Discussion in 'General Electronics Chat' started by imraneesa, Dec 28, 2014.

1. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
I want to check the driver for current. But I don't want to use LED before it spoil. I want to use some kind of test load so that I can use it and adjust the desired current and then finally will use led. Give me a calculation how to make test load for 1w LED. And if it is 500mW LED what will be the test load.

2. ### #12 Expert

Nov 30, 2010
16,298
6,811
P = I E
1 watt = I E
E = IR
1 watt = I squared R
1 watt / I squared = R
Make your load equal to R for 1 watt and 2R for 500 mw.

imraneesa likes this.
3. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
But I don't know how much is I. Then how I will know the value of R.

4. ### #12 Expert

Nov 30, 2010
16,298
6,811
We don't know what I is either, or the voltage the LEDs use, or the voltage you have available to make the test.
You give us almost nothing to work with, not even the color of the LEDs, or a part number or source for where the LEDs came from.

5. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
Iam using two 18650 batteries. So I know the voltage 3.7+3.7. I want to calculate current. So how much load I can put is my question.

6. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,573
2,541
Do you have the LED model number and manufacturer? datasheet?

7. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
It have nothing on it. My friend gave me that. He told to put 350mA. Only that he told.

8. ### eetech00 Active Member

Jun 8, 2013
650
112
Hi

You have to decide on an LED to use then find its recommended operating voltage/current from its datasheet.
Then use the info #12 gave to calculate a resistor to use as a test load.

eT

9. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
Thank you so much guys. I understood finally. God bless you all.

cmartinez likes this.
10. ### #12 Expert

Nov 30, 2010
16,298
6,811
Mouser 1 watt LEDs range from 140 ma to 400 ma, so voltage must be from 2.5V to 7.14V.
Let's assume 140 ma and 2.5V to make a very safe tester.
7.4V - 2.5V = 4.9V
4.9V/.14A = 35 ohms.
Use 39 ohms, 3 watts to start the LED.
That will give an estimate of its proper voltage and we can work at this, bit by bit.

11. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
Can I use normal resistors? Some told me to use 1N4002 diodes as test load. Is that true?

12. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,573
2,541
Thanks for your thanks... you can now calculate R using the simple equations that #12 gave you... but after you assemble the circuit, you will probably find out that it will consume less than the 350 mA that you calculated for, since every LED has a voltage drop, and you won't know until you put your circuit together and make some measurements, the you can recalculate R

Nov 30, 2010
16,298
6,811
Yes, No.

14. ### #12 Expert

Nov 30, 2010
16,298
6,811
1 watt, 350 ma = 2.86 volts
7.4 - 2.86 = 4.54V
4.54V/.35A = 13 ohms, 4 watts

15. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
Why you told 39 ohms and 3 watts. What is 3 watts?

16. ### #12 Expert

Nov 30, 2010
16,298
6,811
3 watts is the size of the resistor.
See post #14 and use 4 watt resistor.
Probably you can only buy 15 ohm resistor.
For 15 ohms, 3 watt resistor will be sufficient.

17. ### imraneesa Thread Starter Member

Dec 18, 2014
184
4
Why you said 2.5v as safe while 7.4v is high. Making tester for high volt is safe right. Also why you substract 2.5 from 7.4.

18. ### #12 Expert

Nov 30, 2010
16,298
6,811
Why you still reading post #10 "How to find resistor with no information except 1 watt" when answer is in post #16? "How to find resistor after you tell us the current".

At this time, I do not have the patience required to type out how to use experimental data to do a decreasing boundary search based on some internet obtained information that might or might not be correct. The fact that I can do the search and find your resistor size is why I answered your question. If you want to know HOW to do the search, that is a different question.
Perhaps that question needs a new thread. The title might be, "How to find resistor for 1 watt LED with no other information provided".

Last edited: Dec 28, 2014
19. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
1,440
368
What is it with people and driving LEDs? Why do they make it so difficult? These LED threads go round and round and round.......

There was a time before the internet when we had no one to bother and instead we would pick up a pen and paper and a calculator and work it out for ourselves. And then we'd wire up some components and check our results. It's just basic electronics, right?

I guess nobody does that any more because it's easier just to make it someone else's problem and avoid the inconvenience of actually bothering to understand anything .

20. ### WBahn Moderator

Mar 31, 2012
17,748
4,797
What kind of driver do you have? Is it a current source or is it a voltage source or is it something else?

You want your test load to develop the same voltage across it when it is dissipating 1W as the LED will have across it, so you need to know what the forward voltage drop of the LED is. If this is a single LED then if you know the color you can make a pretty reasonable guess about the voltage. If it is an array of LEDs, then all bets are off.

To find the forward voltage you can use a simple circuit that will get a small amount of current flowing and then measure the voltage across the LED. For instance, if you take a 9V battery and use a 1kΩ resistor then the most current that will flow when all of them are put in series will be 9mA and the power drawn from the battery will only be 81mW. That should be enough current to get you a reasonable voltage measurement of the LED forward voltage drop since it will not change by much as you increase the current.

So let's say that you do this and you measure a voltage of 3.7V across the LED. That means that when it is dissipating 1W of power that it will have about 3.7V across it (it will actually have a bit more, but not a lot since the voltage will increase by about 100mV or so for every factor of ten increase in current).

So now you want a resistor that will consume 1W with 3.7V across it, which since P=V²/R, means that you want R=13.7Ω and that your current will be about 270mA. Be sure that you use a resistor that is rated for at least 1W (a minimum rating of 2W is better).