I recently bought a car vacuum that came with 12V car power supply (the smoking sign connector). I used the vacuum for about a minute. I noticed that the power supply cable was warm, but still, it held up. As soon as I turned off the appliance, I heard a strange clicking sound.

What I suspected ended up being true, the 9A car fuse for the circuit I had just used had been blown. I found this very odd, since it clearly happened when I turned off the vacuum. I know inrush current, when turning on machines, is a real and serious problem, but I have never heard of an 'outrush current', where there is a spike in current when switching off. Could this be it?

If so, how can I limit this current?


2 Answers 2


A motor is an inductor, which attempts to maintain a constant current. When the motor is switched off, the current that was flowing in the motor prior to turn-off must continue flowing. Since the previous power source (the battery) has been disconnected, the only way for current to continue to flow is for the motor to drive that current with a voltage of its own. The voltage across the inductor (motor) will be whatever voltage is required for the current to continue flowing, and if this current must flow through an open switch (a device designed to prevent current flow), the voltage will be very high.

This alone doesn't blow the fuse. Your typical automotive fuse works essentially by melting from excessive current. But, the motor can't make current higher than it was; it's just increasing voltage to maintain current.

However, there is also overvoltage protection in the car's electronics, most commonly in the form of a crowbar circuit:

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Here, the LM431 and triac normally do not conduct. However, if the voltage becomes higher than a threshold determined by the designer, the LM431 triggers, in turn triggering the triac. The triac essentially shorts out the power supply, as if someone threw a crowbar across the rails (thus the name of the circuit). This brings the voltage down, and the excessive current blows the fuse, disconnecting the power source and isolating the fault.

The solution is to put a snubber across the motor, providing a path for the current associated with the magnetic field collapse that isn't through the protection circuitry. In this application, the snubber can be a simple flyback diode. I'm guessing that your car vacuum was manufactured in China and was very inexpensive. By the sounds of it, they could not justify the cost of a power cord of significant gauge, so I doubt they could justify a diode, or the engineer to tell them that it's necessary. Alternate solution: buy a less crappy vacuum.

  • \$\begingroup\$ Thanks! I'm working on the hardware implementation, see if I can solve this! \$\endgroup\$
    – jhc
    Oct 17, 2013 at 18:51

You have a motor, which usually means it is inductive or has high inductance. The inductor is going to try to fight the removal of current (it's called lenz's law) and spike the voltage up to try to maintain the magnetic flux. Basically the energy is stored in the magnetic field and when the field collapses it has to go somewhere. So it was actually a big voltage spike that then causes current flow (being dumped back into the battery) once the switch is turned off and arcing across the switch contacts back to the fuse. Max current will not be larger than the initial operational current flow. Hat tip to @Andy_AKA.

One solution is to put in a high voltage clamping (over-voltage protection) that will shunt the current locally. But the current flow can be very high so this is not a trivial design.

  • \$\begingroup\$ why would the current flow be bigger when turning off compared to when running. \$\endgroup\$
    – Andy aka
    Oct 8, 2013 at 18:10
  • \$\begingroup\$ Because the rate of change of the magnetic field is much higher. The turn off is nearly instantaneous (relative to a DC supply anyway) and the stored energy responds on similar time scales. The coil that is used in the car ignition operates off the same principle. \$\endgroup\$ Oct 8, 2013 at 18:27
  • 1
    \$\begingroup\$ The current from the back emf of an inductor NEVER gets bigger than the current that was previously flowing. It just can't. It would imply it has stolen energy from somewhere. \$\endgroup\$
    – Andy aka
    Oct 8, 2013 at 18:31
  • 3
    \$\begingroup\$ In your answer, you seem to be saying that the current flowing back thru the fuse is bigger than what was flowing originally and this causes the fuse to blow. This can't/won't happen. \$\endgroup\$
    – Andy aka
    Oct 8, 2013 at 18:44
  • 1
    \$\begingroup\$ I actually don't, but you'd be right in saying it is ambiguous. Thanks. \$\endgroup\$ Oct 8, 2013 at 18:53

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