Eneloop Specifications

[MrAl]

Hello there,



With a little investigating, i found the following data for Eneloop
AAA (here) and AA (scroll down) cells:



AAA size cells:

1. They are basically slightly less than 800mAh at 160ma, and the
capacity comes down to about 725mAh at 1600ma of current. This
is with an ending voltage of 1.0 volts. With 1.1 volts, the rating
is between about 675mAh and 730mAh.
Discharge temperature range is given as 0 to 50 degrees C.

2. With a charge of 800ma, the cells exibit a minus delta V of 10mv.
The charge time is slightly less than 1 hour, which means 800mAh
and the other ratings are very conservative and have to be less
than advertised when charging this way. In other words, you cant
charge a cell with 800ma in one hour and expect to get 800mAh out of it.
Another way to charge is 80ma for 16 hours.
Charge temperature range is given as 0 to 40 degrees C.

3. Long term storage should be between -20 and 30 degrees C.

4. Internal impedance after discharge to 1.0v is roughly 40m ohms.

5. The cells must be charged and discharged a "few" times to get the
full capacity.




The following specs are for AA size cells:

1. They are basically slightly less than 2000mAh at 400ma, and the
capacity comes down to about 1800mAh at 4000ma of current. This
is with an ending voltage of 1.0 volts. With 1.1 volts, the rating
is between about 1600mAh and 1950mAh.
Discharge temperature range is given as 0 to 50 degrees C.

2. With a charge of 2000ma, the cells exibit a minus delta V of 10mv.
The charge time is slightly less than 1 hour, which means 2000mAh
and the other ratings are very conservative and have to be less
than advertised when charging this way. In other words, you cant
charge a cell with 2000ma in one hour and expect to get 2000mAh out of it.
Another way to charge is 200ma for 16 hours.
Charge temperature range is given as 0 to 40 degrees C.

3. Long term storage should be between -20 and 30 degrees C.

4. Internal impedance after discharge to 1.0v is roughly 25m ohms.

5. The cells must be charged and discharged a "few" times to get the
full capacity.

 

[Meterman]

Here you find the Specifications and Typical Characteristics of the Eneloop 2000 as an example.

My point of view:

In the Charge and Discharge graph you can clearly see, that the over and over again mentioned "mAh" (charged or discharged) in reality are of only minor relevance. Only current in combination with voltage is, what counts: "mWh"!

You can see this most obviously at the extremes: if you charge at a temperature of 0°C, voltage is always above 1.5V, so 1mAh charged is 1.5mWh to nearly 1.7mWh for the cell.

On the other hand: when discharging at 4000mA, the voltage is nearly all the time well below 1.2V, down to even 1.0V, so 1mAh only draws 1.2mWh (apart from the very first moments) to tiny 1.0mWh out of the cell.

Considering this, I find it not impossible to (depending on the circumstances) draw more mAh - not mWh - out of a battery than I've charged into it before.

By the way: your provider of electricity bills kWh and not kAh!

Wulf

 

[TorchBoy]

Considering this, I find it not impossible to (depending on the circumstances) draw more mAh - not mWh - out of a battery than I've charged into it before.

Hm... yeah. Sounds quite conceivable. It's not something I've considered before - I learn something new every day, and often on CPF.

 

[MrAl]

Hi again,


The main intent with my post was to provide a short form of reference for how
the Eneloop cells might behave and how they might be used.

With all the NiMh AA cells i have tested i've never once gotten more mAh out of a
cell than was put into it, provided of course that it was totally depleted before the
start of charge. The reason for this is because there are inefficiencies that prevent
all of the charge inserted from being stored and so some of it dissipates as heat.
Apparently this is enough to prevent getting 800mAh out of a cell that was charged
(at the normal room temperature that most people will be charging at) with 800ma
for one hour.
The reader should refer to the term 'Charge Acceptance" to find out more about
the inefficiency while charging a cell.

 

[Meterman]

I know there are complications caused by heat and inner resistance.

To simply verify my consideration, I'll set up the following test.

Preparation: A battery Eneloop 2000 is discharged to 1.00V, connected to an Electronic Load set to constant current 400mA.

Then the loop starts:

1) The battery is connected to an R/C Loader, settings "charge NiMH" at 200mA, charge stopped by limit set to counted 1000mAh, not anything else.

2) The battery is connected to the Electronic Load, discharged at 400mA constant current down to 1.00V.

3) Back to #1.

Charge and discharge currents are the values used for determination of the capacity by the manufacturer.

I myself wonder what happens. If this doesn't prove my expectation, loops with different setups will be necessary until I'm right.

Any suggestion for improvement?


Wulf

 

[eluminator]

The following specs are for AA size cells:
Internal impedance after discharge to 1.0v is roughly 25m ohms.

Interesting. I've always wondered if my method of measuring internal resistance gives the correct amount. Apparently not, although I can discriminate between good and bad cells.

I get something like 35 milliohms when a good cell is fully charged. I get around 85 milliohms when it has been discharged.

 

[Mr Happy]

I think there is no such thing as a single "correct" internal resistance measurement. It must always be stated together with the method used to measure it, and the value depends on the measurement technique.

The value that matters in practice is the one that corresponds to the way that you actually use the cell, for example the voltage drop seen on applying a DC load. When I measure Eneloops this way, I get a value in the region of 40-50 mΩ, so that is the correct value for my application.

 

[eluminator]

Yeah, I guess you can get any value you want depending on the measurement technique.

In any case, it seems to me you have to measure the voltage on the cell rather than the voltage at the cell holder. Well actually I do measure the voltage at the cell holder but I also measure the contact resistance and take that into account.

I find I have to go through a cleaning routine to get the contact resistance to a low and constant value. Otherwise it's easy to get a contact resistance that is greater than the cell internal resistance. Maybe it's not the cell contacts but the contacts of the cell holders I'm using. I haven't figured that out yet.

 

[MrAl]

Hi again,


eluminator:
Dont feel bad, all of my tests always show something like that too.
The 25m ohms is measured with an AC signal source, and the frequency
is set to 1kHz. I never get this low when i do my tests however
so i dont know how they come up with this number.

Mr Happy:
I agree about the internal R measurement, but the 'accepted' method
is to use a 1kHz AC signal source. As i was saying, i never get
as low as 25m ohms myself.

Meterman:
I guess i dont understand you here or something. It appears that
you are saying that when you charge a cell at say 2v and 1 amp for
1 hour (1000mAh total applied) that you can somehow discharge it
for 2 hours at 1 amp (2000mAh supplied) if the voltage happens to
be 1v rather than 2v. Is this correct?

 

[Mr Happy]

If you look at the voltage/time profile when charging and discharging a cell, the charging voltage is higher than the discharging voltage.

So for instance, if you charge an Eneloop at 1000 mA for 110 minutes, that would be 1830 mAh. If the average voltage across the cell terminals over the charging period is 1.4 V, the total energy supplied would be 9.2 kJ. (I've done this experiment and 1.4 V is about right in round numbers.)

If you now discharge the cell at a lower current like 500 mA down to 1 V, the capacity might be 1900 mAh with an average discharge voltage of 1.2 V. This would be a total energy of 8.2 kJ.

So although the mAh might be higher, the energy is lower and there is no violation of physical laws.

Whether this actually happens in a practical experiment is open to testing, and depends on the charge acceptance and charging losses due to heat in a particular cell. But in the case of good quality cells like Eneloops, it is certainly possible that it might be possible to get more mAh out than in. The readings on my C9000 suggest it is a very close call, with near equality in many of my tests.

 

[Meterman]

Mr Happy,
I'm very pleased to see, that you have thoroughly studied and understood what I'm speaking about. Thank you!


MrAl,
of course I do not only think of what you are saying with your given rates. That is never possible because of inevitable losses.
But I think there could be more mAh drawn at the lower discharge voltage than were charged at the higher charge voltage before.

To be as precise as possible in my measurements (I love highest possible precision when measuring) I at first have to build a better battery adapter, as I can't find a really good one in the market. The only one I had hoped to be acceptable showed "Continental US Orders Only" when I wanted to order it. And since my stay in the stroke unit last year all handicraft is a bit slower thanks to less sensitivity in my right hand.

Wulf

 

[UnknownVT]

I think there is no such thing as a single "correct" internal resistance measurement. It must always be stated together with the method used to measure it, and the value depends on the measurement technique.
When I measure Eneloops this way, I get a value in the region of 40-50 mΩ, so that is the correct value for my application.

I did some fixed resistor measurements of voltage and current draw of an eneloop (as well as a Kodak Pre-Charged) AA - I can get 0.5, 1 and 2 ohms vlues - with these I could calculate a number of internal resistance(s) I did this for near fully charged and about 1/2 charged -

Posts #42, #44, #46, #47 on page 2 of thread -
eneloop vs. Kodak Pre-Charged Voltage Maintenance
(long thread summary in Post #50 )
There were 2 ways of calculating the internal resistance see Mr Happy's posts #42, #45 and my post #44.

near full charge -
1 and 2 ohm readings -
ene Rb = (1.400-1.388)/(0.67-1.25) = -0.021 ohms
KPC Rb = (1.360-1.336)/(0.65-1.22) = -0.042 ohms

0.5 and 2 ohm readings
ene Rb = (1.400-1.328)/(2.26-0.67) = 0.045 ohms
KPC Rb = (1.360-1.264)/(2.16-0.65) = 0.064 ohms

0.5 and 1 ohm readings
ene Rb = (1.388-1.328)/(2.26-1.25) = 0.059 ohms
KPC Rb = (1.336-1.264)/(2.16-1.22) = 0.077 ohms
------------
2 ohm readings -
ene Rb = (1.421-1.400)/0.67 = 0.031 ohms
KPC Rb = (1.390-1.360)/0.65 = 0.046 ohms

1 ohm readings (calculated by Mr Happy)
Eneloop = (1.440 - 1.388) / 1.25 = 0.0416 ohms
Kodak = (1.412 - 1.336) / 1.22 = 0.0623 ohms

0.5 ohm readings -
ene Rb = (1.421-1.328)/2.26 = 0.041 ohms
KPC Rb = (1.390-1.264)/2.16 = 0.058 ohms


Approx 1/2 discharged -
1 & 2 ohm -
ene Rb= (1.266-1.247)/(1.17-0.60) = 0.033 ohms
KPC Rb= (1.243-1.220)/(1.14-0.59) = 0.042 ohms

0.5 & 2 ohms -
ene Rb= (1.266-1.224)/(2.09-0.60) = 0.028 ohms
KPC Rb= (1.243-1.176)/(2.01-0.59) = 0.047 ohms

0.5 & 1 ohm -
ene Rb= (1.247-1.224)/(2.09-1.17) = 0.025 ohm
KPC Rb= (1.220-1.176)/(2.01-1.14) = 0.051 ohm
---------
2 ohms load -
ene Rb = (1.294-1.266)/0.60 = 0.047 ohms
KPC Rb = (1.267-1.243)/0.59 = 0.041 ohms

1 ohm -
ene Rb = (1.294-1.247)/1.17 = 0.040 ohms
KPC Rb = (1.267-1.220)/1.14 = 0.041 ohms

0.5 ohm
ene Rb = (1.294-1.224)/2.09 = 0.033 ohms
KPC Rb = (1.267-1.176)/2.01 = 0.045 ohms

 

[Meterman]

Battery holder became ready yesterday night.

As the planned charge/discharge business is fairly time consuming, I then at first only started a series of measuring the inner resistance of a charged eneloop AA at different currents using my electronic load. I checked from 100ma to 1500ma in steps of 100mA, from then on in steps of 500ma up to 4500mA. At 5000mA, the limit of the device, the latter crashed and now must be sent to the manufacturer. :mecry: So the charge/discharge test must be delayed, sorry.

Due to the low resolution for the calculation of only 2 decimals, at low currents the values of Ri (between 33 and 50mΩ) are not as precise as I had expected, caused by the tiny voltage differences under low loads.

Over 1000mA loaded, Ri values were in the range of 42 to 47mΩ, this seems to be realistic.

I habe phoned with the manufacturer in the hope to have him change the calculation to three decimals, as the device itself uses this resolution. (Measurements of U and I are done by the eLoad, values are sent to the PC and calculations of course performed there.)

Wulf

 

[MrAl]

MrAl,
of course I do not only think of what you are saying with your given rates. That is never possible because of inevitable losses.
But I think there could be more mAh drawn at the lower discharge voltage than were charged at the higher charge voltage before.

To be as precise as possible in my measurements (I love highest possible precision when measuring) I at first have to build a better battery adapter, as I can't find a really good one in the market. The only one I had hoped to be acceptable showed "Continental US Orders Only" when I wanted to order it. And since my stay in the stroke unit last year all handicraft is a bit slower thanks to less sensitivity in my right hand.

Wulf

Hi there Meterman and MrHappy,

First off, Wulf, i am happy to hear that you may build a tester for this. I think that
would be very interesting to see...keep me posted about this ok? Thanks.

Meterman and MrHappy:
As to getting more mAh out of a battery than was put it, i think that is going to
be very difficult. The reasoning that the Whr put in can be measured by measuring
the cells terminal voltage and the Whr taken out can be measured by measuring
the terminal voltage isnt exactly correct either, because the internal resistance
is in series with the actual 'cell' when it is being charged AND when it is being
discharged. This raises the voltage a bit during charge and causes a drop in
voltage during discharge. The theoretical internal capacitor that makes up the
bulk of the cell wont change voltage as abruptly. Thus, the terminal voltage
does not represent the watt hours going in nor coming out.

Let me put this another way...
Consider the following:
I connect two cells in series, charge them (in series) with 1000mA for one hour
at 2.8 volts total across both cells. That's 1000mAh into the two cells.
Does this means if i now connect a load that draws the voltage down to 1.2
to 1.4 volts that i can extract 2000mAh from the two cells (still in series)?
If i *disconnect* the two cells and connect them in parallel, *then* i can
get 2000mAh (a little less) from the two cells, but then what i have done there
constitutes a true power converter, and i dont think you can get this kind of thing
to happen with only one single cell. Even regular converters change the circuit
topology (via transistors) when there is to be power conversion.

This in no way nulls out your (Meterman) idea to build a tester of course, as i think that
would still be very interesting and informative and i would like to see this happen.
If you think i can help in some way just let me know too.

More thoughts?

 

[Mr Happy]

More thoughts?

We are not saying you definitely can get more mAh out of a cell than you put in, just that the laws of physics do not apparently forbid it.

The power entering a cell truly is given by the voltage difference times the current, internal resistance notwithstanding. Not all of the power is stored of course; some of the power is dissipated as heat due to resistance losses and other inefficiencies. Exactly how much of the power is lost is what we don't know without experiment.

For the best chance of getting more mAh out than in, the cell would need to be operated in the most efficient manner. That would probably mean finishing the charge at about 80%, before the cell got warm and the -dV signal approached. It would also mean discharging at a low current to minimize losses.

 

[TorchBoy]

... the internal resistance
is in series with the actual 'cell' ...

Just conceptually, though. As far as we are concerned, the cell is what's between the terminals, and includes the internal resistance, with all the inefficiencies it might cause. Anything else I might say here I think has already been said by Mr Happy - got to be fast around here - except that I feel absolutely no inclination to try it. I'm quite happy leaving it as an unknown interesting possibility.

 

[MrAl]

Hi again,

I guess what i am trying to say is that if you put 10 electrons
into a box (and there were none to start with) you can not
pull out 11 electrons. That's just not possible.

 

[Mr Happy]

That's a good point MrAl. I'll have to think about that.

 

[Meterman]

First experiments seem to have proven my expectance. For example:

Charge: Input 1416mWh/1000mAh
Discharge: Output 1313mWh/1070mAh

But there are still quite a few refinements necessary to harden the results in a manner that even those are convinced who have not yet understood the underlying thoughts.

I'd best like to have a few charge/discharge cycles in a single non-stop diagram. But then there'll be another problem: up to now I may not post attachments.

Everybody owning an R/C charger with needed values displayed (or better stored) is invited to do the same experiment.
(Might be even a Powerex MH-C9000 could be used for a test, although I prefer more precise measurements.)

Discharge a battery, charge it up to only half it's capacity (important is, that temperature does not rise at all - that would cause losses) and then discharge the battery. Compare the mAh and, if shown, the mWh of the charge and the second discharge. It's a good practice to put down accompanying parameters for comparison. Good luck!

Wulf

 

[Meterman]

Hi again,

I guess what i am trying to say is that if you put 10 electrons
into a box (and there were none to start with) you can not
pull out 11 electrons. That's just not possible.

Hi MrAl,

can we agree upon counting the amount of electrons put into the box or pulled off the box in mWh? So we'll both come to the same correct results.

You can not move the electrons in or out without the help of tension, otherwise they are too lazy and stay where they are. Thus not only current is what counts, but voltage too, and we come to power in(m)W and from there depending on the time to (m)Wh.

Wulf