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My mobility scooter has a motor controller with regenerative braking - but regeneration seems to take place even at low speeds, well under the maximum! I know it does this because if I reduce throttle at low speed going down a hill, I see the current reverse on the ammeter and the battery voltage start to climb - considerably higher than with no load.

How does the controller achieve this? Normally, charging can only take place when the motor's generated voltage exceeds that of the battery - and this cannot happen at low speed. I ask because I want to build one from scratch.

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The short answer is that regenerative braking is not based only on the back EMF of the motor. It's based on also using the motor's inductance (and/or an external inductor) as the key element of a boost-mode switchmode power converter that can convert the output voltage of the motor (when used as a generator) to a level that can be used to charge the battery.

Since PWM speed controllers already contain the necessary high-current switching elements connected between the battery and the motor, switching between "driving" and "braking" is really just a matter of changing the timing of the control signals to those switching elements.

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  • \$\begingroup\$ Its not so much a "boost mode converter". The rotor is still rotating and energy transfer is occurring. If the speed controller wasn't a 4quadrant controller there wouldn't be any energy transfer UNTIL the speed was high enough that the back EMF > DClink. As an analogy ... sure, but from a motor-control point of view its not complete \$\endgroup\$ – JonRB Dec 16 '14 at 13:27
  • \$\begingroup\$ @JonRB: It isn't merely an analogy. The inductance and the switching element really are working together to boost the battery bus voltage to the point where the current through the battery can be reversed. Otherwise, no power is drawn from the motor when it is operating as a generator, and there's no braking action. \$\endgroup\$ – Dave Tweed Dec 16 '14 at 13:54
  • \$\begingroup\$ I completely dissagree then. You are not energising the stator inductance and opening the switch so the inductance boosts some cap/battery. You do not have to energies the stator to draw energy out of it. Your output inverter provides an out of phase voltage and due to the potential difference current flows \$\endgroup\$ – JonRB Dec 16 '14 at 14:01
  • \$\begingroup\$ @JonRB: Instead of debating me, you should be putting the effort into improving your own answer. I never said we're boosting some "cap/battery". We use the switching element to short the motor briefly so that the induced EMF charges either the motor's own inductance, or an external inductor, with current. When the switch opens, the inductance then boosts the voltage to the point where current flows back into the battery. Note that if you merely left the motor shorted, you'd get braking action, too, but the energy would be wasted. The switching action is what allows the energy to be recovered. \$\endgroup\$ – Dave Tweed Dec 16 '14 at 14:08
  • \$\begingroup\$ I don't see the problem with this analogy. It's actually pretty good. A DC motor will act like an inductor which from a circuit standpoint 'magically' gets charged with magnetic flux (i.e. rotational energy gets put into the inductor as flux). Switching this energized inductor at the right time will create a certain voltage, if the voltage is higher than the externally applied battery voltage, current will flow into the battery and the motor inductance is (partially) de-energized. That is exactly what is happening, and it's a helpful way to view the problem like Dave Tweed stated it. \$\endgroup\$ – user36129 Dec 16 '14 at 14:17

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