Limiting resistor for High Power LED

[Deimos]

I bought some individual 1W (Chinese probably) UV LEDS...
Each one is on a star shaped heat sink. Voltage is 3.8-4.6, current is 350mA.

They will be run off a small 5 volt (2A) LED [power] Supply.
What value limiting resistor (ohms and watts) would I need for each LED?
I am running three of them in parallel off the supply.
Thanks

 

[Bluejets]

Maybe use a 1 to 3W LED driver.
Available from Ebay for a dollar or so.

1-3*1W LED Driver
Wattage：1W
Input Voltage： AC 12V
Output Voltage：DC 2.6.-3.6V

Output Current：300MA
　

3*1W LED Driver
Wattage：3W
Input Voltage： AC 12V
Output Voltage：DC 6-11V
Output Current：300MA

4*1W LED Driver
Wattage：4W
Input Voltage： AC 12V
Output Voltage：DC 3-7V
Output Current：300MA

4-6*1W LED Driver
Wattage：5W
Input Voltage： AC 12V
Output Voltage：DC 12-21V
Output Current：300MA

3*3W LED Driver
Wattage：9W
Input Voltage： AC 12V
Output Voltage：DC 9-12V
Output Current：600MA

 

[Deimos]

Well thanks for the reply, but

1) the supply I mentioned is what I have and that is what I plan to use: 5VDC, 2A supply.
2) I want to know what the value of the limiting resistors should be for the UV LEDS based on the LED Specs and the Supply specs I've listed.

Thanks

 

[Bluejets]

The reason drivers are used is to keep the current constant.
If you want to use a resistor, fine but any voltage variation will change the current and resistors burn up energy, get hot and are inefficient.
To calculate resistor size, subtract led voltage from supply voltage and using basic grade 1 calculation, R=E/I ............where R is resistance, E is the voltage drop on the resistor and I is the expected current.
you get the resistor size. Then by using another I squared R you get the wattage.

 

[Bluejets]

BTW, the above #2 was to give an indication of what is available.
There are 5v units if you care to look.
Here is an example..........

https://www.ebay.com.au/itm/5-35V-L...422554&hash=item51db1d8d68:g:mIMAAOSwI-BWOCXG

 

[kellys_eye]

5V - 4.2V (average LED voltage) = 0.8

0.8/0.35 (volts to drop divided by current flow) = 2.2Ω resistor required.

0.35 x 0.35 x 2.2 (current squared times the resistance) = 0.27 - nearest wattage would be 0.5W.

As above though - if you want longevity and reliability then get the correct driver.

 

[WHONOES]

A word of caution. If you are playing with UV LED's, you need some eye protection in the form of a UV blocker. Even if you point them away from your line of sight, there will still be reflections from objects around you.
UV rays can cause a lot of eye damage and you won't even realise it is happening until it is too late. Because you can't see any visible light, does not mean it's not there.

 

[dave9]

If you really want to get this dialed in right, the easiest way to do that is to use a variable voltage PSU, and a multimeter in series to the LED, so you can see what the current is at any particular forward voltage.

This is just PSU to LED connection, no driver or resistor. You should start the PSU at its minimum voltage setting.

Raise the voltage until your multimeter reads the target current, and do a bit of math at the same time. Multiply the PSU voltage * measured current, to achieve 1W. (V*A=W).

That PSU voltage at the current needed to hit 1W, is your LED forward voltage (Vf) that you would use to calculate the resistor value needed per kellys_eye's equation instead of "4.2V (average LED voltage)".

This is why it's easier to just get a ~300mA current regulating driver, and then put the 3 in series, and it's more efficient too. However you'd usually need higher voltage than 5V.