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-JOE-
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I was wondering since there are 2 batteries do I need to times the amperage by 2.

I ask because my TR-3T6 is only pulling 2.05A. Is that right or is it pulling 4.10A? Or is it 2.05 to each emitter?

Edited by: -JOE- on 05/26/2012 - 01:06
bose301s
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-JOE- wrote:

I was wondering since there are 2 batteries do I need to times the amperage by 2.

I ask because my TR-3T6 is only pulling 2.05A. Is that right or is it pulling 4.10A? Or is it 2.05 to each emitter?

P=VI, Power=Voltage*Current. So, with a 2 cell light you double voltage meaning current will drop in half for the same power. So basically if you want to know the current in a 1 cell equivalent you would double the amperage.
-JOE-
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So what is each emitter pulling then?

scaru
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Assuming a 80% driver efficiency and a vf of 3.0. 2.05*4.2=8.61 watts ,8.61*.8=6.88, 6.88/2.2933, 2.2933/3=.7644 amps to each emitter.

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Hey scaru.. it's 2.05*8.4 = 17.22 watts, he has TWO 18650's.

Which makes it 1.5A per emitter.

scaru
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Oops. That makes more sense. Thanks for catching that.

sontakke
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I am not really convinced that most flashlights which use 2× 18650 cells in series have electronics which to actually drop the voltage without just dumping it as heat. All of those parallel drop down chips used in single cell 18650 use constant current to emitter and the extra voltage i.e. battery volts – emitter volts is essentially dissipated as waste heat.

I would be pleasantly surprised if real power converter circuitry is used 2× 18650 cell flashlights. Anybody knows for sure?

comfychair
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It depends on the driver. Constant current/linear drivers do just discard the excess voltage, buck or boost drivers handle it differently. A buck driver really does take that excess voltage and turn it into current, it doesn't just get wasted.

sontakke
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i just picked up my first 2×18650 (actually 2×26650) light from Meritline. The tailcap shows about 0.96Amps and the light is brighter than all my single cell 18650 lights. Not bad for a cheap light but it could have been driven harder with the two cells.

Bob Loblaw
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how would you calculate this for an mt-g2? Using a trimpot on a tr-0142 driver with 2×18650 i get 5A at the tail

RMM
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Bob Loblaw wrote:
how would you calculate this for an mt-g2? Using a trimpot on a tr-0142 driver with 2x18650 i get 5A at the tail

It should be a little under 5A at the emitter, you always lose some energy through the driver although that number varies widely.

Mountain Electronics : batteries, Noctigon, and much more! What's new?

comfychair
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It's probably not regulating with only 2 cells, so while it's technically a buck driver, without enough voltage overhead it acts like a CC/linear driver. MTG2 and 3 cells will get the driver working as it should be.

The readings at the tail only isn't enough info to really see what's going on. Measure input amps and input volts (under load, while it's running), and compare that to the output amps and output volts. (amps x volts = watts) The difference in watts between input and output numbers is the driver efficiency. Only by checking both sides can you see how a driver is operating.