# Battery question

Discussion in 'Power Electronics' started by rasmus, Dec 8, 2014.

1. ### rasmus

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Nov 6, 2014
Hi

Do battery's in general consume more power when charging compared to the power they output when fully charged ?

2. ### davennModerator

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Sep 5, 2009
power in = power out with minus losses

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3. ### Gryd3

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Jun 25, 2014
Two ways to look at it...
If you're talking about the total capacity, then yeah.
If you're talking about the rate at which it can charge or discharge, that is a different story and will depend on the battery you are using.

4. ### rasmus

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Nov 6, 2014
W
Assume a 12v battery that has a capacity of 100Ahr, charging that with 10A for 10 hours. will it be able to supply that 10A at atleast 12v for 10 hrs ?. Or will the voltage drop under 12v at a certain charge level which means you don't get the full power back ?

5. ### Gryd3

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Jun 25, 2014
There will be losses during charging, which means that not all of the 10A provided will actually be retained... some of it will dissipate as heat.
Likewise for discharging.
Of course, the amount of heat generated depends on a factors such as the battery chemistry and how much current you are providing/taking.

Short answer... if you put 10A into a battery for 10H, you will not get 10A for 10H out of the battery.

*Edit: I should also mention that when you charge the battery, the supplied voltage is not constant... so determining 'power' in Watts will be difficult to do.
This is the same for the output of a battery, as the voltage will typically dip as it discharges. So to determine the capacity in 'power' can be a little more complicated.

6. ### rasmus

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Nov 6, 2014
beside the heat loses, what about the voltage drop that happens during the discharging of the battery, in the words the voltage wont stay at a certain level for the whole charge that the battery have and suddenly drop to zero after the battery reaches 0% charge level.

doesn't that count as a power loss ?

7. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Yes, because they are never 100% efficient. But you need to get your terminology clear.

A battery stores energy, not power. Energy is measured in joules, and one joule is the amount of energy transferred when one watt of power flows for one second.

Take a typical rechargeable cell such as NiMH, rated at 2500 mAh (milliamp-hours), fully charged. For simplicity let's assume that it terminal voltage remains constant at 1.2V during the discharge process.

Because the cell is rated at 2.5 Ah, its "1C" (1 × capacity) charge and discharge current is 2.5A. Nominally, if you charge it fully then discharge it at a steady 2.5A, it will last for an hour before its stored energy is exhausted. At least, that's what the specification claims.

We have enough information here to calculate how much energy it releases when discharged at a rate of 1C. First we need to know how much power it delivers. Power is voltage × current, which is 1.2V × 2.5A which is 3W. So while the cell is discharging, it is delivering a steady 3W of power.

Since we are discharging it at 1C, the discharge time is one hour (its rating is 2.5 amp-hours). One hour is 60 × 60 = 3600 seconds. So the total energy it delivers during its discharge is 3W × 3600 seconds = 10,800 joules, or 10.8 kJ (kilojoules).

Typically, to fully charge a cell like that, you use a complicated sequence of charging stages - precharge, bulk charge, and top-off, at least. During this time, the total amount of energy you need to put into the cell is a lot higher than the energy you will get back out of it. Some of this extra energy is converted to heat during the charge process and lost forever. Some of it is lost as heat on discharge.

A typical complete charge process on an empty cell at a charge current of 1C (2.5A for this cell) might take 100 minutes and raise the terminal voltage to 1.3V on average. So the charging power will be 1.3V × 2.5A = 3.25W and the charge time will be 100 × 60 seconds, so the total charge energy will be 3.25W × 6000 seconds = 19,500 joules, nearly twice the amount of energy you get out. That's pretty typical for the type of cells I'm familiar with. Newer ones like Li-ion and Li-pol might be better.

This is just a very simple description to familiarise you with the quantities involved. In the real world there are a huge number of complicating factors. Here are just two examples: I assumed a constant terminal voltage during charging and discharging, but as you know, this never happens; capacity and available energy depend on the discharge rate (you will get more energy out of a charged battery if you discharge it more slowly - for example, discharging it at half the current will more than double the discharge time).

Sort of. During the discharge process, the terminal voltage does drop, and for most loads, the power will drop as well, but the important quantity when you're calculating battery efficiency is energy, not power.

8. ### BobK

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Jan 5, 2010
That is not correct. You can charge a battery at 1A then pull 10A out, if the battery is capable of 10A.

Closer to the truth is: "energy out = energy in - losses"

And energy = power * time.

Bob

9. ### davennModerator

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Sep 5, 2009
yes I almost reversed it

the point still is you cannot get more out than you put in

10. ### Scotophor

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Oct 8, 2014
Not quite... cells such as this are normally spec'd at close to the optimal discharge rate for getting the most power out, usually something like 10 hours for a full discharge. So if you discharge the cell at a 1 hour rate, you're never going to get the full 2500 mAh out of it, mostly due to internal heating.

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11. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I see. And that's not to mention the fact that often, the capacity specifications themselves are pulled directly out of the ears (or an anagram of it) of the marketing department!

Thanks for the correction!