Do CIM elevator motors overheat?

[MrForbes]

Are you sure about the voltage thing? I don't really understand the whole voltage/current/percent power relationship with speed controllers...but my guess is you're providing it quite a bit more than 1.2 volts if you're commanding 10% power, using PWM.

[AdamHeard]

Originally Posted by MrForbes

Are you sure about the voltage thing? I don't really understand the whole voltage/current/percent power relationship with speed controllers...but my guess is you're providing it quite a bit more than 1.2 volts if you're commanding 10% power, using PWM.

Yes, absolutely.

Model motor as a resistor, you can get it's resistance from the spec voltage of 12V and spec stall current of 133 Amps. V=IR gets you R = .09 Ohms.

Kt is torque/Amp, and therefore stall torque scales with voltage. So, at 10% of spec voltage, you have 10% current draw, which means 10% of spec stall torque.

Stalling a motor in no way leads to failure on it's own. Putting too high of a continuous current into a motor (whether that's at stall, or while rotating) is what fails it (due to heat building up wrecking insulation).

Originally Posted by Thad House

Exactly. If you calculate it so you only need 10% of your stall torque to hold the object up, you only need 10% of the motors stall current to do so. Which you get if you input 10% of the nominal voltage. That becomes 13 amps at 1.2 volts, which is only 16 watts. A CIM can easily dissipate that much wattage, and will have no trouble doing that for the entire match.

You can extrapolate this a tad further and do a BS calc where you say .2 or so of the motors mass is the armature weight in copper, and then do Q = m*C*deltaT, but change Q to Power input and deltaT to deltaT/sec. This lets you get an approximate temp rise per second that's pretty conservative as it neglects ALL heat transfer paths out of the motor.

[Thad House]

Originally Posted by MrForbes

Are you sure about the voltage thing? I don't really understand the whole voltage/current/percent power relationship with speed controllers...but my guess is you're providing it quite a bit more than 1.2 volts if you're commanding 10% power, using PWM.

The old 884's were not linear, so 10% duty cycle was not 10% voltage, but the new controllers are much better, and are almost 1:1.

Adams post above me does a much better job of showing how voltage, current and torque are all related.

[Ether]

Originally Posted by MrForbes

Are you sure about the voltage thing? I don't really understand the whole voltage/current/percent power relationship with speed controllers...but my guess is you're providing it quite a bit more than 1.2 volts if you're commanding 10% power, using PWM.

Everything Adam and Thad said, plus this minor item, which may be key to understanding:

If you're commanding the motor controller via PWM, you're not commanding power, you're commanding output PWM duty cycle.

If the motor controller is linear, then 10% command will create 10% output duty cycle which will effectively cause 10% of the battery's voltage¹ to appear at the motor's input, which -- if the motor is stalled -- will create 10% of the motor's stall current to flow through the motor which will create 10% of the motor's spec stall torque.

The electrical power being put into the motor is Pin = current*voltage, which will equal (10% of spec stall current)*(10% of battery voltage) which will equal 1/100th of the input power at motor stall at spec voltage.

That's why the motor doesn't overheat.



¹Let's assume, for sake of simplicity, that the battery is a constant 12 volt source (same as motor spec voltage).

2The output mechanical power at the motor's output shaft is Pout=shaftSpeed*loadTorque. When the motor is stalled, shaftSpeed is zero so Pout=0. Therefore all the input power Pin is dissipated inside the motor as heat.

[AdamHeard]

Originally Posted by Ether

Everything Adam and Thad said, plus this minor item, which may be key to understanding:

If you're commanding the motor controller via PWM, you're not commanding power, you're commanding output PWM duty cycle.

If the motor controller is linear, then 10% command will create 10% output duty cycle which will effectively cause 10% of the battery's voltage¹ to appear at the motor's input, which -- if the motor is stalled -- will create 10% of the motor's stall current to flow through the motor which will create 10% of the motor's spec stall torque.

The electrical power being put into the motor is Pin = current*voltage, which will equal (10% of spec stall current)*(10% of battery voltage) which will equal 1/100th of the input power at motor stall at spec voltage.

That's why the motor doesn't overheat.



¹Let's assume, for sake of simplicity, that the battery is a constant 12 volt source (same as motor spec voltage).

2The output mechanical power at the motor's output shaft is Pout=shaftSpeed*loadTorque. When the motor is stalled, shaftSpeed is zero so Pout=0. Therefore all the input power Pin is dissipated inside the motor as heat.

Good catch, I overlooked that his question was more related to the motor controller's "10%" output than to the motor itself.