# Brushless motor at constant speed

I'm looking to rewind a brushless motor to run at maximum efficiency for a given voltage and RPM.

let's say given the supply voltage is 12v, I want it optimised to run at a constant RPM of 4000 and drawing no more than 0.5 amps.

how do I calculate the wire size and number of turns?

• You don't. Maximum efficency (minimum power) occurs at minimum current, which basically means the finer the wire (and the greater the resistance) the better. Of course, less current means less torque. Zero current (infinitely fine wire) means zero torque and no rotation. Brushless motors speed is independent of voltage and current, and is determined by the drive frequency. – WhatRoughBeast Jul 2 '17 at 23:25
• I tried thinner wire higher number of turns but it didn't use less current, at the same RPM it used the same current and produced less torque. – P0B Jul 2 '17 at 23:49
• This is perfectly understandable because torque is proportional to current. Also consider mechnical power is the same as electrical power (losses aside) $M\cdot n=U\cdot I$. So, if you want high electrical efficiency, you need low current, but at the same power that means high voltage, which means high speed and low torque. Which means low mechanical efficiency. It's a trade-off. – Janka Jul 3 '17 at 0:18
• @WhatRoughBeast, it is not really accurate to say that brushless motor speed is independent of voltage. While the speed while producing toruqe won't exceed the electric commutation rate, the motor also can't keep up with the commutation once the back EMF exceeds the drive voltage. So the poster's question is not off-base. Designing the motor for the intended usage is perfectly proper; of course it will also need an electronic commutation system to run it at that rate. – Chris Stratton Jul 3 '17 at 1:09
• The first step is to determine how much torque is needed. From that, you can calculate the mechanical power needed (torque * speed). Then you can determine whether your current and voltage targets are realistic. If the mechanical power is greater than 12V * 0.5A, then it is impossible. The mechanical power is easiest to calculate if you use N-m for torque, and rad/sec for speed. – mkeith Jul 3 '17 at 5:10

let's say given the supply voltage is 12v, I want it optimised to run at a constant RPM of 4000 and drawing no more than 0.5 amps.

how do I calculate the wire size and number of turns?

'Iron' loss (magnetic and friction) increases as speed increases, but 'copper' loss (winding resistance x current2) increases as more current is drawn. At idle (no-load, minimum current draw) efficiency is zero. At maximum power output the winding resistance absorbs about half the input power and efficiency is 50% or less. Peak efficiency occurs somewhere between those two points, when iron loss equals copper loss.

Magnetic losses are determined by the physical design and construction of the magnetic path (stator shape, core material, air gaps etc.) Winding resistance is determined by wire diameter and length which is dependent on the space available, but the length required to get a particular rpm is determined by motor dimensions, winding pattern, magnet placement and strength. A large motor needs less turns of thicker wire than a small motor of the same design, and a motor with strong magnets need less turns than one with weak magnets or a larger airgap (but stronger magnets increase iron loss, so...).

If you have a motor that runs on a different voltage than you want, you can rewind it to run on a different voltage by changing the number of turns in proportion, eg. if it does 4000rpm on 6V now then double the number of turns to get 4000rpm at 12V. But that takes up twice as much space so you must reduce the diameter by 1.4 (to get half the area per wire).

However, rewinding will probably not produce maximum efficiency unless you are very lucky. For a given motor, peak efficiency occurs at a particular rpm and power no matter what voltage it is wound for. For example if your motor currently does 4000rpm at 1A on 6V with 70% efficiency, if you rewind it to do 4000rpm at 0.5A on 12V it will still get 70% efficiency.

If you rewind for lower rpm it will have higher resistance and copper loss, whereas if you rewind for higher rpm the magnetic loss will increase. If it is too small then copper loss will dominate due to the smaller amount of copper, but if it is too large then iron loss will dominate because the larger core has higher magnetic loss. So when you rewind it for different rpm the efficiency could get better or worse, depending on the particular motor and where you are operating it on the efficiency curve.

• ok thanks for the insight. ignoring the efficiency curve for a moment, what needs to change to enable it to run at the required RPM/torque with less current? larger rotor with more poles and thicker stator slots for example? – P0B Jul 3 '17 at 12:26
• Power out = torque x rpm. Power in = Volts x Amps. Efficiency = power out / power in. So to get the same rpm and torque at the same voltage with less current, you must increase efficiency. Depending on where on the efficiency curve your motor is operating at present, you may need a larger or smaller motor, more or less poles etc. to match the motor to the load. Highest peak efficiency is achieved by reducing both iron and copper losses, by eg. using thinner silicon-steel laminations in the core and increasing copper fill. – Bruce Abbott Jul 3 '17 at 15:29
• What are the specifications of your motor? What Amps and rpm are you getting now, on what voltage? What is the load? – Bruce Abbott Jul 3 '17 at 15:38
• some of the configurations I've tested with smaller motors: – P0B Jul 3 '17 at 22:07
• load is 100mm dia solid aluminium disk rotor : 24mm outrunner , 12N 14P, 2300KV: 5500RPM, 12v 0.19A (no load) 5500RPM, 12v 0.30A (loaded) 35mm , 12N 14P, 650KV: 4000RPM 12v 0.20A (no load) 4000RPM 12v 0.22A (loaded) – P0B Jul 3 '17 at 22:57