Power from NiMh, etc batteries

[robgraham]

I've just had a nearly heated discussion with my son in law (SIL) about the power from rechargeable batteries and we both agreed in the end that we really weren't that sure of our positions !

Regardless of the battery type, if you have a say a 1.4AH and a 2.5AH battery driving the same device, I'm of the opinion that the actual power output of the motor is going to be same during the 'flat' part of the battery capacity period and that only as the battery begins to reach the end of its charge capacity will the power start to drop off

- true or false ? The voltage will be the same but is the internal resistance and therefore the maximum current output going to be the same at full charge regardless of the capacity ?

Interestingly I discovered, before reading the instructions (!), that my new Ryobi drill Li-ion batteries will switch themselves off before they reach the end of capacity and that I don't have to make a considered judgement on the critical point like my old Bosch with Ni Cd batteries.

Rob

[Peter Parry]

Substantially correct, assuming a 1 amp draw the 1.4AH will last 1.4 hours, the 2.5 will last 2 and a half hours. During these respective time both will deliver an equal current

That depends upon the battery and manufacturer, it isn't related to capacity. It may well be that a larger capacity battery will have better performance (lower internal resistance) but this would be related to improvements in manufacturing process between one and the other rather than intrinsic qualities.

All Lithium batteries do this, as well as limiting charging. If they don't they have some rather interesting characteristics which involve explosions.

[Calvin Sambrook]

I don't know the answer to the underlying question you are asking which I think is "is the internal resistance of a battery related to its capacity?" however in practical usage the power is determined not by the battery internal resistance but rather the load so you are correct the power output of the motor will be the same during the flat part of the discharge curve.

[John Stumbles]

For a given load I would expect a larger battery to have lower internal resistance and therefore to be able to deliver higher current and therefore power. I have absolutely no idea whether the difference is likely to be noticeable or not.

[Dave Plowman (News)]

The peak current output of a battery is nothing to do with its capacity. More to do with the quality of the cells. Cheap cordless drills can often have their performance dramatically improved by changing the battery cells to a better make.

[John Rumm]

Basically true all other things being equal.

The complication comes from the fact that many better quality packs are also higher capacity. So you may also see an improvement in performance with the higher capacity pack. However that is not *because* its higher capacity, only its a better quality[1].

(I have a some DeWalt 14.4V packs - one is 1.3Ah and the others are

2.0Ah - its notable that performance is better from the higher capacity pack in two ways; peak current delivery is better (more torque), and matching is better - so you get a wider "flat" part of the discharge curve with a pronounced "cliff effect" at the end on the bigger packs, than with the smaller one)

Yup, LiIon cells need much more "management" when charging / discharging and have to be treated differently to other rechargeable technologies. Hence why the tools / chargers etc are made with different contacts and fittings from the older technology versions.

[1] Quality being a nebulous term, but implying a number of attributes including lower cell internal resistance, better cell discharge curve matching etc. Note also that different battery chemistries will have differing minimum internal resistances and hence a knock on effect on peak current delivery.

[NT]

It depends on the ratio of battery internal resistance to load resistance. If the load R is much higher than batt R then the same terminal voltage will be present on both batteries under load. This is usually how things work.

But if the battery R becomes a significant percentage of the total Rbattery + Rload, then the 2 batteries woudl give different terminal voltages under load. This occurs when the load current is high, making battery internal resistance significant.

For a given cell design, a larger capacity cell has a proportionately larger conductance, and R =3D 1/conductance. However IRL 2 different batteries can have differing internal details, and their internal Rs may have other than that relationship.

NT

[Dave Osborne]

I agree, this seems to be the underlying question.

however in practical usage the power is determined not by the

Actually, by virtue of the maximum power transfer theorem, maximum power is transferred when the load impedance equals the source impedance.

Assuming a pure resistive load (and an ideal battery) then max power is delivered when the load resistance is equal to the battery's internal resistance.

AIUI, during the "flat part of the discharge curve", the internal resistance of the battery stays about the same. At the end of the "flat part of the curve" the internal resistance rises as the battery is depleted leading to a reduction in available power.

To address the underlying question, the volume of a notional water tank is akin to the Ah capacity of the battery; the height of the water column is akin to the voltage of the battery and the diameter of the outlet tube is akin to the internal resistance of the battery.

As you can imagine, increasing the volume of the water tank does not

*necessarily* change either the voltage or the diameter of the outlet tube - it just (to horribly mix a metaphor) makes the "flat part of the curve" last longer for a given load.

[John Stumbles]

Indeed, and if the system were set up for maximum power transfer then you would only need a much smaller battery, with much higher internal resistance to match that of the load. However with as much power being dissipated in the internal resistance of the battery as is used in the load it would be a toss-up whether the battery would explode due to overheating before it ran flat due to its lower capacity.

In the case of battery-operated equipment you want *sufficient* power delivered to the load in such a way that *most* of the available *energy* in the battery ends up in the load with as little as possible dissipated in the battery itself.

[Dave Plowman (News)]

That's fine in theory - but pretty well everything works best when the source impedance is as low as possible. We're not really interested in efficiency in this case.