Question about tailcap readings for 2x18650 lights

12 posts / 0 new
Last post
-JOE-
-JOE-'s picture
Offline
Last seen: 7 months 3 weeks ago
Joined: 04/25/2012 - 23:54
Posts: 553
Location: Iowa
Question about tailcap readings for 2x18650 lights

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
bose301s's picture
Offline
Last seen: 2 months 1 week ago
Joined: 03/10/2012 - 18:40
Posts: 1224

-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-
-JOE-'s picture
Offline
Last seen: 7 months 3 weeks ago
Joined: 04/25/2012 - 23:54
Posts: 553
Location: Iowa

So what is each emitter pulling then?

scaru
scaru's picture
Offline
Last seen: 4 years 7 months ago
Joined: 03/22/2012 - 13:36
Posts: 6946
Location: Virginia

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. 

Shadowww
Offline
Last seen: 9 hours 27 min ago
Joined: 01/20/2012 - 19:09
Posts: 1340
Location: Northern Europe

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

Which makes it 1.5A per emitter.

scaru
scaru's picture
Offline
Last seen: 4 years 7 months ago
Joined: 03/22/2012 - 13:36
Posts: 6946
Location: Virginia

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

sontakke
Offline
Last seen: 3 years 2 months ago
Joined: 05/18/2010 - 20:43
Posts: 107

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
comfychair's picture
Offline
Last seen: 5 years 2 months ago
Joined: 01/12/2013 - 05:39
Posts: 6198

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
Offline
Last seen: 3 years 2 months ago
Joined: 05/18/2010 - 20:43
Posts: 107

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
Offline
Last seen: 4 months 3 weeks ago
Joined: 07/22/2013 - 04:01
Posts: 104

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
RMM's picture
Offline
Last seen: 1 year 4 months ago
Joined: 07/23/2013 - 13:47
Posts: 4006
Location: USA

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
comfychair's picture
Offline
Last seen: 5 years 2 months ago
Joined: 01/12/2013 - 05:39
Posts: 6198

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.