Electronic – Discharging an inductor

currentinductorvoltage

I wanted to know if it's practically possible to use inductors the same way we use capacitors , in the sense of charging it using a charging circuit then discharge in another circuit.

I thought it could be possible but then i read on the web that when an inductor is disconnected the magnetic field starts collapsing inducing a very high voltage, wouldn't this voltage breaks down any transistors used in the switching from the charging to the discharging circuit? If i understand this correctly, this only happens when there is no way for the current to go so voltage builds up, but then again there is a small split of second between the switching between the 2 circuits, will that cause such a problem?

Edit:

If I used this circuit to charge and discharge the inductor, will there be any voltage spikes?

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Best Answer

I find it helpful to think of capacitors and inductors to be complimentary.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Ideal and imperfect components.

  • Capacitors store energy in an electric field. Inductors store energy in a magnetic field.
  • A capacitor holds energy when open circuit. An inductor holds energy when short circuited.
  • Capacitors lose energy through parallel leakage resistance. Inductors lose energy through series resistance.
  • Capacitors "like" to keep the voltage across them constant. Inductors like to keep the current through them constant.
  • When a capacitor is short circuited the resultant current is very high. When an inductor is open-circuited the resultant voltage is very high.

... when an inductor is disconnected the magnetic field starts collapsing inducing a very high voltage, wouldn't this voltage breaks down any transistors used in the switching from the charging to the discharging circuit?

Yes it would but there's a simple solution:

enter image description here

Figure 2. A simple buck converter. Source: All About Circuits.

In Figure 2 S is the transistor switch similar to that mentioned in your question. When it is switched the inductor tends to maintain current in the direction of I. Since the right side of L is "held" by C and current is to keep going then the left side of L goes negative to try to maintain current. Whe the voltage reaches -0.7 V D starts to conduct and maintains the current through L keeping it "happy" and avoiding a transient high voltage.

You will see this arrangement more commonly in snubber diodes on relay coils.

schematic

simulate this circuit

Figure 3. A typical relay control circuit. Without D1 the inductance of the relay coil would generate a large negative voltage on switch off. This would be likely to destroy Q1 . The diode limits the negative excursion on Q1 collector to -0.7 V.

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