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I am new to electrical engineering aand I need a battery for a project of mine. To power the whole system I need 2.2A of current (that is the max current the motors will need, though they will not be running constantly) on 9V voltage.

Can I even draw that much current from AA batteries? And can you also tell me how the batteries should be allinged to be as least space consuming and what will the capacity of all batteries be?

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    \$\begingroup\$ Well, since you didn't specify how many batteries you want, the answer would be a definite yes. Given an infinite number of cells, regardless of size, you can draw infinite power for an infinite time. \$\endgroup\$ – AaronD Dec 29 '16 at 22:50
  • \$\begingroup\$ @AaronD An infinite number of cells would have infinite resistance in the wiring that connected them. If you used superconductors you would freeze the batteries. \$\endgroup\$ – Loren Pechtel Dec 30 '16 at 3:22
  • \$\begingroup\$ @LorenPechtel If you have an infinite series string, then yes, you'd have infinite resistance. But also infinite voltage. Plug that into Ohm's law, and you don't get an answer for current at all. (Inf/Inf = ???) Instead, take the limit of that as the cell count goes to infinity (1st-semester college calculus), and you get the same max current as if you shorted one cell, or any finite number of cells in series. Replicate these series strings in parallel, and you add that current just like usual. \$\endgroup\$ – AaronD Dec 30 '16 at 3:35
  • \$\begingroup\$ Of course, if you were to actually use a truly infinite number of cells, you might discover something like this: youtu.be/w-I6XTVZXww (true infinity is weird) \$\endgroup\$ – AaronD Dec 30 '16 at 3:36
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For high current drain applications you are best to use NiMH rechargeable cells, which have a lower ESR than most alkaline cells and so can deliver higher currents. 2.2A is not a problem, and is often done in RC cars. This datasheet shows that a 1300mAh cell is rated for up to 3C discharge rate, which is 3.9A. At 2C (2.6A) you should get nearly 30 minutes of use. Higher capacity cells are available.

The only issue is that NiMH cells are only 1.2V per cell, so 8 cells give you 9.6V which may be too much for your motors. Check the spec for your motors to see if they can operate from 9.6V.

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    \$\begingroup\$ Regarding over-voltage, I would say that whenever you're using batteries or another (DC) voltage source that's going to vary with conditions (charge, load, etc.), you should be using a buck-boost converter, not relying on a nominal voltage. \$\endgroup\$ – R.. GitHub STOP HELPING ICE Dec 29 '16 at 20:18
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    \$\begingroup\$ @R.., no. You don't use buck boost converters for motors, you use a high enough pack voltage that you can achieve desired performance via PWM (which is also a form of switching regulation) across the useful portion of the discharge curve. Motors aren't really rated for voltage in the absolute (that would be an insulation rating) but for heating and possibly for current (magnet damage), and PWM effectively controls that. \$\endgroup\$ – Chris Stratton Dec 29 '16 at 23:05
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    \$\begingroup\$ The point of buck-boost isn't to avoid burning up the motor (although maybe I made it sound that way); it's to avoid having the behavior vary with battery charge (or other inconsistent voltage sources). \$\endgroup\$ – R.. GitHub STOP HELPING ICE Dec 30 '16 at 4:52
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It depends on the battery, I think you will be able to get away with it, but barely: here is an graph from an energizer AA discharge graph

from the Energizer E91 datasheet

Notice that it doesn't go all the way to 2.2A (2200mA), but it looks like you should be able to extrapolate the extra bit as it's almost there

Do note that at a ~800mAh capacity, draining at that rate will mean it dies in ~20 minutes or so (or rather the voltage drops under .8*6, or 4.8v)

as for the physical battery placement to minimize space (defined as volume), that would probably be a long chain of 6 all in a "tube", that way you aren't trying to line up circles creating dead space - how practical that is depends on your application, but I would suggest just getting a holder designed for the batteries

For how large the batteries should be, that depends on how long you need them to run and at what average current, which you haven't supplied here, but if you are fixed on a AA size, that doesn't give you much play as far as capacity goes

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  • \$\begingroup\$ 1. OP should also consider the internal resistance for 6 cells in series. 2. OP could use 12 or 18 cells to get parallel combinations and reduce the drain on each cell (and reduce the equivalent series resistance of the whole mess). \$\endgroup\$ – The Photon Dec 29 '16 at 18:23
  • \$\begingroup\$ How does the internal resistance affect the current or the voltage? \$\endgroup\$ – user29519 Dec 29 '16 at 19:03
  • \$\begingroup\$ @ThePhoton internal resistance is already taken into consideration by the graph, as it has a cut-off of .8v - basically if the voltage under load (battery resistance + load resistance) ends up with the voltage drop over the battery being less than .8v due to a high internal resistance relative to the rest of the circuit, it correlates to a lower effective mAh capacity - if there were 2 in parallel (which I wouldn't recommend for reasons that I am sure already exist in a different question), then per battery it would look closer to a 1100mA drain \$\endgroup\$ – user2813274 Dec 29 '16 at 19:07
  • \$\begingroup\$ @user29519 the internal resistance of the battery will lower it's effective voltage output, as a portion of its voltage is lost within the battery itself (it's internal resistance ends up causing an "internal voltage drop" when under load) \$\endgroup\$ – user2813274 Dec 29 '16 at 19:08
  • \$\begingroup\$ The datasheet says the internal resistance of a single AA battery is 180-300 mOhms. What will the voltage output be if I have 6 of these 1,5V batteries in serial? \$\endgroup\$ – user29519 Dec 29 '16 at 19:11
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You can draw 2.2A from AA batteries, but you will need to select a battery that was designed for a 2A load and has low internal resistance to prevent the output voltage from drooping too much.

I would suggest the ENERGIZER NH15-2300 AA battery. It shows a typical run time of just under an hour at a 2.3A discharge rate.

It has an internal resistance of only 30 mΩ, which is 1/10th that of a typical cell (such as the Energizer E91). It is an NiMH type battery so its only 1.2V. If you use 8 of them you can get 9.6V, which is probably close enough. The output voltage for the 8 cell stack at 2.2 A would be 9.6 V - 2.2 A * 8 * 30 mΩ = 9.07 V.

http://data.energizer.com/PDFs/nh15-2300.pdf

You can buy them for $4.20 each on Digikey.

http://www.digikey.com/product-detail/en/energizer-battery-company/NH15VP/N703-ND/1040723



Here is why you can't use a typical AA. Lets use the Energizer E91 battery as a typical example (thanks user2813274).

http://data.energizer.com/PDFs/E91.pdf

The datasheet for the E91 AA battery states that is has an internal resistance of 0.3 ohms and an open circuit voltage of 1.5V when new. You would need 6 of them to make 9V. So the total series resistance would be 6 * 0.3 Ω = 1.8 Ω for the 6 battery stack.

At 2.2A the output voltage of the E91 battery stack would then be 9V - 1.8 Ω * 2.2A = 5.04V. The 5.04V would be for new batteries only. The battery output voltage lowers as the batteries discharge, so over time it will be much less than 5.04V.

The problem is made worse by the fact that the total battery capacity is much less at high currents. For example the E91 battery that has a capacity of 3Ah at 1mA but the capacity drops to less than 0.9Ah at 2.2A. A 0.9Ah capacity discharging at 2.2A would lead to "dead" batteries within 24 minutes.

So a typical cell like the E91 is likely to be discharged to 0.8V each within 24 minutes. At 0.8V point it may not even be possible to get the full 2.2A out of them even with the output shorted.

A typical cell won't last long and due to voltage drop your motor won't receive very much power.

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Yes you should be able to draw 2.2 amps. Assuming your using akaline, they probably wouldn't last more than 5-10 minutes at that current draw. You might get up to 20 minutes from lithium.

Ref: http://www.powerstream.com/AA-tests.htm https://www.duracell.com/en-us/techlibrary/product-technical-data-sheets

You probably want to look at 2s (7.4v) or 3s (11.1v) LiPo.

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It's been my experience that AA batteries produce 1 Amp at 1.5 Volts. They'll produce more Amps than that, but not for very long. If you have 6 AA batteries in a series, you'll get 1 Amp at 9 Volts. Use 12 AA batteries in two groups of 6 to get 2 Amps at 9 Volts. Although I would recommend using a 9 Volt battery or at least 6 D batteries to get that much power. You don't want to have to change your batteries every 5 minutes. Although I haven't tested it, one 9 Volt battery should produce 2 Amps, but not for very long. Maybe 10 minutes.

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