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02-03-2011, 10:38 PM   #1
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## resistor question

good day all.
i am in the process of building an led array. consist of 30X3w XP-E from cree
and 10X3w XP-G

NOW TO MY QUESTION.
each XP-E led runs on 1000ma 3.5VDC
each XP-G led runs on 1500ma 3.5VDC

I have a power supply with 24VDC and 14.6AMPS.
Is there such a resistor that could restrict the power from 24VDC 14.6Amps to 1000ma 24VDC.as i will run the led as a parallel.

02-04-2011, 06:15 AM   #2
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If the load drops 3.5 vdc from a 24vdc supply, what remains is 20.5 vdc to be dropped by the resistor. choose a resistor that will drop 20.5 vdc and contral a 1000 mA current. r=e/i, r = 20.5/ 0.1, or 205 ohms. and, it must be capable of 0.1 a x 20.5v, or 2.5 watt. you're getting into power resistors here, because of the large voltage drop required. and all of this is for one led load, you didn't say what you were putting in parallel.

 02-04-2011, 07:46 AM #3 Member   Join Date: Nov 2007 Location: Nashua, NH, USA Posts: 7,968 Rewards Points: 1,542 Connect the LED units in series strings. Ideally you need seven 3-1/2 volt LEDs in a row to go with 24 volts. In your case seven does not divide equally into the number of LEDs you have (10 1-1/2 amp pieces) and so you might see if it works with ten LEDs in a row. (The 1 amp pieces can be wired as two series strings of 7 and two series strings of 8) But do not mix the 1500 ma LEDs and the 1000 ma LEDs in the same series string. More than the minimum (of 7 here) per series string adds a safety margin since slight variations (sample to sample differences) among the LEDs can result in one getting slightly more voltage than the next in the same series string. For LEDs, each series string should have its own dropping resistor if the counts don't work out evenly (for example a string of 5 where you really need 7). If several series strings (or singles) share the same dropping resistor and one string burns out, the remaining strings will become overvoltaged. Connecting all of the LEDs in parallel and adding one giant dropping resistor is very energy inefficient also. Power is dissipated in the resistor as heat. The amount of heat (here it is wasted energy ) can be measured in watts and equals the voltage drop times the number of amperes. __________________ The good conscientious technician or serviceperson will carry extra oils and lubricants in case the new pump did not come with oil or the oil was accidentally spilled, so the service call can be completed without an extra visit. Last edited by AllanJ; 02-04-2011 at 08:13 AM.

02-04-2011, 07:50 AM   #4
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Quote:
 Originally Posted by TimPa If the load drops 3.5 vdc from a 24vdc supply, what remains is 20.5 vdc to be dropped by the resistor. choose a resistor that will drop 20.5 vdc and contral a 1000 mA current. r=e/i, r = 20.5/ 0.1, or 205 ohms. and, it must be capable of 0.1 a x 20.5v, or 2.5 watt. you're getting into power resistors here, because of the large voltage drop required. and all of this is for one led load, you didn't say what you were putting in parallel.
Your calc is off. 1000mA = 1 amp. It's 1 amp x 20.5 volts or a 20.5 watt resistor for each red array. The green array is even worse. 1.5 x 20.5 = 30.75 watts. That's a lot of heat. You would be better off getting the proper power supply for the LED array.

You'll be wasting more energy in heat than you would save using LED. Probably be cheaper to run regular 24 light bulbs..

Last edited by joed; 02-04-2011 at 05:49 PM.

 The Following User Says Thank You to joed For This Useful Post: TimPa (02-04-2011)
 02-04-2011, 08:49 AM #5 Member   Join Date: Aug 2010 Location: north central pa Posts: 521 Rewards Points: 869 good catch, i was thinking small.

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