# Need some LED help please- battery/resistor?

Discussion in 'The Projects Forum' started by DocOctavius, Jan 29, 2008.

1. ### DocOctavius Thread Starter Member

Jul 19, 2005
12
0
Hi guys-

I'm sure this is an elementary problem, but I'm trying to test a 360nm Fox UV Led for curing purposes, and can't seem to make it work. Here's what I know:

Forward voltage: min 3.6 typ 4.3 max 5.0 Volts at 20mA forward current
DC forward current max : 30mA, 20mA ideal

I'm trying to figure out what kind of battery and resistor to use, tried a 9v battery with a few different resistors, and none would allow enough power to go to it. Since 4.5 volts was ok for the LED I tried putting 3 AA batteries together and running it- it blew the LED out I think. Keep in mind this is for only one LED for testing purposes. Thanks in advance for your help...

2. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
9V batteries, even when new, only output 4mA. They are therefore unsuitable for your test.

LEDs can vary widely in their actual voltage rating from lot to lot; even in the same batch. The voltage drop across them does not change much with an increase or decrease in current. That is why you blew up your LED when you connected it up directly to the battery with no current limiting resistor.

The safest way to test your LED would be:

1) Start with four AA, C, or D cells in series, or a 6V power supply.
2) Your minimum voltage across the diode is 3.6. Let's start with that number. Now to find the voltage remaining to drop:
Edrop = Vsupply - VLED
Edrop = 6 - 3.6
Edrop = 2.4
3) Since your LED has a rating of 20mA, we need to select a resistor to limit the current across Edrop to 20mA:
R = E / I
R = 2.4 (volts) / .02 (amps)
R = 120 Ohms
4) Connect a 120 Ohm resistor in series with the LED across the 6v supply.
Read the voltage across the diode. If it is different than 3.6V, you'll need to re-calculate the current limiting resistor as I did above, using the voltage reading you obtained.

You can avoid such calculations by using a constant current source or current limiter, but a resistor is the cheapest way to go.

Don't try to connect LEDs in parallel using one resistor, as one will ususally carry quite a bit more current than the other(s).

However, you CAN connect multiple LEDs in series, and use just one current limiting resistor for the lot of them.

3. ### DocOctavius Thread Starter Member

Jul 19, 2005
12
0
Thank you so much! One question- how important is the watt rating on the resistor (1/4, 1/8w, etc)?

4. ### Audioguru New Member

Dec 20, 2007
9,411
896
No.
A new 9V alkaline battery has enough power to burn you if it is shorted in your pocket.
Go to www.energizer.com and look at the datasheet for their 9V alkaline battery. It can supply 300mA for almost one hour when its voltage has dropped to 6V.

I have some LED "flash" lights that blink a bright LED 5 times in half a second with 90mA Then a pause of half a second before blinking again. A blue and white LED have a forward voltage of almost 5V at 90mA so I regulate the supply to 5.6V with a low-dropout regulator and use a 9V alkaline battery that drives the flashing LED for about 25 hours.

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5. ### HarveyH42 Active Member

Jul 22, 2007
425
5
Here's a link for a series/Parallel LED wizard, helps design arrays, or eve single LEDs. Useful tool.

http://led.linear1.org/led.wiz

Don't know what you plan on curing, but most of UV LEDs I've seen only put out 80 mcd, not much light. If you find some really bright ones, let me know, been looking for a while. I use a simple 8 led flashlight to find airsoft pellets in my yard at night. Doesn't compete well with streetlights at a little over 3 feet.

6. ### DocOctavius Thread Starter Member

Jul 19, 2005
12
0
The goal is to cure resin in glass a few millimeters from the LED. I've cured the stuff before, but the other LEDs I was using only had a 15 degree viewing angle, and these ones have 90, so there should be a bit more of a spread so I can cure larger sections.

That online tool is awesome- I had found another inferior one, and it didn't seem to be giving me the right info... Thanks a ton guys!

7. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Yes, the power used in the resistors is important.

Take the voltage dropped by the resistor, and multiply by the current through it. You'll then have the power dissipated by the resistor in watts.

You may be able to run multiple LEDs in series, depending upon the voltage of your supply and the voltage drop of the LEDs.

As far as the 15 degree viewing angle vs 90 degree - the intensity of the UV light on the surface corresponds to the viewing angle and the distance of the LED from the surface. Every time the width of the beam doubles, the intensity is 1/4.