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Hello I would like to design protection for HSD, BTS7200-2EPA against inductive load. Spec data for the inductive load is limited. But the PCB was designed for 2.5A peak, and approximate L and R_L values were obtained as 140mH and 9.5 ohm, respectively through measurement. (Approximately 1.4A at 12V)

Refer to HSD datasheet, enter image description here

I got to know that putting a freewheel diode and Zener in series in parallel to load is generally used to protect against reverse voltage from inductive load when switched off.

enter image description here

First of all, can you explain how the concept works during the demagnetization? How fast is the current ramp and how is energy dissipated during the reverse voltage clamp etc.

And next step is how should I select D1 and D2? In consideration to V_Breakdown, Power rating etc.

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The voltage on an inductor is Ldi/dt. Your "L" is 140 mH. Your device can switch "off" in 15 usec, so if your inductor has 1.4 amps flowing through it, the current changes from 1.4 amps to 0 in this period, making the current rate of change (di/dt) -1.4amps/15usec or -93,333 amperes per second. So in an ideal world, the voltage on L would (L*di/dt) would be negative 13,067 volts.

Long before this theoretical negative voltage is reached, something will start to conduct. If you have a high speed catch diode in place, once the diode's forward voltage is reached it switches "on," the diode will start to conduct and the voltage will stop changing. The diode must be able to handle the full 1.4 amperes, and the energy stored in the inductor will be shared between heat in the 9.5-ohm inductor resistance and the diode's forward drop x the instantaneous current.

The energy stored in an inductor is 1/2 * L * i^2 . This amount of energy is absorbed each time the switch is turned "off" so the losses are a function of how often switching occurs. In your case, choose a diode for D1 that can handle 1.4 amps of forward current and has a voltage rating greater than Vs. I generally prefer at least a factor of two for design margin.

D2 is not needed unless you want the instantaneous negative voltage spike to swing more negative than D1 diode drop.

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  • \$\begingroup\$ Thanks for explaination. When switch-off, D1 switches "on". The negative voltage across the inductor forward biases the diode. You mentioned "E" stored in the inductor will be shared between heat in the 9.5-ohm inductor resistance and the diode's forward drop x the instantaneous current. Coil Suppression Can Reduce Relay Life \$\endgroup\$
    – min_c
    Commented Apr 3 at 7:26
  • \$\begingroup\$ E=0.5*0.140*1.4^2 =70mJ. Then wouldn't that cause damage to the load in the long run? Coil Suppression Can Reduce Relay Life \$\endgroup\$
    – min_c
    Commented Apr 3 at 7:51
  • \$\begingroup\$ It would if the coil were able to deliver all its energy at once. But the energy is spread over the coil discharge period. The coil voltage will be -0.7 volts (or -0.3 for a Schottky), so you can once again calculate di/dt and estimate the discharge time. If you use a Zener as in your diagram, the voltage will be higher and the corresponding discharge rate will be faster. This can be useful if you want the coil to discharge more quickly - as is sometimes the case for solenoids when you want them to release right away when power is removed. \$\endgroup\$
    – John Birckhead
    Commented Apr 3 at 17:41
  • \$\begingroup\$ A diode that provides a millijoule rating usually refers to the avalanche mode - when the diode is back-biased and breaks down. This can occur in snubber applications in which case the diode is still discharging a coil when overvoltage is applied to the diode and it breaks down, which happens very fast. \$\endgroup\$
    – John Birckhead
    Commented Apr 3 at 17:47

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