I hope the following will help the people that aren't into electronics all that much. An inductor (relay coil, solenoid, choke, etc.) stores energy in its magnetic field. When the source of current is removed, the magnetic field will collapse back into the core at fast as it can, limited only by current flow, if any. Until current flows the voltage can reach many hundreds of times the actual supply voltage but the polarity is opposite that which created the field since the magnetic field is no longer expanding outward but collapsing inward. This is called the "flyback effect" and it is used to great advantage when you need 6V, 12V, or 200V from a 1.5V penlight battery.
With a diode connected "backwards" in parallel with the coil, the diode will become forward biased when the inductor's field collapses. The energy in the collapsing field is dissipated as heat in the diode and coil winding resistance. If the diode is a silicon diode, then the (negative) voltage of the collapsing field will be limited to the normal forward diode drop (0.7V, for silicon diodes) . There is no need for a high PIV diode since the diode is reversed biased when the relay is energized and that voltage is typically low, say 12V. A 50 PIV diode would work fine. The only real criteria for the diode are that it be able to handle the current from the collapsing field. That's why you don't see signal diodes serving that function. And you don't need a "fast" diode, either. The clunky power diode will be fine. It has a lot of junction capacitance, which makes it "slow", but the capacitance is in parallel with the (theoretical) diode and NOTHING will allow the forward voltage to exceed the diode drop (0.7V, for example). The added junction capacitance only helps "eat" the collapsing field energy as it "charges" the capacitance until the diode clamps "on" in forward bias. So, what happens with a diode-protected relay is that the diode will conduct momentarily when the drive to the coil goes away. A small negative voltage (diode forward drop) will appear briefly until the stored energy is dissipated. The drive circuit sees nothing more than a small negative voltage during that time and thus the circuit is protected. Rick KC0OV ______________________________________________________________ Elecraft mailing list Home: http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/mmfaq.htm Post: mailto:Elecraft@mailman.qth.net This list hosted by: http://www.qsl.net Please help support this email list: http://www.qsl.net/donate.html
