In a H-bridge, suppose a brushed DC motor has been provided +24V for some time, such that the winding current has reached +1A and the back-EMF has reached 23V (R=1Ohm in this example and no other losses).

When we supply -24V across the windings from that point, the motor is even more effectively braked than if it had been short-circuited in a freewheeling state: the current decreases rapidly to -(23+24)/1=-47A, so the motor is in generator mode. However, these 47A are also provided by the power supply (same loop) whereas the kinetic energy of the rotor is supposed to decrease...

I understand from this example that the net power is still negative (2.2kW lost in the windings, 1.4kW provided by the power supply), but is there a way to understand intuitively what is the provided power doing/going?

It is difficult to explain, it just sounds counter-intuitive that in order to decrease energy, we need to provide some still, and if so: where is it going?


2 Answers 2


If the voltage supplied to the motor is reversed, that is called plug reversing, or plug braking if the reverse voltage is removed before the motor actually reverses. If the power supply will accept reverse current, energy from the rotating mass will be returned to the power supply. If the power supply will accept the energy, but not control it, a high current will flow and a high braking torque will be produced. That will likely be bad for the health of the motor.

With electronically-controlled motors, the controller may contain a braking resistor that dissipates braking energy. If energy is supplied from the supply during braking, at least part of that energy may be dissipated in the braking resistors. That is also an undesirable situation.

Neither of the above two braking schemes are really good design approaches, but plug braking can be implemented inexpensively with motors that are not electronically controlled, so if it can be implemented without harming the motor, it might not be a bad scheme.

The torque vs. speed diagram below illustrates plug-reverse braking. At steady-state, a motor operates at the intersection of the motor's torque vs. speed capability curve and the load's torque vs speed demand curve, point 1 on the diagram. When the motor is reversed by swapping the voltage polarity of the armature power supply, the resulting torque vs speed curve as a reverse motoring curve that is the red curve rotated about the origin by 180 degrees to the negative speed and negative torque quadrant. It is then extended through the positive speed, negative torque quadrant and shown as the orange curve in that quadrant.

Since the speed and direction have not changed, the operating point moves from point 1 on the original motor curve to point 2 on the new motor curve. However that not a stable operating point because it doesn't intersect the load curve. That intersection is in the negative speed, negative torque quadrant. Therefore motor brakes the load moving the operating point along the orange curve to point three. There we assume that the motor is stopped. If the motor is not stopped, it will accelerate in the reverse direction into the negative speed and negative torque quadrant.

Note that the load torque adds to the torque generated by the motor when decelerating the load.

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  • \$\begingroup\$ +1 for putting the traditional name to it while I was commenting. \$\endgroup\$
    – user16324
    Commented May 4, 2020 at 13:05
  • \$\begingroup\$ Thanks, it's clear and I have some reading with this term. \$\endgroup\$
    – user42875
    Commented May 4, 2020 at 20:04
  • \$\begingroup\$ I have studied the specific case of a brushed dc motor with a separate field supply or permanent-magnet field in greater detail and revised my answer. I think I have the basics described properly assuming that nothing catastrophic happens. I have not considered armature reaction or permanent-magnet demagnetization. \$\endgroup\$
    – user80875
    Commented May 4, 2020 at 22:31

Because you are actively braking.

If you simply provided 0V (i.e. short circuit the motor) it generates 23V across its winding resistance giving practically its stall current (23A vs stall current 24A).

As you are supplying -24V, there will be (momentarily) 47V across the winding, providing 47A, approx double the stall current. The generator and PSU are in opposition (where else would 47V come from?) with only the winding resistance in between them.

Half of this power (24V * 47A) comes FROM the power supply, providing mechanical torque to generate acceleration (greater than the stall torque) in the reverse direction by increasing the current above the stall current.

It should go without saying that this is a brutal thing to do to the PSU, the motor, and the mechanical load (or passengers!)

As for where the energy goes; look at the heat dissipated in the windings : approx 4x the dissipation when stalled. (assuming an ideal motor controller : a real one will dissipate its share, proportional to its own device resistance). Both the mechanical power from deceleration and the electrical power from the PSU end up as heat.

If you don't cut the supply when the motor stops, it will reverse and then the power flow will be easy to understand.

  • \$\begingroup\$ @ChrisStratton I think you're confusing relatively gentle regenerative braking which is not the subject of the question, with plug reversing. In regenerative braking you reduce the PSU voltage (to less than the back EMF) instead of reversing it as here. That extracts mechanical power to store in the PSU (where the PSU is bidirectional) - or failing that, in a braking resistor. \$\endgroup\$
    – user16324
    Commented May 4, 2020 at 13:03
  • 3
    \$\begingroup\$ What you are still getting wrong is your claim that power comes from the supply "providing mechanical power to generate acceleration in the reverse direction". That cannot happen until rotation passes through zero, as rotational inertia is not relative. Physically, you cannot brake by adding mechanical energy, but only by removing it. \$\endgroup\$ Commented May 4, 2020 at 13:10
  • \$\begingroup\$ @ChrisStratton +1 on last comment. Rephrased; the PSU supplied current is providing mechanical torque, but (as the answer already said later on), the power it provides only appears as heat. \$\endgroup\$
    – user16324
    Commented May 4, 2020 at 13:18
  • 1
    \$\begingroup\$ With Charles's clarification that this is an unsustainable scheme I can accept that energy is not returned to the supply. But your attempted explanations which involve adding mechanical energy are erroneous. Any energy added would have to transfer from the electrical realm to that of heat without doing mechanical work, as the mechanical work done in braking is negative. This presumably is why such a scheme is unsustainable - there's no good place for the energy to go unless you build some sort of fluid cooling system into it. \$\endgroup\$ Commented May 4, 2020 at 13:38
  • 2
    \$\begingroup\$ With the edit changing wording from "providing mechnical power" to "providing mechnical torque" it now possibly works, since a torque against rotation is one which absorbs power: it is providing mechanical torque to remove power \$\endgroup\$ Commented May 4, 2020 at 13:49

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