I have a battery carrier for a flashlight that is in 3P1S configuration. Using a single Panasonic NCR18650BD (10A) battery, I am able to get a measurement of10+ A when I shorted the terminals using a multimeter, regardless of which of the 3 battery bays I put the battery in. This agrees well with the theoretical maximum discharge current of the NCR18650BD.
However, if I load in 2 or 3 of the same battery into the battery carrier, I am only able to get a slight increase in current, at about 11.4A for 2 batteries and 11.9A for 3 batteries, in parallel. This was a result I did not expect. I would have thought that putting the batteries in parallel should get me up to 30A for 3P1S or 20A for 2P1S configuation, since each battery bay is capable of passing 10A through.
Could anyone advise what could be going on here? Why am I not getting the higher current I am expecting?
Edit: Found out that resistance decreases a little in parallel circuit while voltage remains the same , so current only increases a little due to the slight decrease in resistance.
Don’t short the batteries. You will cause damage to the batteries, your multimeter or your environment.
The batteries are able to deliver more than their nominal maximum current, but will heat up quickly while their voltage drops. The limiting factor of your test is your multimeter. For proper current tests you need a clamp meter. But please, don’t short the batteries. The short circuit current won’t tell you anything.
Does your light draw more than 10 amps — Are you using a clamp meter ( if you are make sure the jumper wire is big enough to handle more amps)— If you are using a meter with Leads, they have to be Big and as Short as possible — Best way to go is a clamp meter---- A picture of the carrier would help – also are the springs in the carrier Bi-passed
That won’t be enough because the multimeter uses a shunt resistor (basically a thick wire with known resistance), measures its voltage drop and then calculates the current. This is enough to get completely “wrong” values.
Yes my LED and driver is capable of drawing currents up to 20A. The springs in the carrier are all bypassed.
Noted that a multimeter with leads isn’t necessarily the ideal way to measure current, but given what I was testing, the absolute accuracy of the reading is not that important. What’s more important is the relative increase in current as I increase from 1 cell to 3 cells in the carrier. Even if the accuracy of my readings is off by 30%, the relative increase in current (if any) should still show clearly.
The minimal increase in current wasn’t what I expected initially, but after going back to basics, I realised that with the voltage of the cells fixed at 4.2V, the discharge current is basically dependent on the resistances of the multimeter and battery carrier. As my battery carrier is already highly bypassed, the decrease in overall resistance due to adding more cells, is quite low, hence the small increase in current.
I guess the only way to provide more current to my light is to change to even higher-drain batteries, but I am not sure if the increase would be significant, since the difference in total resistance of the setup (due to lower internal cell resistance) would only be in the order of tens of milliohms…Please correct me if I am wrong about this.
Is it possible that IF you were actually feeding the current from the cells to an actual LED through the multimeter wires, the skinny size of the multimeter wires could be limiting the amount of current?