I am looking for ways to make a mobile power supply for 10 units of MG996R servo motors, each of which can draw up to around 800 mA to 1 A and operate between 4.8 V to 7.2 V.
I have 12 Eneloop AA batteries (BK-3MCCA8BA), and from Panasonic Eneloop BK-3MCC (4th gen) – where I can find maximum discharge current?, I gathered that each of my Eneloop AA battery can discharge up to 6 A, which means that I should be able to power up to 6 MG996R's with a 4 or 6 AA battery pack.
This appears to agree with my setup where I am trying to control the 10 servo motors using the Adafruit 16-channel 12-bit PWM driver with a 1000 uF capacitor.
Once I begin to control over 5 servo motors, I get jitters, which I assume is due to the insufficient current.
However, I also read at How do I determine the maximum amp output of a battery pack? that I can model the batteries as a Thevenin-equivalent circuit, where the current is V_Th/R_Th. If I had 4 1.2 V AA batteries in series and assumed an individual internal resistance of 100 mOhms, I should be getting 4.8 V/ 0.4 ohms= 12 A.
This current should be sufficient for 10 of my servo motors? However, this is not the case.
What am I missing here? Also, could I put 2 packs of 6 AA battery holders in parallel to provide sufficient current?
Best Answer
12A with 0.4ohm series resistance would use up all the battery voltage leaving nothing for the motors. 12A * 0.4 = 4.8V.
To determine the maximum current you can take out of the batteries you first need to know what minimum voltage you need for the load.
For example if the minimum voltage you need to drive the servos is 4V that allows 0.8V drop across the internal resistance.
0.8V across 0.4 ohm is 2Amps.
Yes, you can put strings in parallel to increase the current. From the previous calculation each string would give 2A. To get 10A you would need 5 strings in parallel.
You can't some extra headroom by adding an additional series cell.
For example if you had 5 series cells in each string you could take 5A before the voltage dropped from the 6V @ no-load to the 4V minimum voltage.
With this in mind you could get 10A peak with just 2 strings of 5 cells or 10 cells total. The lifetime would be less as the capacity would only be for two batteries in parallel.
Be aware that the voltage of the cells is not constant. When fresh the cells are probably about 1.5V then drops as the batteries are used. Cells are normally reckoned to have no usable energy when they have dropped to about 0.9V.