Determining motor amps based on rpm and time

Thread Starter

Chris Jones (cjonesy20)

Joined Jun 20, 2019
3
I'm taking a coursera/University of Colorado Boulder course on Motors and there's a question that none of the material covered and the instructors are MIA.

I'm not looking for a straight up answer, but some help in finding the right equations I would need to calculate this. "Final" current seems pretty straight forward at 5V/2ohm = 2.5 amps but I'm not sure how to use the 0.009 per rpm constant to figure out what it would be at t=20ms. Any help would be appreciated!

Suppose the motor resistance is 2 ohms, motor inductance is 5 milli-henries, external voltage is 5 volts, and the back EMF time constant is .009 per RPM.
As the motor accelerates, what is the motor current in milli-amps at time t = 20 milliseconds when the motor is running at 500 RPM. Ignore any oscillation of the back EMF voltage due to commutation. Type an integer value of milli-amps.
 

Thread Starter

Chris Jones (cjonesy20)

Joined Jun 20, 2019
3
I took a stab at it and figured maybe 0.009 per rpm is voltage. So 500 rpm * 0.009 = 4.5V. So that means that 5V supply - 4.5V back emf = 0.5 V / 2 ohm = 0.25 amps. so 250 milli-amps. Why provide the 5 mH if it's not needed?
 

MaxHeadRoom

Joined Jul 18, 2013
19,730
The off off load current of a DC motor is virtually dependant on the inertia and friction (load) of the armature, as voltage increases, the supply-BEMF difference is fairly constant, so the current curve is fairly flat from 0 to max rpm.
This has been found by empirical testing using a 2.5hp 120v DC motor.
Max.
 

Thread Starter

Chris Jones (cjonesy20)

Joined Jun 20, 2019
3
My experience with some worm-gear permanent magnet DC motors is that there's a huge spike in current at 0 rpm and then it tails off from there. Definitely not flat. I had to put supercapacitors in place to prevent overloading the power supply at startup (didn't want to put in any acceleration ramps).
 

MaxHeadRoom

Joined Jul 18, 2013
19,730
That is when the motor rated voltage is applied instantaneously, I misunderstood and thought that you were applying power gradually.
Max.
 
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