Some M@g led questions?

Question one: If you put a standard m*g dropin led into a 3 cell mag, and one into a 6 cell mag (both are standard 3 watt dropins) will the 6 cell be brighter than the 3 cell?

Question two: What is the best,and brightes drop in led I can get for m*g (by drop in I mean just replace the bulb with it)

While I would prefer to go with the above, if I cant get what I need from this I am prepared to go another direction.

Question three: Suggestions on getting this result> A m*g with either 1, 2, or 3 leds. At least 400 lumen. A minimum of 1.5 hours run time ( I prefer nimh, but if needed I can go li-ion) and this light MUST THROW! My budget for this light would only be about 100$ max so any help is appriciated, also if you have liks for part sources that is great.

As always Thanks in advance for all your support cpfers.

UPDATED: Ok guys thanks for all the info. I have decided what to do but I still need your assistance. Help me build this light.

I will cut down and rethread a m*g to a 1 D size so it will run on one protected li-ion cell. Emitter will be p7 heat sinked.

Question one: Do I have to use a driver board with this?

Question two: Will the KD aluminum reflector (camless) that I have work for this?

Question three: Where can I get a p7 emitter, and the heat sink for mag D cell?

Question four: If anyone has done this or similar is there anything I need to know about installing theese parts that will be a problem?

First thing to notice is that there are 3 specific stock M@g LEDs. 2-cell, 3-cell and 4-cell. They are (imho terribly) designed for specific voltages. Hence there is no 6-cell stock dropin made by M@g. If you put one into a 6 cell I suspect nothing good will come of it.
The LEDs they use are 4+ year old technology, Luxeon IIIs, so they will be half as bright or less than what is currently available. I personally wouldn't recommend buying one. I do not believe there is much difference in brightness between the 2, 3 and 4 cell versions.

The Best dropin you can get for a M@g is a Malkoff. You definitely get what you pay for here, even if it is 3X as espensive as most other dropins.
Some lower cost aturnatives are from Terralux. They even have a 3 emitter setup that will get plenty bright.
Another place to get bright M@gmods is from Elektrolumens. Wayne makes some crazy bright stuff, but often only designed for direct drive... no regulation circuit, so only for sue with 3 cell setups (or a single LiIon cell).

Beyond that you get into more hardcore modding with soldering irons and circuitboards, or buying custom mods. It's crazy around here. :devil:

Hope this helps

Check this thread.

Maglite never made a LED drop-in for the 6cell. The ones that they have will burn out at those voltages. The highest output MagLED seems to be the 3 cell one.

If you want lots of throw, go for a single LED drop-in(the Malkoff or the KD aspheric drop-in, the Elektrolumens P7 drop-in/mod for the 3D also works). Multiple LEDs will have a smaller reflector area and less throw.

Please see updated Questions. IE help me build this light. TY

1) I think that would be a good idea. You can use up to 20v with this driver but it does cost a good bit. This KD one should work for 1 higher capacity Li-ion along with this one too if you want modes. Not sure how this would work.
2) The stock reflector works fine, but you will get a doughnut beam at full focus. That won't be a problem with a textured reflector(you can self texture a smooth reflector with some spray products). The KD ones are the cheapest textured Maglite fitting ones I know of.
3) You can get the emitter at a bunch of places. KD and DX being the cheapest but check the Marketplace if you want a specific bin that isn't offered by KD or DX(probably faster shipping the either too). DX Seoul P7 one, two. KD Seoul P7 one, two. Litemania has heatsinks.
4) There are quite a few Seoul P7 Maglite Mods in the Modding section. Here's one. This could help too.

Thanks Gunner, and to others that have posted so far. I have found this into to be useful, a little more info and I will be ready to order some parts. Please if any one has more useful info, tips, ideas, or places to get parts let me know.

Also one more Question Please: Could someone explain this "Bin" thing to me? I gather that is has somthing to do with quality, but what else? What is the best, and worst "Bin"?:thinking:

Bin(might not be the full technical term) is what LED companies use to sort their LEDs.

At a minimal(from what I've seen), it sorts a LED to the color(red, blue, white etc.), flux(output/efficiency), and tint(warm, greenish, yellowish etc.). Some add the LED's Vf(forward voltage) and a few other bits of info too to the bin code.

This thread might help.

I was just wandering if someone could explain the math behind this to me.

On KDs web site I found this driver borard http://www.kaidomain.com/WEBUI/ProductDetail.aspx?TranID=5296

They say that it will run a P7 emitter @ 3.0A for 3+ hours on 2 18650s (2400mah I believe)

I understand how to get incadecent mah ie. eneloops are 2000mah my light draws 2a so my runtime is 1 hour.

But I do not understand how they are equating 2x2400mah batteries to run a light drawing 3a to 3+ hours?

Also will p7 emitter be putting out full 900 lumen at 3amp?

Thanks always for your support cpf.

vestureofblood said:

I was just wandering if someone could explain the math behind this to me.

On KDs web site I found this driver borard http://www.kaidomain.com/WEBUI/ProductDetail.aspx?TranID=5296

They say that it will run a P7 emitter @ 3.0A for 3+ hours on 2 18650s (2400mah I believe)

I understand how to get incadecent mah ie. eneloops are 2000mah my light draws 2a so my runtime is 1 hour.

But I do not understand how they are equating 2x2400mah batteries to run a light drawing 3a to 3+ hours?

Also will p7 emitter be putting out full 900 lumen at 3amp?

Thanks always for your support cpf.

Click to expand...

The voltage of 18650's is 3x higher than AA's. What the LED draws is not the same current as what the battery draw is. Two 18650's in series providing 3A will only be using approximately 1.5A each

Marduke said:

The voltage of 18650's is 3x higher than AA's. What the LED draws is not the same current as what the battery draw is. Two 18650's in series providing 3A will only be using approximately 1.5A each

Click to expand...

I see what you are saying I think. Li-ions provide the rated mahs at 3.7 volts ie 1x18650 is 2400mah at 3.7 volts. So using 2x 18650 each battery provides only half of what is needed 1.5a each. However still this does not add up to me. 2x 2400mah= 4800mah divided by 3.0a = 1.6 amp hours no? Not 3+ hours.

It seems as though I am still missing somthing? :thinking:

In series, you add voltage, not capacity. Both voltage and amperage come into play, so it can be easier if you convert power usage into watts. 3 hours does seem optimistic though, 2 hours seems much more reasonable.

Could you show me the conversion to get every thing to watts. I think I saw this before. Where both the draw from the emitter is converted and power from the batteries. If you know this formula please give examples. Say useing one li-ion then same thing for two. If its not too mutch trouble. I think once I understand this formula I will be better prepared to build my light.:thumbsup:

volts X amps = watts

in series, batteries add voltage, not mAh

in parallel, batteries add mAh, not voltage

Say you have a emitter running at 3.7v drawing 1 amp. you have two 18650's powering it in series, each at 3.7v 2400mAh. the power draw is 3.7v X 1A = 3.7watts. The power available is 3.7v X 2 X 2.4Ah = 17.76 watt hours. First off, a rough estimate says you can power your emitter for about 4.8 hours (17.76 / 3.7 ). The draw from your batteries, since they have double the required voltage, is only half the current, which is .5A

Confused yet??

i have been looking for that exact explanation. you freakin rock! :rock:

Marduke said:

volts X amps = watts

in series, batteries add voltage, not mAh

in parallel, batteries add mAh, not voltage

Say you have a emitter running at 3.7v drawing 1 amp. you have two 18650's powering it in series, each at 3.7v 2400mAh. the power draw is 3.7v X 1A = 3.7watts. The power available is 3.7v X 2 X 2.4Ah = 17.76 watt hours. First off, a rough estimate says you can power your emitter for about 4.8 hours (17.76 / 3.7 ). The draw from your batteries, since they have double the required voltage, is only half the current, which is .5A

Confused yet??

Click to expand...

No I think you have made this clear to me now. Lets find out.


If I have 2 D cell li-ions (5000 mah) in series, my driver boar regulates a constant 3.0a draw. Is 3.7v x 3.0a=11.1 watt draw. My power available is 7.2 volts x 5.0a=36 watt hours available. 36 divided by 11.1=3.24 hours runtime. Correct?

vestureofblood said:

No I think you have made this clear to me now. Lets find out.


If I have 2 D cell li-ions (5000 mah) in series, my driver boar regulates a constant 3.0a draw. Is 3.7v x 3.0a=11.1 watt draw. My power available is 7.2 volts x 5.0a=36 watt hours available. 36 divided by 11.1=3.24 hours runtime. Correct?

Click to expand...

Except for the terminology, yeah. Your driver doesn't "draw" 3A, it provides 3A. It will draw different amounts from your batteries, depending on the voltage. It "draws" closer to 1.5A from your batteries.


Also, in reality there is some circuit inefficiencies which suck up some runtime. So, using that knockdown factor, I would expect closer to around 3 hours runtime.

Thanks Marduke, And to all others who have been posting.

My favorite part about learning somthing new is understanding the science,and or principals, so that I can apply it (I feel like I have begun to do this here) and hopfuly contribute to the advancement of somthing new.

IE Give a man a fish, and he will eat for a day
Teach that same man how to fish, and he will eat for a lifetime.

You'll probably enjoy this

https://www.candlepowerforums.com/posts/2561827