3 star mounted luxeon rebels- lenses unavailable?

[Jarl]

I'm looking to build my own lights, and I was thinking to use one of these, with a buckpuck, a few switches and a heatsink.

At the bottom of the page, however, it says "Standard Fraen and L2 lenses are NOT compatible with this LED. Lenses are not currently available." I'd be looking for about 10'-15', or possibly an oval type of lens. Anyone know any bodges or official lenses which would suit this?

edit: just found these. Anyone know if they fit?

 

[Gunner12]

Since the tri Rebel is a after market component and the rebels are spaced so close together, I don't think there are any optics that will fit that star. The buckpuck has a v-out of about 3.7v and the tri Rebel star needs somewhere from 10-11v to operate.

Why do you need 3 high power LEDs? From my experience, one high power LED is more then enough for most tasks.

Welcome to CPF!

 

[IdiotsAbound]

Why do you need 3 high power LEDs? From my experience, one high power LED is more then enough for most tasks.

Welcome to CPF!

Can't believe you asked that on CPF, a forum where the brightest wins lol.

But from what i understand with these 3-Rebel boards, one can easily set it to light up just one, two or all three LEDs with just a click of the switch.

 

[frank777]

Why do you need 3 high power LEDs? From my experience, one high power LED is more then enough for most tasks.

This is perhaps the most disturbing statement I have ever read on CPF. Having a flashlight that is "too bright" is impossible. It is like being too good-looking, or having too much money. Personally, I want to hold the sun in my hands.

 

[bondr006]

Why do you need 3 high power LEDs? From my experience, one high power LED is more then enough for most tasks.

Gunner, Gunner, Gunner my friend....Do you not know that you can be banned around here for such subversive talk????

 

[Gunner12]

I was just trying to be somewhat realistic. The Shark driver should work with the tri Rebel star.

 

[Luminescent]

A three way star like this can't be used at the focus of ANY single standard optic (lens or reflector), because the spacing between the emitter dies is HUGE (several mm) compared to thier size (1mm), so it's impossible to get anything but an ugly three lobed pattern unless you completly defocus.

If you want a totally flood only with more than 90 degrees coverage, then you can use this type of star with NO optics. This will flood half of a normal sized room pretty nicely.

There are small plastic combiner optics designed for RGB LED clusters that may work with this star and give a slightly tighter beam (maybe 20 - 40 degrees), but that's still not very tight.

To get perfectly clean circular pattern with good throw, try 3 separate rebels with 3 separate 15mm to 25mm reflectors in a clustered arrangement.

For even more fun, 7 rebels with 15mm reflectors could clustered togeather in a hexagonal grouping [one in the center] inside the head of a maglight, and would pump out an incredible 1260 lumens at about 14 to 15 watts of drive.

This output would match some HID lights while still providing close to 50,000 hour LED reliablility (if the heatsinking is handled properly), and better overall efficiency. On throw, the HID would still have the edge because the HID discharge presents all it's lumens as a small point source that can be focused by a single larger reflector, but the Rebel light would be more practical for most uses, because too much throw in a pencil beam can limit the applications.

As far as drive goes, assuming you run the 7 LED's in series with an 85 to 90% efficent boost converter, and have an 8 AA cell 9.6 volt NiMH pack (fits very nicely in a 2 D-Cell Maglight host), you would get a runtime of over an hour.

Sounds like a nice light, I'm surprised that someone hasn't built one already.

 

[Jarl]

Ok, understand what people are saying, and now i've come up with this:

power supply (battery, 2/3D's if i go parallel, or 10 v. high capacity AA's for series) to driver (haven't decided on this... something to give me variable output (350,700,1000 type thing)), to LED's, to battery again, possibly throw in a resistor if it's needed.

Does this sound alright? Also, opinions on parallel vs. series would be good; parallel seems easier, but i'm confused by drivers. If I go parallel, would I buy a (say) 2100ma driver to get 700ma to each LED, one 700ma driver to get 700ma to each LED, or 3 700 ma drivers wired in series with the LED's?

At the moment i'm leaning towards parallel, because I can dim and turn off the LED's without wasting energy, but if I need to buy lots of drivers, this would take away the advantage, as drivers are fairly expensive.

I'm planning on making this a bike light, which doubles as a powerful flashlight that I can take if I know that i'll be going outside in the dark, hence the 3 rebels. Also, 3 seems a nice number as a compromise between silly amounts of light and decent burn times (as long as the batteries are as stated, I should get 3 hours on 700ma to each LED- 2700 mah AA's gives me a fair bit of leeway)

Anyone know how much heatsinking i'll need? The same website that does the rebels and the lenses also does heatsinks, though they're a bit small- link and i'm not sure if that will be sufficient, having done no LED projects before. Most the time they will be run at 700ma, in cold conditions and with air going past, though it'd be nice to know that I could run them at 350ma, at warmer ambient temperatures, with no ventilation and have no problems with heat buildup.


edit: been checking the instructables (thanks!) and it seems none of them use drivers... what's the advantage of a driver?

 

[Luminescent]

Ok, understand what people are saying, and now i've come up with this:

power supply (battery, 2/3D's if i go parallel, or 10 v. high capacity AA's for series) to driver (haven't decided on this... something to give me variable output (350,700,1000 type thing)), to LED's, to battery again, possibly throw in a resistor if it's needed.

Does this sound alright? Also, opinions on parallel vs. series would be good; parallel seems easier, but i'm confused by drivers. If I go parallel, would I buy a (say) 2100ma driver to get 700ma to each LED, one 700ma driver to get 700ma to each LED, or 3 700 ma drivers wired in series with the LED's?

At the moment i'm leaning towards parallel, because I can dim and turn off the LED's without wasting energy, but if I need to buy lots of drivers, this would take away the advantage, as drivers are fairly expensive.

I'm planning on making this a bike light, which doubles as a powerful flashlight that I can take if I know that i'll be going outside in the dark, hence the 3 rebels. Also, 3 seems a nice number as a compromise between silly amounts of light and decent burn times (as long as the batteries are as stated, I should get 3 hours on 700ma to each LED- 2700 mah AA's gives me a fair bit of leeway)

Anyone know how much heatsinking i'll need? The same website that does the rebels and the lenses also does heatsinks, though they're a bit small- link and i'm not sure if that will be sufficient, having done no LED projects before. Most the time they will be run at 700ma, in cold conditions and with air going past, though it'd be nice to know that I could run them at 350ma, at warmer ambient temperatures, with no ventilation and have no problems with heat buildup.


edit: been checking the instructables (thanks!) and it seems none of them use drivers... what's the advantage of a driver?

The problem with that triple star, is that the three Rebels are too close together to use the little clip-on optics you posted the link to at led-tech.de (or the separate reflectors I suggested), and at the same time way too far apart to use a single reflector or lens without getting an almost useless triple beam pattern.

Without a reflector, even three rebels will be marginal as a bike light, and with a reflector, in order to smooth out the beam you are going to have to either use a diffuser or throw the thing way way way out of focus.

If you can find an old halogen light with a very highly dimpled "orange peel" reflector, that will help combine and smooth out the beam.

How you do the heatsinking is dependant on how you are going to put the light togeather mechanically. The little heatsinks you showed will only do the job if they are outside in free air, not inside some kind of enclosure.

Most custom flashlight builders take a different approach and start with an off the shelf metal shell light, like a D or C cell Maglight.

The Maglights outer shell is very heat conductive aluminum, and makes a great heatsink all by itself, so with a Maglight you just use a slab of aluminum or copper to mount the LED's and couple the heat into the body of the flashight.

Some folks have cut the 2 D-Cell maglight down and shortened it by removing a section that extends from just behind the threads (keeping all the threads plus about another 1/8 inch, on back to the point after the switch (loosing the switch and a few inches of the 2 D-Cell battery tube).

Then the two sections are rejoined with a large cylindrical plug of aluminum and the seam is hidden under the area where the head is threaded back on. Since the threads on the head still work, focusing is still potentially possible, and the missing pushbutton is replaced by some kind of 'twisty' activated switch inside the head.

A cut down 'shorty' 2 D-Cell Mag light running on 4 NiMH AA's driving 3 Rebels would make an extremely cool bike light, and a standard 'flashlight clamp' can be used to mount it on your bike.

The discharge curve of the NiMH batteries is pretty flat, so you can just use a separate resistor for each LED.

3 ohms with 4.8 volts = approx 500ma.

2 ohms with 4.8 volts = approx 750ma.


Personally, if you are planning to run all three LED's on a star, I would stay with the 3 ohm resistor, and see how that works out first.

The resistor can dissipate about a watt so I would either use a 2 watt power resistor to be on the safe side or plan on running several 1/2 watt resistors in series or parallel to get the correct value.

An easy way to stack up the current without worrying about a lot of complicated power calculations is to just get a handful of 10 ohm 1/2 watt or 10 ohm 1 watt power resistors at Radio Shack.

With 4.8 volts (4 x 1.2 volt NiMH battery) each 10 ohm resistor you parallel in the series bundle feeding the LED will increase the current by 150mA.

1 x 10 ohm = 150 mA.

2 x 10 ohm (in parallel) = 300mA.

3 x 10 ohm (in parallel) = 450mA.

4 x 10 ohm (in parallel) = 600mA.

5 x 10 ohm (in parallel) = 750mA.

The note above 'in parallel' obviously only means that the resistors are all placed in parallel with each other then the parallel bundle is placed in SERIES with ONE Rebel LED.

Each of the other 2 Rebels in the star are wired to their own separate parallel bundle of resistors.

So to wire up all three Rebel LED's and drive them at 600mA for example, we would need a total of 12 resistors (4 x 10 ohm resistors in parallel, then the one of these 4 resistor parallel bundles placed in series with each of the 3 Rebel LEDs)

I know that sounds like a lot of resistors, but 1/2 watt or 1 watt resistors (either will work fine), are pretty cheap.

EDIT:

Just to avoid any confusion about the parallel and series arrangement needed to stack simple 1/2 watt 10 ohm resistors, here's what it would look like set up for a full 5 x 10 ohm (5 x 150mA = 750mA).

PLEASE NOTE:The 750 mA arrangement is only shown for TEST purposes (because you mentioned wanting to use that drive level in your post), but this drive level is probably only ok for SEPARATELY mounted Rebels. If you still insist on using the one piece star arrangement you showed earlier, there is probably no way you will be able to drive them safely at that high of current level due to the bad heat concentration problems inherent in that design. I would try to hold the current down to about 450mA per emitter when mounting them in this arrangement (450mA. would only use 3 of the 5 parallel 10 ohm 1/2 watt resistors in each string).

 

[jwl]

Jarl - Take a look at some prototypes that McGizmo has done. Post #20 of that thread is AWESOME! There is also information on where to get some of the parts that should be of interest to you.

 

[lexina]

Jarl - Take a look at some prototypes that McGizmo has done. Post #20 of that thread is AWESOME! There is also information on where to get some of the parts that should be of interest to you.

Thanks for the link. McGizmo has done it again!

 

[Jarl]

Thanks for all the help guys, I really appreciate it.

Now i'm thinking to get 4 rebels on PCB boards and mount them on heatsinks (one on each), then put the lenses on the rebels, glue the heatsinks together (probably 2x2), mount a handle of some sort (so I can hold it when the heatsinks are hot), squirt slicone gel or something on the PCB and around the lenses to seal, permanentise and waterproof it, then wire up the batteries. The heatsinks would be exposed to the air, and the whole thing would be REALLY small (1 inch x 1 inch x 1 inch or so). The battery pack would go somewhere on my frame.

The would be in a parallel/series circuit- two sets of 2 LED's in series, making 4 LED's, so i'd need about 8 volts (9.6 from 8 AA's would do the trick and allow some play with resistors).

I'm planning on 2 switches- on and off, for each set of LED's in parallel, so I can run 2 lights for places where I don't need so much power, and 4 for where I do.

I'm very happy with that (in theory), now I just need to work out all the stuff like how big are the resistors (graphs say for about 700ma draw, i'd need about 3.7v for each string, so 7.4 total volts, giving me 2.2 volts to play with with resistors and drivers, so after it goes through the driver it wouldn't be too hard to calculate the size of the resistor needed.

edit: I've made a diagram to help visualise my plans, here. Comments?

 

[Jarl]

quick question about drivers, as I can't get my head round them. If i have a 350ma driver in series with 2 LED's, do they both get 350ma with the driver outputting 350 ma, or do they each get 175 ma with the driver outputting 350ma, or do they each get 350ma with the driver outputting 700ma?

If it's the first, then if I had a 350mah battery (for example..), the battery would drive the LED's for 1 hour (in theory...), if it's the second then the battery would again last one hour, and if it's the last the battery would last 2 hours. Which one of these is correct?

 

[BentHeadTX]

Jarl,
To understand series and parallel, it goes something like this.

Series is when the LEDs are wired together with the + from the regulator hitting the + of the first LED and the - from the regulator attached to the - of the last LED. Take the - from the first LED and attach it to the + of the second LED. You can wire up large strings of LEDs this way.

Parallel is when both + connects of both LEDs are wired to the + output of the regulator. Both - connects of both LEDs are wired to the - connections of both LEDs. This can cause problems unless the LEDs are balanced and have identical forward voltages.

Series LEDs are great because they all run at the same current level measured in milliamps (or amps for the crazy folks) The problem is as each LED is added in series, the voltage required to drive them increases. For example

Luxeon Rebel 3.3 volts at 350mA drive level
Two series Rebels 6.6 volts at 350mA
Three series Rebels 9.9 volts at 350mA
Four series Rebels 13.2 volts at 350mA
100 series Rebels 330 volts at 350mA

Parallel is a larger pain in many ways than series when calculating LEDs. The voltage stays the same at 3.3 volts but as each LED is added, it cuts the current available to each LED. For instance (in theory to understand)

Luxeon Rebel 3.3 volts at 350mA
Two parallel Rebels 3.3 volts at 175mA (each)
Three parallel Rebels 3.3 volts at 117mA (each)
Four parallel Rebels 3.3 volts at 87.5mA (each)
100 parallel Rebels 3.3 volts at 3.5mA (each)

The complications of parallel connections when using the current driven LEDs add a bunch of additional problems. As the current level drops to LEDs, the voltage required to run them at that level also drops. This is good but all LEDs are not identical and weird things start to happen.

The forward voltage of a Rebel to run at 350mA varies so current will vary between the LEDs when running in parallel. Just taking a wild stab at this to illustrate the point.

Three Luxeon Rebels under test to calculate actual forward voltage at 350mA.

Rebel 1 3.30 volts at 350mA
Rebel 2 3.22 volts at 350mA
Rebel 3 3.41 volts at 350mA

Actual current driving the LEDs when configured in parallel off a 350mA current regulator (wild guess)

Rebel 1 117mA
Rebel 2 190mA
Rebel 3 43 mA

Parallel has different drive levels because the LEDs have different voltages they operate. The above example would have one bright, one a little dimmer and the last one much dimmer in operation. They are not "in balance" The way to get them in balance is to add resistors to each one so they all run at the same current. Say 100mA each with resistors (resistors will consume some power) As LEDs age, they will drop a little in forward voltage so the drive level will every so slightly drop again with the resistors.

As far as your battery question, you can calculate runtimes this way.

Take the voltage and mAH rating of the battery to get watt hours. Say a 3.6 volt battery at 1,000mAH. That would equal 3.6 watt hours (3.6 x 1.000) Then take the level you would like to run the Rebels (350mA) and multiply it by the forward voltage they will operate. (3 Rebels = 3.3 x 3 = 9.9 volts) 9.9 volts x 350mA = 3.465 watts

Before you jump in and say that would be an hour runtime, there is another power sucker than needs to be calculated. The regulator! The regulator will eat at least 20% of the available power trying to boost the battery voltage and regulating the current. 1 divided by 0.80 (percent of efficiency listed by the regulator) is 1.25. Multiply 1.25 times the power the Rebels need (3.465 watts) 1.25 x 3.465 = 4.331 watts.

The 3.6 watt hour battery divided by the 4.331 watt load is 0.831. Multiply .831 times 60 minutes for an estimated runtime of 49 minutes.

Confused yet? I think I am and a beer is calling me

 

[Jarl]

I think i've got it! so if I use a 1000ma puck, with 2 LED's in a parallel circuit , each LED will only ever see 500ma. If I take out one of the parallel LED's (with a switch), however, the remaining one will get 1 amp each cool.

Would I be able to run 4 3 volts LED's in series with 9.6V using a puck to sort out voltages, or would they all end up with 2.4 volts each (assuming no voltage drop). If so, I may go down that route with a 700ma puck, though it'd be nice to have the 1000ma just for the "naa, look how bright these are" effect... An adjustable 1000ma would be good....

edit: just realised I can't, so looks like it's back to the parallel/series.