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Is my RGB LED broken, or am I just dumb?

I have a 10W RGB LED like this one: http://dx.com/p/10w-1000lm-rgb-light-9-led-module-silver-white-9-11v-168507

It's rated to 9-11v, so I hooked it up to a 9v battery. The LED has 4 leads:
1) +9v
2) Red -
3) Green -
4) Blue -

The green and the blue work as expected, and I can combine them to get cyan.
The problem is with the red:
1) It won't combine with any other color.  If the red is switched on, the green and blue switch off.
2) It works on both polarities.... It turns on both with +9v and -9v.

Am I doing something wrong?  Am I missing something?  Am I going crazy?  Am I just stupid!?

Please help!

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Looking closely at the tiny little wires in this image,
http://img.dxcdn.com/productimages/sku_168507_1.jpg
I think I see how it is wired. 

There is a series string of 3 red LEDs,  a series string of 3 green LEDs, and a series string of 3 blue LEDs.  If you want I'll draw you a diagram picture of what I imagine I am seeing here.

Note that red, green, blue LEDs do NOT have the same characteristic forward voltage. 

If you wire all three strings in parallel, then this forces them to have the same voltage, and that  voltage is probably going to be the voltage of the string with the lowest forward voltage; i.e. the red ones, just the red ones turn on. 

Ideally you should have some way of providing a controlled current to each string. I think that is the way this thing was intended to be driven.

Probably the easiest way to do that is put a resistor in series with each string, although resistors are not a great solution for high powered LEDs.  I don't know if you actually wanted to drive this thing at its full rated brightness?  I mean 10 W?  What is that, 3.3W per string?  1.1W per individual LED on the die?

Anyway, I think putting some resistors in series with each string would be an easy way to prove you can turn on all three strings at once.

Also note, I am expecting the blue string to have a total forward voltage drop of approximately 3*3.6 = 10.8V, so I cannot explain how the blue string it is turning on at all with just a 9V battery.

So let me amend that to suggest: a 12 V battery, and some resistors in series with each (red,green, blue) string, and then you should at least be able to see something; i.e get R, G, and B all working at the same time.
I'm not sure wher eyou got this from but I have some RGB LED's here.

Theyre from Spiratronics Part Number LH3-018 and they work like any other LED, they jsut have a common and separate anodes. RGB all at the same time too.

These are the pins I have and the recommended resistors:

With the pins down, and the flat of the LED on the left:

1-> R-> +9v
2 -> Common -> 470 ohm resistor (yellow purple, black, black banding, ignore the silver/gold stripe i.e. the 5th)
3 -> B -> +9v
4 -> G -> +9v

In order of PIN LENGTH from longest to shortest:

Common
Blue+
Green and Red+ are the same
Morgantao (author)  asluyters3 years ago
That *would* work, but a 10W capacitor is not exactly practical.
It's expensive, and it's HUUUUUGE!
Morgantao (author)  Jack A Lopez3 years ago
You nailed it pretty much on everything...
* Yeah, there is a series string of 3 LEDs for each color.
* Yeah, after researching a bit on high power RGB LEDs, I noticed the Red LED has a lower forward voltage, of about 7v, as opposed to the 11v of the Green and Blue LEDs.
* Yeah, current goes where there's least resistance, Making the Red LED the preferred path.
* Yeah, they are intended to be used with a driver. Problem is, the driver costs more than the LED.

Driving the LED at full brightness makes resistors a bit impractical, as a 10W resistor is expensive and almost as big as a driver.

What surprises me most is that the Red LEDs aren't burnt to a crisp, as they lit up at more than max current with both forward and backward voltage applied.
Maybe a cheapo LED, but a resilient one so far :D
How can you expect there to be a total blackout of the string of LED's?

Remember ohms law. V=IR? Or in this case you need to refer tothe triangle explaining this:

V
----------
I R

Voltage drop isn't as straight forward as 3 x 3.6 in your case. You need to take into consideration the resistance of the circuit (As a whole) and the current it draws.
Vyger3 years ago
Yep, yep, probably not, and maybe-- its a relative thing, as compared to a door knob?
Morgantao (author)  Vyger3 years ago
I think you lost me.... Wait... Yeah, you lost me...
frollard3 years ago
A 9v battery can't supply enough current to run a 10W rgb led. Red diodes have lower voltage (about 2v) requirements, so any individual colour will work, but when you use red in conjunction with another the voltage drop is too high (because of the current draw) - the voltage from the battery drops below the required voltage to light the blue or green (usually 3.5v each plus)...

2 important things:
1) never use 9v batteries, they are garbage.
2) Get a proper constant current driver for your led or it will burn out.
9V batteries are not all bad, for this reason:

I think the high series resistance of that little battery may have prevented OP's LED from drawing too much current, thus saving his LED from being instantly overcurrented and destroyed the first time he tried to use it, by um... naively connecting it directly to a battery.
;-)
+1 and +1 to you both :)
iceng iceng3 years ago
What did you pay for your RGB 10W wonder and where ?
Morgantao (author)  iceng3 years ago
Got it as a sample at work.
Should be around $6, if Deal Extreme is any indication.

I wouldn't call it expensive, then again, I also wouldn't call it high end.
Short of very low current and high-ish dc voltage requirements of something like a smoke detector, they are deplorable. :)

I was even going to mention just that; the crappiness of the battery saved him his expensive rgb led :D