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I read an article from wikipedia about freewheeling (flyback) diode which is used to drive a relay. I understand it but I have four questions. Would you give me a hand, please ?

1- Why does the coil change its polarity ?

In figure 1 the positive terminal of the coil is the upper terminal (at node 1), but in figure 3 the positive terminal of the coil is the lower terminal ( at node 2). that means it discharge in the opposite direction. If it was a capacitor the polarity would not change.

enter image description here

2- If I apply AC to a coil, Does the voltage spike happen each period when the voltage is zero ?

I read that the coil makes a voltage spike when we turn off the supply. does that mean the coil make spikes at the arrows of the following figure?

enter image description here

3- Why don't we put a diode for each inductive load such as motors in children toys .. etc. ?

4- Can I put an LED rather than a diode ?

Thank you very much,

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  • \$\begingroup\$ The spike is caused by the resistance becoming very high, not just by the current becoming zero. An inductor "tries" to keep a constant current flowing through it; smallish current times very large resistance = large voltage. \$\endgroup\$ – user253751 Sep 9 '15 at 23:58
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1) An inductor's current remains flowing in the same direction when open circuited and the only way this can be achieved is by reversing its polaity so you can't have both.

2) When the voltage is zero the current flow is at a maximum but this energy feeds back into the voltage source hence no spike - the spike only happens with non zero-ohm voltage sources with discontinuous signals.

3) Generally there are diodes but they are probably on the circuit that drives the motor and might easily be inside some chips.

4) No an LED won't withstand much in the way of reverse voltage (maybe 5V) so any motor supply greater than 5V might destroy it but ostensibly it will work.

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  • \$\begingroup\$ thank you very much for your answer, it is very useful but i still don't understand question number 2. Would you explain it in more details, please ? \$\endgroup\$ – Michael George Sep 9 '15 at 22:18
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    \$\begingroup\$ A voltage source is what it is. It produces a perfect voltage and nothing will disturb that hence no spike. But even if it had several ohms source impedance, the terminal voltage across the inductor would still look sinusoidal and so would the current (but shifted a quarter wave with respect to voltage). \$\endgroup\$ – Andy aka Sep 9 '15 at 22:31

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