5
\$\begingroup\$

I sometimes see that "A perfect inductor would be with superconductors" and so on. What properties of "Perfect inductors" makes them different from regular and what advantage would they create towards electronic circuits?

\$\endgroup\$
5
  • \$\begingroup\$ This is a fairly subjective question, so it may draw some downvote attention (the 'advantage' part, at least). \$\endgroup\$ Commented Jun 13, 2013 at 2:06
  • 1
    \$\begingroup\$ You need to add some context because that statement doesn't make sense alone. I'm pretty sure the term you're looking for is ideal inductor. \$\endgroup\$
    – Matt Young
    Commented Jun 13, 2013 at 2:13
  • 1
    \$\begingroup\$ If you want to see a superconducting inductor, visit an MRI or particle accelerator facility. They call them "superconducting electromagnets", but they are inductors. \$\endgroup\$
    – Phil Frost
    Commented Jun 13, 2013 at 3:05
  • \$\begingroup\$ A theoretically perfect inductor might have infinite inductance and zero resistance. You can approximate this simply by having two pieces of wire that are not connected. An infinite inductance would take an infinite period of time before it would pass a measurable amount of current. Since we don't have a infinite amount of time, a broken connection will have fairly equivalent behavior for the time-scales on which we can make measurements. \$\endgroup\$ Commented Jun 13, 2013 at 4:56
  • \$\begingroup\$ @ConnorWolf: You make an interesting point. The amount of stored in an inductor which is passing a certain amount of current is proportional to the inductance, but the amount of energy that can be added to an initially-empty inductor by applying a certain voltage for a certain time will be inversely proportional to the inductance. \$\endgroup\$
    – supercat
    Commented Jun 13, 2013 at 15:45

5 Answers 5

6
\$\begingroup\$

Ideal or perfect inductor would be/have, in my book: -

  • Zero DC resistance (unless requiring inductor with defined-peaking characteristics in a tuned circuit or winding a solenoid that naturally suits having a DC resistance)
  • Zero core loss (eddy current loss) unless requiring an EMI suppressor
  • Zero hysteresis loss
  • Linear i.e. has no saturation (unless you are requiring a saturable reactor or desiring to create 3rd harmonic distortion)
  • Zero capacitance and hence no self resonant frequency
  • No change in L as temperature changes
  • No curie point (applies to non-air-cores I believe) unless by design you need one.
  • Zero disaccomodation factor (no change in permeability with mechanical shock)
  • No flux leakage unless building a transformer.

Hopefully you can see that some "ideal" or perfect requirements do not suit other applications.

\$\endgroup\$
10
\$\begingroup\$

The perfect inductor has reactance without any resistance. In other words, the real component of its impedance would be zero. Loss of power as heat within the inductor is thus also zero.

The perfect inductor presents no impedance to a constant current (i.e. DC), yet opposes any slightest change of current. Any non-superconducting material can not meet this condition, as it is bound to have some resistance.

Hence, a perfect inductor would need to be made of superconducting material.

Advantages at the trivial level would be elimination of any wasted power through resistive heating in an inductor, of course. Beyond that, one enters the realm of speculation: There may be many advantages, but also design challenges.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ One advantage beyond elimination of wasted power is achieving a high Q for tuned LC circuits. \$\endgroup\$
    – Kaz
    Commented Jun 13, 2013 at 6:31
  • \$\begingroup\$ @Kaz Agreed: That falls into the realms of speculation for me: What the inductor is being used for would determine such advantages... After all, intentional detuning for lower Q and wider pass band is a real requirement for some designs :-) \$\endgroup\$ Commented Jun 13, 2013 at 6:35
  • \$\begingroup\$ It would also have the ability to store an infinite amount of (magnetic field) energy, create an infinitely strong magnetic field and most probably destroy the universe. Fortunately the perfect inductor like the perfect answer doesn't exist. \$\endgroup\$ Commented Jun 13, 2013 at 10:16
  • \$\begingroup\$ @JImDearden Actually, no: The perfect inductor in and of itself does not specify any given core. Air cores, ferrite cores, and others, will each have their own saturation points, and therefore will be the limiting elements. \$\endgroup\$ Commented Jun 13, 2013 at 11:52
  • 1
    \$\begingroup\$ @JImDearden: I don't see how an inductor which could store an infinite amount of energy could threaten the universe, absent a source of infinite energy with which to feed it. It's hardly beyond the realm of imagination that superconductors could create inductors that were, within certain operating limits, "perfect" to many significant figures, nor is it beyond the realm of imagination that such things might have practical uses. Their abilities, however, would not include collapsing the universe. \$\endgroup\$
    – supercat
    Commented Jun 13, 2013 at 15:49
4
\$\begingroup\$

A perfect inductor, at first glance, would have the following:

  • zero series resistance
  • infinite permeability
  • infinite saturation flux density
  • zero core loss
  • infinitessimal volume

I wouldn't want to try switching one in a regulator, though. The voltage induced when the magnetic field collapses would be a sight to see :)

\$\endgroup\$
2
  • 3
    \$\begingroup\$ What would you expect to see, really? A superconducting inductor doesn't have infinite inductance; it just has zero series resistance. The induced voltage will still be defined by the impedance you present to the inductor as it prevents instantaneous changes in current, just like with ordinary inductors. \$\endgroup\$
    – Phil Frost
    Commented Jun 13, 2013 at 3:04
  • \$\begingroup\$ @PhilFrost: I think the biggest observable difference would be that because the efficiency of adding energy to a conventional inductor drops with the amount of energy already added; the total energy that can be stored using a given drive voltage is thus finite. A superconducting inductor might be able to hold a lot more energy than could be practically stored in a conventionally-resistive inductor of the same value. \$\endgroup\$
    – supercat
    Commented Jun 13, 2013 at 15:54
4
\$\begingroup\$

A "perfect inductor" (or "ideal inductor") would be a two-terminal device with the following voltage - current relationship:

\$v_L = L \dfrac{di_L}{dt} \$

Note that this implies that the voltage across the device is zero when there is a steady (constant) current thus, it would necessarily be the case that such a device would have zero resistance, i.e., it would not dissipate energy but only store or deliver it.

It's really the lack of properties like, for example, resistance, capacitance and the associated self-resonance frequency, etc. that distinguishes a perfect inductor from an "ordinary" inductor.

A perfect inductor would be simple in that it would possess the property of inductance period. This would certainly be an advantage in a circuit in that you wouldn't need to take into account the non-ideal properties of a real, non-perfect inductor.

In other words, a "perfect" inductor is a fantasy. It doesn't exist except in the abstract world of ideal circuit theory.

\$\endgroup\$
0
\$\begingroup\$

A perfect inductor that had it's 2 leads connected by zero ohms would maintain a current flow through itself forever, would it not? Assuming we could get current started. Regardless of the value of inductance. However any form of measurement would rob energy and cause the current to be reduced.

It would act as a flywheel spinning in free space. Interesting, unobtainable, stuff.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.