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There is unbelievable amount of contradicting information on the internet. The textbooks don't explain it in enough depth, so I have to ask here.

I have an induction heating system which has a coil through which there is applied an AC voltage which frequency I can regulate and I use cooking pots as a load.

I'm interested in the power consumption of different cooking pots. I'm most interested in why aluminium pots can't be heated at the same frequencies as pots with higher permeability.

One reason could be the hysteresis losses, but many papers say that hysteresis losses only contribute up to 10 % of total power losses inside the pot. If this is true, this couldn't be the reason.

They then say that the losses increase if the resistance of the pot increases, but that doesn't make sense because then plastic pots would have the highest losses because of high resistance. My thinking is that if the pot has a low resistance, then the currents will be higher which will increase the losses. If a pot has high resistance, less current will flow and the losses will be lower, but this is not what we observe.

What is the reason for this? Low resistance materials induce poor losses and high resistance materials induce poor losses. If resistance is not what determines the losses, then what is?

EDIT for mkeith: I am more interested in how the eddy currents are induced inside the material and how it leads to losses, than an actual product.

To answer your question, induction cookers can only heat aluminium pots at high frequencies (around 100 kHz), while steel can be heated equally well at 18 kHz and lower (they use 18 kHz+ to avoid noise in the audible range.)

Inductance with steel pots is indeed a lot higher, and so is the resistance felt by the coil. What happens if we put on an aluminium pot? The resistance and inductance decrease. This affects the resonant frequency of the resonant LC circuit. The resonant frequency increases. Also because of the decrease in resistance, the current increases (if we increase the frequency to match the new resonant frequency.) This would mean that because we have a low resistance, the current drawn would be very big and because of so low resistance the heating would still be smaller. Big current increases the negative losses because of internal resistance of the coil and the switching elements can only take a limited amount of current.

This explains why we can't heat aluminium at low frequencies. We increase the frequency to increase the resistance of aluminium pots because of increased skin effect which allows us to have lower current running through the switching elements for the same amount of heating power.

Now I understand this, but the real question I'm interested in is the following. If we have a current source that maintains 10 A at a fixed frequency, then in one case we have an aluminium pot causing 0.3 ohms increase in resistance over primary and in the other a steel pot which adds 5 ohms of resistance over primary. In both cases the internal resistance of primary is neglectable.

We see that aluminium pot is taking 30 W of power while steel pot is taking 500 W. How does this make sense from the following perspective?

The current is the same in both cases, so the electromotive force created in both pots is the same. Because the pots have different electrical resistances, the eddy currents inside of them should be different as well.

The losses are calculated by current flowing inside the pots squared times the pots' resistance.

Because the resistance of aluminium is lower, the currents inside is is larger, which means more losses should be created in the aluminium pot, but the actual result is just the opposite. This is the main part of my confusion.

JRE
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MaDrung
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    High enough resitance is neccessary but of course only until some limit (which is still so low that only materials we call "conductive" will work at all). It is similar to the maximum power transfer theorem: if resistance is too low or too high only little powre is transfered. In insulators no power is transfered because there is no current; in good (or super-) conductors no power is transfered because there is no/too little voltage. And: of course conductivity is only one factor. Another one is permeability. – Curd Oct 24 '17 at 13:16
  • I am interested in what causes that limit. Nobody ever explains that. :) – MaDrung Oct 24 '17 at 13:17
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    I've read it all numeros times. And what you say is wrong. Eddy currents Will be induced even if the material is not ferrous. What the permeability Will do is decrease the skin depth, increasing the effective resistance. But increase in resistance decreases the losses because it decreases the current which contributes more to the losses than the resistance. The only way this could make sense is if the increase in resistance because of decreased skin depth would not effect the ammount of current induced. But I think this is not the case. – MaDrung Oct 24 '17 at 13:23
  • @Bimpelrekkie You can induce eddy currents in any conductor (even semi conductors). However induction cookplates are optimized for ferrous materials. – ratchet freak Oct 24 '17 at 13:25
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    It is not that they're optimized, but that they can not heat non ferrous pots at low frequencies. The losses are just too low. Which is weird for me since the induced currents should be very high if the pot is very conductive since emf generated trough the pot is the same no matter it's resistance. If resistance is low, it should induce very high current, which should induce more losses. yet the effect is the opposite. – MaDrung Oct 24 '17 at 13:28
  • If I understand this correctly, non-ferrous materials don't get heated as efficiently because they don't "capture" the (varying) magnetic field, while a ferrous material gets a lot more of the magnetic fields inside it? – Oskar Skog Oct 24 '17 at 13:29
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    The reason ferrous materials work better is because of hysteresis. The pot becomes magnetized in one direction, then alternatively in the other direction, and if you look at the B-H curves in a transformer core, you can see that there is a loss on each cycle. This energy loss appears as heat. Try a Google on "hysteresis loss." – John Birckhead Oct 24 '17 at 13:35
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    Many papers say that in induction cooking, hysteresis losses represent only up to 10 % of total power loss. This is not the case then if this is correct. – MaDrung Oct 24 '17 at 13:36
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    The heating can only come from two sources: hysteresis or resistive heating. If the majority does not come from hysteresis, it must come the induced current combined with the material's resistivity. The material's impedance must be "matched" to the induced current and frequency. Too high resistivity (plastic) results in little or no current; too low (aluminum at lower frequency) means less resistance and low induced voltage. If you get the frequency up high enough, the aluminum skin effect will increase and the apparent resistivity will increase. – John Birckhead Oct 24 '17 at 18:28
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    Good question. For sure the magnetic permeability is they key, not resistance by itself. If it were only resistance, then pot thickness would be critical, and you could heat a thin aluminum pot just as easily as a thick cast iron pot. I think what happens is that the frequency needs to be adjusted when heating ferrous metals because they change the coil inductance (due to magnetic permeability). There may also be a capacitor to tune the resonant frequency of the work coil. Changing coil inductance will change the resonant frequency and power transfer. Just a theory, hence a comment not answer. – user57037 Oct 24 '17 at 20:07
  • It might be interesting to measure the coil current with different frequencies and different materials. If you can put a current probe on the wire leading to it. Also try heating aluminum foil to positively rule out the resistance idea. – user57037 Oct 24 '17 at 20:23
  • @John Birckhead as far as I understand induced voltage has nothing to do with resistance. Induced voltage is fixed, but the current magnitude depends on resistance. Now if you decrease resistance, the losses Will only get better because current contributes more to losses than resistance if you look at P=i^2*R. – MaDrung Oct 25 '17 at 05:42
  • The effective resistance of the pot must be matched to the effective resistance of the power supply. Too much resistance=low power transfer. Too little resistance=low power transfer. If you had an infinite power supply with zero impedance, you could heat aluminium pots. You don't, and you can't. In this respect, the same as any other power supply circuit. – david Oct 27 '17 at 03:00
  • @david, it is a common mistake to think that you want to match resistance between source and load for maximum power transfer. But, actually, unless you want the source to get very hot, this is probably not what you want. What you want to do is make sure the source resistance is much lower that the load resistance. In this case, it will be the load that gets hot, not the source. But the situation is more complicated here because there is also a transformer, and perhaps a reactive matching network (we don't really know). – user57037 Oct 27 '17 at 03:12
  • @MaDrung, I think perhaps it is time for you to provide a bit more information. Please edit your original question to explain a bit more about your setup. What frequencies have you found to work with aluminum? What frequencies work with steel? My prediction is that steel requires lower frequencies because the input inductance is higher. Comparing steel and aluminum at the same frequency, steel will have higher inductance, and therefore lower current at the applied voltage. So the input power will be lower, and heating will be less. The reason you have to use different frequencies is purely... – user57037 Oct 27 '17 at 03:19
  • to increase input power to the coil when using steel. – user57037 Oct 27 '17 at 03:20
  • I have added an edit to my original post and I think I have worded my question realy clearly at the end of the edit. – MaDrung Oct 27 '17 at 11:21
  • Good edit - I can see your confusion. When you expose a conductor to a changing magnetic field, a secondary field is generated in the conductor that opposes the original field. In a perfect conductor, this secondary field would completely counteract the primary field. This is why a magnet levitates over a superconductor. In imperfect conductors, eddy currents are generated from difference of these two opposing fields. It is these eddy currents that heat the conductor, not the primary induced current (which is only theoretical since it doesn't exist by itself). – John Birckhead Oct 27 '17 at 21:14
  • After reading your edit, my main observation is that I don't think EMF in the steel pot will be the same as in the aluminum pot because the steel pot is ferromagnetic. If we use the analogy of a transformer with shorted secondary, the steel pot is a steel-core transformer with shorted secondary, and the aluminum pot is an air-core transformer with shorted secondary. I am not enough of an expert to be sure this is what's going on, but it seems right. – user57037 Oct 28 '17 at 04:45
  • Oh, and if you start by assuming that EMF is the same, the logical thing would be to calculate loss using V^2/R rather than I^2 * R. But as I said, I think V would be higher in the steel pot. – user57037 Oct 28 '17 at 04:49
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    @MaDrung Please stop writing will with capital w. – winny Oct 28 '17 at 07:17
  • @mkeith If I thought that impedance matching was a complete answer, I would have written an answer, not a comment. I am aware that Edison's great insight was that you didn't want impedance matching in a power supply. But in answer to a question about why you want matched impedance power systems, it's an observation, not a misunderstanding. – david Oct 30 '17 at 02:11
  • @winny I'm not doing it. This site has some weird autocorrect I can't switch off. – MaDrung Nov 02 '17 at 07:18
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    @MaDrung No, it's on your device. I've fixed it now along with some spelling and grammar errors. – winny Nov 02 '17 at 11:25

5 Answers5

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I am considering your edited case, where the frequency and current are held constant. You observe that the EMF should be the same in both pots, therefore the pot with a lower resistance should heat up more. However, I believe you are assuming something that is not true. The EMF will not be the same in the steel and aluminum pots because the magnetic flux density (B-field) will not be the same.

The primary coil, driven at constant current, will give rise to a time varying magnetic field (H-field) which will be the same in both pots. But due to the high magnetic permeability of the steel pot, the magnetic flux density (B-field) in it will be much, much higher than in the aluminum pot.

As you know, Faraday's law relates the rate of change in the B-field to EMF. Since the steel pot will have much higher B-field, it will also have a much higher EMF, and thus a much higher power dissipation despite its high resistance.

If the EMF in the two pots really were the same, then you can easily see from the equation P=V^2/R that the pot with lower resistance would have higher dissipation.

To be just a bit more numerical, note that the permeability of steel is around 100x higher than aluminum, so the EMF in steel is actually around 100x higher, and V^2 is 10,000x higher. Aluminum is a better conductor than steel, but not 10,000x better, so the net result is that the steel gets much hotter under the same H-field magnitude.

user57037
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  • I have another question. If we use 2 pots that both have permeability equal to that of air and we have a current source that maintains 10 A. The only difference between the two pots is their resistance. We would then expect that the same EMF will be induced in both pots. This would mean that inside the pot with lower resistance, there will be more losses because less resistance means more eddy currents. But if you calculate the losses on the primary, because the pot has a lower resistance, voltage required is smaller to maintain the current. this means that losses should be smaller, not bigger – MaDrung Nov 02 '17 at 09:43
  • Is it possibly, that because the eddy currents inside the pot increase because its lower resistance, that those eddy currents create bigger counter voltage to the coil, which means that the voltage of the current source actualy had to increase instead of decrease after the transient period? – MaDrung Nov 02 '17 at 11:29
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    We may be abusing transformer equations somewhat. Also, I am not an expert on transformers. It is an interesting question to be sure. Basically what you are saying, I think, is that if we transform the load impedance back to the primary, a lower resistance should give rise to a lower voltage for a fixed primary current. Thus power in is less. Yet it is heating up more. So what gives? I don't know. – user57037 Nov 02 '17 at 19:09
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I think that a different perspective is required. We need to see it from a power transfer point of view.
There are at least 3 components to the "induced" power transfer. 1 - transformer action, 2 - hysteresis, and 3 - resonance.
Most of the power is transferred by transformer action. The metal pot acts as a shorted, "1 turn" secondary, as well as the transformer's core.
As is well known, iron has a better permeability than other metals (aluminum), therefore it is better at transferring the primary power to the secondary.
Since the secondary (the core) is "shorted," the power transferred gets dissipated in/by the core.
The second component, hysteresis, has already been mentioned and contributes an additional 10%.
Resonance is also an important factor because different metals (materials) will have different resonant frequencies, and the resonant frequency is very important to achieve maximum power transfer.

Guill
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A induction heater is really a transformer where the secondary is the thing being heated. One important advantage of this is that there can be isolation between the power input and the thing being heated. Magnetic fields can pass thru electrical, thermal, chemical, and other barriers.

As with any transformer, there is a equivalent impedance presented at the secondary. For maximum power transfer, the load has to match this impedance. Power is voltage times current. At 0 impedance, the voltage is 0, so the power is 0. At infinite impedance, the current is 0, so the power is 0.

The induction heaters you are describing are apparently set up to deliver maximum power to something of iron or steel. Aluminum is more conductive, and apparently has too low impedance to work efficiently with the induction heater you have. Plastic won't work at all because the impedance is very high, effectively infinite for practical purposes.

Olin Lathrop
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    Can you read trough my EDITed part to answer my question? Also how come if plastic has very high resistance, the resistance of the primary in presence of plastic is also not very high? It seems that the primary just ignores the materials that aren't good conductors. What effect causes opposition to current in primary then? – MaDrung Oct 27 '17 at 11:37
  • Whoever downvoted this, what exactly do you think is wrong, misleading, or badly written? – Olin Lathrop Oct 27 '17 at 21:30
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    @Olin Lathrop Hi Olin. I am not the guilty party - I never downvote, but I do think both answers have the same problem. Inductive heaters are not analogous to transformers; they are more like poorly designed transformer cores in a transformer with no secondary. Does this make sense? – John Birckhead Oct 27 '17 at 21:43
  • I am not the guilty party either, but I don't really understand your answer. What two things are you saying should have their impedance matched? – user57037 Oct 28 '17 at 05:02
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If we have a current source that maintains 10 A at a fixed frequency ... In both cases the internal resistance of primary is neglectable...

... We see that aluminium pot is taking 30 W of power while steel pot is taking 500 W. But how does this make sense from the following perspective?

The aluminum pot is reflective. It reflects most of the energy. Since that reflected energy is not absorbed by the Aluminum pot, it is mostly not taken from the 10 Amp low-impedance power supply you postulate. So the voltage on your theoretical 10 Amp power supply is lower: if you create that with a switching supply connected to the mains, that will take less current from the mains. If you create it with a linear supply connected to the mains, the linear supply will get hot.

The mechanism of reflection involves the induced currents, so you can get there from here, but to do so you have to accept the observable facts: lots of current flows in the aluminum, not much power is lost in the aluminum.

The aluminum is in the near field of the induction coils so you can't call it 'radiation', so you don't have 'radiation' and 'reflection', but at the atomic level you have EM field which, through interaction (mostly) with electrons is either transferred to thermal energy or retransmitted. In the case of aluminum, the mechanism for transfer to thermal energy at low frequencies is not efficient, and the mechanism for re-radiation/reflection is efficient.

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david
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  • What would happen if you had 2 pots with permeability equal to air with 2 different resistances and a current source that kept 10 A trough the coil. I would say that the power is greater in the pot with lower resistance, since it induces more eddy currents because of the lower resistance. But because of lower resistance, the voltage required to run 10 A is smaller, so the power should be smaller as well for a fixed 10 A current. This contradicts itself. – MaDrung Nov 02 '17 at 12:21
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There are several existing answers (including one that is accepted), but none talk about the role of the leakage inductance between the drive coil and the pan (though it is mentioned in a comment by John Birckhead).

Leakage Impedance

In a traditional transformer, there is a high permeability core threading both the primary and secondary coils, which loops back on itself to form a "magnetic circuit".

In most designs this leads to near unity coupling (to the 1st order) between the two coils. This means the primary effective impedance is turns ratio related to the secondary load, and shorting of the secondary effectively shorts the primary.

However in some designs (such as traditional magnetic neon sign transformers, some microwave oven transformers, and transformer based stick welders) there are magnetic "shunts" which provide a bypass path, allowing some of the primary flux to loop back without passing through the secondary coil.

The shunt path can be modelled as a "leakage inductance" in series with the primary coil, and this leads to a leakage impedance at the transformer operating frequency. This becomes the effective supply impedance.

If the secondary is shorted (say by a negative dynamic impedance plasma in a neon tube, or welding arc) the leakage impedance limits the short circuit current, with the primary flux preferentially flowing through the shunts rather than the secondary coil.

Load Matching

Maximum power transfer occurs when the load impedance matches the effective source impedance produced by the leakage inductance.

For instance, if you have a "10kV 50mA" neon sign transformer (10kV open circuit voltage, 50mA short circuit current), then the effective impedance is 200kΩ (10kV/50mA).

If the secondary is open circuit, no power is delivered. If it is shorted with a 0.1Ω load, 50mA will flow, but less than 1mW will be delivered to the load (though secondary winding resistance will probably lead to at least a few watts being dissipated in the secondary). However, if you apply a matched load of 200kΩ, then 25mA will flow and 125W will be dissipated in the load.

Leakage impedance between induction stove and pans

The relevance to the induction hob and pan of the OP, is that the coupling between the primary coil and the pan base (which is both the secondary coil and the load) is "core-less" (i.e. no high permeability core threading both the drive core and the pan base). This means there is a low coupling coefficient, a high degree of flux leakage, and therefore a high primary leakage impedance.

As with the example of the neon sign transformer, this means shorting of the secondary (say with a thick, highly conductive aluminium or copper pan) will lead to very little power delivery.

To maximize the power delivery you need a pan base with an effective impedance which matches (or at least approaches) that of the leakage inductance.

Use of ferro-magnetic pan bases

This matching is normally achieved through the use of a ferro-magnetic material (iron or some steels), which due to a very high-skin effect, show a high effective resistance (the eddy currents are contained in a thin surface layer, of what is already a relatively high resistance metal).

As noted in another question, you can also get limited heating of thin aluminium foil, because like with the ferro-magnetic skin effect, the thin section leads to a relatively high resistance.

Tim Williams
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colintd
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