Brushless motors come in many different winding choices. These winding choices are critical to the specific application that you are looking to place a motor in to. However, did you ever wonder how a brushless motor is able to perform equally in terms of power when Kv is changing?
Continuous Power Output Limitation
The first topic to dig in to is, what truly is limiting the power output of a brushless motor? The true electrical limitation to a brushless motor is heat. Too much heat as we’ve discovered in another article, will destroy the brushless motor.
Brushless motors do not operate with 100% efficiency. If they did, we would not have to worry about heat in any way. However, a typical brushless motor will operate between 80 to 90% efficiency.
We are going to follow an example throughout this article. Let’s take a look at a 1750w motor where this power can be delivered continuously. If we assume an efficiency of 90%, our 1750w motor will be delivering 10% of 1750w in heat. This roughly equates to 175w. If you have ever touched a 100w old school incandescent light bulb while it was on, you would know what about 90w of heat feels like. Our motor has to be able to dissipate 175 watts of power in order to not over heat. Anything more and the motor should not be rated for 1750 watts.
A larger motor simply is able to dissipate more waste heat. If we look at an inrunner motor, the motor is capable of removing heat by the surface area of the case. Increase the surface area by increasing the size of the motor and you are able to remove more heat allowing a higher amount of power to be produced.
The Heat Producing Parameter of Brushless Motors
We have also learned from a previous article that heat is produced by current. Voltage does not directly impact the amount of heat that a motor will expel. The more we load our brushless motors, the more current we should expect to draw from the battery pack.
If current produces heat in our brushless motors, how is it possible that the high Kv motor (draws more current) is able to have the same output as the low Kv motor? Surely, we would expect that drawing such a high amount of current would contribute to significant waste heat.
The answer is in the data chart below. Motor one is on the top, and motor two is on the bottom.
|Continuous Wattage (Watts)||Kv Value (RPM/v)||Voltage||Current||Rm (ohms)||Io (Amps)|
Waste Heat produced by each Brushless Motor
We know the limitation to power output is waste heat. In order to understand how both of these motors will end up producing similar waste heats, we can use the above data set. Two forms of waste heat produced is known as copper losses and iron losses. Copper losses are found in the windings of the motor while iron losses represent the power consumption to keep the motor rotating.
To calculate the Copper Losses, we take the square of the current and multiply this by the Rm. To calculate the iron losses, we take the voltage and multiply it by the Io.
|Motor||Copper Losses||Iron Losses||Total Losses|
|Motor 1 / 2600 Kv||117 watts||65 watts||182 watts|
|Motor 2 / 580 Kv||133 watts||41 watts||174 watts|
As we can see, both motors end up producing similar waste heats. These values closely match the waste heat calculated by taking 10% of the total wattage of the motor. Now the point here is not how closely we can get back to the original number we were expecting. You may try this on a motor and your results may be significantly different.
The point here is that motors running at higher voltage will have a higher internal winding resistance. The higher resistance results in significant waste heat. Most would expect the higher current running through the first motor would result in a lot of waste heat. However, it’s not as bad as the motor with a much higher kv has the While on the other hand, the high no load current of motor one running at a lower voltage still produces wasted power leading to heat.