Modeling Optima 300 Brushless DC Motor 2208-1100KV

In summary, the speaker is seeking help with finding a transfer function for their brushless DC motor for a robotics project. They have conducted various tests and have obtained data on the motor's RPM at different duty cycles and steady state voltage. They have also attempted to determine the mechanical and electrical time constants, but are unsure if their calculations are correct. They have ended up with a transfer function of \frac{74500}{s^{2}+60s+500} but are not sure if it is accurate. They are looking for advice on how to further test the motor's response in the time domain.
  • #1
Fronzbot
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This is not homework but I would consider it to fall in this category. I am working on a robotics project and wanted to simulate a PID controller before implementation. I need to find a transfer function for my motor for this to work.

Homework Statement


As mentioned above, I need to find a Transfer Function for my brushless DC motor (Optima 300 Brushless Motor 2208-1100KV). I have done quite a bit of testing and so the data I have available is:
- RPM of motor at various Duty Cycles (controlled via microcontroller PWM)
- Steady State Voltage at various Duty Cycles
- Voltage across motor (not sure what it is though- I probed the left most wire and grounded it on the right most wire- red and black respectively. Not sure what I am ACTUALLY measuring on the motor, though I believe it is my applied voltage after conversion from my ESC)

Homework Equations



[itex]G(s) = \frac{K_{m}}{\tau_{m}\tau_{e}s^{2}+(\tau_{m}+\tau_{e})s+1}[/itex]


The Attempt at a Solution


I used a plot of Shaft Velocity in rad/s over Applied Voltage whose slope I said was Km

I said my [itex]\tau_{m}[/itex] (mechanical time constant) was 0.02s by saying it was equal to [itex]\frac{R_{a}}{K_{m}\bullet K_{e}}[/itex] where [itex]K_{e}[/itex] is the slope of my applied voltage vs. Shaft Velocity (I think this is wrong) and [itex]R_{a}[/itex] is given in the datasheet as 22[itex]m\Omega[/itex]. This gives me a [itex]\tau_{m}[/itex] value of 0.02s.

This seems reasonable to me.

The next step was to find the electrical time constant, [itex]\tau_{e}[/itex]. I guarantee this is not right but I was hoping I would be approximating it close enough. I said [itex]\tau_{e} = \frac{L_{a}}{R_{a}}[/itex] where [itex]L_{a}[/itex] is the armature inductance which I assumed to be 3mH. This gives me [itex]\tau_{e} = 0.1s[/itex] (approximately).

When all is said and done I end up with a transfer function of:
[itex]\frac{74500}{s^{2}+60s+500}[/itex]

The denominator seems ok, but the numerator appears to be WAY to high. If I multiply [itex]K_{m}[/itex] by my [itex]K_{e}[/itex] I'd have a numerator of 462 which is far closer to what I would expect. My only problem is that I have no idea if I did any of this correctly. Nothing, besides the numerator at least, jumps out at me as being wrong... but that doesn't mean I'm right.

Worst case I'll just have to build the robot and tune everything (the PID parameters) manually, but I'd MUCH rather have some simulations to refer to.

Hopefully someone out there can help me out!
 
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  • #2
I can't see that any of your tests have involved the time domain. All you have done is vary the average DC applied, and noted the new steady state speed. Nothing to do with time or frequency response there.

In case you're wondering why I don't suggest what tests you could do, it's because I'm not sure. Can you impose a small sinewave on the driving signal? The ac in the voltage from the tachometer on the shaft would then indicate the motor's response.
 
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Related to Modeling Optima 300 Brushless DC Motor 2208-1100KV

1. What is the Optima 300 Brushless DC Motor 2208-1100KV?

The Optima 300 Brushless DC Motor 2208-1100KV is a type of electric motor that uses a permanent magnet and electronic commutation instead of traditional brushes to generate motion. It is commonly used in applications that require high efficiency and precise control, such as in drones and RC cars.

2. What are the main advantages of using the Optima 300 Brushless DC Motor 2208-1100KV?

The main advantages of this motor include high power-to-weight ratio, low maintenance, and high efficiency. It also offers precise and smooth control, as well as the ability to operate at high speeds.

3. How is the Optima 300 Brushless DC Motor 2208-1100KV different from a brushed motor?

The main difference is in the way they are powered and controlled. A brushed motor uses mechanical brushes to transfer power to the rotating shaft, while a brushless motor uses electronic commutation to control the flow of electricity to the motor's windings. This results in a more efficient and precise control of the motor.

4. What are the key specifications of the Optima 300 Brushless DC Motor 2208-1100KV?

The key specifications include the motor's KV rating, which indicates its speed, and its voltage and current ratings, which determine its power output. It also has a maximum power output and weight, as well as dimensions and a shaft diameter that may vary depending on the specific model.

5. How can I determine if the Optima 300 Brushless DC Motor 2208-1100KV is suitable for my application?

You will need to consider the motor's specifications, such as its KV rating, voltage and current ratings, and power output, and compare them to the requirements of your application. You may also need to take into account the motor's weight and dimensions to ensure it can fit in your system.

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