I am making a robotic device with the following servo motors. How do I go about calculating the battery capacity? It is supposed to run for 10 minutes. I am bit confused as to use the no-load current or load current or the stall current to add up the amps.

Thanks Ro

  • HS-422
  • HS-485HB
  • HS-645MG
  • HS-755HB
  • HS-805BB
  • HS-85BB
  • HS-785HB
  • $\begingroup$ To get a reasonable answer, you need to give more details. What will be servo load? Will all the servos run continuously? Is losing power critical to safety of people and the device, or will it be just an inconvenience? $\endgroup$
    – mactro
    Commented Jan 5, 2017 at 13:14
  • $\begingroup$ Hi, The servos are used in a robotic arm. The load is coins. They will be running continuously together. It is for a school project competition so losing power is not ok. $\endgroup$
    – Dharmesh
    Commented Jan 6, 2017 at 5:32
  • $\begingroup$ For a robot arm (stationary), it is easier to get a desktop or bench power supply. You can get a cheap one that provides power at a fixed 5V or 6V, and it will deliver however many amps are needed until it hits a heat limit (probably 5A, which is enough for a small arm). Personally, I bought one that I can adjust from 2V to 30V and delivers up to 10 amps. The constant power makes testing and debugging MUCH easier because you wont need to constantly disconnect and reconnect, or wonder if the battery is depleted. Servos can get weird on low power. $\endgroup$ Commented Oct 20, 2017 at 18:30

2 Answers 2


This typically requires a bit of testing with your actual payload. You can try to model it mathematically, but it's easier and more accurate to just plug it in and measure the current draw.

Current draw in a motor increases with the amount of torque required. No load current is how much current the servo will draw if you move it with nothing attached to the shaft at all. The stall current is how much it will draw if you clamp the shaft in place (say, with a vice grip) and tell it to move. Keeping that in mind, your real-world current draw will be somewhere in between those two numbers.

A reasonable approximation for an arm like you're building would be to measure the average current for each time you pick up and move your payload and how long that (averaged out) current draw occurs for per motion. If you know how many of those motions per minute you need to do (or more conservatively, are capable of doing), then you can figure out how much power you're going to draw over the course of your 10 minute runtime.


First, can you add as much battery as you wanted? Meaning, is there a technical or score-wise drawback to the amount of battery? Or mass?

Does the servo load increase with the increased battery amount? (Probably not).

The above considerations would require a precise battery sizing. Otherwise, taking a rough estimation and then buying one size bigger and smaller batteries would be the fastest and most pragmatic road to success.

If you do not want to purchase unnecessary items, you can calculate the energy consumption using power utilization as below:

1) divide the total operational time into phases. Like, phase 1. startup and waiting Phase 2. Competition run Phase 3. Wait for Xxx Phase 4. Run-2 Phase 5. Finished and waiting

2) In each of these phases determine the power drawn from the battery, and multiply that with the amount of time of that phase.

3) And when you find the exact number, add 20% for uncertainties. This would help select the required battery.

Notes: 1) temperature is ignored (assume the competition is indoors). 2) validating this method can be done by literally testing it 2-3 times for especially the operational cases (competition runs).

  • $\begingroup$ As a side note: if you can get extra points for decreasing the servo size, try designing a mechanism such that the servo loads are minimal. (Springs, counterweights etc). $\endgroup$ Commented Jan 7, 2017 at 9:50

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