Deenergizing Inductor

by user3048782   Last Updated November 09, 2018 16:25 PM

I am reading through "Practical Electronics for Inventors" and I came upon a chapter about deenergizing inductor. I understand the concept that when you cut the current going into inductor, it will generate high voltage drop, since dI/dt goes to infinity. How come this is not an issue when the inductor "starts"? The voltage drop should be the same, right? Why this isn't an issue with capacitor? When we "turn" it on/off, there should be infinite current (minus internal resistance) according to laws. Shouldn't high current be a bigger problem than high voltage? What am I missing?

Answers 1

Think of inductors and capacitors as complimentary components.


simulate this circuit – Schematic created using CircuitLab

Figure 1.

  • (a) The inductor current is limited at start so no problem at switch on as it will rise more slowly than an equivelant DC resistor would. As you have identified, the problem is at switch off.
  • (b) The capacitor is the opposite. There is a big problem at switch on as it behaves as a short-circuit so current is only limited by any inherent resistance in the circuit. At switch off there is no problem.
  • (c) The solution for the inductor is to maintain current after switch-off using the snubber diode.
  • (d) The standard solution for capacitors is to limit the charge with a series resistor. This may be bridged out when the voltage has reached a certain level. (This is done in high-powered switched-mode power-supplies or variable frequency drives, for example.)
November 09, 2018 16:14 PM

Related Questions

Updated August 10, 2015 17:11 PM

Updated October 02, 2017 06:25 AM

Updated March 06, 2018 15:25 PM

Updated December 14, 2018 17:25 PM

Updated October 06, 2018 03:25 AM