So I'm still new to electronics, and I've been taking a look at Boost Converters and such (just learning power supplies and the different types)...which got into explaining inductors. Needless to say it was a little bit to take in. Inductors seem quite complex for such a simple component.

  1. Just so I have this straight, inductors resist change in current, so if the current is lowering it will "create" a higher voltage to try and make up for this according to Lenz's law. (Is this correct? ....does anyone know how this happens?). When it is creating this voltage is the current being lowered or just drained quicker?

  2. In a schematic such as this:

    Lets pretend the diode wasn't there. What would happen? Would the inductor just keep building up energy with nowhere to go? Would it just dissipate in the air? In the Wiki article it said it would arc over to the next wire. Is there a limit to how far it can arc (like What if the wires were FAR away: would the inductor melt, or would the energy just dissipate in the air?

  3. What determines how much energy an inductor can store? The number of turns? Or does the size of the inductor actually matter as far as "rate" of storage.

  4. Unrelated Sort-of, but are there any "cool" experiments I can do with them to just kinda see how they work? I saw this one on youtube essentially he just has a switch that he turns on and off and you can see the voltage jump up super high. Im assuming this is how a boost converter works.

Sorry for the multiple questions, just trying to grasp the magic of inductors. They seem so simple (A coil of wire) but do so many crazy things.


3 Answers 3


Yes, a inductor sortof resists current changes, just like a capacitor resists voltage changes. In fact, inductors and capacitors are current/voltage mirrors of each other. The way I like to think of inductors in circuits is that they give inertia to current. They don't of course, but it seems a useful conceptualization technique.

In the schematic without the diode, if everything starts out at 0 and the switch is closed, the current will be a exponential decay toward Vs/R. Initially all the voltage is accross the inductor, and in the steady state there is 0 voltage accross it.

The interesting stuff happens when the switch is opened. At any one instance, the inductor will maintain its current constant. This includes the instance the switch is opened. Without the diode, there is no obvious path for the current. The inductor voltage will increase to whatever maintains the current thru it.

A mechanical switch works by touching together two conductors. When the switch opens, the conductors move away from each other. This can't happen instantly, so when the switch first tries to stop the current thru it, the contacts will be very close together. It won't take much voltage to cause arc over. Once the arc is started, the gas between the contacts becomes a plasma, which has high conductivity. The arc can therefore continue for a while as the contacts move farther apart. During this time, the voltage accross the switch isn't zero, so the inductor current decreases. As the contacts move further apart, the arc voltage increases, decreasing the inductor current more rapidly.

Eventually the current is low enough that it can't sustain the arc and the switch finally opens for real. At that point there is little energy left in the inductor. The only place for that current to go is onto the inevitable parasitic capacitance accross the inductor and other parts of the circuit. Every two conductors in the universe have some non-zero capacitance between them. This capacitance is small, and therefore the voltage will rise quickly. This also decreases the current in the inductor rapidly. Eventually a peak is reached where the voltage on the capacitance actually starts to push the inductor current the other way. In a perfect system, all the energy on the capacitance would be transferred to the inductor as current, but this time in the opposite direction. Then it would charge up the capacitance again in the opposite direction, and the whole cycle would repeat indefinitely. In the real world there is some loss, so each swing back and forth will be a little lower in amplitude as energy is lost as it is being sloshed back and forth between the inductor and the capacitance. Voltage plotted as a function of time (as a oscilloscope does) will show a sine wave with amplitude decaying exponentially towards Vs.

  • \$\begingroup\$ I think I got most of that, except the part where your talking about reverse direction. If the switch was large enough (like a larger cap between the two conductors (and fast enough), the Voltage I assume Voltage would build quicker? But what your saying is the Voltage/Current Would eventually just decay in the Inductor itself if the switched was left open? \$\endgroup\$
    – user3073
    Commented May 30, 2012 at 21:37
  • \$\begingroup\$ @Sauron: It's not clear what you are really asking, but if you made a switch that could turn off faster, there would be more energy left in the inductor and the therefore more energy left to ring with the inevitable capacitance. Enough energy would require a high enough voltage that it could arc elsewere. After all arcing stops, you are left with the capacitor/inductor system holding the energy, which will ring with a decaying envelope as the energy is lost in resistance as it sloshes back and forth between inductor and capacitance. \$\endgroup\$ Commented May 30, 2012 at 23:28
  • \$\begingroup\$ @RussellMcMahon: Perhaps your first "Capacitor" was intended to be "Inductor?" \$\endgroup\$
    – Shamtam
    Commented Jun 5, 2012 at 11:12
  • \$\begingroup\$ @Shamtan - No, but thanks, one did need changing. - second capacitor is inductor - it has DC continuity. \$\endgroup\$
    – Russell McMahon
    Commented Jun 5, 2012 at 14:18
  • \$\begingroup\$ Here's a water model: Resistor = pipe with variable length or dia. || Capacitor is a rubber sheet across a tank with entry and exit either side of sheet. Vary tank size and sheet stiffness. || Inductor is a pipe with a rubber walled section. Pressure makes the wall bulge and retain current and build pressure. Not a perfecyt model but gives some feel. \$\endgroup\$
    – Russell McMahon
    Commented Jun 5, 2012 at 14:19

(1) Yes, inductors resist change in the flow of electrons. Lenz's law, Maxwell's laws, and the equations in any electronics textbook or physics textbook a b c d e work great for calculating the relationship between current, voltage, inductance, magnetic field strength, etc., much like Ohm's law works great for calculating the relationship between current, voltage, and resistance.

As any of those textbooks will tell you, for any short amount of time dt, the change in current through an inductor will be very small (di), and can be exactly calculated as

di = v dt / L

where v is the average voltage across the inductor during that short amount of time and L is the inductance.

The larger the reverse voltage across the inductor, the faster the current drops to zero.

(This is still true whether we are forcing the voltage across the inductor to be some particular voltage by putting a battery across it, or whether we have some load resistance across the inductor and the voltage is somehow being caused by the inductor itself).

When we apply a voltage across an inductor, the current slowly rises, and energy goes into the inductor, stored in a rising magnetic field inside and outside the inductor.

When we disconnect the inductor from the power source, leaving some resistance connected between the ends of the inductor, the current slowly falls. Meanwhile, and the energy comes out of the mysterious, invisible magnetic field (g) and into whatever is connected to the inductor.

(2) Olin gives an excellent answer.

(3) As any of those textbooks will tell you, the energy e stored in an inductor at any instant is

e = (1/2)L i^2,

where i is the current at that instant. This energy (magnetic field energy) is the same as the amount of electrical energy that would come out of a battery (it doesn't matter what voltage) connected to that inductor during the time it takes to ramp the current from 0 to that same i.

With any given physical inductor (so we are given some fixed L), the amount of energy that I can store in that inductor is generally limited by the maximum current rating of that inductor. High-power inductors generally use thicker wires and better ways of getting heat out of the wires, but exceeding the current rating causes those wires to melt and fail. This is a maximum energy rating, not maximum power rating -- many designers fill inductors (and also transformers, for the same reasons) with energy and then dump it out again thousands or millions of times a second, in order to get more power through the system than if they only did it 60 times a second.

I find o'scopes excellent for "seeing" what goes on in circuits with inductors. Perhaps you might enjoy building some kind of switch-mode voltage regulator such as the Roman Black +5v to +13v boost Converter.

  • \$\begingroup\$ An Inductor won't do anything in a D/C Circuit correct? like it wouldn't be up any voltage or anything im assuming? \$\endgroup\$
    – user3073
    Commented May 30, 2012 at 22:23
  • \$\begingroup\$ Yes, in a DC circuit (i.e. where the current through the inductor is constant), the inductor will appear to be a short circuit; it will have no voltage drop across it. \$\endgroup\$
    – Jason R
    Commented May 31, 2012 at 2:13
  • \$\begingroup\$ @Sauron - Note that having a DC power supply it's not a DC circuit when you close the switch! It's DC in steady state. \$\endgroup\$
    – stevenvh
    Commented May 31, 2012 at 10:19
  • \$\begingroup\$ So in a D/C Circuit if you had an inductor and a switch.....opening the switch wouldn't cause it to Arc then? \$\endgroup\$
    – user3073
    Commented May 31, 2012 at 11:26
  • 1
    \$\begingroup\$ @Sauron - Yes it will. The inductor's current doesn't stop immediately, and it will act as a voltage source to keep it flowing. The higher the resistance, the higher the voltage it will generate. The energy comes from the magnetic field that it built up. So the voltage across the switch can become very high which will the arc to persist longer. \$\endgroup\$
    – stevenvh
    Commented Jun 1, 2012 at 4:11

This is very interesting question. Just for clarification, I will reword it. For ideal inductance with non zero current, zero capacitance and ohmic components, what happenes when DC path is destroyed with lossless switch ? No thermal dissipation, no ringing is allowed, no DC either, as there is no switch. The energy conservation law must be met completely.

I certainly understand that even with all things ideal, there is a physical materially measurable gap which will allow the current to continue flowing even through vacuum. But what if vacuum is perfect insulator ?

There is no real correct answer, as even arithmetical infinities and zero propagation times, infinite speed of light etc. will not help.

But say, if all abstraction still allow material charge particles to be involved, the conductor will violate electroneutrality and will loose the cloud of electrons, which will continue travelling with some inertia away from conductor. The magnetic field will turn momentarily from being toroid into cylinder, then culon force will return particles back into conductor. Repeating forever, it will be ringing, but with volumetric (or as you wish electrostatic) capacitance of of coil body (not the parasitic capacitance).

Hmm. Still problem with having non-ideality. If wire is infinitely thing, then there is no capacitance, the frequency will be infinite, higher than gamma. It is like a big bang all over again but with limited total energy.

The answer: with all thing ideal the magnetic pulse produces will be Dirac Delta Function, infinitely high and infinitely narrow pulse with integral of 1. (or any particular total integral depending on initial total energy).

The closest practical device is studied in Los Alamos http://en.wikipedia.org/wiki/Explosively_pumped_flux_compression_generator


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