LED Project

[Electriceye]

Hi all, I have procrastinated far too long now on this project. I have managed to wire a few together and make them light up, and show them off...however I want a proper solution.

I have these 2.4-2.8VF 700Ma, Colour LED. I need to get them powered by whatever PSU I have laying around. Currently I have to work with 32V @ 925MA PS. 3 x 5V @ 700MA.

I am using this nifty calculator "LED series/parallel array wizard" on wiring LED'S in series, and it tells me what size resistor and how much power it will dissipate based on my build.

We have 13 LEDS x 2.4VF @ 700Ma. Which consumes 31.2V, since my PS is 32V...I need to drop off 0.8V. It reccomends a 1w 1.2Ohm resistor.

I also would like to limit the current to the LEDS to 650, how would I go about getting this ?

I really need some help, I am scared to just plug and play here....

 

[MikeMl]

You need to allow for a larger fraction of the available power supply voltage to appear across the current limiting resistor. The whole idea of driving leds is to drive them with a nearly constant current. If you only drop 0.8V out of 32V, you are not coming close to inoculating the leds against thermal variations. This is especially important for high power leds.

You really should be using a constant-current led driver type of power supply...

 

[audioguru]

Your LED Wizard is very stupid and it doesn't know that if all the LEDs are the maximum voltage of 2.8V then 13 in series will not light from your 32V supply because they need 13 x 2.8V= 36.4V plus more voltage for the current-limiting resistor.

Use common sense intead of a stupid Wizard:
1) Do not operate LEDs at their absolute maximum allowed current. You are smart to consider 650mA.
2) Connect 9 LEDs in series. If they are all 2.4V then they use 21.6V. The current-limiting resistor is (32V - 21.6V)/650mA= 16 ohms. Use a 10W 16 ohm resistor if you can find one.
3) If all 9 LEDs are 2.8V then they use 25.2V. Their current is (32V - 25.2V)/16= 425mA.
5) If all 9 LEDs are halfway at 2.6V then they use 23.4V. their current is (32V - 23.4V)/16= 538mA.

You might not notice the small amount of brightness difference.

An LM317 IC plus one resistor makes a constant current source for 10 of those LEDs when powered from a 32V supply.

 

[Electriceye]

Thats what I thought the resistor was doing My intentions were on providing constant current.

With your calculations in 3 and 5, you are using the 10w 16ohm resistor ?

Thanks for your input all, I will make the constant current with an LM317, found a good instructable. I will come back if I have any questions.

 

[MikeMl]

Since it will be dissipating something close to 10V*.6A = 6W, you will have to put the LM317 on a heat sink. If it blisters your wet finger, make the heat sink bigger

 

[audioguru]

With your calculations in 3 and 5, you are using the 10w 16ohm resistor?

Yes, both my calculations show the 16 ohms resistor.

I will make the constant current with an LM317, found a good instructable. I will come back if I have any questions.

DO NOT look at the circuit of an Instructable because many are defective and were designed by little kids who know NOTHING about electronics. Instead look at TI-National Semi's datasheet for the LM317 (National Semi invented it).