Internal Resistance

[electrobent]

I have two battery space 39.6V NiMH packs. One never was able to charge continuously even at .9A at above about 65F which it normally is here. I can put the pack in the freezer and then charge it right up from where it was when it over heated. After having problems with discharging the pack as well as charging it, I ordered a second pack and it immediately took the charge fully with no over heating problems. Running the packs in parallel has worked well for me even though they have slightly different fully charged voltages but I have been wondering why they are different.

I work at an engineering college and they sometimes have some neat trash. There was an older shaker from a chem lab out where they put things they want to disappear and I could see that it had a monster 50 Ohm power resistor and pretty big AC motor. Well the monster resistor is actually 51.2 Ohms according to my meter but I used it as a load for my batteries.

I measured the voltage with no load, put the load on an measured again. I used this second voltage and the 51.2 to calculate Current from V = IR.

I then plugged the resultant I value (Current) and the two voltages into Ri = ((V0 - VL)/I) and got .021 ohms per cell for my problematic set and .007 for the non-problematic one. I dug into the specs for the Batteryspace.com D cells of which my packs are allegedly comprised and see that they are supposed to be .007 or less so I feel pretty good about the one set but now am wondering if it could be one or two bad cells or if the whole pack is compromised .

Does anyone have any experience with this sort of thing or know anyway of determining this without taking the whole pack apart?

Thanks!

 

[Link]

Other than feeling to see if there are a few hot spots as opposed to the whole pack being warm, there's no real way to tell without taking the pack apart. Not that I can think of, anyway. :?

 

[electrobent]

good idea--I can zap it with my Harbor Freight Infrared Thermometer. I think only 6 are surrounded.

Thanks!

 

[olaf-lampe]

A common problem of batterypacks is bad contacts between cells. (Spotwelded on a hangover monday morning)

Also a singe bad cell can ruin the whole performance.

Take the 50R load and measure the single cell voltages if possible.

-Olaf

 

[deardancer3]

There are several potential problems inside your pack, all of which need you to take the pack apart. (remove covers and pack shrink rap and construction paper)

Me and a couple other folks repair NIMH packs for other people. Some problems we often see are: Last cells in the string, where the output +and - wires attach, are connected inapropriately. Like using a slow a soldering iron and over heating the cells. so the end cells are the most commonly failing cells from certain manufacturers.

bad/loose/marginal cell interconnects.

Bad cells that are not up to the other's performance wise, dragging the pack down.

To take the pack apart takes some safety precautions, they are posted around in several places, please find them and use them. (basically no metal jewlry, or even glasses, no metal surfaces, no metal tools(use plastic knives and scissors) and a second person to help if you can.

cells on the outside of the pack sometimes have gotten some trauma, cells on the inside sometimes run just a bit hotter than the ones on the outside. both cause premature failures.

after inspecting for obvious wiring and mechanical faults, and cells that look suspicious,
test pack with no and full load and monitor individual cell voltage. discharge pack about 50-75% of rated capacity and check cell voltages under load.

order identical or slightly higher AH cells with DUAL tabs attached for replacing the problems you see. order extra. charge new cell to pack charge level before inserting in pack. float/balance charge ~1/30c after full charge for 6 hours to rebalance. (do this monthly)

carfully reassemble after bench testing passes.

There are some pretty good folks around that might also give you some more tips.

best

d

 

[electrobent]

Thanks Dear Dancer:

I have just a couple of questions:

order identical or slightly higher AH cells with DUAL tabs attached for replacing the problems you see. order extra. charge new cell to pack charge level before inserting in pack. float/balance charge ~1/30c after full charge for 6 hours to rebalance. (do this monthly)

By dual tabs you mean one on each end or two on each end?

And how does one do a "float/balance charge to rebalance if they are all connected. Since min are 10 Ah this would mean a 1/3 Amp charge through the whole pack for six hours?

I see I am going to need a more sophisticated charger.

And what about providing some distance between the cells when I rebuild the pack? Would that help with the cells in the middle getting baked?

 

[deardancer3]

Yes, solderable tabs on both ends of the cell. resoldering will be tabs to tab, not tab to cell.

you might be able to use your present charger, just insert an appropriate resistor in the + line.

or use a 50 volt power supply with appropraite resistor to generate 100- 300 miliamps.

there are other steps to take, but this is the basics

without knowing hour your pack is put together, hard to advise. one row of middle dellls is ok, but 4 rows is troublesome.

d

 

[monster]

how is your pack arranged?

when i had a problem with a pack i discharged it and then made wee little holes in the insulation to probe with my meter to check individual cell voltages. no need to disassemble for diagnosis! the bad ones should show up as 0 volts.

 

[ZapPat]

I measured the voltage with no load, put the load on an measured again. I used this second voltage and the 51.2 to calculate Current from V = IR.

I then plugged the resultant I value (Current) and the two voltages into Ri = ((V0 - VL)/I) and got .021 ohms per cell for my problematic set and .007 for the non-problematic one.

While this method will give you a good idea of the internal resistance, I think it would be more precise if you took current and voltage readings for 2 different loads, and avoid using the floating (no load) voltage in your equation. So it would be:

Ri = (V1 - V2)/(I1-I2)

 

[electrobent]

Time to hunt down another power resister . . .