Build a lipo discharger

[izeman]

I know this is an old story, and i posted some comments on that topic in various threads as well.
But i need something different.
I don't want to discharge the lipo, i want to build a IR meter of some kind - very basic, as i need it only once a year.

Connect an icharger to the balance port and start recording. This gives me accurate cell voltage of every cell, every second.
Connect a massive load (made of a resistor network of eg. 100W 5Ohm) to a FET switch that can be triggered by a button switch.
Then connect the full charged battery to the switch and press the button for 10s.
This will give some sag voltage and with the current known i can calculate the IR.
The question now: can i directly connect a wire resistor to the battery? Last time i did, it sparked and i quit the project.
Would i need to put a light bulb in series? Or would this limit the current?
I plan to see around 20A for a single 6s lipo.
Thanks



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[izeman]

i just calculated the resistors. i have 5x 100W 5Ohm resistors. if i put them in parallel i get 1Ohm total and 500W.
a 6s lipo is 25.2V fully charged and 18V fully discharged.

U = I * R = 25.2V = I * 1 -> I = 25.2A
P = U * I = 25.2V * 25.2A = 635W

so this does not match the 500W total load. what will happen?
voltage will drop to something below 25.2V instantly with a 5C load, but still ...

 

[Hillhater]

Your calc are wrong.
With that 10 ohm resistance, you will only have 2.5 A in the circuit.

 

[Ianhill]

^^ 1ohm 500watt misread it i think.
I'd say your resistors will start glowing up the wattage or reduce the ohms.
Some of the ichargers measure IR same way as u mentioned with their internal discharge resistor not sure if all models do this but i got the IC 1010b+ and when its at start menu i hold forward button it measures cell voltages then i hold back and it IR checks the cells individually and in total ive never used my PC lead, My laptop wouldn't connect to the charger but the program itself looked nice, served me well for laptop cell reclaim checking capacity and ir.

 

[Hillhater]

Nope !
With 10 ohms and 22.5 v, there will only be 2.5 amps ,...0.5 amps through each resistor.
Those 100W resistors wont even get warm :lol:

 

[Punx0r]

Not sure where 10ohms came from... It's 1 ohm...

Izeman, the current through the resistors will change depending on the voltage and the voltage will change depending on the current. Just use a volt/amp meter to measure both at the same time.

 

[izeman]

@hillhater: again: you MISREAD it. it's 1 Ohm and not 10 Ohm. if you take 5x 5 Ohm and put them in parallel it sums up to 1 Ohm. correct?
so it's ~20A load.

but the main question is: may i just put those resitors DIRECTLY to the battery? or will something bad happen?

the resistors are screwed to the back of a massive heat sink (200x100mm 25mm fins) and 2 90mm fans blowing at it.
but this isn't important at all, as the resistors will see load only for 10s or so during measurement.

 

[Hillhater]

Ahh ! ..i see...5 ohm not 50 ohm .. :lol:

i just calculated the resistors. i have 5x 100W 5Ohm resistors. if i put them in parallel i get 1Ohm total and 500W. ...

..but why do you want so much current for measuring IR ?
Just use two of those 5 ohm resistors in parallel and get 10 amps (ish) and 100w in each resistor.

 

[izeman]

whatever the more amp you have the more precise the measurement will be. if you read 4.200V and apply a 5A load the voltage may sag to 4.100V. if i draw 20A it may sag to 3.950V. makes the delta much bigger, and evens out any little measurement readings.
and it doesn't matter if i take 1,2 or 5 resistors in parallel. every additional amp would lower the resistance of the total network. and the load for each resistor stays the same.

but still: what happens if i connect the resistor to the battery? will it short circuit the battery? 1 Ohm resistance is almost a short circuit. or am i wrong?

 

[Ykick]

1 Ohm resistor across a 4V battery cell will draw 4A current dissipating 4W of power. It will get hot but a 5W resistor should handle the load.

I’ve played around a great deal with DC IR testing RC Lipo. I just don’t find much practical difference with IR results using 1C or 5C load.

But if you really want a powerful discharger, I suggest scavenging appliance heater elements and parallel ‘em to your hearts content until you achieve the current draw you desire.

 

[Hillhater]

No its not a short...its a 25 amp load...but will give one big spark.
..but a good reason for using less current for the test

 

[izeman]

No its not a short...its a 25 amp load...but will give one big spark.
..but a good reason for using less current for the test

ok. so if a spark is to EXPECT, then i'm fine with it, because this is what happened when i tried it some time ago i will need a inrush limiter (2x IRFB 3077 in parallel) to switch the load on/off. so there will we no spark then. if i remember correctly there where some major sparks with the 6 light bulbs (2s3p with 12V100W bulbs). they worked great but the light was SO bright that it was not really useable

 

[izeman]

But if you really want a powerful discharger, I suggest scavenging appliance heater elements and parallel ‘em to your hearts content until you achieve the current draw you desire.

that's easier if you live in 110V AC countries. with 220V here you need a lot more devices to get some decent load. you need a lot of heaters, hairdryers or what ever. and it takes a lot of room. the 5x 100W resistors fit onto this one heat sink.

and i did see some difference dependent on the load. if you count mW, and your device has some percentage wrong reading, the lower the reading, the more prone it is to some rounding errors. and a weak cell will be much easier to identify with a high load. if you discharge the battery at 1C it all may look good, but it will be a different picture at 5C or even more.

i now know pretty well which ones the bad cells in the packs are. they can be identified quite easily on the charge/discharge curves logged by logview or dataexplorer. and it once again proved the "let 'em sit fully charged on a shelf for a day or two" theory. the very same cells that performed bad on the charge/discharge test showed significant voltage loss during the "shelf test". all cells charged to 4.20V, and after 2 days, 5 are at 4.18V and one is at 4.14V or so. pretty easy. after the test the pack is discharged to 3.85V again, and before repairing the cells they are discharged to 3.00V, as they are pretty harmless at that stage - in case something goes wrong during soldering.

 

[Ianhill]

Nice little read, I guess that why my Icharger is not that accurate at IR readings, as it only has inbuilt 30w discharge not enough to test larger cells its OK for single 18650 but thats about it, when I test my lipos I get between 3-5mohm and it fluxates guess I need more oomph.

 

[izeman]

Nice little read, I guess that why my Icharger is not that accurate at IR readings, as it only has inbuilt 30w discharge not enough to test larger cells its OK for single 18650 but thats about it, when I test my lipos I get between 3-5mohm and it fluxates guess I need more oomph.

i guess that's the reason why this reading is so super inaccurate. measuring IR was why i bought the icharger, but i should have read forums before where they said that it was a feature you can't really trust.
every time i measure i get different values. it can be as bad, as the worst cells show the lowest resistance. useless.
and i do this measurement off all my packs maybe once a year to sort out the ONE cell of the pack that limits the whole big pack's capacity. and i want to be 100% that i replace all cells that need replacement, and none that are still good.

 

[izeman]

i just finished the project, and i must say it work exceptionally well
for the first test i took a 4-year old 40% charged 6s lipo and had the window open in case something bad would happen.
the little 2 3077 IRFB FET inrush limiter does a great job. a small microswitch to turn it on, and automatic turn off when you let the swtich go. that way i must sit next to it to operate it.

i also made a little video to show the massive sag of this old 20c zippy pack at 5C. and you can clearly see from this little video what the bad cells are. there is no need to attach the icharger. but i will do it to get some real values instead of guessing, and something to feed my spread sheet with.






[youtube]kiOfEnyO1TY[/youtube]

 

[Ykick]

Very clean and professional. Nice work!

 

[izeman]

this is a screenshot of dataexplorer in monitor mode. only drawback is, that i can't monitor CURRENT. so i have to note that down from the inline current meter. but as i charge every battery to full 4.20V, and then do the test, the current is always the same. it's 6x 5 Ohm resistors in parallel now, so the current is around 30A.



as you can see sag is between 12mV and 14 mV. current is 30A. so IR between 4,1mOhm and 4.8mOhm per cell.
i will now do this with all my 8x 6s and the compare the results in the spread sheet.

this is the data from the spreadsheet:

 

[Ianhill]

Thats really nice how are you pulling the data from it ?

 

[izeman]

Thats really nice how are you pulling the data from it ?

thanks. the icharger is in MONITOR mode and then connected to my pc running DATAEXPLORER (java app for M$/linux/MAC) or LOGVIEW (M$ only).

 

[izeman]

the discharge works really nice as well. the resitors get REALLY hot. they reach 150C / 300F at the surface. the heatsink is around 100C / 200F. but they have no problem with that.
a single 6s 5000mAh pack is empty in under 10min.

 

[DAND214]

the discharge works really nice as well. the resitors get REALLY hot. they reach 150C / 300F at the surface. the heatsink is around 100C / 200F. but they have no problem with that.
a single 6s 5000mAh pack is empty in under 10min.

SO, is the pack 5000mah or?

Dan

 

[Ianhill]

Thats really nice how are you pulling the data from it ?

thanks. the icharger is in MONITOR mode and then connected to my pc running DATAEXPLORER (java app for M$/linux/MAC) or LOGVIEW (M$ only).

Silly me I couldn't pull data off my icharger to laptop it wouldn't recognize it maybe it was because I did not use the monitor mode I had the drivers installed and the mini disc it came with etc but I'm far from computer wizz.
Does the resistance change with the high temperature reached in the resistors ? don't it scew the results at all.

 

[Hillhater]

Neat test set up.
But normally you would want the voltage (and current) under load to stabilise to a reasonably constant value befor you take readings for calculating IR.
However, your voltages are still changing, which would suggest either not enough time before you stopped the load , or the cells cannot hold a stable voltage with that 30Amp load.
Also, with a resting voltage up at 4.1-4.2 v, you are not going to see a "normal" discharge performance as the voltage drops rapidly before a normal dischage rate is established below 4.0 v
Have you tried repeating these results at different states of charge ?.... IR does vary noticeably with SOC.
Did you see this thread... https://endless-sphere.com/forums/viewtopic.php?f=14&t=73701&p=1168548&hilit=Dcir#p1168548

 

[izeman]

hillhater, i hear you, and you're absolutely right. i know that sag is highest for a fully charged battery. and this is why i chose to do it with fully charged batteries. just to cleary see which cell sags the most. sag at 4.00V is less, and cells behave more evenly.
as is said: those cells that have been identified as "less than optimal" cells all show the same signs:

.) they have the biggest sag at any voltage
.) they store less energy
.) when charged their voltage raise is faster than for the other cells
.) they loose more voltage faster when sitting on the shelf

what i needed was some method to quantify that. so some numbers for a spread sheet to be able to compare absolut numbers. and this was accomplished by the resistors connected to the icharger.

what i don't agree with is, that the cells haven't reached a stable voltage while doing the test. they DID. what you see in the graphic is the normal discharge of the cells. a single 5000mAh cell discharged at 30A will be completely empty in less then 10min. so even with a linear voltage decrease from 4.20 to 3.00 or 1.200mV in 600s this is 2mV/s. what you see here is around 4mV/s. and as you said: sag will be less with lower voltage.
but it will never stabalize. you can clearly see the cliff when 30A load is engaged, and this is what we need to calculate IR. please correct me if i'm wrong.