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Suppose I have a simple circuit with a 9V battery that is connected to a DC motor, causing it to spin. Then I suddenly squeeze the motor shaft between my thumb and fore fingers, and bring it to a sudden stop.

How can understand what happens to circuit voltages and currents immediately after this event, as well as longer term? For instance, if I hold the motor for a long time, will the battery continue to feed power into the system? If so, will the power be more or less than when the motor was spinning? Would the difference in power between the two scenarios be completely different magnitudes (i.e would something start burning? if so, would it be the battery or the motor)?

Immediately after I stop the motor, will there be a large spike in something (current?)? How would I calculate the magnitude of the spike based on the current prior to the event, number of windings, physical dimensions, time scale of the event, etc.

I just don't have a good mental model of how to think about motor behavior when they are in a "failure" mode.

Thank you!

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    \$\begingroup\$ The back EMF collapses and the windings become a dead short. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 9, 2016 at 16:55
  • \$\begingroup\$ Thanks -- the part about windings is clear (and makes sense...) -- what is "back EMF"? \$\endgroup\$
    – lilinjn
    Commented Oct 9, 2016 at 17:02
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    \$\begingroup\$ @lilinjn The back electromotive force is the voltage present at the motor terminals when nothing is connected to them. It's a result of the motor acting as a generator as it spins (so at rest it's zero), and its magnitude is purely proportional to the rotation speed on permanent magnet motors. When you supply e.g. 10 V to a motor from an external power supply the motor speed increases until the back EMF approaches 10 V, at which point the supply voltage and back EMF largely cancel each other out, the current decreases (which proportionally reduces torque) and the motor stops accelerating. \$\endgroup\$
    – jms
    Commented Oct 9, 2016 at 17:18

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When you stall a dc motor you get stall current and this is usually several times full rated load current. Because the motor is no longer spinning, the natural back-emf isn't present and the full 9 volts is applied across the armature coil resistance. This resistance is undesirable (it represents a loss) when normally spinning hence it has a low value and stall current can be high.

If the battery is regarded as a source of voltage then power into the armature resistance is stall current squared multiplied by R. It can easily burn a motor but, for a feeble little 9 volt battery, there's usually not a problem .

When you release the armature the current flow tries to maintain itself because of the armature coil inductance and this can generate a voltage spike. You can calculate this if you know the value of rotor inductance and snubber values.

I'm not going to extend my answer into AC machines because this would take too much time.

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    \$\begingroup\$ Thanks -- I removed the bit about AC to keep the question more focused. \$\endgroup\$
    – lilinjn
    Commented Oct 9, 2016 at 17:31

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