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I've been getting some amazing info on this site so thanks for all the info so far!

Onto my question.

I know I can brake a dc motor by shorting the terminals but I also know how much current that can generate. I need to be able to disconnect my speed controller before shorting the terminals as i'll probably cause a failure or worse, a fire!

I've attached 2 drawings...one of my current setup and one with a DPDT switch that will disconnect the motor and at the same time, short the terminals. (I think I've got the wiring right?!) My concern is that the current generated by the braking will still damage my SPDT relays that are used for polarity switching. Is there another way around this?

It's for an electric car. Motors are 12vdc pulling approx 4a no load. SPDT relays rated at 40a. enter image description here enter image description here

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  • \$\begingroup\$ If you are going to design circuits, even simple circuits like this, you need to learn how to draw schematic diagrams. Pictorial connection diagrams like the ones you provided are too difficult to follow. In this case, your text description is sufficient to indicate that you have not designed a proper braking scheme. \$\endgroup\$
    – Charles Cowie
    Apr 28 '20 at 17:35
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It is never a good idea to short motor terminals. For a permanent-magnet DC motor with a commutator, you need to disconnect the armature from the power source connect a resistor across the armature terminals. Rather than using just a fixed resistor, better braking performance will be provided by a PWM braking circuit that will regulate the braking current.

The ohmic value of the resistor should be chosen to draw no more than 150 percent of the rated armature current at the rated armature voltage. The power rating needs to be based on the motor plus load inertia, the load torque and the duty cycle. The total dissipated energy in the braking resistor for a braking cycle will be the energy stored in the rotating mass minus the energy dissipated in the load and in the motor losses.

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    \$\begingroup\$ Hi, Charles. In the last sentence do you mean, "The total dissipated energy in the motor for a braking cycle ...:? \$\endgroup\$
    – Transistor
    Apr 28 '20 at 17:23
  • \$\begingroup\$ @Transistor; I meant the braking resistor, but neglected the motor also. I revised my answer. Thank you. \$\endgroup\$
    – Charles Cowie
    Apr 28 '20 at 17:28
  • \$\begingroup\$ I think I get the gist of your answer. RE my diagrams, I'm not gonna learn to draw schematics as I don't do this often and for me it's easier to understand by actually drawing the circuit as I would put them together. \$\endgroup\$
    – Sal
    Apr 28 '20 at 19:05

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