# What determines the maximum discharge rate of a battery and how can I create a high current power supply from batteries?

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?

• Assume may be a mistake. It could be many things including interference between motors. Please post your schematics, and measure the voltage on the battery as more motors are started and chart the voltage droop if any. Mar 22, 2017 at 2:40
• I tried power my servos from a power supply, and I noticed that the voltage drops to around 1.8 V and the curerent to 0. My schematic is identical to that found on learn.adafruit.com/16-channel-pwm-servo-driver/hooking-it-up except that I have 10 servos instead of 1. Mar 22, 2017 at 2:59
• The eneloops probably have less than 100 mOhms resistance. But they are 2 Ah batteries. When you drain them at 12A, that is 6C. You are asking a lot from the battery. Voltage will drop very rapidly. I would suggest that for best results you try to keep yourself to 2C, or use Lithium batteries. High discharge lithium ion batteries can support 6C discharge fairly well. Mar 22, 2017 at 3:30
• Since they showed the curve at 25'C, it would be safe if you ensure the 1Watt or so dissipation with an enclosed battery stays at 25'C 6A^2*25mOhm=1W Also you can expect the capacity reduced >50% and the life cycle rating reduced the same. The datasheet also says capacity is rated at 16h using 400mA. So in conclusion your choice of battery pack is seriously over-loaded. Mar 22, 2017 at 4:54
• datasheet goo.gl/Xpcg59 Mar 22, 2017 at 4:56

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.