Testing methodology for untrusted Chinese LFPs

[Maast]

Buying LFP batts from Aliexpress/Alibaba/(shudder)globalsources/etc is pretty much a crapshoot.

You really dont know what you're gonna get so they should be tested prior to use. IMO at the individual cell level if you can.

I've got a routine that I've got but if there is a better one I'd love to hear it.

What I've been doing is cc/cv charging to 3.65V, disconnect and then I let the cell sit for 72 hours exactly and then measure the cell voltage.
Then charge back to 3.65V and discharge at .1C, then charge/discharge at .5C then at 1C. I'd do a 2C run if I could but my load bank can't pull than many amps from a 3.2V cell.

Thoughts?

 

[Craig]

I have no clue but am looking to discharge some cells at 400 amps as well pretty much to do same as you and verify quality and capacity. I'm wondering if something like this would work. https://www.amazon.com/DealMux-a110...ords=1000+watt+resistor&qid=1572658759&sr=8-3 Ive used small wire wound resistors to test 18650's mkaybe a larger one would do the same?

 

[Craig]

ANother thing i have thought about is if you had some cells that you had say 5 cells pretested and knew their capacity. you could add one more and subtract the original 5's capacity from the whole 6 after testing 6 in series.

 

[SCClockDr]

To determine the load (R) needed to create a specific current (I) the voltage of the battery bank is needed.
A 20 ohm resistor would draw 400 amps provided the source voltage was 8,000 volts (E).
For a 3.6 volt cell you would need a 0.009 ohm Resistor capable of dissipating 1.44 Kw.
For a 48 volt battery you would needed a 0.12 ohm Resistor capable of dissipating 19.2 Kw. (Neither is very practical) I'm sure it's why Will uses an inverter for a load in his videos.
The formula is E=I*R or I=E/R or R=E/I
For watts (P) P=E*I, E=P/I, I=P/E

 

[Craig]

To determine the load (R) needed to create a specific current (I) the voltage of the battery bank is needed.
A 20 ohm resistor would draw 400 amps provided the source voltage was 8,000 volts (E).
For a 3.6 volt cell you would need a 0.009 ohm Resistor capable of dissipating 1.44 Kw.
For a 48 volt battery you would needed a 0.12 ohm Resistor capable of dissipating 19.2 Kw. (Neither is very practical) I'm sure it's why Will uses an inverter for a load in his videos.
The formula is E=I*R or I=E/R or R=E/I
For watts (P) P=E*I, E=P/I, I=P/E

That's all good but we need to test 1 cell at 3.6 volts. So an inverter will not work. Have any practical ideas? Is it even possible to get 500 amps from 3.5 volts. I mean I know it is theoretically possible. But what about the real world.

 

[Maast]

I have no clue but am looking to discharge some cells at 400 amps as well pretty much to do same as you and verify quality and capacity. I'm wondering if something like this would work. https://www.amazon.com/DealMux-a110...ords=1000+watt+resistor&qid=1572658759&sr=8-3 Ive used small wire wound resistors to test 18650's mkaybe a larger one would do the same?

That'd work, but you couldnt adjust it. Make one of these, its called a "liquid rheostat". I put this one together out of PVC pipe, some galvalume roofing and a brewing barrel I had laying around. I think I have about 30 bucks into it and thats mainly the tinned marine wire and welding lead connectors I had to buy.

Basically you have two plates submerged in a conductive fluid (NOT saltwater) the most common fluid is sodium carbonate (washing soda) and you can vary the resistance by how far apart the plates are. In this one the plates are rings and I can slide the inner ring up and down the center pipe and then clamp it in place when I hit my target amp draw.
My minimum resistance is .5 ohms (ish) with the rings as close as they can get - about 3/4". To get that I had to mix up the washing soda solution and keep adding it until no more would dissolve, then I added some sodium hydroxide (which is much more conductive) until I got my target resistance.
The solution gets warm but I've never boiled the 30 gallons of it. Use it in a well ventilated place as it does produce a little hydrogen and oxygen - but not as much as you'd think.

The drawback is that the resistance goes up a little as the water gets warmer so you dont have a completely steady amp draw but it doesnt really matter because the amp-hour meter just keeps recording it as it varies. You also have to keep in mind the amps also go down as the voltage droops on the battery but again it doesnt really matter. A little water does evaporate but I only have to add maybe quarter inch every 7th or 8th testing run.

 

[Maast]

Thinking about it, you could do the same thing with two copper pipes in a rubbermaid tub and just vary their distance from each other. It takes an huge amount of energy to heat water so you could still dissipate 10-15kw that way and with the large surface area of the tub it'd self-cool from evaporation and no matter what it'd never go above 212F as it boils. At that temp the tub might start to deform though.