Need Technical Assistance from the pros..

[Zman]

Hi everyone...I have been thinking about doing a luxeon star project for a while. This weekend I scored a 4v, 700 ma transformer for only $2. I also bought a nice aluminum heatsink. Can this be used to drive a 1 watter? I know the basic V=IR, but how do you calculate the wattage capacity of the resistor? Any hints, tips, or help would be greatly appreciated. Once I figure this part out, its time to look for some dirt cheap green/bluish lame bin luxeons!! Thanks everyone!!!

 

[RGBrainbow]

Hi Zman,
is this only the transformer (4V AC) or a powersupply (4V DC)?
What colour is Your 1 watter (different Vf for different colours)?

Vresistor = Vsupply - Vf

R = Vresistor / I

Presistor = Vresistor * I

If it's AC You'll have to add a bridge-rectifier (and perhaps some 220 to 470µF to avoid flickering).

And please always have in mind:

You're dealing with High Voltage. THIS IS DANGEROUS !!!

greetings
Joachim

 

[Zman]

Thank you so much for the assistance so far and for teaching me a few terms Joachim /ubbthreads/images/graemlins/smile.gif Its a power supply...4v DC. I dont have the emitter yet. I was hoping to buy a few white ones with poor color on the forum once I figured out the electronics side of it.

 

[Doug Owen]

IMO the next thing you need to buy is not a handful of LEDs to fiddle with but a meter so you know what you're doing.

The next thing you need to do, calculate the needed resistor, depends on knowing the real voltage from your power supply. Tenths of volts count in these ranges. You'll also need to be able to measure the actual current (and maybe Vf) once you're 'up and running'.

We've seen suitable DMMs (for Harbor Freight on sale) for as little as $3 in past months, but you'll probably end up paying $10 to $15 after looking around a bit. Be sure the one you get has a high amps range (typially ten amps), lower ranges often have too much series resistance for some of the hotter LED uses.

Doug Owen

 

[Zman]

Ok the sunpro digital multimeter is reading 5.9V!!! Sounds like I need to check with another meter to be sure, but maybe its just plain mislabeled...bummer.

 

[Darell]

Those power supplies run high about 99% of the time in my experience. I'm surprised it is only 5.9V, actually. I have about six 12V nominal supplies here that read almost 19V.

 

[eluminator]

5.9 volts sounds about right if there is no load connected to the supply. The more current you draw, the lower the voltage. It should drop to 4 volts when it is supplying 700 ma, according to the specs you provided.

Your supply is unregulated, no doubt. You don't often buy a regulated supply for $2. It should work fine, but you will have to experiment a little to find the right resistor to give you the current you want through the LED.

 

[PlayboyJoeShmoe]

I agree with Darell. Most of the wall units I have acquired read WAY higher than the rated voltage.

If I meter the charger for my 2.4V M7 I read some 11 odd volts. I read a WHOLE lot more than 3.6V on my Stinger charger.

The 19V for a Wagner cordless drill reads nearly 22V....

You get the drift of this eh?

 

[Zman]

Yes fellas, I'm starting to understand. I guess now its time to do some calculations to determine resistance and the max current I want to run through the led. I read the luxeon III's put out great light at 700ma. Would one of those be ok at 4v 700ma?

 

[whiskypapa3]

Get some 10W 10Ohm resistors from Radio Shack. Hook one across the WallWart output. Measure the voltage with this load. Get a sheet of Quadrule paper (or just draw a big tic-tac-toe set of squares) and mark one axis as volts (0-6V)and the other mA (0-1000). Plot the open voltage/current (5.6V,0mA)then the voltage/current for 10Ohms (the current will be V/10). Do the same with two 10Ohms in series and again in parallel (series. current = V/20 and parallel, current = V/5). Connect the dots with a smooth curve.

Find the drive current you want to use on the current axis, move up to the smooth curve and read the voltage off the other axis. If it is higher than the Vf of the LS you may go to next step else trash WallWart and try another.

Subtract Vf (typ 3.5V)from the voltage found on the other axis and divide the result by the current you want (typ 350mA), this is the required resistance. It will probably be somewhere near 3-5Ohms, make up a resistor out of 3-10Ohm Radio Shack 1/2W resistors in parallel (power in resistors = current squared times resistance, here approx 1/4W). This should get you near enough to the current you want without harming anything. If it is much different try other combinations of series/parallel resistors to get what you need. You can make a precision resistor out of fine nylon coated stainless steel fishing leader if you want later...

 

[Doug Owen]

[ QUOTE ]
Zman said:
Yes fellas, I'm starting to understand. I guess now its time to do some calculations to determine resistance and the max current I want to run through the led. I read the luxeon III's put out great light at 700ma. Would one of those be ok at 4v 700ma?

[/ QUOTE ]

Yes, it *should* be (given reasonable heatsinking), but you can't just 'go there' directly in a single shot there's too many variables (your supply the the Vf of the specific LED for instance).

I suggest you start by chosing a resistor for half that current or so. Say 5.9 less 3.5, 2.4 Volts divided by 400 mA, for six ohms. Not all that common a value, I'd go with two ten ohms in parallel (for five ohms), plenty close enough for a first cut and still on the safe side by a wide margin. Ten ohm half watt resistors are common enough, although marginal in power, go with ones or use four 20 ohm (more or less) in parallel if you can only get half watters. Then *measure* the current you're actually using with that value and 'trim' it by adding more resistors in parallel with the first two. Another ten ohms will give you about 50% more current (a bit less, but a good rule of thumb). Do a 'cut and try' to get to whatever value you think is right.

Doug Owen

 

[eluminator]

The electrical circuit is easy. The proper heat sinking is more difficult. If you haven't done your homework here and just want to see some light come out of the LED without a heatsink, I'd suggest no more than 50 ma or so. You could probably run 100 ma for a short time if you keep your finder on the back of the LED and shut it down about the time you get burned.

I'm no expert on the Luxeon and I may be off by a factor of two or so, but I just wanted to let you know that you could let the magic smoke out of it quite easily with that power supply unless you had a good heatsink.