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I know the mathematical reason behind this. Given the inductor I-V characteristic we know that:

enter image description here

I understand mathematically that a function which is differentiable must be continuous, but what is the physical reason behind this?

We know that current creates magnetic field and that the energy stored in that magnetic field is:

enter image description here

I guess it is pretty much related to the energy somehow, but then what's next, and what is the explanation for this?

The same question applies to the voltage of capacitor and energy stored in the electric field in case of a capacitor.

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    \$\begingroup\$ I think this may fall under Physics SE rather than EEE SE unless you're happy with a very simplistic explanation such as "the current/voltage represents the energy stored in an inductor/capacitor and you can't instantaneously change the energy in a step amount because it would require infinite voltage/current". Similar to how kinetic energy of an object is stored in it's velocity and you can't dump that energy instantaneously or achieve a step change in velocity without infinite force to decelerate it an infinite amount. \$\endgroup\$
    – DKNguyen
    Apr 25 at 21:19
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    \$\begingroup\$ Well, instantaneous change in current means infinite di/dt which means infinite voltage. Irrespective of whether the mathematical function is differentiable etc. Infinite voltage will exceed the dielectric breakdown strength of all real materials. Infinite voltage in a vacuum gap will cause electron ejection and acceleration so that won't work either. \$\endgroup\$
    – mkeith
    Apr 25 at 22:41
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    \$\begingroup\$ Related question on Physics SE. \$\endgroup\$ Apr 26 at 6:57
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    \$\begingroup\$ @gvg be aware of the new comments under the answer you had accepted. In a nutshell, the back-emf of an inductor does not define, or shape or influence the current that flows. Just as the voltage developed across a resistor due to current flow does not seek to restrict that current in any way. There is a lot of faulty answers for inductors to be found on the internet so be cautious here. \$\endgroup\$
    – Andy aka
    Apr 26 at 12:22
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    \$\begingroup\$ @Andyaka unless you are aiming for a pedantic argument where everything influences everything else and there is no causality, you seem to be simply wrong, to me... the voltage developed across a resistor does act in opposition to the current flowing through it. \$\endgroup\$
    – user253751
    Apr 26 at 17:30

6 Answers 6

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Here's the real reason.

A moving charge (current) produces a magnetic field.

When a current is applied to an inductor (coils of wire), there is a magnetic field that builds up within the inductor.

The change in magnetic field (flux) produces an EMF across the inductor. The EMF opposes the change in current through the inductor.

So you have an EMF fighting the change in current.

The higher the rate of change in current, the higher the EMF across the inductor opposing it.

That's why the current doesn't change instantaneously.

Key point to remember is that the energy storage mechanism of an inductor is a magnetic field.

The magnetic field (energy) also can't change instantaneously. It's physically impossible to instantaneously change the energy in an inductor (or capacitor).

The fields build up or collapse with respect to time.

Have a look at Lenz's and Faraday's laws. https://en.m.wikipedia.org/wiki/Lenz's_law https://en.m.wikipedia.org/wiki/Faraday's_law_of_induction

One important equation is this:

EMF = -N*dΦ/dt

Look at the EMF equation in terms of magnetic flux.

See that EMF is proportional to the rate of change in flux?

Looks similar to the equation in terms of current doesn't it :-)

Here are some more examples along with a simple picture: https://andropoide.blogspot.com/2017/02/inductor-lenz-law.html?m=1 enter image description here

Hope this helps.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Apr 26 at 19:05
  • \$\begingroup\$ @mrbean Could you also please give us the proof of why voltage cannot change instantaneously in a given capacitor? I'm curious, because in that case there will not be any back-emf. \$\endgroup\$
    – gvg
    Apr 26 at 21:41
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    \$\begingroup\$ @gvg Good question. For a capacitor voltage to change, charges need to be moved and stored across the plates. An electric field is created by the charges stored at the plates. Energy in a capacitor is stored in the electric field. That energy can't change instantaneously. Energy can't change instanteously in a capacitor (or inductor) because God said so :-) \$\endgroup\$
    – mrbean
    Apr 26 at 22:30
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    \$\begingroup\$ @mrbean I liked your argumentation. Especially the last one. ;-) \$\endgroup\$
    – gvg
    Apr 27 at 18:59
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Why can't current change instantaneously in a given inductor?

Plausible reasons as I see them: -

  • Current cannot change instantaneously in an inductor without creating infinite voltage and that ain't happening in the real world. It's all in the \$\frac{di}{dt}\$ becoming infinite.

  • Voltage cannot change instantaneously in a capacitor without creating infinite current and that ain't happening in the real world. It's all in the \$\frac{dv}{dt}\$ becoming infinite.

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I guess it is pretty much related to the energy somehow

Since the energy stored in the inductor is proportional to the square of the current, changing the current requires either increasing or decreasing the stored energy.

Therefore you can't change the current instantaneously without delivering (or absorbing) infinite power to (or from) the inductor.

That's basically it. Infinite power sources don't exist in the real world, or really even in the ideal world, so therefore the current through the inductor can't change instantaneously.

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You have it backwards.

The other answers talk about the physics and the maths explanations. They are not wrong. But they are not fundamental.

The only real reason is that when we do the experiment, we notice that current doesn't change instantaneously in an inductor.

This 'noticing' can be quite involved. For instance, Newton noticed that 'When a body is acted upon by a force, the time rate of change of its momentum equals the force'. Before he did careful experiments that reduced friction, people thought that 'things slow down and come to rest'. Once he had invented the concept of 'momentum', and defined 'force', he could express his Second Law.

As it turns out, real inductors tend to be made with wire that has resistance, which complicates its behaviour. When the experiment is done carefully enough, we find that 'when an inductor is acted on by a voltage across it, the time rate of change of the current through it is proportional to the voltage' (any similarity to Newton's Second Law is totally intentional). Instantaneous change of current would imply infinite voltage.

So having noticed this fact about the world, we sought to explain it. Hence inventing the language of voltage, current, conservation laws, and describing the behaviour with mathematics.

We could attempt to describe the behaviour with many different concepts and equations. We keep using the ones that describe what we see when we do the experiment, and reject those that don't. As we come up with deeper theories, we use the test of whether they ultimately describe what we see to decide whether they are worth keeping or not.

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  • \$\begingroup\$ Obviously, this is correct. We can boil the question down to a point where noone can answer it anymore. Like what is magnetism anyway? youtube.com/watch?v=luHDCsYtkTc. My intention was to give him some fundamentals so that when working with a circuit or an inductor he can get a basic idea of what is happening. Both physically and with simple math concepts. \$\endgroup\$
    – mrbean
    Apr 26 at 21:35
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Another simplistic explanation, but energy-based...

With capacitors and coils, to make changes, we'd have to remove some EM energy from, or add some EM energy to, each component.

In wires, what is the rate of electrical energy-transfer? That's the watts, the joules-per-second rate of energy-flow, proportional to V * I.

If we wanted to make an instant change to a capacitor or coil, we'd have to transfer some of their field-energy INSTANTLY, meaning, we're needing a few joules to be transferred per ZERO seconds, meaning infinite joules/sec rate, or infinite wattage.

In other words, with coils/capacitors, the speed of any change is roughly proportional to the watts, to the rate that the field-energy is being moved, in joules/sec. Higher watts gives faster changes to their internal fields. Infinite watts could give instant changes. To create near-instant energy-transfers, we'd need either immense volts, enormous amps, or both.

Knowing this, we can step back down to consider the actual values of volts and amperes.

With a capacitor, the voltage is (somewhat) constant, while the rate of energy-transfer at any point in time is proportional to the current (to watts = Vconst * I = joules/sec rate of energy-transfer.) In capacitors, since voltage is partially fixed, the current begins acting like the energy-flow rate (even though it's not.) So, as others here point out, we'd need infinite current, if we wanted to create infinite watts, to move some of the capacitor's energy instantly, which then steps the capacitor's voltage instantly to a new value.

And with coils, the current is (somewhat) constant, and the rate of energy-transfer is proportional to the voltage. In coils, since current is partially fixed, the voltage begins acting like the energy-flow rate (even though it's not.) To transfer energy instantly, we'd need infinite watts = V * Iconst, which requires an infinite voltage, if we wanted to move some of the coils energy instantly, which then makes the amperes step instantly to a new value.



Misconceptions: isn't current a bit like energy-flow? Nope. The path for current is in closed loops, in complete circles. Charge doesn't move from one component then into another, instead it circles around like a rotating flywheel, or like a closed-loop drive-belt. The charge is everywhere in the conductors, and during a current, it rotates. Obviously, current is not energy-flow. (Voltage isn't energy-flow either, voltage is a measure of the electrostatic fields extending between pairs of conductors.) To cause electrical energy to flow to a different place in a circuit, we need a transfer of electro-magnetic energy ...of e-field energy, or of magnetic energy, or both.

The e-field energy (the capacitor-energy) moves in proportion to the current, while the b-field energy (the inductor-energy) moves in proportion to the voltage, and both together: the net rate of energy flow, involves both at once, V * I.

Heh, if you're not confused yet, go see my article... IN CIRCUITS, WHERE IS THE ENERGY FLOWING?

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Proof by True Assumptions (for both L & C)

  • Every conductor has resistance including inductors (Rs or DCR)
  • Every capacitor has an effective series resistance (ESR) due to the electrode interface
  • All time response changes are non-zero for both current flow in inductors and charge value and voltage in capacitors.
    • They have an exponential time response with \$\tau_{_L}=\dfrac{L}{DCR}~ [s]\$ and \$\tau_{_C}=ESR*C~[s]\$

Conclusion

Since all material and components must have some resistance , they must have a finite rise time.

Example

  1. Consider a PCB power trace inductance that is 1mm wide with 1oz Cu (35um) with no ground plane
  2. DCR = 3.333 mΩ/cm and L = 1.8971 nH/cm thus τ=0.5392 μs (max) for one (1) cm.
    • Yet adding more resistance speeds up the risetime, but makes it more lossy.
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  • \$\begingroup\$ Thanks for your answer, you have described the real world accurately. I just wasn't precise enough in my original question, so in my question I was assuming ideal inductor and ideal capacitor. Even the V=L*di/dt equation applies only for ideal inductor. For a real inductor it would only be valid for its ideal inductor part. Related to your answer just one tiny correction: ESR means Equivalent Series Resistance not Effective Series Resistance. \$\endgroup\$
    – gvg
    Apr 26 at 21:21
  • \$\begingroup\$ TY @gvg but those terms are effectively equivalent. (lol) maximintegrated.com/en/glossary/definitions.mvp/term/… \$\endgroup\$ Apr 26 at 21:25
  • \$\begingroup\$ Right, thanks for the clarification! :) \$\endgroup\$
    – gvg
    Apr 26 at 22:12

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