Surefire 6P White Luxeon Module.

Joined
Mar 15, 2001
Messages
598
Location
Ohio
I've been working on it for a while and it's finally time to get some outside opinions.

I'm doing this conversion for someone who contacted me when I had my DB24 on eBay. He's a frequent CPF reader, but not a member yet (I think).

Anyway, it's a lamp module for a Surefire 6P (old model?) that is smooth all around instead of having the hex pattern on the head.

It it a direct drop in replacement for the P60 lamp @ 6 volts.

It draws about 340mA using the 2 123 lithiums I have and barely gets warm when used ouside the 6P.
It uses 3 - 22 ohm in parallel for about 7.3 ohms resistance.
I'm hoping to have time to lower the resistance and get more output.
I'm thinking the 6P body will help to disperse the heat.

Appreciate any comments.

No, I don't know what to charge for them yet.
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And now for the pics...
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If you'd like to view the album (or if the pics don't load) ... http://community.webshots.com/user/bajagadget
 
Excellent job!
The only bad thing - more than 40% of the battery power is wasted. Thats why 2x123 setup NEEDS step-down. OR, you could connect (with a little wiring, soldering, and taping) the batteries in parallel, and use the Peter's LS step-up. Wait... It is not available yet... I was day-dreaming.

Addition: If I'm day-dreaming at 3:27AM, shouldn't it be considered night-dreaming?
 
The lost power bothers me, but it was an economical tradeoff.

I'm wondering... Do diodes generate heat when used to drop the voltage? I would think they do since they have a voltage drop and current going thru them.

I don't know how I would incorporate a step-down without getting one of the NatSemi LM3477 step-down eval boards for $11.
 
You know, I've read that Step-Downs can be much more efficient than Step-Ups so I'm thinking about giving them a go sometime. Resistors are just WAY too wasteful and primitive of a solution to suit me anymore. Of course the big drawback with Ups or Downs is the cost, but when you are in pursuit of the ultimate flashlight, concerns of cost just seem to melt away somewhat. Plus there is also the HUGE added benefit of having a constant brightness level until the end of the battery life that resistors just can't provide.
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Only problem with the step-down I can use (LM3477 for $10) is the size.

The darn Eval board is 2.4 x 2.6 inches. Even trimming away some of the board is gonna be over an inch square.

What else can I use to make an efficient step-down?
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i had the same question (unasked)
about using diodes as the voltage drops, because most of them drop 1 volt (sounds like a set voltage), of course they put out heat as a byproduct, and would be no more or less effective than a resister.
so i tested with what i had here, the diodes didnt drop one volt, ????
aparentally the drop is one volt at some effective current, and my testing was based on a 20ma led.
so it would take the RIGHT diode properties, just like the right amount of resistance.
 
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Go, Go Gadget Flashlight:
I'm wondering... Do diodes generate heat when used to drop the voltage? I would think they do since they have a voltage drop and current going thru them.
<HR></BLOCKQUOTE>

I used a diode to drop the voltage on a LS module on one of my first prototypes for a drop in replacement for Koehler / Wheat mining lamps that run off of a 4V lead acid battery...

I read in a couple of places tht the voltage drop on most diodes is .7V - so it was a natural:- drop .7v from 4V and you get 3.3 volts which the LS likes.

I used a fairly small diode - it worked fairly well - and dropped the voltage to the right range. But that diode got HOT! hotter than the LS heatsink and hotter than a resistor would have, in fact it melted the heatshrink I had around it. (still ran for several hours though)

Don't know specifically what kind of diode it was, one out of the "pulled parts" box, but it was pretty small... I did get some beefier diodes to see if they would work and stay cooler, but haven't gotten around to trying it yet.

I don't know what kind of efficiency you would get, but I'm pretty sure that you could build a step down circuit based on a LM317 voltage regulator on a 1 inch square (or even 3/4" X 1 1/2") PC board without having to get into surface mount components with just a three resistors and a couple of caps--- (I have been looking for a ready made one like that (or with a different VR chip) since I would prefer not to get into manufacturing circuit boards too..)

I did rip apart one of those multi-voltage car lighter plug adapters thinking that THAT might work- It sort of does, although the voltages come out either too high or too low, (just replacing resistors would probably let you get the right range, but again, I havent tried it yet) The adapter I pulled apart just had a 317 VR chip and 7 resistors. Not even a capacitor... which I would think would be necessary for some sort of power stability.. This method could even give you a switch for adjusting for battery drain....

ANYWAY, I don't know if this route would be more efficient than using a resistor or diode to drop voltage, but it COULD be done in quantity pretty cheaply... I have seen the 317 chips selling on eBay for about 40 cents each in quantities of 25.
 
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by JoeyL:
oooh...

I want one.
Please post if you're taking orders.

Thanks.
<HR></BLOCKQUOTE>

Sorry, JoeyL - I was here first !!!
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Well, second, (after Glow Bug)
Glow Bug, welcome aboard - nice name too
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Form an orderly queue, chaps ......

Gadget !
PK congratulated you !!


lightlover
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For reducing the voltage to a LED the use of resistors, diodes, or linear regulators (like LM317, or a low drop out regulator) will all result in wasting the same (large) amount of battery power.

There are two pretty efficient things you can do. You can use a switched capacitor circuit - which generally would not provide much amperage for an LS module, or you can use an inductor based switching supply.

The benefit of the inductor based switcher is that you actually SAVE a significant amount of battery energy. Watch this.

Say you want to power the LS at 3.3V and 300mA which is 1 watt of power. If you use a 6 volt battery with a resistor (or diode or linear regulator) you're taking 6V @ 300Ma or 1.8 watts from the battery - wasting nearly as much as you are able to use for light. If you use a switching supply that's 85% efficient you'll be taking 192mA instead of 300mA. (1 watt)/(6Volts)=167mA/85%=192mA. Much more economical, plus you'll be able to run the battery down to probably 4V extracting every last electron you paid for with an evenly bright light output till the battery is nearly dead.

(saving up for an LS module....)
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Silviron: The problem you had with the diode getting too hot is that diodes are rated to take different amounts of wattage. You just need a diode that can handle the amount of power that you are pushing through it is all.
 
Thanks for all the praise and info guys. Appreciate it.
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It seems the only way to do this "correctly" will be with a step-down regulator. I have been reviewing the info on the LM3477 and it looks promising. Problem is, it's pretty big for an SMD board and I doubt if I can fit it in to a circular space of 3/4" diameter.
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Any other suggestions on alterative "tiny" circuits that will regulate current output with varying input?
 
I know that Maxim has many step down chips, but so far as making a recommendation, I've only begun to explore the option of using a step-down, so someone else would probably be able to give you a more definitive answer. I do know though that a step up/down is definitely the very best way to go these days for anything that is battery powered.
 
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Secret:
Are there any super efficiant resisters?<HR></BLOCKQUOTE>

I don't think so, KenB!
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Nice try.
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I think the only efficient way to do it is to "chop" the output thereby effectively reducing the voltage the LS gets.
 
Gadget -

As you probably know ... the Linear 1308b SEPIC circuit design might give you what you want in output specs. Too bad it's such a booger to build.

Well, my PCB building skills weren't up to the task. You'd do better with ten thumbs and no wrists.

If I remember right, the completed board measures about 3/4" square.

Too bad someone can't crank out the circuit boards for that solution ... a bunch of folks on CPF would likely buy them.

Mark
 
Mark,

Thanks for the tip.
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I just found and read the specs on the 1308. Looks the same. For 1.2v IN, you can only get about 110 mA out.
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I'm still searching for another altenative...
 
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