# how to calculate current

Discussion in 'General Electronics Discussion' started by docb, Apr 17, 2010.

1. ### docb

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2
Feb 11, 2010
I think a 9v battery is about 600ma
If AAs are 1200ma

If I use 6 AAs in series to get 9v, is it still a total of 1200ma?

In other words, do AAs get me about twice the usable time as a 9v?

EDIT. I meant to wrote AA's in all my messages here, not AAA.

Last edited: Apr 18, 2010
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Are you talking about capacity (how much energy is stored in a battery)? This is roughly the thing that tells you how long a circuit will operate when using that battery.

Or are you talking about the maximum current that can be drawn from a battery (the maximum instantaneous power)?

I suspect it is the former, although current (and the units you use) reflect the latter.

The appropriate units of measure are mAh (milliamp hours). This is a unit that combines the rate of discharge (mA) with the time the battery will last (hours).

In practice, this is done for a specific current, say 900mAh at 25 mA. If you look here you will notice that as the current increases, the effective capacity of the battery tends to drop. This is partially caused by power being lost in heating the battery.

However, you are perfectly correct in that placing batteries in series causes the total capacity in mAh to be unchanged (placing them in parallel allows you to add them -- but at the expense of not being able to add the voltage).

Your conclusion is probably true. The 6xAA cells will last around twice the length of time that the single 9V cell will. In fact they will probably last longer than that as the current at which they are rated to give that capacity is probably higher than the one used to calculate the capacity of the 9V battery. The same relationship shown above means that the capacity will be higher at a lower current (although not dramatically higher).

If the batteries are the same type (e.g. alkaline) then the difference in volume between 6 AA cells and 1 9V battery is also indicative of their relative capacity. In fact, if you open up some 9V batteries you'll find they have 6 AAAA cells inside.

3. ### docb

131
2
Feb 11, 2010
Yes I was just talking about time that you can run a device.

I have a keyboard that uses 6 AA's and wondered how much less time I'd get with a single 9v.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The keyboard probably uses very little current. I have wireless keyboards with 4xAA cells that last too long to remember when I changed the cells. Presumably a 9V battery would last about a third of that time.

The only risk is that the keyboard may have short, high current demands to operate a transmitter and this may not work as well with the smaller 9V battery. A capacitor in parallel with the battery (which will help with this potential problem) may have more leakage than the average current required by the keyboard.

5. ### docb

131
2
Feb 11, 2010
Why would it last a third of the time, and not half the time?

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The AH rating of the AA cells is probably based on a higher current. At lower current they would probably show higher capacity.

My gut feeling is between 2 and 3 times as long for AA cells, but that is based on their volume, not their rated capacity.

There are heaps of variables. You have considered only one.

Try it and find out.

7. ### Resqueline

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Jul 31, 2009
The Ah capacities of primary cells vary a lot between brands, not to mention between technologies, and also a lot depending on how they are used (current & rest times).
A typical alkaline AA has around 2000mAh and a 9V has around 500mAh, so there's a factor 4 right there. I don't know why you refer to AAA's first and then AA's..?
The longer time you use to discharge a battery the higher Ah number you'll get (due to ESR & chemistry). This is also a factor to consider when estimating run times.
Like steve says, there's only one way to find out (the exact number for your application).

8. ### docb

131
2
Feb 11, 2010
Ok, thanks! I guess I'll just need to test.