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NiMH battery pack mystery

HANKMARS

Jul 28, 2019
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I use 6 - AA NiMH cells together in a 6 x AA battery holder. My application typically uses a timed on state drawing 50mA to 2A. In this application, I have sufficient power to drive the device for 90 to 120 minutes. These times should improve as I refine my circuit values. Each cell has a rating of 2800mAHrs. That rating seems accurate. A dead short across the output of this 6 pack can momentarily deliver around 11 amps. Using an off the shelf charging module, which is designed to charge NiMH and Li batteries, and charging the 6 cells as a single unit, the 6 pack battery delivers 7.xx amps. If I disconnect battery from charger and provide a rest period and recharge, my available current out does increase. The highest charge is reached if I charge each cell individually but that is impractical with this application.Thinking that I could achieve a greater use time from my device if I use 8 cells, I loaded up an 8 x AA battery holder. Even if I charge each cell independently my maximum current out is 6 - 7 amps. If I charge the 8 pack as a unit, my max current (using a dead short across my meter or driving a 1 ohm load ) is 3.5 amps. I have ohmed out each spring connector ( which connects each battery to the next ) and have found no discrepancies. A milliohm meter may give me more accurate readings but the best I can tell is that each connector internal to the 6 and 8 cell holder units, are nearly identical. I have mixed and matched my NiMH cells. I have tried multiple battery holders. I have independently charged each cell to maximum capacity but this discrepancy in max current out still lives. Why? Somebody blurted out "Threshold of material," which I think I have a vague understanding of but from a quite different application. I am inclined to believe that trying to push 11 amps thru an extra 2 cells might not be achievable at the voltage of 8 NiMH cells ( typically 1.2 V per cell ). I am also inclined to believe that I may be overlooking some obvious characteristic. Any thots or info that you can provide. My next stop will be battery theory but even a logical starting point would be much appreciated.
 

kellys_eye

Jun 25, 2010
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Test each cell individually for max current - with them connected in series your max current flow will be equal to the 'worst' cell. If you want more current then you need to series/parallel them.

Max current flow is a function of the cells internal resistance and connecting them in series only exacerbates the problem.
 

Bluejets

Oct 5, 2014
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The batteryuniversity site has a lot of detail on all types of cells.
Might be a good stop over and do some study.
To begin with, many of these cells are overrated in output capacity if (most likely) they come from China.
Then any capacity may well be a 10 hour discharge rating or "C" rating as sometime referred to.
The output curve is not linear.
 

HANKMARS

Jul 28, 2019
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The batteryuniversity site has a lot of detail on all types of cells.
Might be a good stop over and do some study.
To begin with, many of these cells are overrated in output capacity if (most likely) they come from China.
Then any capacity may well be a 10 hour discharge rating or "C" rating as sometime referred to.
The output curve is not linear.
I most likely will make a stop at battery university. The current output curve of NiMH cells is nearly flat in relation to V out.
 

HANKMARS

Jul 28, 2019
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Test each cell individually for max current - with them connected in series your max current flow will be equal to the 'worst' cell. If you want more current then you need to series/parallel them.

Max current flow is a function of the cells internal resistance and connecting them in series only exacerbates the problem.
When connecting cells in parallel, the V out will be at or nearly at the voltage of the lowest voltage cell. I currently do not believe that cells in series will drop their current flow to the lowest "charged" cell. If a cell is faulty because of a manufacturing error or because of damage, then yes, I believe the faulty cell may impede A out. I do believe that the internal resistance is key to my problem here.If that is the case, I have proven that 7.2V overcomes the internal resistance of 6 cells but that 9.6V is not enuf voltage to overcome the internal resistance of 8 cells. What I will do is to series connect 12 cells and see if they produce an expected current out of approximately 20A.
 

kellys_eye

Jun 25, 2010
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Imagine a series chain of resistors. The current flow is a function of the sum of the series resistances. Current flow can only increase if one (or more) of those series resistances is lowered but will always be a function of the HIGHEST resistance even if all the other resistances drop to zero. You can't 'force' the resistance lower....

Replace those resistors with batteries (and their equivalent internal series resistance) and you have the very same issue. the battery internal resistance will drop as a function of its charge status but only to a defined minimum resistance according to their chemistry and that 'minimum' determines the maximum flow when in a series circuit.

The current flow will always be proportional to the sum total series (internal) resistance which is why, if you want higher current, you connect batteries in parallel (as per paralleling resistors allows a greater current flow as a function of the reduced equivalent resistance).
 

HANKMARS

Jul 28, 2019
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Imagine a series chain of resistors. The current flow is a function of the sum of the series resistances. Current flow can only increase if one (or more) of those series resistances is lowered but will always be a function of the HIGHEST resistance even if all the other resistances drop to zero. You can't 'force' the resistance lower....

Replace those resistors with batteries (and their equivalent internal series resistance) and you have the very same issue. the battery internal resistance will drop as a function of its charge status but only to a defined minimum resistance according to their chemistry and that 'minimum' determines the maximum flow when in a series circuit.

The current flow will always be proportional to the sum total series (internal) resistance which is why, if you want higher current, you connect batteries in parallel (as per paralleling resistors allows a greater current flow as a function of the reduced equivalent resistance).
With this theory it seems that you could almost connect enuf cells in series to eventually have no current out. I did series connect 12 cells resulting in a current flow of 9+ amps across a 2 ohm resistance. Test lead resistance explains some loss in current. I recognize internal resistance in a cell but I am guessing that, for whatever reason, there is also a PN (diode) type of current blocking also happening here, too. The voltage to overcome this junction is not linear but more of a stair step form. The R2R schematic is not an exact analog of my theory but may help to convey my theory. Replace the Rs with a diode junction and you might get the point that I am working towards
I will peruse battery university and see if I can find characteristics related to this matter pertaining to NiMH cells.
 

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Harald Kapp

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With this theory it seems that you could almost connect enuf cells in series to eventually have no current out.
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.
 

HANKMARS

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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.
As the voltage increases across a fixed resistance, current will increase also. Using a 1 ohm resister, the voltage across the resister reads the same as the current value. 6 cells, 1 ohm load reads 10 amps. 8 cells, 1 ohm load reads 5 amps. What happened?
 

HANKMARS

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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.
R2R A TO D 1.3.png
 

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Harald Kapp

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6 cells, 1 ohm load reads 10 amps. 8 cells, 1 ohm load reads 5 amps. What happened?
10 A @ 1 Ω is equivalent to 100 W. Is the resistor rated for that much power?
When increasing the voltage (8 cells instead of 6 cells), one would expect current to rise proportionally. But at such high power dissipation the resistor will heat up and the resistance will increase , thus current will actually decrease.
Compare this to the current vs. voltage vs. time characteristic of a conventional tungsten light bulb: When cold, a high inrush current will flow, heating up the filament and thus decreasing over time.


As to post #10: What is the intention of this circuit diagram? It doesn't show cells in series. What it does show is a R-2R ladder network with R replaced by diodes? What is this for?
 

kellys_eye

Jun 25, 2010
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6 cells, 1 ohm load reads 10 amps. 8 cells, 1 ohm load reads 5 amps. What happened?
Yes, what happened - TO THE RESISTOR?

Measure the volt drop across the resistor while measuring the current flow.
 

HANKMARS

Jul 28, 2019
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Yes, what happened - TO THE RESISTOR?

Measure the volt drop across the resistor while measuring the current flow.
I do that and consider my voltage reading, across a 1 ohm resistor, to be the value of my current flow, per ohm's law.
 

HANKMARS

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Yes, what happened - TO THE RESISTOR?

Measure the volt drop across the resistor while measuring the current flow.
Oops. Misunderstood your instructions. I predict a significant drop in current due to my meter in series with circuit.
 

Harald Kapp

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I predict a significant drop in current due to my meter in series with circuit.
Nope. The voltmeter across the resistor is in parallel and will have a truly negligible influence on the current.
Unless I misinterpreted your reply?
 

HANKMARS

Jul 28, 2019
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10 A @ 1 Ω is equivalent to 100 W. Is the resistor rated for that much power?
When increasing the voltage (8 cells instead of 6 cells), one would expect current to rise proportionally. But at such high power dissipation the resistor will heat up and the resistance will increase , thus current will actually decrease.
Compare this to the current vs. voltage vs. time characteristic of a conventional tungsten light bulb: When cold, a high inrush current will flow, heating up the filament and thus decreasing over time.


As to post #10: What is the intention of this circuit diagram? It doesn't show cells in series. What it does show is a R-2R ladder network with R replaced by diodes? What is this for?
My 1 ohm resistor is rated at 100W. Aluminum body, Size of a baby carrot stick. Little bigger. Circuit diagram messy yes trying to convey idea that along with internal resistance, there may be a sort of barrier that has a bias voltage required to allow it to conduct. Let's say that a 1,5 voltage is created by the active components within the cell. Perhaps, for whatever reason, the physical barrier between cell components, requires 0.3 volts to allow current flow from pole to pole. Output V = 1.2V. As you series cells, barrier breakdown V virtually increases. 0.6V, 0.9V, 1.2V, 1.5V etc. I have not laid out the math so that notion my be moot. If it has any validity, I suspect the barrier breakdown voltage to be a much higher % of voltage produced internally. More like a 1V Vbd and 2.2V internal actual. A hyper biased barrier will conduct easilier than a barrier bias just barely above its Vbd.
 

HANKMARS

Jul 28, 2019
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Yes, what happened - TO THE RESISTOR?

Measure the volt drop across the resistor while measuring the current flow.
I read your instruction to mean for me to measure voltage across 1ohm WHILE simultaneously measuring current flow, with a meter, I presumed.
 

HANKMARS

Jul 28, 2019
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The question is. is 1 Ω still 1 Ω when 10 A are going through?
If temp is finely regulated, I say Yes. In this particular case, I can connect 1 Ω load across 6 cell batt, read V across resistor at 10.xxV. Remove 6 cell batt, connect 8 cell batt read 5.xxV, switch 6 cell back in and read 10.xxV. Just that quick. If resistive element internal heats to a temp of an incandescent bulb as rapidly as a light buld does, then that would explain some. I also do testing with a 2Ω load, 2 100W, with nearly identical results so for now I will disregard resistance variance due to temp change as low current cause.
 
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