This is most likely a dumb question, just for some reason I have just forgotten the fundamentals about how charging and discharging work.

Essentially my question is for this simple circuit: enter image description here

When the switch is closed the inductor keeps charging and when it is opened it discharges through the diode. Now here is my question. Why doesn't the inductor keep charging as the current is entering back through the inductor? This is the fundamental knowledge I have forgotten.

Is it due to the diode voltage drop? Hence charge out is always less than charge in? (I know its a bit of a dumb question but I would like to just clear my confusion).

Also if the diode wasn't there and say the inductor had charge in it already (the switch is open at this time) and say the diode shorted completely, no charge will be lost will it?

****In short what causes an inductor to discharge again and how does it discharge through itself (when the switch is open). Does it discharge through itself due to internal resistance or no? ****

  • \$\begingroup\$ The stored energy, in the form of flux in the core of the energy, builds up, and then declines. The polarity of dFlux/dTime sets the polarity of the inductor's voltage. When that switch opens, the source of energy has been disconnected, and the flux must begin to reduce; the very start of reduction causes dFlux/dTime to change polarity. \$\endgroup\$ – analogsystemsrf May 15 '19 at 3:32
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    \$\begingroup\$ Dunno if this'll clarify or not: the inductor is not charged with charge. It's charged with flux in the core (or the air in the middle of the coil). So it's not about accumulated charge -- it's about built-up magnetic field. \$\endgroup\$ – TimWescott May 15 '19 at 4:49

When you send current through the inductor, the inductor will use the energy in that current to maintain a magnetic field. When the power source that was supplying the current through the inductor dissapears, the inductor tries to keep the current flowing through it the same and it does so using the energy stored in the magnetic field.

The inductor uses the energy in the magnetic field (which corresponds with the magnetic field collapsing as its energy is consumed) to maintain the current as close as possible to the same current levels as when the current source disappeared.

The inductor does this by becoming a power source using the energy stored in the magnetic field to produce a voltage as high as necessary to push that level of current through any obstacles so that the current flowing through the inductor is the same. If the easiest path is across the air gap of the open contacts of a mechanical switch or relay, the voltage required to spark across that air gap will be very high so that's the voltage that will be produced. That requires a lot of energy so the current will collapse to zero quickly.

If the easiest path is something like a diode then then the voltage (i.e. the diode's forward voltage drop) required to push that current won't need to be so high and the energy stored in the magnetic field lasts longer so the current drops more slowly.

  • \$\begingroup\$ Ok thanks, so essentially is it due to the diode voltage drop and with time that the source (inductor) since it only has a set charge, it has to become less? Also what causes the inductor to not keep charging itself if current is entering back through it? \$\endgroup\$ – Student May 15 '19 at 3:33
  • \$\begingroup\$ If somehow, the discharge current was less than the current entering back through after the diode, would the inductor charge? \$\endgroup\$ – Student May 15 '19 at 3:34
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    \$\begingroup\$ In an ideal world, you could swap the power source across an inductor with a perfect wire and current would continue to circulate through forever. It would would circulate indefinitely (or until you did something to draw energy from the system such as through heat or radio waves) since no energy was lost. It would be like a perpetual motion machine (the kind where energy is not lost, not the kind where you can draw infinite energy from). Think about why we use superconductors in MRI machines. \$\endgroup\$ – DKNguyen May 15 '19 at 3:37
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    \$\begingroup\$ Yes, in the case with the diode, the diode is a device that supplies no power, but has a voltage drop across it and current flowing through it, which means it is dissipating energy. Same thing with resistors, except that the voltage drop through a resistor drops as the current drops so the time it takes for the current and magnetic field has different characteristics than if it were a diode \$\endgroup\$ – DKNguyen May 15 '19 at 3:40
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    \$\begingroup\$ That's why LC circuits oscillate. The inductor's magnetic field collapses as the energy is used to push current into the capacitor, storing energy in the cap. But then the balance swings back the other way and the charge in the capacitor produces a voltage sufficient to drive current through the inductor which restores the energy back into the magnetic field. The energy sloshes back and forth this way causing oscillation. \$\endgroup\$ – DKNguyen May 15 '19 at 3:42

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