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In many manuals (for example for 3D printers), one can find the warning not to connect or disconnect the stepper motors from the electronics while powered on because that will very likely damage the drivers.

  1. What is the actual process behind the damage occurring?
  2. Is ist just costly to design the driver in a way that it would survive this event or is it actually not possible for some technical reason?
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    \$\begingroup\$ I can't think of a good reason a stepper motor driver shouldn't be able to tolerate the motor being disconnected at any time. Maybe if the diode clamping was left out, but that hardly saves any money, and is a bad idea for other reasons. \$\endgroup\$ Dec 14, 2017 at 16:10
  • \$\begingroup\$ I think the reason is that they want to cover their assess if the driver fails for whatever reason. \$\endgroup\$ Dec 14, 2017 at 16:12

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Because it is possible that one of the lines will disconnect first, the inductance on the line still connected can lead to dangerously high (or low) voltage levels and the other side of the free wheeling diode has nowhere to recirculate current through. This can cause a very quick voltage spike that can blow out the output MOSFET(s).

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    \$\begingroup\$ Draw out a circuit, and you should be able to see current paths for the inductive kickback. If the phases are being driven with H bridges, then each transistor will have a diode around it, or at least the body diodes will kick in. Even if simple low-pulling each phase with a common to power, there really should be a diode to power for each phase. All it takes is a few diodes in the right places, and there should be no problem. \$\endgroup\$ Dec 14, 2017 at 22:49
  • \$\begingroup\$ @OlinLathrop I agree with you, however, on many of these 'stepper-driver-in-a-chip' things you see all over the hobbyist sites, they are not robustly protected. I have seen these chips burn out many many times while hot-plugging. \$\endgroup\$ Dec 14, 2017 at 23:38
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Because of the inductive kick back. If there is a current passing through windings at moment t=0 and you disconnect it, then at the very next moment the current is the same. How come? The induced voltage of the winding will force the same amount of current to continue to flow unchanged. The induced voltage will be so much high, as much is needed, hundreds,kilo,.. volts, this voltage will break down your driver.

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    \$\begingroup\$ Whoever downvoted this, could you please explain why? It's basically the same as the other answer, and equally correct... A downvote without a comment is basically a "go **** yourself" instead of anything constructive. \$\endgroup\$
    – jms
    Dec 14, 2017 at 22:06
  • \$\begingroup\$ @jms If it is supposed to mean the same, it does so in a much less understandable way. If I only had this answer: Why would the current be the same if I disconnect something? What is "t=0" in this instance supposed to mean, other than demonstrating technobabble? (Yes, I suspect what it might mean, but in which reference time frame?) And as a non-native speaker, "will be so much high" doesn't parse either. From my point of view, the other answer is better (even though I still don't fully understand the problem). \$\endgroup\$
    – vwegert
    Dec 15, 2017 at 6:32
  • \$\begingroup\$ @vwegert Q: Why would the current be the same if I disconnect something? A: Because the inductivity has this property to not allow rapid current change. Q: What is "t=0" ? A: It's a common EE term to represent a switching event with a timestamp. All further timing events explanations are referred from this specific time event. \$\endgroup\$ Dec 15, 2017 at 9:23

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