B-H Curve uestion

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

I believe it?s the third option. (looks like a smaller BH curve) The BH curve doesn?t change unless the material changes. The core still has the same hysteresis.

Another note, Your magnetizing the core. so At 'A' the core would be in a neutral state. ( or core 'reset', B=0) If the core is not reset, well that's usually bad ;) Some high power stuff has circuitry to keep track of which way the core is magnetized.

Cheers

I would say a curve rather like the second one you drew.

Sorry; for clarity i should have said "the second answer" or the bottom-most curve.

Sorta kinda the second "or".

From "Electric Machines: Steady-State Theory and Dynamic Performance", Mulukutla S. Sarma, Wm. C. Brown, 1985:

It's similar to the first picture, assuming the curve is traversed very slowly. H changes, B doesn't, i.e. the remanence of the material rules. In practical B-H curve generators, an AC signal is used, and the magnetization has to slowly penetrate the material, so there are skin-depth effects, and those would cause phase shift and give rise to open loops WITHOUT the material hysteresis; that kind of practical experiment might give rise to more complex traces like your second picture.

That doesn't look like the classic remanence problem.

See

If you are thinking about 'stopping' as cutting the current at a zero crossing and being left with the remanence, the stopping points would be at Br and -Br.

If you are going to modulate your current waveform as you suggest, it is a time dependant problem. You can't just make a 'right angled turn' quickly, as that would imply di/dt --> infinity with all that implies. If you follow the B-H trajectory slowly, stopping at the B = 0, H = Hc point and then drop the current (H) to zero as B 'relaxes' along the horizontal axis, once you pass H = 0, going negative, the B-H trajectory will follow one of the inner B-H loops (along a slope approximating an ideal permeability) until the material begins to saturate and develop a reverse remanence (-Br).

That's my guess.

I'm trying to Spice model a magnetic field sensor. My waveforms are getting quite close to lab... I'm just getting into the fancy details ;-) ...Jim Thompson

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=A0 =A0...Jim Thompson

=A0 =A0| =A0 =A0mens =A0 =A0 |

=A0 | =A0 =A0 et =A0 =A0 =A0|

=A0|

=A0 =A0 =A0 |

LTspice? are you going to share?

LTSpice has a non-linear inductor. See Help/Contents/LT Spice/Circuit Elements/Inductor. It describes two types available and shows the B-H curves.

Cheers, John

On Wed, 24 Aug 2011 12:56:44 -0700, bg wrote: (top posting fixed)

I rather suspect that what Jim is trying to model is the undesired residual magnetism in his sensor when the external field is removed or reduced.

HAH! Lotsa luck doing that starting with a B-H model. The common 'demagnetization' of a tool with slowly diminishing AC excitation actually magnetizes the bulk in layers, onion-like, so that the EXTERNAL field is nil, but it's fully magnetized still, at the micro level. Nonlinear and heavily geometry-dependent real fields inside real inductor cores and such is a BIG problem, and SPICE can only maybe handle its baby brethren.

Materials considered to be magnetically "soft" , do not retain their magnetic field (B) when the magnetizing force (H) is removed. Therefore the plot of "B" traces right back over itself as H is reduced. The BH curves for soft materials are narrow in the horizontal directions. You find these types of materials in applications where "retaining" the flux is not wanted such as motors, solenoids, etc.

When retaining the magnetic field is desired such as magnetic recording, tape, disks etc. "hard" magnetic material is used. The BH curve for those materials becomes wide in the horizontal direction, in other words as H is reduced, the value of B hangs in there, so the two curves are somewhat displaced. The right most curve is produced as the value of H is increased and the left most curve is produced as H is decreased. B is the result of H and whatever flux was stored in the material. Oersteds is one of the units used to measure the stored flux. That's all I can remember, but if you need some equations or specific figures, I can probably look them up.

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

Where is the *stop* ? At zero B or part way up the B curve.

Assuming part way up he B curve...

What is the impedance of the current driver when you *stop*?

I think that will affect the angle of the return to zero H.

Slight droop... dependent on supply stiffness.

Upper and lower left quadrants are also interesting to ponder.

Mikek

The arrows are drawn backwards. Approaching A from above, if you then reverse dH/dt, you stay on the right side of the curve for small signals, For large signals, you hit the saturation horn where history is erased.

Any Spice. It's a subcircuit. Waveform shortly, the .SUBCKT later or tomorrow, Skype from Hong Kong coming up. ...Jim Thompson

=A0 =A0...Jim Thompson

=A0 =A0| =A0 =A0mens =A0 =A0 |

=A0 | =A0 =A0 et =A0 =A0 =A0|

=A0|

=A0 =A0 =A0 |

Are you serious? The whole point of the curves is that they represent the location that the function(in this case f(B,H) =3D 0) is bound to. Both of your possible answers are invalid since they break the constraint. (it maybe possible given certain circumstances that you did not delineate)

This seems rather basic but the point of such a representation is that while you are on the one "branch" of the graph you must stay on it. So the point A must stay on the right side of the graph, along the curve, no matter how it moves... else it would be pointless to show such a graph the way it is. When we reach a branch cut it then is possible to move to another branch in which case we are then constrained to that part of the curve. The only time you can move from one curve to the other is when they intersect at the branch cuts.

As far as the practical aspects the it is more complex. The real hysteresis curves is much more complex and it may be possible to setup a circuit so that both answers are right. This is not the normal behavior of magnetic though as B and H generally have a straightforward relationship, at least ideally.

Your question is a bit ill-posed and doesn't make much sense. You talk about changing the "trajectory" and give to answers that are nonsensical unless there is more going on. You also say "stop at A" and "change direction" but you don't give us the initial direction(unless you mean A is going up on the right branch by those arrows).

In any case the answer is that if you "change direction" the point A MUST move in the opposite direction ALONG THE CURVE!!! You cannot just make up your own trajectories willy nilly.

How it looks...

Will play on most any modern Spice. I'll post the .SUBCKT tomorrow. ...Jim Thompson

Jim:-

See linked image. Consider also how degaussing works.

Best regards, Spehro Pefhany

Whoa! Way more than a Spice model can handle. I settled on just providing "memory" to get hysteresis, since this is a "regulaly" driven sensor. The end user is happy ;-) ...Jim Thompson