Losing 0.255Ah somewhere!

[PB]

I put my Watt meter between the battery charger and the battery being charged and it shows me the current flowing into the battery, the battery voltage, the Wattage the battery charger is supplying and the accumulated Ampere hours the charger is replacing in the battery.

Ok, so I have a 3s 3000mAh battery which shows 50% on the battery checker and I attach it to the charger via the Watt meter. When the battery charger beeps to tell me that the battery is fully charged I would expect the accumualted Ampere hour reading on the Watt meter to be 1.500Ah which is the 50% of the full charge that needed to be replaced, but it doesn't read that it reads only 1.245Ah. The battery checks 100% charged on the battery checker after charging so where is the missing 0.255Ah? Is it just down to the battery checker and Watmeter accuracy or am I missing something?

[Biggles' Elder Brother - Moderator]

Probably lots of factors at play here PB - not least of which the fact that the battery checker can only ever be an approximation - it can't measure the capacity left in the battery, it can only estimate it from the cell voltages. So "50%" could easily be "55%" in reality - and better for the checker to err on the safe side of underestimating the remaining capacity rather than overestimating it.

BEB

[Frank Skilbeck]

Plus over use, especially at the high currents we use, the batteries lose some capacity, I can see this on the telemetry, you can see the amps at max throttle dropping off later in the flight but the mah monitor is still saying that batteries should have more left, but comparing the the mah used from the telemetry and then with the mah put back in with the charger there isn't much difference.

Also the telemtry would be telling me that I have 25% left in the battery, but the battery checker would read somewhat less.

Based on your readings I would say your 3,000 mah battery is effectivelly a 2500 mah battery

[PB]

That's inetersting thanks. Yes, I suppose accuracy toleranaces across the various bits of kit in use could account for it, especially the battery checker. I hadn't looked at it in that way though Frank, that if 1.245Ah represents 'about' 50% of the battery capacity then it's 'about' a 2.5Ah battery, but I guess that must be right.

Thanks for your thoughts on it......

[Depron Daz]

I have several chargers and several cell checkers, and once a battery is charged on say my genuine IMAX B6, I can check it on the other chargers and checkers and they'll all tell different readings! I put it down to the cheap components used. As a general rule of thumb my B6 is what I go by, but then who's to say that that is accurate???

[Shaunie]

I suspect that the battery is down on capacity.

Also remember that both the charge and the discharge process are not 100% efficient, some enrgy is lost in both conversions.

The battery checker is only estimating remaining charge from the battery voltage and as battery chemistry will vary slightly from one manufacturer to another it is never going to be that accurate. The accumulated Ah reading will be closest.

As a trainee (Electronics Design) I was instructed "never give the user more than one way of measuring anything, as they will never match and the customer will pick you up on it" .

You can measure capacity 3 ways, energy in, energy out, or remaining capacity estimated by voltage and they will never be the same. Choose one and be consistent in using it is my advice.

Shaunie.

[Sparks]

As Shaunie says, the charging process is not 100% efficient, some of the 'energy' will be lost as heat.

Most 'hobby' battery checkers estimate battery capacity by simply measuring the open-circuit voltage, which varies with cell temperature and is not factored into the measurement results. So don't treat the readings from these battery checkers as gospel.

[Simon Chaddock]

And to add the final nail.

Just because its says 3000mAh on the label doesn't mean it actually ever was!

You can get better and worse batteries that are other wise identical.

Edited By Simon Chaddock on 25/10/2013 11:01:48

[Paul Marsh]

Also you get resistance in the wires. When current flows, resistance tries to slow the electrons down, causing friction - heat. This heat requires energy and some would be from your lost 225mah.

For example, one client was having a problem, as the landlord had a multimeter in his switch room, and the customer had his own meter in his building. Both were set the same, but over a year, it made out that he insisted that was paying too much. There was a slightly higher current usage upstream - in the switchroom, to what the customer had. Takes it that the customer was paying £21 per year more that he thought, but after losses in the cable underground (about 50m) both reading will never be the same, although very minute.

[Vecchio Austriaco]

The capacity of a battery is only valid if you "empty" the battery with a certain current. Normally this current is rather low - much lower than the current we use in our models. The rated capacity is normally based on a 20h discharge time. So I am sure if you have a few LEDs as a load on your battery the capacity (of a new battery) will be as written or even higher than on the label. if you take out for instance 25C it will be much less.

Who was flying electric 30 years ago will be very happy with the performance of todays batteries. So better to have a spare battery and change over early than to fight for the last mAh.

VA, happy LiPo user

[Shaunie]

VA,

Lead acids are rated at C20 (20 hour rate) as you say but I don't think any LiPo manufacturer gives details about how the Ah rating is measured.

Lead acids are rated this way as their performance drops off rapidly with increasing current, those of you running LiPo chargers from lead acids take note.

As already mentioned, how good was the battery in the first place? I think many cheap LiPo's would best be rated in "fibs".

Shaunie.

[Peter Beeney]

With the greatest respect, Paul, I’m afraid I’m entirely convinced about your synopsis. An ammeter is in series with the circuit to measure the current flow, and therefore it does’t matter where in the circuit you measure it, it will always read the same. The heat radiated from the wire doesn’t affect the current, the heating effect is caused because of it; and in the case of the battery, the heating effect caused by the charging will be equalled by that when it’s discharging anyway. Also, because the resistance of the wire is going to be low, the I squared R losses are going to be pretty low, too, as well.

Relating to your unbalanced supply, I’m assuming this to be mains electricity? If the power used was less, as measured, then this would point to some sort of leaky line, or much more likely, a neutral/earth fault, which would not normally be apparent until some other situation arises that disconnects the neutral in the right, or perhaps that should be wrong, place. When I said leaky line, I meant a current flow to earth, I’d have thought any capacitive losses at mains frequency in 50 metres of cable (2 x 25m?) would have been negligible. So this will be one caveat to what I’ve said above, if you use two ammeters and one is only reading part of the circuit, then they are bound to give different readings. Or maybe the meters just want testing and calibrating.

The discrepancy that PB mentions in the OP might easily be a variation between the instruments; and trying to use a voltmeter measuring the open circuit voltage as an estimate of capacity must be really, really fraught! A good example of this happened very recently. An electrician club mate gave me 8 10Ah 12V SLA batteries, recovered on an ME (Maintenance Exchange) basis from a UPS, (Uninterruptible Power Supply), to check out. The open circuit voltage was around 12.6-7, which is not bad for lead acid batteries that have stood around doing nothing for a few weeks. However, on a discharge test at a mere 500mA the best result was 1.4Ah capacity and the worst 30 milliamps before the voltage hit a low of 10.8! Attempting to recharge them then caused the charger to say it thought they were very quickly recharged. This was because the internal resistance was relatively so high. As I remarked to him later, if these had been required to work in anger the UPS would have instantly fallen over, so that would have been a double failure. Lack of a little TLC at the right time, I’d have thought.

As I found out a long time ago, very often this ’lectrikity stuff ain’t always quite what it appears to be…….

PB

[PatMc]

Posted by Paul Marsh on 25/10/2013 16:22:56:

Also you get resistance in the wires. When current flows, resistance tries to slow the electrons down, causing friction - heat. This heat requires energy and some would be from your lost 225mah.

For example, one client was having a problem, as the landlord had a multimeter in his switch room, and the customer had his own meter in his building. Both were set the same, but over a year, it made out that he insisted that was paying too much. There was a slightly higher current usage upstream - in the switchroom, to what the customer had. Takes it that the customer was paying £21 per year more that he thought, but after losses in the cable underground (about 50m) both reading will never be the same, although very minute.

The current will be the same at any point in series, therefore the AmpHours figures will be identical wherever measured.
Your example is about a power measurement, i.e. WattHours, which isn't the same.

[Paul Marsh]

I put it down to laymen's terms - don't want to sound pedantic!

Basically the two reading were the same, but over a year, there was a difference.

Nothing is 100% efficient, as stated above. you put 100 in, but get 90 out. Unless you start taking about Zero Point Energy, but if were could use Zero Point Modules in our planes, then we would never need to recharge in our lifetime...

[PatMc]

Paul, the point is that PB is refering to AmpHrs whilst your example is about WattHrs. AmpHrs is a linear measurement, WatHrs is a measurement of volume.

[Dave Bran]

With regard probable equality of cell capacity, the below pic shows several batteries from one supplier of allegedly the same capacity.

[Biggles' Elder Brother - Moderator]

Paul, I'm afraid its not "layman's terms" at all and zero point molecular efficiency has nothing to do with!

Pat and PB are correct - its a consequence of the law of conservation of charge - the current is the same everywhere in a closed circuit. The electrons cannot just dissapear! Inefficiencies may slow their flow yes, but that effect is felt throughout the circuit not just a some points!

BEB