Why for an inductor the average voltage across it must be zero? - Page 1
Forum section: electronics / beginners

author: markusz
author_rank: Contributor
timestamp: January 13, 2022, 11:46:57 pm
content: Hi experts,I'm reading the Art of Electronics and do not get this.Description and my evaluation to the point that blocked me:I simply don't get how to go from there. I then watched a youtube video but it confuses me more:In the video, around minute 2, the professor conveniently puts i(0) = i(T), which is kinda cheating IMHO...for my circuit, I guess the fast flipping does something but the description of the problem seems to indicate a general rule (not specific to this diagram).

author: Benta
author_rank: Super Contributor
timestamp: January 14, 2022, 12:23:19 am
content: I get your frustration, that video is lousy.The mathematics aren't that difficult, but getting them settled in your brain are.I have it like this: "Capacitors are to voltage what inductors are to current."A capacitor will change voltage over time according to the current supplied to it. If you keep supplying it with a current, its voltage will rise.An inductor will change current over time according to the voltage applied to it. It you keep applying a voltage, its current will rise.Capacitors and inductors do the same thing, but with either voltage or current. They're complementary.

author: rstofer
author_rank: Super Contributor
timestamp: January 14, 2022, 12:40:47 am
content: The last line of your work shows V headed toward infinity as di/dt heads toward infinity. That is correct because the basic equation is:V(t) = L di/dtand a step function input has an extremely high di/dt. It is the principal that makes ignition coils work. If the current isn't changing, the only voltage across L is from the internal resistance and that is simply V=I*R and is usually ignored in this context.V(0) thru V(T) covers one cycle in an AC waveform. Consider a sine wave centered at 0V. The waveform starts at zero, goes positive until returning to 0 at pi radians and then goes negative until completing the waveform at 2*pi radians. The AVERAGE of this sine wave is 0. It starts at 0 and it ends at 0 and spends equal periods above and below 0VThe same thing with KatKim's triangle wave except there is a constant offset and the waveform is asymmetric. The average of the triangular portion IS 0 but finding the value of the average can be messy. The steady offset isn't changing so the inductor isn't reacting to it. There is no di/dt in a DC voltage (once it is established).I love her videos about control systems. I haven't watched the ones about power systems but that is her specialty.

author: TimFox
author_rank: Super Contributor
timestamp: January 14, 2022, 12:47:39 am
content: Quote from: Benta on January 14, 2022, 12:23:19 amA corollary to the discussion above is that the mean (DC) current through a capacitor is zero, and the mean (DC) voltage across an inductor is zero. Note that the actual two-terminal device you buy is not a pure inductance or capacitance (parasitic ESR, reactance, etc. complicate the practical circuit).

author: Manul
author_rank: Super Contributor
timestamp: January 14, 2022, 01:00:47 am
content: This is not a direct analogy, but maybe it will help you to grasp the concept intuitively.Imagine that you travel from your home any direction, any course and come back home. Your elevation gain and elevation loss will exactly add up to zero. Does not matter what mountains you climb and how many hills you roll down. Simply because you start and finish at the same spot.For ideal inductor average voltage is zero, because otherwise current will fly away positive or negative and never come back "home".For ideal capacitor average current is zero, because overwise voltage will fly away positive or negative and again never come "home".

author: IanB
author_rank: Super Contributor
timestamp: January 14, 2022, 01:12:57 am
content: Mathematically, you can look at it like this:V=LdI/dtNow, over a finite time interval, the average value of dI/dtis ΔI/Δt(you can see this if you integrate the derivative).Furthermore, over a sufficiently long time period, the value of ΔI/Δtmust tend to zero, otherwiseIwould be increasing without limit, which is impossible.SinceVis proportional to dI/dt(Lis a constant), and the average value of dI/dtis zero, it follows that the average value ofVis zero.This is true even if you posit a DC source that endures forever, since you will eventually be dividing by infinite time, and dividing by infinity will make everything go to zero.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 14, 2022, 01:21:05 am
content: Note on notation: you've written I(T) and I(0), implying these are functions evaluated at the values given; rather, they should be subscripts, which says that I at time T is [that value], and I_0 is the initial condition. This is a differential equation, so we need some sort of boundary condition to define an answer -- so the latter is equivalent to the "plus a constant" of an indefinite integral. If T is a variable, then it will indeed be I(T) "current I as a function of time T".Something's missing between the left and right steps: \$\overline{V}\$ is not simply VT; in general it is the time integral of V(t). (This being the flux applied to the inductor. Neat fact: flux isn't just something floating around in space (like B (flux density)), it's a more concrete thing, at least, I like to think so(?); it's a waveform we can measure at the terminals of the inductor!*) In the case that the applied voltage is constant for some period of time, we have Phi = VT. If voltage is a square wave, we need to sum the two fluxes together. And notice V is the voltage applied to the inductor, not the supply voltage.*An ideal one, anyway. Real ones have resistance, the voltage drop of which adds to the EMF measured. And we don't measure flux on the 'scope directly, but we can integrate voltage easily enough. And if we know the current flow and resistance, we can subtract the resistance part to get the ideal EMF again.The key insight is that, under quiescent conditions, I is constant, or at least not drifting up or down over time. (In the setup shown, it will cycle around a mean value, but as long as we measure it at a consistent point during that cycle, we can measure the baseline, and if it's drifting or what. Or just subtract out the cycling altogether.) For this to be true, dI must be zero over a cycle, and so we can simplify things further:One more pre-simplification: notice that, for straight line waveforms (square voltage and triangular current), the differential equation can be rewritten as the difference equation,\[ V = L \frac{\Delta I}{\Delta T} \]for some (finite nonzero) deltas.If we take two instances of this, labeling them with voltages V1 and V2, for respective durations T1 and T2, and we and chain them together forever (for a cyclic period T = T1 + T2), we must have:\[ \Delta I_1 + \Delta I_2 = 0 \]That is, current returns to its initial value after one cycle. And so, substituting the above into this and rearranging, we have\[ \left( \frac{V_1 \Delta T_1}{L} \right) + \left( \frac{V_2 \Delta T_2}{L} \right) = \Delta I_1 + \Delta I_2 = 0 \]and you should be able to see V1 DT1 = -V2 DT2 from there. In other words, the fluxes must balance.And then, as for what that does to the circuit -- notice this is independent of (DC) load current; it may take some time for that current to stabilize, but once it's there, the flux balance condition holds, cycle upon cycle.As for the value of the capacitor, and other particulars (switching frequency, risetime; control methods?!), that's another matter. It might perhaps be better understood as a signal filtering operation (which, can probably be hand-waved easily enough -- but is best understood with help from the Fourier or Laplace transforms). And also connects with actual engineering matters (because, the output isn't going to be perfectly stable DC, there will be some AC ripple on that output, corresponding to the capacitor's reactance, and the ripple current through the inductor; and, well, we can't have it exactly zero, so how much can we actually tolerate in a given application?), and, engineering problems are less well defined (or incorporating many more variables), so I will leave it at that -- hints of where we might go from here, but all in due time, of course!Tim

author: golden_labels
author_rank: Super Contributor
timestamp: January 14, 2022, 01:38:11 am
content: An average is a measure of position. It tells where in some space a set of values is located. Arithmetic average does that by indicate “center of mass” of all those values.(1)So, having all possible values the waveform takes over a period, we may tell that “center of mass”. As you may expect, the values will be on both sides of that average location.Now, the part skipped in the video. If the average is zero and we have values on both sides, clearly there must be both positive and negative values. Right? And there are. If the current increased, there was corresponding positive voltage. But we know that it must decrease to go back to its initial value. In an inductor the only way to decrease current is by applying opposing voltage. Here you have your negative part. Since they must balance exactly to get back to the same current, the average must be zero.If that’s not clear yet, consider a mechanical analogy: a vehicle traveling in space, powered by rocket engines. To avoid confusion: forget that I used words “position” and “space” above — they were used in a different, non-mechanical meaning. Now, the analogy is:Current is represented by vehicle’s movement (speed).Inductor’s opposition to change in current is vehicle’s inertia (opposition to change in velocity).Voltage is the force obtained by firing thrusters.Imagine the vehicle is traveling at some speed. You observe that the speed increased. That means a thruster must have been fired. If the vehicle has to be brought back to the initial speed, what must be done? There is only one solution: using thrusters on the opposite side. Both thrusters may be fired many times and deliver different forces, but to get back to the initial speed they must balance exactly. The average value will be zero.(1)Which isn’t in any way special, privileged, or having some natural ontological meaning: it’s just a convention that happens to be very useful and convenient.

author: IanB
author_rank: Super Contributor
timestamp: January 14, 2022, 02:40:32 am
content: There is a sense in which "the average voltage across an inductor must be zero" is a meaningless statement.Consider the fundamental equation:V=LdI/dtWe have one equation, three unknown variables:V,LandI.But what this equation really says is this:V(t) =LdI(t)/dtThere are two degrees of freedom to remove. We can fixLas a constant, and we cannot really fixIas it is a state variable. So we choose to defineV(t) as a forcing function, perhaps from a signal generator.But if we are fixingV(t), what has the inductor got to do with it? If we makeV(t) a forcing function, then its average is defined by that function. So there is nothing we can really say about the inductor.If instead, we posit that the inductor is part of a bigger system, andV(t) is just another voltage in that system, then still the average value ofV(t) is defined by the system as a whole.So what we can say is that the average value of all voltages in a system is determined by the inputs to that system. These averages may or may not be zero according to how the system is driven.If the system is left unperturbed and allowed to settle to a resting state, then all time derivatives will settle to zero.Either way, the statement that "the average voltage across an inductor must be zero" is a non-sequitur.Bite me!

author: sclafa
author_rank: Newbie
timestamp: January 14, 2022, 03:51:53 am
content: Here's a better picturehttps://youtu.be/A78yP8oApqk?t=1h17m54s

author: golden_labels
author_rank: Super Contributor
timestamp: January 14, 2022, 09:23:45 am
content: IanB: you have missed one constraint: that’s a cyclic sygnal.The current must repeatedly go back to the same value. Of course if we consider V(t) to be the driving force it may go anywhere and not average at 0, but in all those cases the current on the inductor wouldn’t be cyclic.Looking at it from a more mathematical perspective: since it’s a repeating sygnal, it is a sum of sines and some fixed offset. All sines have average value of 0 by definition, so the sum of the averages is also 0. Waht remains is the constant. But that must also be 0, as any non-zero value would produce current that grows to infinity, which contradicts the initial assumption that it’s cyclic.

author: CaptDon
author_rank: Super Contributor
timestamp: January 14, 2022, 03:19:26 pm
content: Want a super simple answer without all the math....It doesn't have to be zero and often never is. Mainly due to internal resistance, but often due to the very way it is used. Take for example the old 'electro-dynamic' speakers from way back when. The field coil was used as a smoothing reactor for the D.C. plate supply voltage and due to the D.C. current the reactor formed an electromagnetic pole piece that the voice coil worked against. The rectified d.c. had a varying voltage and current component which the field coil acting as an inductor helped to smooth out but the sum result did not return to zero. It had an offset. Offsets in most inductors and transformers are bad as they can lead to saturation.

author: CaptDon
author_rank: Super Contributor
timestamp: January 15, 2022, 03:17:53 pm
content: In SMPS units that utilize 'flyback' topology it is absolutely required to return the flux to zero. They use 'reset' windings and often even the snubber circuit is part of the flux reset operation. Leaving flux in the core causes a huge transient current when the driver mosfet/BJT turns back on. So a person needs to look at the circuitry surrounding the inductor. In many cases a designer does indeed want to insure the flux returning to zero. In other cases like that of a 'Saturable Core Reactor' once common in older power supplies and welders a certain amount of flux density is held or induced in the core. The higher the residual flux the more the reactor tends to look only like its D.C. resistance and less like its inductive reactance. It was one way of regulating voltage or current flow but not very common anymore since we have SCR's, Triac's and I.G.B.T. devices capable of thousands of volts and amps. So there isn't really only one inductor flux rule for returning to zero.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 15, 2022, 04:43:11 pm
content: Flybacks can be CCM just fine... it's not often done, just because other types are preferred at power levels where that would become desirable.One-switch forward is superficially similar but does however require a reset winding or clamp circuit. Two-switch forward is very common these days; by switching both ends of the primary, it can be clamped against the primary supply voltage directly, saving the leakage inductance of a separate winding, and recycling the power that would otherwise have to be dissipated in a clamp (or recycled with a more complicated circuit).Tim

author: TimFox
author_rank: Super Contributor
timestamp: January 15, 2022, 10:14:59 pm
content: Another interesting circuit to analyze is a superconducting solenoid electromagnet.Since the resistance is literally zero (when operated within maximum field and current density), the shorted inductor will have a persistent current with no external voltage connected after the current flow is established.(In practice, it decays very slowly due to imperfections in the circuit, etc.)To "charge" the magnet, the external constant-current supply connects across a short section of the superconducting wire, which can be heated above the transition temperature to act as a normal resistor (relatively low value, but not zero). When the external supply current is ramped up from zero, there is a self-induced voltageV= -LdIL/dt, whereILis the current through the superconducting coil. Some of the charging current, however, flows through the resistor due to that voltage across it. When the ramp of the external current supply stops at the desired value, dIL/dtgoes to zero and there is no voltage across the resistor. At that point, the heat is removed and the resistor goes cold and superconducts, and the external supply can be disconnected.This can be calculated by elementary differential equations, or simulated in SPICE (simple model of L and R in parallel driven by a ramped current source).

author: radiolistener
author_rank: Super Contributor
timestamp: January 16, 2022, 04:59:53 am
content: Quote from: Benta on January 14, 2022, 12:23:19 amcapacitors are storing energy in the electric field. Inductors are storing energy in the magnetic field.

author: golden_labels
author_rank: Super Contributor
timestamp: January 16, 2022, 09:59:55 am
content: You people do realize the topic is about calculations regarding an ideal, lumped-element model inductor?