I decelerate the velocity of my robot very quickly by applying a common voltage (eg. 24V to both the terminals using Motor driver) to the DC motors. Given, the motor is running at a very high speed and suddenly a common voltage of 24V is applied to the motors. I suppose the kinetic energy should dissipate in the armature resistance. Is it safe for the motors or external braking is advisable?

  • $\begingroup$ Both the terminals are applied the same voltages. $\endgroup$ – DVen Jan 4 '20 at 1:45
  • $\begingroup$ No. The motor driver I'm using has a feature of locking which applies same voltages to both the terminals. $\endgroup$ – DVen Jan 4 '20 at 1:47
  • 1
    $\begingroup$ how is the ground connected to the motor when both terminals are connected to the same voltage potential? $\endgroup$ – jsotola Jan 4 '20 at 1:51
  • $\begingroup$ When same volatges are applied to both the terminals, we can think of a virtual short. No potential difference between the terminals resulting in stopping of the motor. $\endgroup$ – DVen Jan 4 '20 at 1:54
  • $\begingroup$ You will find they’ll probably get hot real fast, depending how much energy needs to be dissipated, on top of that as the magnetic energy drops as its short circuits in your system, your breaking won’t be that fast. External breaking would be advisable. $\endgroup$ – morbo Jan 5 '20 at 21:47

You can do various Bad Things to a motor by doing this -- but whether it's bad for your motor in your machine depends on what you're doing.

When you short a motor like that, it'll generate a lot of current, and in addition to heating up the armature, that current can damage the commutator or brushes. On older motors it could demagnetize the magnets, but that's probably not an issue if your motor has rare-earth magnets.

You can compute the current from the motor's speed, it's speed vs. voltage constant and its armature resistance. Compare that to the motor datasheet's maximum current. A good motor datasheet will have not only a maximum current, but some sort of instantaneous current that gives you an idea of how much of an overload you can give it for how long.

You may also want to consider how much torque the motor will be absorbing during this, and ask whether you'll be exceeding your mechanism's strengths. You'll know the current, so use the motor's torque vs. current constant to calculate the torque, and compare that to the capability of whatever you have the motor attached to.

  • $\begingroup$ One important information: 24V 24V to the motor is given by means of a motor driver. $\endgroup$ – DVen Jan 7 '20 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.