thru the LED...  It will be really bright. (for a while)

Keith Cress
Flamin Systems, Inc.- <http://www.flaminsystems.com>
jimkirk (Electrical)
12 Apr 06 7:31
Thanks for the catch, logbook.  Yes, 425 ohms, or 430 for an E12 series value.  At least I got the power correct. smile

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