NiMH battery pack mystery

[HANKMARS]

Not at all. As the resistance increases, so does voltage. And by Ohm's law I = V/R, the max. current is the same for any number of cells in series. Limited by the cell in worst condition.

The problem may still be the material or connections in the battery holders themselves. If I had 8 dummy batteries, I would be able to most likely observe a discrepancy in conductivity between holders, 6 cell and 8 cell, with my course meter reading. 200Ωs is the finest scale I have right now. I use to have a Fluke meter that had a 2Ω scale, which was nice for verifying small resistances. Also, I suspect V at battery terminal drops to 1 V or less with only 1Ω load. I will check that. Otherwise, how can I produce 10A from a max 7.2Vots?

 

[kellys_eye]

If the current flow changes (drops) between using the 6 cell pack and 8 cell pack then the 8 pack cell has a faulty cell in it.

You can only produce 10A from a 7.2V source if that source has a sufficiently low internal resistance to permit that level of current.

 

[HANKMARS]

If the current flow changes (drops) between using the 6 cell pack and 8 cell pack then the 8 pack cell has a faulty cell in it.

You can only produce 10A from a 7.2V source if that source has a sufficiently low internal resistance to permit that level of current.

I can take a 6 cell battery of fully charged cells and draw 10-11 amps. I can take another 6 cell battery of fully charged cells and draw 10-11 amps. Take any 8 of those cells, put them in my 8 x AA holder and I draw 5-6 amps max. ( 1Ω load ) Various 6 cell holders, various 8 cell holders. It may be caused by the 8 cell holders physical construction but currently can not prove it with equipment on hand. Open to procedures for experiments that would prove current loss due to holder.

 

[HANKMARS]

BIC lighter, a single 1Ω 100W resistor, 2 - 1Ω 100W resistors in series, a 6 x AA battery holder, an 8 x AA battery holder, and 4 big honkin - 0.39Ω resistors just for fun.

 

[HANKMARS]

That was not the point.
Output capacity decreases as one increases the discharge rate.

I am not sure I get your point. Output capacity decreases even if discharge rate is constant or even if discharge rate decreases. Any current flow produced by a battery is going to be decreasing available energy. The battery is discharging any time that it is in use. The available energy is decreasing ( charge depletion ) but of course battery capacity is mostly unaffected. Please clarify.

 

[bertus]

Hello,

Does this article help?
https://batteryuniversity.com/article/bu-501-basics-about-discharging

Bertus

 

[kellys_eye]

Got two 6 cell holders? Try two in series and see what you get.

Occams Razor - the 8 cell holder....

 

[Alec_t]

Open to procedures for experiments that would prove current loss due to holder.

Measure the individual cell voltages actually at their terminals while in the holder and maximum current is being drawn by your load. Does the sum of those voltages equal the single voltage across the cell pack ends? If not, the interconnections within the holder have a high resistance.

 

[HANKMARS]

Hello,

Does this article help?
https://batteryuniversity.com/article/bu-501-basics-about-discharging

Bertus

Interesting article. I keep my Amax test as short as possible. Approximately 1 second. Seldom over 2 seconds unless need be. Good idea to have a percentage buffer on charge and discharge ends. Avoid undue stress.

 

[HANKMARS]

Measure the individual cell voltages actually at their terminals while in the holder and maximum current is being drawn by your load. Does the sum of those voltages equal the single voltage across the cell pack ends? If not, the interconnections within the holder have a high resistance.

This is the direction that my conclusion is headed. This is a difficult or rather involved test. Having only generic VOMs on hand, moving around to the various test points takes way longer then I am willing to short ( 1Ω ) the battery. This would be a definitive test, however. Great idea. I may well check ebay for a used or inexpensive milliohm meter..Or maybe it is called a micromo meter. Even with that meter checking each internal connector, I could easily miss a bad connection.

 

[HANKMARS]

Measure the individual cell voltages actually at their terminals while in the holder and maximum current is being drawn by your load. Does the sum of those voltages equal the single voltage across the cell pack ends? If not, the interconnections within the holder have a high resistance.

I can do this test at less than Amax, say maybe 1 or 2 amps. May very well give good indication.

 

[HANKMARS]

Got two 6 cell holders? Try two in series and see what you get.

Occams Razor - the 8 cell holder....

Two 6 cell batteries in series gave me an expected Amax out. ~ 20A. I am leaning towards the idea that the 8 cell holder may be the culprit.

 

[Bluejets]

Output capacity decreases even if discharge rate is constant

Not unless the cells are defective.
Point was, if you read any of the batteryuniversity data on NiMh or Nicads or whatever you would see the capacity is rated at a certain discharge.
This may be called C10 or C5 whatever the manufacturer decides. (C5 = 5 hours)
So if a cell is has a certain capacity at C5, and you discharge in 2 hours, the capacity is now a lot less.
This new capacity now depends on the discharge curve from the manufacturer.

 

[HANKMARS]

Not unless the cells are defective.
Point was, if you read any of the batteryuniversity data on NiMh or Nicads or whatever you would see the capacity is rated at a certain discharge.
This may be called C10 or C5 whatever the manufacturer decides. (C5 = 5 hours)
So if a cell is has a certain capacity at C5, and you discharge in 2 hours, the capacity is now a lot less.
This new capacity now depends on the discharge curve from the manufacturer.

This particular thread is specifically concerned with NiMH cells. Are you using the term "capacity" to mean battery capacity as well as % capacity, or do you mean that the amount of energy storage of a cell is affected by its discharge rate?

 

[Bluejets]

I gave you the link, go read it.

 

[HANKMARS]

I gave you the link, go read it.

Dear Bluejets, I scrolled thru this thread 3 times finding no link sent by you. You mentioned a site name of battery university but "link" and "site name" are not interchangeable terms. Thanks anyway.

 

[Bluejets]

Some people still cannot use Google........

https://batteryuniversity.com

 

[HANKMARS]

Not unless the cells are defective.
Point was, if you read any of the batteryuniversity data on NiMh or Nicads or whatever you would see the capacity is rated at a certain discharge.
This may be called C10 or C5 whatever the manufacturer decides. (C5 = 5 hours)
So if a cell is has a certain capacity at C5, and you discharge in 2 hours, the capacity is now a lot less.
This new capacity now depends on the discharge curve from the manufacturer.

Two "completed" projects where "definition of terms" was of no concern.

 

[Harald Kapp]

Good technical data on battery holders seems to be hard to find. I found one for an 8 cell AA holder similar to the one shown in post #24 where a peak current of 2 A ist stated! Far from the more than 8 A you want to draw.
The springs in these holders are made from steel (not the best electrical conductor) with a thin coating of nickel. Feeding high current through these springs will dissipate a lot of energy.
This discussion supports that theory. @Alec_t suspected this phenomenon in post #28 but you never came back to to his request.

For high current you will need special high current battery holders, potentially not easy to acquire.
Better yet: use a spot welder and nickel strips and build a battery pack.

 

[HANKMARS]

Good technical data on battery holders seems to be hard to find. I found one for an 8 cell AA holder similar to the one shown in post #24 where a peak current of 2 A ist stated! Far from the more than 8 A you want to draw.
The springs in these holders are made from steel (not the best electrical conductor) with a thin coating of nickel. Feeding high current through these springs will dissipate a lot of energy.
This discussion supports that theory. @Alec_t suspected this phenomenon in post #28 but you never came back to to his request.

For high current you will need special high current battery holders, potentially not easy to acquire.
Better yet: use a spot welder and nickel strips and build a battery pack.

I did send a reply to Alec t in post #30 and post #31. He recommended a test procedure that could very well pin point this problem. I now have 2 projects affected by this problem. One project uses a current of only 0.5 to 2A. The other project needs short bursts of high current to rapidly heat a tiny coil of nichrome wire to ~ 1000 degrees F. I was opting to go from 6 cells to 8 cells in order to have longer operating times. When the nichrome coil stood alone and not in its fixture, 6 cells brought it to temperature in ~ 2 seconds. The 8 cell pack resulted in a considerably slower elevation of temperature and was unable to bring the coil to even a cherry red color. Again with the coil unencumbered by the heat sinking effect of its fixture, 6 cells easily and rapidly took the coil to a bright orange color and probably beyond if I allowed continuous current flow. If I were to try to drain the batteries at this high rate of current, my battery holder would be destroyed.