# Testing new LED, higher power than I'm used to

Discussion in 'General Electronics Chat' started by ke5nnt, May 4, 2011.

1. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
I ordered a sample of an 8mm 100mA LED that produces 90K mcd, which I'm super excited to test out. I thought I was pretty good at determining power dissipation and resistance, etc, but I'm getting different info from different sources and I wanted to ask here just to be sure I don't blow up my sample, or a resistor for that matter.

Vf = 3.0V
If = 100mA
Vcc = 9V battery (just to test the diode, not a permanent power source)

So, 9V - 3V = 6V drop across the resistor at 100mA = 60 Ohms. Don't think that's a standard value, but we'll just use that for now.

So, as for power dissipation across the resistor for testing, my math figures 6V x 100mA = 600mW. I'm not getting any discrepancies there... However, let's say I want to do 5 diodes in series with a 16V power supply. 5x3Vf = 15 V, so 1V drop across the resistor at 100mA for the string... wouldn't this result in a power dissipation for the resistor at 100mW? After all, 1V x .1A = .1W? The reason I ask is because using a resistor calculator with this info in place gives me a power dissipation across the resistor of 1.3W, which I believe is wrong but has left enough doubt in my mind to want to ask.

Thanks.

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
It is OK do drop the current a few ma. 64Ω is a standard value, and you can always parallel resistors to compensate.

9V batteries (if that is what your using) are notoriously fickle, and not very stable under loads. Have you measure the source voltage while trying the LED?

If the resistor is dropping 6V, and running around 100ma, the power would be 0.6W. You would need 1W resistors.

You are running into the problem most online LED calculators have. You need more than a volt to reliably regulate the current with a resistor. Call it headroom.

3. ### marshallf3 Well-Known Member

Jul 26, 2010
2,358
201
or 4 15 ohm resistors in series.

Just a warning: A fresh Alkaline 9V battery should be able to put out 100mA without the voltage starting to drop, a tired battery or a "General Purpose" one will have trouble keeping up.

I've got some 100 mA LEDs I'm working with right now, specs say 3.4 - 3.8V but you want the current level right, they only show the voltage figure to get you started.

I bought one of those Radio Shack holders for 4 AA cells, it's fully enclosed, has an on/off switch and wire leads.
http://www.radioshack.com/product/index.jsp?productId=2062254 - \$1.99

For my 100 mA LEDs I put two 15 ohm 1/2 watt in series - too much but enough to get 86 mA at 3.1V through the LED. ideally I'd want to use something between 22 and 26 ohms - whatever gives me the proper 100 mA maximum current but running them off this battery pack is for testing only.

Oddly enough you can usually buy 4 AA alkalines for the same price or less than a single 9V and they hold a lot more energy than the 9V. (except if the 9V is one of the new Lithium ones)

PS: This is one handy battery pack for \$1.99, especially since it's fully enclosed so the batteries cant fall out not to mention the on/off switch and the wire leads.

These LEDs I got from Goldmine are incredible, at 265,000 mcd and a 7,000*K white they will light up a wall 20' away.
http://www.goldmine-elec-products.com/prodinfo.asp?number=G16642

These will knock your socks off too for a red LED:
http://www.goldmine-elec-products.com/prodinfo.asp?number=G16750

Last edited: May 5, 2011
4. ### Audioguru New Member

Dec 20, 2007
9,411
896
Is 100mA the tested current when the light output was very high or is 100mA the absolute max allowed current and any more will burn it out?

Is the LED made by a name-brand manufacturer or made by somebody in their garage in a country you never heard of who lies about the brightness and beam angle??

5. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
Specifications:
Lens Type - Water Clear
Life - 100,000 Hours
Forward Current - 100mA max
Forward Voltage - 3.0 typ, 3.4 max
Reverse Voltage - 5v
View Angle - 140 Degrees

I can't say as if I know who the manufacturer is, but the supplier does state that these LEDs are made in the U.S. Not that saying that necessarily proves anything, but hey...it's a sample. If it doesn't work out well, then oh well I guess. I can however say, I've bought over 400 5mm super-bright LEDs from the same supplier, and I've never had a problem with them, nor have I ever been disappointed with the product. I guess we'll see.

On another note, I've decided to test the LED using a 12V source through a 9V linear regulator instead of using a 9V battery. We'll see, if I blow something up (doubtful)... I'll clean off the soot and try something different.

6. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
What is your resistor configuration?

7. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
Well I stopped at Tanner's today on my way to work and picked up a 62Ohm 2W resistor. Basically, without the neatness of a proper schematic...

Vcc-->9V regulator-->diode-->resistor-->ground

I did include a 220uF 25V cap from source to ground, Vcc is a regulated 12V supply that outputs an even 11.99V. I know there are brighter LEDs out there, but the 5mm super-bright 13,000mcd LEDs I'm used to don't compare to this thing... IT'S BRIGHT! The photos don't do it justice.

8. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
I like that style of case. Looks like fun.

9. ### Audioguru New Member

Dec 20, 2007
9,411
896
Using a 12V supply to light a 3.3V LED is throwing away a lot of power.
Instead of using a 12V supply, a 9V regulator and huge resistor why not use a 6V supply and an LM317 regulated current source that uses one small resistor?

10. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
Well, the short answer is because this was nothing more than a test, and for the 30 seconds or so that the LED was on, power waste didn't mean that much. When I get around to actually using the LEDs for something, I'll use a more practical design to power them.