so my question is regarding the diode used across the relay coil. My A level textbook (the section from the textbook is given below) says that its purpose is to protect the op-amp from a back emf induced when the current in the coil is switched off. But this explanation is not adequate enough for me. I don't understand how this will protect the om-amp.

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My problems are:

1)I don't understand, how the emf is induced in the relay coil when the current is switched off.
2)I can't visualize how the current will be moving in the flyback diode when the current in the coil is switched off.
3)I don't understand how the diode will protect the op-amp, given that the first diode will stop any current from flowing back to the op-amp.

Please explain in layman terms. My qualification is just UK Cambridge international A level physics. Hence I don't understand the other complicated explanations given about flyback diode in this site.

thank you very much!


In layman's terms, a current through inductance cannot change suddenly, so even if supply voltage is cut off, the energy stored in the inductor would still like to keep the same current flowing. It must generate voltage over the inductor to do that, and this can be limited with the diode over the inductor. The current has a path via the diode and voltage is limited to the diode forward voltage until the current has ramped to zero.

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