# Question about LEDs relative to voltage

1. Aug 6, 2016

### Fizzman1

Back story : I have cheap 12v LED 5050 strip from China.
Question: how do LEDs respond to 13v and 14v in terms of % intensity with higher voltage neglecting lifespan and assuming proper heat dissipation.
Details: the LEDs at 60/m rated 14.6w at 12v. They are typically 3led in series with a resistor per set x3 and the rest in parallel. I have 60led in parallel with a fan about 1w. The total current I measured with a 13v power supply is 1.6A. I am thinking about putting it on a 14v dc supply since they should be able to handle it since it was designed for car batteries which is about 14v with alternator. How much is the led actually brighter versus lost in efficiency in heat at the resitor. I don't know the register value in the strip but could find out if it's pertinent for the calculation.

2. Aug 6, 2016

### Fizzman1

I can probably calculate the power each led produce with the changes in current relative to the voltage, but how much of it actually translate into light intensity versus heat generated at the LED. I think these same LEDs are used in 24v strips with just a different value resister to regulate the current across the actual diode. The ultimate question is it worth it to run 14v vs 12v in terms of light intensity vs heat. +5% light +10%heat, worth it! +1% light +99% heat not worth it!(the application is a cheap dying lighting for coral aquarium 1.5gallons) And also, I guess I have a question about what excess current does to a resistor. Obviously failure at some point but by what mode? Purely heat or possibly other causes of failure.

Last edited: Aug 7, 2016
3. Aug 7, 2016

### NTL2009

A little hard to answer with specifics - the "5050" refers to the package size of the LED, the specs of the different LEDs sold in that package size can vary widely.

But, looking at the first few data sheets I found:

https://d114hh0cykhyb0.cloudfront.net/pdfs/5050-WW6000.pdf
http://www.wayjun.com/Datasheet/Led/5050 SMD LED.pdf
http://www.cree.com/~/media/Files/Cree/Chips-and-Material/Data-Sheets-Chips/CPR3ER.pdf

It looks like the brightness increases fairly linearly with current increase, no sharp knee or anything that I saw. But not a 1:1 increase, maybe .6/1? But you would need to know specifics of your LEDs, and the R value to get specific values.

I'd try experimenting. But first, are you sure these are rated for 14V? I think I've read that these can overheat with car voltages? If you can read/measure the R value, you can make a first cut assumption that almost all the extra 2V will be across the R ( ~ 0.1~0.2V per LED, so 0.3~0.6V to the LED and 1.4~1.7V to the R?). If you can measure the current at 12V and 14V that will tell you how much extra power they draw. I'd expect it to be > 14/12 ratio, the LEDs will clamp and cause it to be a non-linear draw.

Probably OK to run at 14V, but watch the heat, keep them well ventilated, get data sheets on the LEDs if you can.

-NTL2009

4. Aug 7, 2016

### CWatters

I think white LEDs have a forward voltage of around 3.2V so three in series is about 9.6V leaving 12-9.6V = 2.4V across each resistor.

If you increase the voltage from 12V to 14V the extra 2V will all appear across the resistor.. 14-9.6 = 4.4V

Since the resistor controls the current I would expect the current to increase by a factor of 4.4/2.4 = 1.8 times. Almost a factor of two.

A quick look at the data sheet for a 5050 led chosen at random by google had a recommended current of 60mA and a max of 90mA. If the resistors in yours had been chosen to produce 60mA at 12V then they would push 1.8 * 60mA = 108mA at 14V. That would exceed the recommended 90mA.

Without data on the LEDs they actually used it's impossible to say if 14V is going to damage yours. They might have designed it to operate on 14V or they might not.

Relative Luminous Intensity isn't quite proportionate to current. So increasing the voltage from 12 to 14V might increase the current by a factor of 1.8 and give you an increase of say 1.5 in brightness.

Overall the voltage is increasing by a factor of 14/12 = 1.17 and the current by 1.8 so the power will increase by 1.17*1.8 = 2.1. Twice as much heat.

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