Amp draw calculation.

[mhertz]

When wanna calculate true amp draw of a cell, then am I correct in the following:

With an inline-voltmeter, and a mech with a 0.25ohm build and a fresh cell, then lets say that the resting voltage is 4.16v and then under load it sags to 3.8v.

Of course the VD(battery-sag-mostly from IR, but nonetheless) can be calculated to 0.36v, but what I was interested in getting confirmed was that the actual true amp draw of said cell is calculated from the voltage under load i.e. 3.8v, and not the resting voltage of 4.16v, correct; i.e. 3.8 / 0.25 = 15.2a.

Please note, i'm not talking about safety here, which always is calculated from full voltage i.e. 4.2v, but I just wanna get confirmed that it's correct that when talking true amp draw then it's calculated from voltage-under-load(of course I know the meter also is about 0.4% imprecise).

 

[Chips177]

(3.8vx3.8v)/.25Ω=57.76w
57.76w/3.8v=15.2a

The true amperage is voltage under load. The safe way is full voltage (4.2v). Dependent on the battery, you're technically safe. Also, remember the inline voltmeter draws a small current to give you that reading, making the calculations slightly inaccurate.

 

[mhertz]

Thanks for your reply, I appreciate it...

My question was purely theoretical, and based from my feeling that the above was correct, even though it's sometimes disputed i've seen on occasion through googling alot, with reasoning of that the cell tries to put out the full voltage and hence that is the true amp draw seen from the battery, but to me that didn't make sense seeing as the voltage is conditioned by the full circuit connected to the power source(cell) in terms of what is drawn, and so the cell's own IR + mod and coil, determines what's drawn out...

Thanks again for conformation!

 

[duc916]

the actual true amp draw of said cell is calculated from the voltage under load i.e. 3.8v, and not the resting voltage of 4.16v, correct; i.e. 3.8 / 0.25 = 15.2a.

Correct. Read a little on Thevenin equivalent theorem.

 

[mhertz]

Thanks alot duc, i'm gonna look up that theorem! Appreciated!

 

[Baditude]

Also keep in mind that many battery brands advertise false or misleading specifications.

Efest, Imren, and several other brands often advertise the pulse discharge rate and list it as a "maximum discharge rate", not specifying whether it is the "continuous" discharge rate or some other made up spec. If you look over Mooch's battery tests, you'll find that many of the advertised specs for 30 - 60 amps are only 20 amps continuous or less when bench tested. Buyer beware. 18650 Battery Safety Grades -- Picking a Safe Cell to Vape With

Pulse discharge ratings are pretty useless, and misleading to consumers. There are no industry standards for "pulse" ratings, and every "manufacturer" has a different definition of what a pulse rating is -- anywhere from a millisecond to a few seconds. The continuous discharge rating is an accepted industry standard and makes choosing between different brands and models of batteries more useful.

Battery pulse ratings are useless!
There are no 18650 batteries with a genuine rating over 30A!

 

[mhertz]

Thanks mate and fully agreed!