# Gave it a go, but ultimately I'm too thick.

Discussion in 'General Electronics Discussion' started by qwertyface, Apr 24, 2013.

1. ### qwertyface

1
0
Apr 24, 2013
Hi all,

I'm trying to hack together a rudimentary circuit to drive a high power LED. I have a mere basic understanding of electronics and electrical theory and I am turning to you brighter chaps for support with my calculations as I don't want to switch my circuit on and watch it smoke! These high power LEDs ain't cheep...

I waffle a bit to begin but only because I want to demonstrate that I've tried and not just posting on here for a quick answer without giving it a go first!

I hacked apart an old 300W computer PSU putting a dummy load between the PS_ON and GND wires. I've done this so that I can use the 5v/12v rails.

My LED has a Vf of 3 to 3.7V and forward current of 700 to 750mA. I would like it to take 730mA (to be persnickety!) but because I expect to use a number of these at a later date I have opted to use the 12V rail and figure therefore that I'll need a substantial resistor in the interim.

On the back of some paper my circuit is simple: Vcc --- LED --- R --- GND.

To my understanding KVL states that the sum of all voltages across components is equal to Vcc. I'll use 3V as my LEDs Vf, then the voltage drop across the resistor is 12 - 3 = 9V. I'd like to operate the LED at 730mA, so the wattage of the resistor is going to be 9 * 0.73 = 6.57W, but doubling up as per convention means I'm going to need a 12 to 15W resistor. AFAIK its value will need to be (12 - 3) / 0.73 = 12.32R, ~13R. So a 13R 12W resistor... a ceramic one comes to mind. And I have one, but at 10R 20W.

The problem:

LEDs are sensitive as you well know and so I need to determine what effect this resistor value is going to have in terms of both the new voltage across the LED (I am assuming that as the resistor I have is a lower value then what I require based on the previous calculation I'm guessing it will not sink the voltage I require it to and therefore the voltage drop over the LED will increase?) and the current through the LED (note, my DMM says the resistor is 10.14 ohms)

1. Am I correct in calculating the current through the LED with the following: (12- 3) / 10.14 = 0.888A, or ~890mA ?
2. I don't know how to determine the voltage, how would I do this?

General questions:
Thanks
QF.

Last edited: Apr 24, 2013
2. ### GreenGiant

842
6
Feb 9, 2012
Generally you put the resistor first, but it doesnt matter that much, and ideally you would want a larger resistor (less current) than you "need" though at this current and voltage you may be better off getting a current regulator, you are going to be losing so much power in the resistor (almost 3 times the power used by the LED!!).

If you don't want to do that you can get a 1 watt 2 ohm resistor and put it in series with the resistor you have now and that will bring your current down to 750mA (for a 3V drop on the LED) or 690mA (for a 3.7V drop on the LED)

3. ### BobK

7,682
1,688
Jan 5, 2010
Your calculations look correct. And you are using the low end of the Vf, which is the correct way to be safe. If you want to use the 10R resistor, use two 1A or more diodes in series. This will drop and addtional 1.2V or more.

Shield your eyes from the LED, it is going to be bright! I temporarily blinded myself for minutes when I connected a high power LED with the wrong voltage. My sight recovered fortuneately, but the LED did not.

And be sure to heat sink the LED. Hopefully it is on a star, but that will not be enough the for the 3W or so it wil be running at.

Bob

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
Yeah, I agree with most of the above. Especially shielding the LED when you power it on. I've placed paper over them, or deliberately pointed them toward the bench rather than toward me.

There is a long discussion about driving LEDs in the tutorials section that may be worth a read. Generally I wouldn't advise using a resistor for a high power LED, but that's not absolute as long as you're aware of the power dissipateion in the resistor (which you are).

I also presume the LED has a heatsink...

I would tend to err on the side of operating the LED at a lower current. If you calculate 13 ohms, see what 15 ohms gives you -- or follow the excellent advice which precedes mine.

5. ### cjdelphi

1,096
104
Oct 26, 2011
I'd opt for PWM output rather than relying on the a huge watt resistor alone...

6. ### CocaCola

3,635
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Apr 7, 2012
Last edited: Apr 26, 2013

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Apr 12, 2013