Battery Level and Laser Performance

[Cyparagon]

Yeah, sounds like the steel and wiring accounts for the extra voltage drop you're seeing. Stick to IMR cells, and/or use larger and shorter wire to connect everything.

 

[Arc]

I see, the voltage drop is occurring further down the current path where the leads of the volt meter panel connect to. I didn't understand enough about electricity when putting this together. My thinking was that if the wire doesn't heat up at all during use, then it's plenty thick enough to carry the current without limiting power to device. But now I'm realizing that I should have used a much thicker wire for the main wires that go from the battery terminals to the drivers which are connected up in parallel.

Also I did get in contact with the eBay seller supersports600 and he presented me with the following info:

"well talk about the 700mAh aw18350s as they will sag less
I have repeatedly measured 14 milli-ohm internal resistance, that is consistent
and accurate. So if you measure the sag at the battery before the contacts
(as the contacts and wires have resistance) it will be the following
2 cells
V^2 = PR
WHERE P = IV = 8 X 2 = 16WATTS
SO
16 X (0.014 + 0.014)
= 0.448 = V^2
So V drop will be 0.67 Volts across the 2 batteries"

Now that seems correct to me but earlier in this thread we had,
Quote:
Originally Posted by Arc View Post
Ri=(VNL-VL)/IL=(7.4V-5.9V)/2Amps=0.75 Ohms

That is correct, assuming you've measured everything properly. I would expect <150mΩ per cell.
End Quote:

So based on that we would get the following for the AW IMR cells from supersports600,
Ri=(VNL-VL)/IL
Hence, (VNL-VL)=Ri*IL=0.014*2=0.028V of sag

Any idea why these numbers aren't in agreement?

 

[Cyparagon]

Because you have that steel strip in the middle, and also wire on either end of the cells.