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Hysteresis losses in an inductor - resistance proportional to frequency?

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Silversonic
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Apr15-12, 09:25 AM
P: 130
1. The problem statement, all variables and given/known data

Consider an inductor, with a core of ferromagnetic material, connected to an a.c. power supply. Explain why the contribution of the hysteresis losses to the resistance of the coil is theoretically proportional to the frequency of the oscillator.

3. The attempt at a solution


The energy lost per second, i.e. the power, is simply the amount of times the hysteresis loop of the material is 'completed'. So the power used is proportional to the frequency. However, how do I then show from this that the resistance is also proportional to the frequency? It feels like I'm over-simplifying it, but is it simply because I^2 = P/R and since I is not proportional to the frequency, this means R is? I've looked hard, and it may be simple. A text book of mine says to consider representing the hysteresis losses as a resistor in parallel with the inductor - why parallel? I would have assumed the correct approach would be to consider it in series - after all, in a circuit diagram we consider the resistance of an inductor as being placed in series with the inductor.
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rude man
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Apr15-12, 03:33 PM
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You got started off right - the loss is proportional to frequency since frequency = no. of times the loop is traversed per second. The energy dissipated per cycle is proportional to the area within the loop. The additional increase in R with frequency, on top of the above loss mechanism, is due to skin effect, eddy currents, etc.

As for R in series vs. in parallel with L - it can be either way. In either case you get, for a given frequency, a real component plus a positive reactive component. With series it's R + jwL, with parallel it's wL/√(1 + w2L2/R2)exp{j(π/2 - tan-1wL/R)}. The argument in the exponent is always positive.
Silversonic
#3
Apr16-12, 04:15 PM
P: 130
The additional increase in R with frequency, on top of the above loss mechanism, is due to skin effect, eddy currents, etc.
Hey thanks for the reply. I've taken note of the skin effect and eddy currents too, but how would you show that this increase in R with frequency is proportional?

Also, doesn't the skin effect only effect the currents in the core, but not the circuit? I know that the skin effect cancels out the field in the interior of the core, and thus lowers the inductance.

rude man
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Apr16-12, 08:31 PM
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Hysteresis losses in an inductor - resistance proportional to frequency?

Quote Quote by Silversonic View Post
Hey thanks for the reply. I've taken note of the skin effect and eddy currents too, but how would you show that this increase in R with frequency is proportional?
As I said, it's the area under the B-H curve, as defined by the H end-points (proportional to excitation current). It should be intuitive that the more times the curve is traversed, the more the power loss, since each traverse represents so many Joules of energy. If the energy per traverse is Q, then power loss = fQ.

Also, doesn't the skin effect only effect the currents in the core, but not the circuit? I know that the skin effect cancels out the field in the interior of the core, and thus lowers the inductance.
Eddy current power loss is proportional to the square of the frequency. Also covers eddy currents a bit.

For eddy currents see http://en.wikipedia.org/wiki/Eddy_current

For the skin effect see http://en.wikipedia.org/wiki/Skin_effect. The skin thickness δ is inversely proportional to the square root of the frequency. This means the resistance is inversely proportional to δ.

Power inductor and transformer cores are constructed in laminations isolated electrically from each other to minimize eddy currents. Skin effect is in the wires, not the core, far as I know, but I cold well be wrong - anybody? Low-signal inductors sometimes use powdered iron to effect the same thing. Other core materials are inherently electrically non-conducting.
Silversonic
#5
Apr17-12, 12:46 PM
P: 130
Thanks for the help Rude_man. I apologise but there is one thing that is not intuitively obvious to me;

As I said, it's the area under the B-H curve, as defined by the H end-points (proportional to excitation current). It should be intuitive that the more times the curve is traversed, the more the power loss, since each traverse represents so many Joules of energy. If the energy per traverse is Q, then power loss = fQ.
I understand completely that the power less is proportional to the frequency of the oscillator and that was what I originally understood. What is not intuitive to me is how this means the resistance is proportional also. Maybe because it's so simple that it essentially goes without saying. I would be fine if the answer is because the current is completely independent of the frequency and thus (R = P/I) the resistance is proportional. I believe this is true.

Power inductor and transformer cores are constructed in laminations isolated electrically from each other to minimize eddy currents. Skin effect is in the wires, not the core, far as I know, but I cold well be wrong - anybody? Low-signal inductors sometimes use powdered iron to effect the same thing. Other core materials are inherently electrically non-conducting.
This is what I have been given for the theory explanation of this experiment;

"At higher frequency the more effectively do the induced currents cancel the field in the interior of the core. This leads, at high freq., to the skin effect where the field and eddy current are confined to a surface layer whose thickness is δ. The variation of the inductance is associated with the skin depth".

Although it does not state where this skin effect occurs, the core or the circuit, it specifically mentions that the eddy currents, which form in the core, witness the skin effect as opposed to the currents in the circuit. I have here no mention that the skin effect has any effect on the resistance of the inductor, only its inductance. Although if the skin effect DOES have an effect on the current of the circuit then it WOULD make sense to me how the resistance of the circuit/inductor would be affected. Later it goes on to mention that the variation of resistance is associated with energy dissipation (with no mention of the skin effect).
rude man
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Apr17-12, 08:58 PM
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Quote Quote by Silversonic View Post
Thanks for the help Rude_man. I apologise but there is one thing that is not intuitively obvious to me;



I understand completely that the power less is proportional to the frequency of the oscillator and that was what I originally understood. What is not intuitive to me is how this means the resistance is proportional also. Maybe because it's so simple that it essentially goes without saying. I would be fine if the answer is because the current is completely independent of the frequency and thus (R = P/I) the resistance is proportional. I believe this is true.
For constant current, which implies constant H, power dissipation goes as R, right? P = i2R. If you understand that power loss is proportional to frequency in the hysteresis curve, then it follows at once that R is proportional to frequency, right? That's just basic logic, if A=B and B=C then A=C.

However, this does assume constant current. If you assume constant voltage, then H drops with frequency (since L goes up and so does reactance wL) and the entire hysteresis curve takes on a new shape.


This is what I have been given for the theory explanation of this experiment;

"At higher frequency the more effectively do the induced currents cancel the field in the interior of the core. This leads, at high freq., to the skin effect where the field and eddy current are confined to a surface layer whose thickness is δ. The variation of the inductance is associated with the skin depth".

Although it does not state where this skin effect occurs, the core or the circuit, it specifically mentions that the eddy currents, which form in the core, witness the skin effect as opposed to the currents in the circuit. I have here no mention that the skin effect has any effect on the resistance of the inductor, only its inductance. Although if the skin effect DOES have an effect on the current of the circuit then it WOULD make sense to me how the resistance of the circuit/inductor would be affected. Later it goes on to mention that the variation of resistance is associated with energy dissipation (with no mention of the skin effect).
I have no quarrel with your text - obviously it's a wider source of info than me. But I know this: skin effect results in resistance increase in wires! Look at the website I sent you. I quote from a section of it: " The skin effect causes the effective resistance of the conductor to increase at higher frequencies where the skin depth is smaller, thus reducing the effective cross-section of the conductor. The skin effect is due to opposing eddy currents induced by the changing magnetic field resulting from the alternating current. "

It should be easy to be convinced that if you reduce the cross-sectional area of a wire, that the resistance increases!

http://en.wikipedia.org/wiki/Skin_effect

P.S. resistance is always associated with energy dissipation. Pretty hard to get around i2R!


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