Patriot

I decided on building an LED light using a drop-in unit from DealExtreme. It is a unit for a Sure-fire flashlight.

My light bodies are metal, but are insulated with rubber from the rest of the bike. The ground spring on drop-in (along with the reflector) will come into contact with my light body.

Is this a big deal?

How should I make sure I build this thing as not to fry it?

https://www.dealextreme.com/details.dx/sku.11836

Patriot

A couple more questions.

1. Do I need to add a resistor to control current?

I have a 12.1Vdc/4.2A constant supply from a DC converter, so I was thinking it may not be necessary. Although 4.2 Amp is alot. I heard some guys just drop these into Xenon flashlights with no other mod. How could this be without ruining the LED?

2. is the center spring the positive?

Just want to make sure. I'd hate to hook up the polarity backwards and ruin the LED.

Unknown Cyclist

The module you have requires "Input Voltage Range: 3.7V~18V"

Therefore it's likely that the circuit board can only reduce the voltage to suit the LED.

You need a basic circuit: power - switch - and the module.

What exactly are you putting the module in ?

The circuit board in the module is there to supply the LED with the correct current.

Your DC converter may be capable of supplying 4.2A but will only do so when connected to a suitable sized load.

Connect your module to a supply of between 3.7V to 18V and the module will draw as much current as it needs (for any given voltage) to supply the LED.

From the design, the spring would sit on the positive terminal of the battery. All the modules I've seen work like this. I guess you could hook it up to a low voltage initially to be cautious.

Additional information about the light you are fitting the module to would increase your chances of getting helpful advice.

Patriot

Deal extreme doesn't have much other info.

The module has two springs. One outer and one inner.

I assume the inner spring is the positive terminal.

When I watched this video, the guy pulled off the outer spring, and dropped it into his Sure-Fire without it. I assume the batteries he installed are a pair of 18650's with the positive on the center spring?

Also, what would be the best spot to connect my GND lead to?

https://www.youtube.com/watch?v=JgsUpRdgsp8

Ok, I know my converter will put out up to 50w (12v/4.2A), but wasn't sure if the LED would control current on its own. I assume it is, but wanted to make sure.

Unknown Cyclist

What are you putting the module in ?

Patriot

I'm installing it in a homebuilt metal light body.

https://www.bikeforums.net/electronics-lighting-gadgets/459817-project-2x20w-headlights.html

Basically, a piece of chrome plated thin walled pipe with a collar on the front to install the reflector.

Unknown Cyclist

The module is likely to get warm in use and unlike halogens it needs to get rid of the heat.

Try to make sure that either the body or the reflector makes good contact with your tube.

The more heat you can conduct away from the module the better.



ps. have a look at my HID project....

Patriot

I wish I could attach some aluminum cooling fins to the side of the reflector somehow.

The front of the tube is where the most contact will be. It will be in full circle contact with the collar that holds it in.

Would that be enough?

Unknown Cyclist

All depends on what current rating the module is.

A lot of the modules will reduce power to the LED if it gets too hot.

You'll have to let us know how well it works.....

Patriot

I think one reviewer said it drew 0.4amps at 12v.

So, I'm wondering what 0.4 amps will do.

Get warm would be a minimum. But, what about extended use for cycling? Would that harm it?

One can only wonder.

Unknown Cyclist

The circuit in the module regulates the current TO the LED, without you measuring the current at the LED there's no way of knowing what the regulator is outputting. It's likely to be 750mA, but could be 1000mA. It's also possible on higher (input) voltages that the LED is driven harder.

Voltage x current at the LED = Watts consumed by the LED, some of this will be light output, some will be heat. The input voltage and current are important as the regulator may well have a sweet spot - there is likely to be a specific input voltage somehwere between 3.7v and 18v where the regulator is working most efficiently.

You might find that running it from one 18650 (3.7v) produces less light, but at the same time the regulator does little work so consumes little power and produces little heat.

You might find that running it on two 18650s (7.4v) produces full brightness and moderate heat.

It's even possible that running it on three 18650s (11.1v) would produce full brightness, but so much heat that runtime is no better than on two 18650s (7.4v).

Basically you have Watts in = Watts out - less the losses from the regulator.

I have an Ultrafire AA/14500 torch when running on a single AA the input voltage is 1.2v and the capacity is 2800mAh, the torch puts out good light and runs for hours on a fresh battery.

When I use it on an 14500 3.7v 900mAh - the battery is almost exactly the same capacity, but the light output is much, much higher, probably doubled, but the torch gets too hot to hold and runtime is about 1/3.

I'm not saying that your regulator isn't capable of behaving perfectly over a wide voltage range, just that it isn't likely - and it is likely that there will be a sweet spot where V x A in gets you best efficiency - light out/power in.

Patriot

^^ I'm aware of that. The power issue is one of my concerns. I would think that if the other reviewers say it runs at 400mA at 12v, then I would tend to think that is about right. That would mean it is using about 5w, which would seem about right for the rating of 250 lumens.

If it was using 700mA-1000mA, then that would mean it's using 10-12w of power at 12v. That would be about the same power consumption/lumens as a Halogen bulb, would it not?

Or, am I just totally off in my thinking here?

I'm hoping that the spot where it connects to the light body at the rim will allow enough heat to dissipate off to prevent damage on longer rides. I was even digging around ebay, looking for a 30mm 12v cpu fan to mount on the back of the drop-in module to keep it cool while running. I could drill some holes on the bottom of the light body for air ventilation. I would prefer not to have to do this though.

Unknown Cyclist

You are confusing power in with power out, they are different because the module is not 100% efficient.

The important thing is what the regulator is supplying the LED and the efficiency of the regulator.

If it's drawing 0.4A @ 12v then yes, the power consumption is 4.8W.

If it's supplying 750mA @ about 3.5v the power consumed will be 2.6W.

All of this is conjecture, I'm simply suggesting that checking the efficiency of the module throughout it's input range could yield better results - IF the amount of power used by the light is of concern to you.

Patriot

True, the efficiency matters a great deal. The less power converted to light, then the more current needs to be removed as heat.

As a guess (if I remember), doesn't it take about 3.5w to produce 250 lumen? (in a 100% ideal circuit)

If so, I can see the other 1w getting lost as heat, giving it an overall power to light conversion efficiency of about 75-80%. This is about what good quality LED's provide if I remember right. I could be totally wrong though, but I read somewhere that should be about right.

My point was, if it was using upwards of 700mA, then it would only have an efficiency of about 35-40%, which is very poor for an LED. That sounds more like the efficiency of a halogen.

Anyway, all this is rather moot.

I just purchased the drop-in unit, and a larger 35mm reflector to make it fit in my light body.

What I'm wondering now, is where would be a good spot to attach my -/GND wire?

Unknown Cyclist

The voltage that any given LED requires to run at a set current will vary from one LED to another - check the LED spec sheets.

It's not the efficiency of the LED that is in question, it's the efficiency of the regulator that is.

You have to take both into account and while you can't change the efficiency of the LED, you can at least check the efficiency of the regulator through it's voltage range.

You'll have some power turned into heat inside the LED, but you may well lose a greater amount of power in the regulator.

The regulator may have good efficiency at one point in the input voltage range and be very poor elsewhere, it's not uncommon, some of mine do.