# DC Fan & Motor Battery Configuration

What is the best battery configuration for a cooling fan and 48000 rpm dc motor, 9volts, about 20 watts of power consumed. Short run cycles approximately 30 sec per cycle. I need maximum performance for each cycle and with the total number of cycles equal to 30. There is also a weight constraint of under 1 pound for the battery pack. I'm trying to decide if 6 1.5 volt AA's are sufficient in series or would another configuration work better. My first Post so be easy on me if this question is not suitable for this forum.

• will it really go 48000 rpm with full load or is this no load? – Sunnyskyguy EE75 Feb 13 '18 at 0:22
• I haven't actually run the motor at this time but that was the spec maximum limit. I assume that is no load. – Ricky Raynor Feb 13 '18 at 14:11
• Did you understand my analysis why AA is no good? Use the same method to analyze any battery choice such as 3 18650's with a current limiter or voltage controlled current source or PWM controlled and you will get better results. basically is your load is 2.2A you need at least 2.2Ah batteries to get 1C ratings which are easy in LiPo and not possible in Alkaline do better than 1C and your design is >4C so the battery heats up due to ESR. – Sunnyskyguy EE75 Feb 13 '18 at 17:10
• Would the 30 sec cycle heat the batteries too much? There would be up to an 83 second pause between each cycle where the batteries could cool. Then the cycle would go again. There would be 6 series of 5 cycles to get the total 30 cycles. Batteries could be exchanged to cool between each series. – Ricky Raynor Feb 13 '18 at 19:29
• You would never get 9V for 30s at 2.2 A due to ESR dropping voltage.. read my answer again and aH capacity extrapolates to zero – Sunnyskyguy EE75 Feb 13 '18 at 20:29

The maths is easy:

20 W x 30 s x 30 cycles = 18000 Ws (watt-seconds) and since there are 3600 s in one hour that's 18000 / 3600 = 5 Wh.

But since you are likely to find your batteries rated in mAh we can do it that way too:

The current, I = P/V = 20 W / 9 V = 2.2 A (or 2200 mA). You're going to run for 15 minutes or 0.25 h so you will need to supply 2.2 A x 0.25 h = 0.55 Ah. (Note that if we multiply this by the voltage we get 5 Wh as in the first calculation.)

Now lets pick an alkaline AA datasheet at random. First hit was the Energiser which is a respectable brand.

Figure 1. An extract from the Energizer AA datasheet.

At the bottom we can see that they don't even quote for 2200 mA and that there's an implied limit of 500 mA discharge current. Note also how the capacity drops to half capacity of the 25 mA discharge rate.

We'll have a look at another aspect too. The nominal internal resistance (IR) is 150 to 300 mΩ when fresh. (It will be worse as it discharges.) The internal resistance will mean that the voltage will droop as you increase the load. We can calculate this too: from Ohm's Law we know that V = IR so the drop will be 2.2 A * 0.25 Ω = 0.65 V per cell. That's 2.2 V from your 6 V pack leaving only 3.8 V.

I'm trying to decide if 6 1.5 volt AA's are sufficient in series or would another configuration work better.

You could try it but it won't be good and you could chew through batteries fairly quickly. I won't recommend a battery but I think you have enough information to work out the performance of any other battery from the datasheet. You're looking for one with a rated discharge current of > 2 A and a capacity of about 1 Ah (giving a margin of nearly 100%).

# Good question, but maybe bad assumptions

## Impressions

• You want a motor to drive a coiling fan and 40 kRPM motor Pd=20W @9V for 30x30s periods or 1/4 hour. So total energy = 20W/4 = 5 Wh

Ideally you want a water cooled motor and you should study about Arrhenius Law about MTBF. Its a chemical reaction that causes material fatigue typically 50% loss in life for each 10'C rise. You dont want batteries to get warm to be efficient and this battery has a short life to begin with.

## How hot is each AA cell?

I estimated ESR from their optimistic curve of V-I for a short term drop in voltage to be 0.13 Ohms. But the worst case spec is 0.3 Ohms (new cell)

Pd=I²ESR per cell = 2.22²* 0.13 Ohm = 0.64W per cell
A quick comparison with hollow ceramic resistor about the same size rises 325'C/20W = 16.2'C/W ,so 10.4'C if it were hollow. But it's not hollow so expect >20'C rise and <25% of the capacity.

If you extrapolate the Battery graph below to 2.22A , you may see the intersection of mA to reduce down to 0.

So back to the drawing board.