If I keep the shaft of a DC motor in place while it is running does this harm it in any way? Is it bad for it at all?
Electrical – Is resisting DC motor bad for it
dc motormotor
Related Solutions
What's important is to not overheat the motor or any of its parts (windings, brushes, bearings, etc.)
The heat in the motor comes from current-squared times resistance.
If the motor is rated for 1.95 amps max, then you have three options to avoid overheating it:
- Make sure the load is never so high that the motor stalls or overloads.
- Make sure that the voltage is so low that the current through the windings will never be higher than rated.
- Use a current controller to drive the motor that can limit the current at the given maximum. (Typically, average current is limited through PWM.)
The reason your "2A power supply" didn't do the limiting is that it wasn't built with "continuous current limiting" as a feature. There exists power supplies that have this feature, but they are generally more expensive as it's usually harder to build that feature than either an unlimited power supply (that you can destroy by overloading it) or a intermittent-limiting power supply (that turns off entirely on overload or overheat.)
The amp rating for a power supply is generally how many amps it can safely deliver -- NOT a limit or exact number, like the voltage rating. The reason is that you don't "push" amps into a load; the load "draws" amps based on the voltage you supply and its internal construction (resistance, impedance.)
My recommendation for you is to get a 12V/2A power supply and see how much current is drawn if you stall the motor. If < 2A, great! If > 2A, reduce voltage even more, until the current drawn when stalled is acceptable.
If that's not good enough, then get a power supply that lets you establish a current limit, where the response of that power supply is to reduce voltage until current is under that limit. You can buy these as "components" from places like Jameco or Digi-Key, or you can buy a cheap 18V/3A benctop power supply from Amazon (which will also come with handy digital read-outs.)
A constant current means, for an ideal motor, a constant torque. This is approximately true for real motors. It doesn't matter what you attach to the motor, or how fast it's turning.
What you seem to be missing is Newton's second law of motion. It states that force is the product of mass and acceleration:
$$ F = ma $$
The constant current you supply to the motor is one force. The weight opposes that force. The difference is the net force, \$F\$ in this equation, and \$m\$ is the mass of the weight, plus the mass of the rotor and the string and everything else the motor must move.
You set current to be sent to the motor so that the torque applied is 10 in-lbs without any load.
Not possible. There is nothing for the motor to "torque against". This is the mechanical equivalent of trying to develop 10 volts across a dead short. The motor will rapidly spin at its maximum speed, and the back-EMF will rise to the driving voltage such that your driving electronics are unable to supply enough voltage above the back-EMF to make enough current to have that much torque.
Let's just say you determine how much current is required for 10 in-lbs of torque, and you drive your motor with a constant-current supply set to that.
What happens when the torque from the weight/load is 5 in-lbs?
Assuming that the rotor and the string are massless and frictionless, the weight will be accelerated upwards by the net 5 in-lbs of torque (motor's 10 in-lbs, less 5 in-lbs from the weight). The rate of the acceleration is determined by the mass of the weight and Newton's law above.
As the speed of the motor changes (the weight is accelerating), the back-EMF also changes. Your constant-current supply to the motor will have to apply an increasing voltage to maintain the same current. Electrical power thus goes up, as does mechanical power.
What happens when the torque from the weight/load is 10 in-lbs?
Motor torque balances weight torque. However fast the weight is moving (if at all), it keeps doing that. Newton's first law applies.
What happens when the torque from the weight/load is 15 in-lbs?
The weight will accelerate downward, overpowering the motor. However, it won't be a free-fall. The motor cancels some of the force of the weight, resulting in a slower acceleration downwards.
If the weight overpowers the motor, then eventually it can get the motor to run backwards, relative to the way it would run if there were no load. When this happens, the back-EMF now adds (instead of subtracts) from the voltage you apply to the motor. At some point, your controller, which is attempting to maintain a constant current, must apply a negative voltage to maintain that current. In other words, the back-EMF is sufficient to create the necessary torque on its own: your controller must oppose it.
This is perfectly symmetrical with the first case, where the motor was overpowering the weight. In that case, electrical and mechanical power went up (without bound, if you let them). In this case, electrical and mechanical power go down (negative, if you let them). Energy is conserved because you are changing the gravitational potential of the weight.
The need to resist the back-EMF usually means storing electrical energy in a capacitor or battery, or using it to heat a resistor. If you can't do this fast enough, then the motor will create more torque than your desired 10 in-lbs, and you have hit the limits of your "constant current" driver.
Further reading:
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Best Answer
In general, yes, because the stall current for a motor can greatly exceed the rated current, and exceed the continuous current rating of the motor's windings, brushes and commutator, and burn out the motor.
In some motors it won't instantly kill the motor but heat it - the motor will survive short overloads but can't dissipate the heat from a continuous overload. If you've been making heavy cuts with a saw or a drill it's often good practice to run the motor unloaded for a minute afterwards so the built-in fan blows cool air through it.
This excessive current under heavy load is a necessary consequence of keeping the winding resistance down to keep the motor's efficiency high under normal (high speed) operation.
Cheap motors tend to have incomplete data but for a reasonably complete specification see this datasheet and note (the first column)
"Maximum continuous current" 6A. (The nominal rating for 100% duty cycle)
"Starting current" 105A. (This is also the stall current).
Typical motors like the Mabuchi RS550 are designed for lower efficiency so the stall current may only be 6-10x the rated current (here 83A vs 10.8A at max efficiency, max continuous current is not specified).
In neither case should you mistake the stall current for the rated current : the RS550 surely cannot survive 83A at 9.6V (about 800W) for very long!
However, in your case (a motorized fader) the motor is small, low powered, probably has quite a high winding resistance and low efficiency, and may be able to survive a fairly prolonged stall. This is a deliberate design choice to limit its stall torque rather than injure a sound engineer's fingers! Alternatively its drive current may be deliberately limited. Over and above that, its controller apparently detects its drive current to detect stall or manual override, and cuts off the power before any damage can be done. It is completely safe to stall this motor by hand.