I am following this example of a sensored BLDC motor.

  • This post assumes conventional current flow.

I am trying to understand how the diodes in this circuit protects the rest of my circuit from back EMF. If my understanding is right, when the motor produces back EMF, the back EMF will flow from the motor to the positive terminal of my battery because the top diode makes a direct path for the back EMF to flow to the positive battery terminal. Please let me know if my understanding of how current will flow is incorrect.

If my battery is 36 volts and if the back EMF flows directly to the positive terminal of my battery will that damage my battery because of the voltage spikes? If so how do I protect my circuit from back EMF?

  • \$\begingroup\$ What kind of battery??? Lead acid would eat those spikes up happily. Some others won't be so tolerant \$\endgroup\$
    – Kyle B
    Aug 18, 2022 at 4:31
  • 1
    \$\begingroup\$ The battery impedance should look like a brick wall to the flyback energy. You’ll see a similar circuit arrangement on most ESCs. Note that the circuit in the link is not a good reference - the mosfet drive is poor. \$\endgroup\$
    – Kartman
    Aug 18, 2022 at 4:41

1 Answer 1


Back EMF isn't a lightning strike. I think a lot of poor quality references get zapped (literally, shocked) by inductive kick/flyback, without understanding what conditions define the effect. And then, after not understanding it, assume any other case of bEMF causes the same. Which it doesn't.

To wit: here, a motor only produces bEMF proportional to RPM. If you spin it faster than it can go at 100% battery voltage (pushed by an external force), sure, it can act as a generator, charging the battery. Pushed hard enough, or for long enough, the battery might get overcharged; or the inverter isn't designed to handle such current and overheats.

There is the additional effect of winding inductance, which is not coupled to motion, but due to the change in current at the terminal. This EMF is also well constrained by the inverter clamp diodes, and current is never higher at turn-off than the peak current at turn-on.

What conspires to make a high voltage, is a sudden change in load impedance. An inductor charged by a battery, switch or transistor, is held to a low impedance; when the connection is broken or the device is turned off, the impedance suddenly goes very high. The same current continues to flow, and since the impedance is high, the voltage drop is high.

A clamped circuit (as an inverter, SMPS, etc.) simply maintains the low impedance, and EMF is well defined at any current.

  • \$\begingroup\$ "What conspires to make a high voltage, is a sudden change in load impedance.".This isnt true.What conspires to make a high voltage is the rate of change of current. \$\endgroup\$
    – Miss Mulan
    Aug 18, 2022 at 11:58
  • \$\begingroup\$ @MissMulan Since \$V = L \frac{dI}{dt}\$, the voltage and current rate are equivalent. A clamped circuit can be described by either measure. An unclamped circuit permits a high voltage = high dI/dt, by presenting a high impedance to the current (~ V = Z * I). \$\endgroup\$ Aug 18, 2022 at 21:35

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