In the past few days, I've encountered an application note on the Maxim Integrated website about rapid demagnetisation of an inductive load:
https://www.maximintegrated.com/en/design/technical-documents/app-notes/6/6307.html
There was the technique of using a Zener diode in series with the classic freewheeling diode, and I am interested in knowing how to calculate the Zener voltage and power rating e.g. for a 24 VDC/400 mW relay coil.
The higher voltage you allow the coil to spike at, the faster the current will turn off.
Something, usually a transistor, is driving the inductive load. You have to keep the maximum spike voltage below what will kill the transistor. Different transistors have different breakdown voltages, consult the data sheet for the one you're using.
The Zener has to conduct the same current that the coil does when energised, at a power given by the product of its voltage and the current. It has to absorb the same amount of energy as heat that the coil has stored when on. You rarely find single pulse energy ratings for zener diodes, but generally if they will handle the power and current from a small coil, they'll handle the energy. If it's a big coil, and a small zener (for certain values of small and big) you might have problems.
Just about any zener will handle the 17mA (decreasing rapidly to zero) current from the coil. An ordinary 400mW type would be fine. If you're insistent on using the smallest SMT type you might want to investigate transient thermal resistance etc.
The voltage should be as high as possible given the capabilities of the driver transistor. Since the load is inductive, you should look at safe operating area as well as simply looking at breakdown voltage, especially for small geometry BJTs. If in doubt, a larger transistor is usually a good idea.
The Zener voltage needs to be selected such that the Vdc+Vzener+Vf is less than the breakdown voltage of your high-side switch.
Instantaneous power of the zener needs to be greater than Vzener*Ion; and continuous power of the zener needs to be high enough to deal with the switching duty cycle of the relay.
Choose safety factors for your voltages and powers that suit your needs.
The Zener voltage is chosen to keep the voltage on the drain of the MOSFET below its breakdown voltage.
Example: VDC = 330 Volt Vbreakdown = 650 Volt Vflyback = 200 Volt (Voltage developed by the coil)
Vmargin = Vbreakdown - Vflyback - VDC
Vmargin = 650 - 330 - 200 = 120 Volt
Vzener must be smaller than Vmargin.
In this particular case I would pick a 100 Volt Zener diode.
Note: In my formula I omitted the leakage inductance of the coil in order to keep my answer simple.
If the zener has the capability of more watts than the relay coil normally consumes, there will be no way to damage the zener, because there's no way to put more energy into the zener than first goes into the coil.
