I've come across a circuit that has a diode in parallel with a power switch that's controlling a BLDC motor circuit. I was told that it's a flyback diode to prevent voltage spikes from an inductive load, but that doesn't really make sense to me. I've always seen a flyback diode placed in parallel with an inductive load, like with a relay coil, or in a buck converter. The two ways I understand it are that:

a) the diode provides a path for the current in the inductive load to continue to flow

b) when the current though the inductive load is suddenly stopped, the voltage, which is V=L*di/dt is a very large negative number, so V_LOAD goes negative. However the diode is then forward biased, so it clamps V_LOAD at the forward voltage of the diode

With the circuit in question, neither of those conditions hold true. The current could have to instantly change direction for the diode to conduct (which would probably cause a lot of other issues), and V_LOAD would have to increase in voltage in order for diode to be forward biased.

Has anyone seen this before and know what the purpose of this diode is? It kind of seems like someone didn't know what they were doing by putting the diode in that position, but I might be missing something super obvious.

Switched Inductive Loads


3 Answers 3


Is the circuit of reputable origin?

I am fairly certain it is due to someone seeing two similar things:

  • A flyback diode anti-parallel to an inductive load
  • RC snubbers or TVS diodes in parallel with a switch

and then trying to combine them the two (probably due to convenience of connecting across switch rather than the motor or motor driver) without truly understanding what is going on.

Either that or they thought "inductive loads arc and get a diode across them and the switch also arcs when it opens" and it was more convenient to connect across the switch so that's what they did, mistaking the source for the symptom.

It also doesn't make sense because it's not just a motor, but a complete BLDC circuit which already contains measures to deal with flyback.

  • \$\begingroup\$ It's from a large consumer electronics company that should know what it's doing - though I've seen some other questionable things which makes me think it was added due to a lack of understanding like you said. So you agree that the diode placement doesn't make any sense? \$\endgroup\$ Oct 16, 2019 at 21:54
  • 1
    \$\begingroup\$ @newothegreat Still doesn't make sense. Are you sure it's a regular diode and maybe not an ESD diode? \$\endgroup\$
    – DKNguyen
    Oct 16, 2019 at 21:55
  • \$\begingroup\$ @newothegreat What is this schematic of exactly? I'm leaning towards just braking when the switch is opened. Are there separate control signals for the BLDC driver? Or is it just apply power and go? \$\endgroup\$
    – DKNguyen
    Oct 17, 2019 at 13:45
  • \$\begingroup\$ There's a separate controller (not included in the diagram) that's always on, regardless of the switch position. The controller probably takes care of the braking when the switch is opened. \$\endgroup\$ Oct 17, 2019 at 13:55
  • \$\begingroup\$ @newothegreat It might not be able to since the driver loses power when the switch is open, but the presence of an external controller does reduce the chances the diode is for braking. \$\endgroup\$
    – DKNguyen
    Oct 17, 2019 at 13:58

Is there a circumstance where the motor is rotating against the torque demand, like say doing dynamic speed control or motor field braking?

A BLDC under these circumstances is called an alternator and can raise the bus voltage to well above normal, this diode prevents that by recovering the energy back into the battery.

  • \$\begingroup\$ But it would be ineffective if the switch was closed...which wouldn't make sense for dynamic speed control. \$\endgroup\$
    – DKNguyen
    Oct 17, 2019 at 13:43

It's possible the BLDC circuit is capable of regenerating energy (from a spinning motor + load) back into the supply and the diode limits the BLDC supply voltage to one diode drop above the battery voltage.

  • \$\begingroup\$ The thought crossed my mind but then thought it's strange because it would be ineffective if the switch was closed. \$\endgroup\$
    – DKNguyen
    Oct 17, 2019 at 13:33
  • \$\begingroup\$ @DKNguyen It's unnecessary when the switch is closed because the battery is right across the supply. I have seen this cause occasional but repeatable failures depending on usage. Very annoying for the manufacturer (and customer, obviously). \$\endgroup\$ Oct 17, 2019 at 13:35
  • \$\begingroup\$ This sounds plausible. If the motor is spun externally when the switch is open, it would probably start to charge up the electrolytic cap across the bus terminals. The diode would prevent the bus voltage from rising too high and exceeding component voltage ratings. Let me see if I can find more info. \$\endgroup\$ Oct 17, 2019 at 13:50
  • \$\begingroup\$ @newothegreat The voltage produceD is based on RPM and won't continuously climb. It only exceeds the voltage ratings if the motor is oversped beyond normal operating RPM. \$\endgroup\$
    – DKNguyen
    Oct 17, 2019 at 14:00
  • \$\begingroup\$ @DKNguyen If it's doing regenerative braking then the voltage should climb. It uses the energy from the rotation (assuming it was spinning when the switch is opened) and boosts it via motor inductance. \$\endgroup\$ Oct 17, 2019 at 14:05

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