My understanding is that the L in the inductor I-L graph represents dynamic inductance for a small signal variation, so cannot be used to directly see the total energy. The formula E=L*I^2/2 is an approximation in case L is constant, but in the general case where L is function of I, let's call it L(I), the energy can be calculated with an integral E(I)=\int_0^I L(I)*I*dI that for a small current increment can be approximated as: E(I+dI) - E(I) = L(I)*I*dI so this means that on saturation when L(I) becomes very small, the increment of extra energy stored will be very small, however the energy previously stored is still present, there is no lost energy just inability of storing more energy.
Good riddle! I need to think about it. At first glance i think a good clue here is that even saturated inductor when goes out of its saturation region recovers all the energy pumped in so nothing was wasted.
Interesting riddle I think energy is there in the magnetic core not going anywhere. When core saturated it means most of the core magnetic dipoles are aligned so energy is there working to maintenain this state. So what is lost is not energy but the ability to oppose the current (reluctance). Whic means no more energy can be converted to magnetic field but original energy is still there.
In that case the only valid way to calculate the stored magnetic energy is from the B, H curves. The formula which applies is E = 1/2*int(B*H*dV), where V corresponds to the volume of the geometry. The physical explanation is that the magnetic energy will not be reduced for high currents but its increasing rate will abruptly be reduced and theoretically will reach absolute zero (dE/di =0) for 100% saturation of the core.
The energy of a saturated inductor isn't lost! Although the differential value of the inductance isn't constant, (depending on the actual current value) there are no losses in the system. It looks different, if you introduce such effect as hysteresis and finite resistance of the ferrite core.
In my opinion: 1. When inductor saturates its inductance at the saturation point goes to zero (tangent to B-H curve) 2. To saturate an inductor an energy is required to create a magnetic field within its core. 3. When inductor is saturated (inductance is zero) then the voltage at inductor terminals drops to zero 4. No voltage on the inductor = no voltage on the current source = no power (no energy delivered to the inductor) 5. Answer is: inductor is still charged, energy is not changed, energy stops to be delivered as the inductor voltage is zero.
Hi Professor Ben-Yaakov. The way I see it, the energy stored in the magnetic material is "full", so there is a maximum energy stored in the material, after which the stored energy is in the "air". So the energy stored keeps increasing, it's just that the rate of DE/DI is slowing considerably
E = Integral of power, E = Int I V dt, but V = L dI/dt, (L is defined instantaneously by the derivative of I and voltage), so by substitution E = Int I L dI, thus the energy is the area under the curve, the stored energy remains mostly stored in the magnetic domains in the high inductance part and the remainder of the energy is stored in the region around the inductor as the loss of magnetic permeability cause the magnetic lines to no longer be tightly contained exclusively within the core.
To answer a riddle with a riddle, the energy is hiding in plain sight :) The energy stored in the magnetic field has 2 parts (Energy stored in the core and Energy stored in air gap/leakage (even if there's no airgap)) The total energy stored in the field is integral B^2/(2*mu) * dv The rate of energy rise slows down after saturation (since the B field in the core is increasing much much slower). But the B field in the air keeps increasing. and the energy keeps increasing (At a slower rate) Since the rate of rise of magnetic flux has decreased in the core, the voltage drop across the inductor reduces. The above analysis assumes a constant current source. It's also interesting to consider a constant voltage source and some DCR.
Dear Prof. Sam Ben-Yaakov, I think as the core saturates, coil acts as coil without ferromagnetic core, something like choke in vacuum. Energy will still increase, but with different slope. Energy stored in core will remain there. In that area formula you are using will work better, because magnetic circuit will be more linear. Generally, energy stored in choke is proportional to area up the flux curve and flux will be still increasing in our case. More interesting for me is supplying choke with current source driving current lineary increasing. It means that in linear choke voltage should be constant. In case of nonlinearity, voltage somehow decrease as the inductance decreases...
@@sambenyaakov There is some theory about atoms of ferromagnetic material represented by current loops mounted flexible in material. Flexibility is important because it store energy like a spring. As this loops changes orientation in direction with magnetic field, they are storing energy. When all are oriented in same direction of field, they are not able to store more energy. But I can not proof it mathematically. Thought about experiment. Let's assume you use h Bridge to energise inductor. 2 transistors, 2 diods, charged dc cap, simple 2q invertor. It is able to conduct 2 polarities of current, only one polarity of voltage. At one instant you turn on both transistor, starting energise inductor. Cap voltage decreases as energy is moved to magnetic field of choke. Cap voltage decreases. As you turn off transistors, choke hold the direction of current open diodes and move energy back to the dc cap. Ideally, you will get same cap voltage like at beginning. In reality, you loss some energy in conduction losses and one switching cycle.
Why do they never show us these graphs? I've never seen an inductance graph in tandem with its corresponding permeability graph in any electromagnetics course. That's a killer question. I hope no student ever has to face it when examined (unless of course it was part of the material covered!).
Energy stays the same because the I^2 term goes UP as the L goes down... Since the energy was stored in the low permeability air gap before saturation, this energy will still be there too, but also more energy stored in the core because its permeability has gone down because of the saturation. When extracting the energy back, the hysteresis losses in the core may absorb some of that energy.
The equation e = 0.5 L i^2 is only true if the inductance is constant ( linear inductance). To calculate the energy in a non linear inductance I would use the flux linkage
@@sambenyaakov The energy storage in an inductor is increasing with the current even when the inductance drastically drop down. When the inductor is saturated the incremental energy is marginal but exist. The problem is in the way that the E2 was calculated because that equation is only true for a linear inductance. To calculate the energy in a saturated inductor we need to calculate the integral
Hi Professor Ben-Yaakov. There isn't a loss of energy because what is getting lower is the differential or incremental inductance, not the total inductance. At the beginning you define the relative permeability as the slope of the B-H curve, but in fact that's the differential relative permeability, the total relative permeability is just the ratio B/H, and it is always increasing. The same thing happens to the inductance, the voltage at the inductor terminals will drop as differential inductance drops, but the stored energy so far is still there, it just is more "reluctant" (winkle winkle) to store any more. Am I close??
@@sambenyaakov yes! Sorry, after I wrote the message I realized it eventually reaches B/H=mu_0 (permeability of the vacuum) and it stays that way, but the main issue in the problem is about considering total inductance and not differential. BTW, I LOVE your channel, thanks for posting such nice videos with such a good content!
My understaning is that inductors with a core have a limited energy storage capacity. The drop in inductance after saturation is such that the energy is a constant, which also means the product of the current*inductance equal a constant number, which is why the inductance drops.
Energy went to magnetic domains? I mean permabilty is changed with number of domains aligns to external magnetic field. I think this is mechanical change in crystal structure that "magazine energy" like spring. I don't know if this phenomena is simmilar to piezoelectric efect, but I will glad to know if there are related :)
energy doesn't go anywhere and doesn't appear out of nowhere. the expressions given in the middle of the video are valid separately from each other for two different inductors that have a constant inductance over the studied current interval in them. In reality, the energy storage process can be divided into several stages( for ease of integration): 0) energy consumption in heat from the winding resistance ( including eddy currents and skin effect) and core heat for the remagnetization of dipoles 1) the current increases from zero to the inflection point of magnetic induction, energy accumulates in the dipole moments of the core as long as the dipoles are able to rotate freely in a certain direction, then different behavior is possible for different materials, two or three inflection points are possible, etc. and different slopes between them 2) after the inflection point, there is a long section of current that ends with constant magnetic induction through the core(only through it), at this point all the dipoles are aligned along magnetic lines. During this entire interval, the energy accumulation in the inductance formed only by the wire( an inductor without a core, only air) also continues. 3) then there is accumulation only in the inductivity formed by the wire up to the theoretical limit, if there is one. I have never asked myself whether there is a maximum magnetic induction for a particular volume of space On the interval 1-2, the inductance remains in the "nominal" on the interval 2-3, the inductance "decreases" according to the law depending on the core material, which can be divided into its own sub-intervals on the interval 3-infinity, the inductance is equal to the inductance of the coil winding without a core using splitting into these intervals, we can use the formula from the middle of the video separately for each interval to calculate the energy by summing the energy of each interval. As a consequence of the non-decreasing sum of the series, we have E1
@@sambenyaakov maybe I really misunderstood the question. just based on the formula V = -L(dI/dt) which precedes the energy formula E= (LI^2)/2 which is obtained by integration, and is only valid for constant inductance, which is not true in our case( since L(i)). So I wrote my message above, about how to calculate the inductance energy at any current, without integration(piecewise integration). The graph will be non-decreasing, as I wrote above, so there will be no "loss" of energy, and the stored energy graph for your inductance will look like a transition between two parabolas with different focal lengths(determined by the initial and final inductance).Just the formula for calculating the total energy will be different(not LI^2/2), and to get it, we need to integrate L(i)*i*(di/dt) , then find the constant C and it will be far from the standard, while graphically it is solved quite simply. Or do you claim that there is a loss of energy? then in my understanding of the process of energy storage, there must be an error somewhere, but I can't imagine where(I hope you will clarify this in your video response) . I have not described anything about magnetostriction, because I have not studied this topic much, but I doubt that the energy stored on the deformation of the material will be sufficient. Although the young's modulus for ceramic materials is quite high
This is a small riddle about physics and energy stored in bodies. "You have a metallic spring (maybe steel) with k constant. Then this spring is compressed and it stores energy given by kx^2/. You bond it compressed with a small cannabis string or fit inside a plastic box to mantain it compressed. Then submerges this the bonded spring in acid and it disolves completely. Where this kx^2/2 energy go?" Cheers.
You have to define things in your question more clearly. At first, I thought you meant the string or the box when you said "it disolves(sic) completely". But that didn't make for an interesting question, so I'm guessing you mean the steel spring dissolves. I would think it ends up as additional heat in the acid. The structure of the metal is under stress. As bits of it dissolve away, pieces of the crystal structure are going to have their strain released. Some will go into kinetic energy of the just-dissolved bit being kicked away, some will go into kinetic energy transfer from the bit that has had it's tension released kicking some molecules of liquid away.
WEll, why not more close to home: you feed a current to an inductor with no losses and short it. Current will continue flowing and energy is stored. And the you submerged it in acid - where does the electrical energy go?
Hello Professor! Not sure about the answer, my guess would be the rest is stored in the air or free space, as magnetic domains are full. Can you also explore magnetic switches and pulse compression networks sometime in the future. Thanks!
Answer is not exact. Magnetic switches and pulse compression are excellent subjects will try to prepare a video on that. See comment by Power Max and my response.
I'm nervous with this one. My intuition says that from V=L*di/dt, at t=0 and I increasing, V_L is small and positive (inductor absorbing energy). When L decreases, dL/dt is negative, so V_L is negative and the inductor now acts as a generator. So the current source absorbs this deltaE. After reading other comments, I should have known better than to trust you with that deltaE. Free space has it's own permitivity, even if the inductor is saturated. Should just take less power for the desired di/dt after I2
I think there are some assumptions when you have the formula E=LI^2/2. This formula suposses that B is proportional to L and I always (The energy is stored in the magnetic field). But now, with current dependent inductance ( non lineal magnetic field dependence of the current) you will need to integrate to obtain a correct formula.
You can think that the infinitesimal increase in energy stored by an inductor given an increase in current by dE=LIdI, If L goes to 0 (in reality goes Lu0/u where u0 is the vacuum permeability and u is the initial permeability) then you cannot store more energy with an increase in current.
What about the nonlinearity here? Only for linear inductors the energy can be defined as 0.5Li^2, otherwise not (only in the case of linear inductor we can claim that half of the energy provided by the source is accumulated and another half used to magnetize the inductor). It would be correct here to define energy as the integral of i*v over time, where v = d Fi/dt = L*di/dt + i*dL/dt. Anyhow, in this situation where the inductor is driven by the current source, the difference of energy is "eaten" by the inductor during the L-transition caused by saturation.
The simple answer is that this is a trick question, as the given energy expression does not apply to varying or non liner inductance. There should be another term involving L^2 and i. Coil energy is derived by integrating instantaneous power vi, but L is a function of i, so derivative should be according to chain rule using two parts... This is similar to energy stored in a moving coil, both i and L are functions of the displacement parameter (x or theta). In this case, some of the energy dissipates as core losses, due to the change in current leading to dipole movement... x
I'll take a wild stab at it: I suppose for the inductance to be zero, the magnetic properties of the cores material have vanished. So, for a ferromagnetic core, I suppose the domains should no longer be aligned and if they're not aligned then they're probably going all over the place, in other words, randomly oriented and in the process of randomly re-ordering their orientation (through the electrons' orbits shifting their axes) they produce heat. Just don't ask me to explain why this would happen though!
@@sambenyaakov actually... now that I'm thinking about it, what we need is the plot of the square of the current on the same graph as the inductance. Then, we'll be able to tell if the inductance is dropping fast enough for the product of inductance with the square of the current to be decreasing or not - because if it doesn't then we have no energy loss to account for. I suppose the premise of the question is that it is in fact decreasing faster than the square of the current...
Hi Sam, how are you? I think that the answer is laying in the inductance - it is not remaining constant. As a result of that, the inductor voltage form contains a part of phrase that is dissipative.
@@sambenyaakov L=L(t) v=d/dt (L•i)=Ldi/dt+idL/dt =v1+v2 where v2 = i•dL/dt = k•I You can see that part of the voltage (noted as v2) is relative to I- hence this part is lossy.
My 2 cents: The real inductance is only a function of geometry and must be measured in the linear portion of the B(H) plot, i.e. when H is small. The fact that the inductance seems to drop at high current values (high H) is just a math artifact which just reflects the fact that the inductor is no longer able to store more energy. Whatever energy it was stored before saturation it's not lost.
For the absolute inductance to actually fall, wouldn't the slope of the B-H curve have to go negative? that is to say it would have to be non-monotonic? Apparently there is actually a practical use of induct or core saturation, you can use it to amplify a signal! sparkbangbuzz.com/mag-amp/mag-amp.htm
Maybe I'm wrong, but let's try it to explain what I think in a few words: When the µr decreases the current rises and the stored energy remains almost the same. Does that make sense?
@@sambenyaakov When you increase the current you store more and more energy in the magnetic field. This limits the current. Coming into saturation this effect is no more working. The inductor becomes more like a resistor with its dc resistance remaining. So the current can increase. And you can no longer store more energy in the magnetic field.
The energy equation doesn't hold for non-constant inductance. We should integrate the energy for each inductance separately or come up with some "RMS" equivalent value for the overall inductance.
