# LED specific voltage for specific wavelength

Hello

I am trying to design a precise circuit for LEDs. I have a datasheet from a manufacturer and it says an LED is capable of running 3.0v at 640nm and 3.4v at 660nm. Is there a way to tailor the power (voltage) so that the LED will run exactly or at least approximately at 3.0v or 3.4v?

Will the LED accept a specific voltage of 3.4v with multiple LEDs in series?

Do i have to maintain 700mA to light up the LED?
Can run it at 350mA but same voltage? Two LEDs connected in parallel in series connection with other LEDs?

Thank you!
Ivan

vk6kro
Could you post a copy of the data sheet?

That would be a red LED and they normally have a voltage drop of about 1.5 volts and the output is not a single wavelength like that.

They don't normally use 700 mA either. More like 20 mA for the common size ones. But you may have something unusual like a multicolor LED.

Assuming it is a normal LED, you can choose the current through it by putting a resistor in series with it and then varying the supply voltage, or you can start with a fixed voltage and vary the resistor. Resistors values like 300 ohms are typical.

Either way, the voltage across the LED is determined by the current through it. This voltage varies according to the color of the LED.
There is a chart of these voltages here:
http://en.wikipedia.org/wiki/Led

You can put LEDs in series, but if they are placed in parallel, they need their own series resistor.

You can calculate the resistor value like this:
R = (voltage across the resistor) / (current through the resistor)
where (voltage across the resistor) = the difference between the supply voltage and the LED voltage.
and (current through the resistor) = LED current required as long as it doesn't exceed the maximum rated current of the LED.

eg R = 12 volts - 1.5 volts / 20 mA (0.02 amps) = 10.5 / 0.02 = 525 ohms

#### Attachments

• 3w 660nm.pdf
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SO. There are two ways to do it:
i got 14 LEDs(x3.4v=47.6volts) in series my power supply is 48volts. target wavelength is 660nm

1. choose the current through it by putting a resistor in series with it and then varying the supply voltage - how do i vary a supply voltage if i got 14 of them at 48volts supply voltage? will the resistor precisely adjust the voltage to the LED, lets say 3.4v which is 660nm on the data sheet? how can i make sure it draws 3.4volts instead of 3.0volts?

2. "you can start with a fixed voltage and vary the resistor. Resistors values like 300 ohms are typical." If i got one LED and one power supply capable of a certain voltage i might just get away with the approximate voltage range of 3.4v. But if i have 14 LEDs in series how can i adjust them to just work at 3.4v to give me the 660nm wavelength?

Thank YOU!
Ivan

vk6kro
That is a high powered red LED. Very bright and it would get hot without a heatsink.

The wavelength is given in an unusual way. The lower extent of the band is shown as 640 nm and the upper extent is shown as 660 nm but you don't get to choose between them. It is a band of wavelengths between those limits and it doesn't depend on the voltage.

Also, the graph of voltage vs current shows a range of 2.05 volts to 2.65 volts, yet the text says 3 volts to 3.4 volts because the 3.4 volts is at 700 mA.

You can get dedicated driver modules for these high powered LEDs and these are a lot more efficient than just using a resistor. Try Ebay.

there definately will be a heatsink

So the 640 to 660 it merely depends on a batch made? so some LEDs will draw 3volts and some 3.4v?

dedicated supply could you please write me a few words on how it works? does it choose the needed voltage itself? It seems like a good idea if i do not have to use resistors.

Thanks
Ivan

i think i misuderstood. dedicated you meant specifically designed for LEDs? :)o the 640 to 660

does it depend on a batch made? some LEDs will draw 3volts and some 3.4v?

vk6kro
A LED does not behave like a laser which produces just one output frequency. It produces a band of frequencies.

That 48 volt supply makes the design difficult. I'm not sure if you could get a power supply module for 48 volts. These provide a fixed current output and help to stop these LEDs from destroying themselves.

If you had to use the 48 volt supply, you could take two strings of 7 LEDs and calculate a resistor to go in series with all of them.
I make it about 39 ohms and about 15 watts actual dissipation so you would need two 20 watt 39 ohm resistors. I don't know if you could buy them.

Have a look on Ebay for LED regulator modules. I have seen lower voltage modules.

LED driver 100w 4.3amp 24v(11$) LED constant current driver 700mA DC6-50v 35w(9$)
LEDs: 700ma 2.0-2.4v 3.0-3.4v 3.6-4.0v

using the first driver i could divide 4.3amps into 6 parallel circuits of 716mA and drive 12 LEDs at 2.0v at one circuit. Is this correct? Do i calculate the amount of LEDs for it lower voltage end of 2v or 2.4v? Can it be done this way?

using the second driver is more expensive, and only 700mA is used and LEDs in series. will this work better than the first method?

What do you think?

vk6kro
That 700 mA is an absolute maximum rating, so it is probably good practice to stay well away from 700 mA.

As these are expensive devices, I think you could start with 500 mA as a goal. You don't want a few LEDs to suddenly stop working just because you get a hot day.

So, could you have a look where you saw the 700 mA constant current drivers and see if they have any for 500 mA?

At 500 mA, the LEDs should have 2.7 volts across them, according to the first graph, so 7 would drop 18.9 volts.

Depending on the rating of the current driver, you would probably need two drivers to supply two strings of 7 LEDs.

This costs more than using resistors, but the power saving should be large.
Two strings of 0.5 amps from 48 volts (with resistors) would take 48 watts from the power supply.
Two constant current drivers supplying 7 LEDs each would be supplying 18.9 watts and (depending on their efficiency) would possibly take only 22 watts from the power supply.
The constant current drivers would also be a lot more immune to mains voltage fluctuations.

Very interesting.

In the datasheet it shows that the peak current is 900mA, does that mean that 700mA is already with the safety?

With a constant current driver, does it mean the voltage will automatically adjust to its optimal levels?

Can i measure the voltage drop exactly when i turn on the circuit? I would like to figure out the maximum possible number of LEDs in a circuit so i can make it as more efficient for the driver as possible.

vk6kro
Very interesting.

In the datasheet it shows that the peak current is 900mA, does that mean that 700mA is already with the safety?

No, peak current means it will blow up even if this 900 mA happens very briefly. 700 mA means that you might JUST get away with this current. It would have no safety factor at all.
A bit like putting 300 tons on a bridge that is known to fail at 301 tons.

With a constant current driver, does it mean the voltage will automatically adjust to its optimal levels?
Yes.

Can i measure the voltage drop exactly when i turn on the circuit? I would like to figure out the maximum possible number of LEDs in a circuit so i can make it as more efficient for the driver as possible.
Yes, you can. The output voltage will rise to the maximum level if there is no load.
This information would be given in the data sheet for the current regulator too. You can request this information before you buy anything.
The main factor is the 48 volt supply. You have to get a current regulator that can handle this voltage with some safety margin.