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Just wondering, a flyback diode is typically used to protect the circuit.

Could the magnetic field generated by the flyback be captured to drive a reed switch and protect the circuit the same way a flyback diode does?

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  • \$\begingroup\$ Could you post a rough schematic for this proposed active clamp, please? \$\endgroup\$ – Nick Alexeev Jul 3 '13 at 5:43
  • \$\begingroup\$ @NickAlexeev: A pretty crude contraption really. What I had in mind was to have a fine coil wrapped around the reed switch with the switch itself connected to the relay/inductor coil supply. Can't figure out how to make it into a schematic though ... \$\endgroup\$ – Everyone Jul 7 '13 at 16:55
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The diode must (and can) respond very fast. A reed switch, being a mechanical device, will be much too slow (even when you could set it up to switch at the correct moment, which would be difficult if possible at all).

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    \$\begingroup\$ This is highlighted by the fact that there are "fast" type diodes specifically for this task. \$\endgroup\$ – John U Jul 3 '13 at 9:03
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A coil that is carrying current and is then open-circuited will want to sustain that current until the energy in its magnetic field is zero. This continuation of current can't be diverted into the coil of a reed relay because the reed relay's coil will oppose the sudden surge in current and might as well become open-circuit for a few milli-seconds. This will generate a spark/high voltage that will inevitably harm semiconductors.

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Short answer NO

Even if the switch was fast enough you would need to find some means to synchronize it precisely with the turn off spike. As Wouter says - its a slow mechanical device.

A diode automatically conducts and shorts out the reverse spike and costs a lot less than a reed switch.

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  • \$\begingroup\$ If one drove a large coil with a current-limited supply, would there be any particular problem shutting off the coil by first shorting it with a reed relay and only cutting off power once the reed relay was fully engaged? The relay would have to be able to switch the "on" current of the coil, but would there be any other problem? \$\endgroup\$ – supercat Jul 3 '13 at 14:43

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