Some power supplies and regulators don't like to have current flowing back into the output terminal. It's something I've seen casually mentioned in various answers here and it sounds reasonable enough. This has never been a problem for me until now because my designs have always had a net outflow of current, but now that I'm designing a motor controller, I need to make sure that my power supply will be able to accept potentially large transients flowing back into the supply from the flyback diodes.

Common regulators like the 7805 (not that I would use one anywhere near motor control) do not indicate a reverse current limit on their datasheets, so how do I look for a regulator that will tolerate these transients without self-destructing?

(My motor is a little 24V/35W BLDC to clarify scale.)

  • \$\begingroup\$ You haven't said but I presume it's being fed by a H bridge. \$\endgroup\$ – Andy aka May 13 '14 at 21:24
  • \$\begingroup\$ You want it to clamp the voltage or to tolerate overvoltage? \$\endgroup\$ – Spehro Pefhany May 13 '14 at 21:28
  • \$\begingroup\$ Yes, the regulator powers a 3 phase H bridge. I would prefer clamping as much as possible. \$\endgroup\$ – BB ON May 13 '14 at 21:30

Synchronous buck regulators are pretty useful at dealing with current in reverse. After all they do it all the time if the current is discontinuous - the lower transistor has to spend a certain amount of time dragging the current backwards through the inductor. OK normally the current is all forward current but in discontinuous mode the average current could be zero.

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Added bonus - you are also recovering some of the energy from the motor into the inductor.

  • \$\begingroup\$ Synchronous converters will happily soak up backfeed current, which can be dangerous if there is nothing to absorb the recovered power at the input side. Connecting a battery at the output of a synchronous buck converter and then setting the setpoint below the voltage of the battery may cause the regulator to overvolt itself. Don't ask how I learned this... \$\endgroup\$ – jms Jun 3 '15 at 11:11
  • \$\begingroup\$ Conceptually it may help to notice that if you flip the schematic left to right it looks just like a boost converter (with Q2 being the switch and Q1'd body diode being the flyback diode). \$\endgroup\$ – pericynthion Jul 23 '17 at 18:53
  • \$\begingroup\$ @pericynthion yes that is exactly what happens in a buck boost converter - in effect, a H bridge is formed. \$\endgroup\$ – Andy aka Jul 23 '17 at 18:59

Generally, I've found that BJT-based regulators may dislike being back powered, whereas MOSFET-based regulators use a P-channel MOSFET that contains a body diode that conducts sufficient current to the input side that dangerous back voltages don't exist. This is perhaps not a universal rule, but I've been lucky with it so far.

Also, why do you worry about back propagation into a 5V regulator when you're powering a 12V motor? I presume that only control logic runs on the regulated voltage, and your motor (and the power rail of the H-bridge) draws power directly from a DC power source. If so, there's no back power to worry about.

If you're powering the motor through a 12V regulator, then A) make sure you have sufficient cooling, and B) couple a reverse diode (something like a 1A Schottky) from the output to the input, so that a dangerous negative voltage is not allowed to exist across the regulator -- the higher voltage on the output side will effectively be shunted to "input" for the regulator, minus the diode drop.

  • \$\begingroup\$ The 7805 was just an example of a common regulator. \$\endgroup\$ – BB ON May 14 '14 at 13:21
  • \$\begingroup\$ Which is why I answered with the actual distinction between BJT regulators (like the 7805) and MOSFET regulators (like more modern devices.) \$\endgroup\$ – Jon Watte May 15 '14 at 15:47

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