Help me understand the role and effect of an inductor used in the following electrical circuit.

The circuit in question is a power supply circuit for around 50 servers in a data center. The circuit uses a MOSFET transistor for commuting the power provided to the servers. In order to protect the transistor, the apparatus must be able to evaluate the connected charge in case it might be a short-circuit or too high of a capacity, as this will destroy the transistor.

For this purpose, an inductor is connected in series with the circuit, but I fail to understand how this addition fulfills the intended role.

Circuit schematic

  • 3
    \$\begingroup\$ It's a mystery to us as well. Can you provide some sort of schematic? \$\endgroup\$ Commented Jul 18, 2018 at 13:31
  • 2
    \$\begingroup\$ And where is the circuit? \$\endgroup\$
    – PlasmaHH
    Commented Jul 18, 2018 at 13:31
  • \$\begingroup\$ Added a quickly made schematic of the part near the inductance. The left side is connected to the power supply circuit, and the right to the servers. \$\endgroup\$ Commented Jul 18, 2018 at 13:44
  • \$\begingroup\$ THe inductor can delay the rise in fault current in sufficient microseconds for a fast transistor with current sensing to quickly shutdown . The diode switches only rapid shutoff current to ground that would otherwise result in a large negative voltage spike. \$\endgroup\$ Commented Jul 18, 2018 at 14:24

1 Answer 1


If a MOSFET drives a voltage into a short circuit or a circuit with an impedance that is too low it will exceed its ratings and become damaged. If a small value inductor is placed in series with the MOSFET, it will not cause a rapid short circuit but a ramping-up of current. This is due to the formula for an inductor: -

$$V = L\dfrac{di}{dt}$$

That ramp-up of current is proportional to voltage AND, if that current ramp is monitored (via a small series resistor), it can be quickly ascertained that either the incoming supply voltage or the load are beyond acceptable limits before the current reaches a damaging level to the MOSFET.

  • \$\begingroup\$ Does this imply that a high capacity = low impedance? \$\endgroup\$ Commented Jul 18, 2018 at 13:48
  • \$\begingroup\$ I don't know what your definition for "high capacity" is. \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2018 at 13:48
  • \$\begingroup\$ If there is a discharged capacitor of a relatively high capacity directly connected to the circuit output, can it possibly force it to output a too high of a current compared to the MOSFET ratings? \$\endgroup\$ Commented Jul 18, 2018 at 13:54
  • \$\begingroup\$ @AlexisSorokine yes this could happen and the inductor circuit would likely detect this but, that isn't too bad because once the capacitor is charged a little (by the first attempt at closing the MOSFET), the MOSFET can try again within usually a few microseconds and advance the capacitor voltage a bit more. This of course requires a control loop but, within a few tens of milliseconds, the server capacitor could be close to full potential and the current it now takes is much smaller and more manageable. \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2018 at 13:58

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