Magnetic Flux Density Depending on Current

Mar 25, 2007 11 Replies

I don't know what it is about magnetics..it just never sinks in.. :(



Here's what's on my inductor: Voltage: 240Vpp 40% duty square wave @ 100khz Current: +/-100mA triangle wave hovering around 1.5amps. L=1.3mH u=125



Problem:



1) What's Bmax
2) What's delta B

I thought about solving this is terms of voltage using Bmax=(Vrms*10E8)/4.44fAN But the Vrms is supposed to be from a sine wave...I have squareish.



However, it looks easier to solve in terms of current. Neglecting the ripple current.. (low impact of accuracy) and just calling it 1.5Amps of DC current and using:



magnetizing force H expressed by: H=(0.4*pie*NI)/l N=turns I=peak current Amps l=magnetic path length cm



along with u = B/H (u=125)



B can be solved. (Gauss) Hopefully to be less than the datasheet core sat.



Delta B would be from the +/-100mA or 200mApp.



200mA is put into the B=uH equation. I'd use that to look up the core loss on the datasheet. (The more Bchange the more core loss.)

Then I found this eqn: Bdc=(L*Idc*10E8)/(N*Ac) L=inductance Idc=DC amps N=turns Ac=cross section cm^2 Which should give the same results.



Have I goofed up somewhere? D from BC


Magnetics datasheets are a pain. For things like pot cores and toroids, for a given material, why don't they give us gauss per ampere-turn? That, plus Al which they usually *do* supply, would be a big help.

John

Pick one terminal of inductor as reference so that for flux balance you must have VH*0.4+VL*0.6=0 and VH-VL=240 making VH=240*0.6=144V and VL=144-240=-96V. Then you have 144V=N*d/dt(phi)=N*d/dt(Aeff*B) so that dB=144*dt/(N*Aeff)=144*0.4/(N*Aeff*100E3). Then Bmax=Bdc+dB where Bdc=N*Idc*u(1.5A) where u(1.5A) denotes permeability of core at 1.5A dc bias.

Actually that's a bit conservative. Since flux balance requires that

+dB=|-dB|, or equivalently, Bdc=(Bmax+Bmin)/2, then since Bmax=Bmin+db and vice versa, you have Bmax=(Bmax+Bmin)/2+(Bmax-Bmin)/2=Bdc+dB/2.

B units of Tesla, area units in square meters, inductance units of henries - knocks out the 10E8 multiplier from this formula, (if it is the correct multiplier with the original units).

DC bias has no frequency relation, so the earlier sinusoidal formula is misapplied in the average static case, regardless of its potential suitability elsewhere..

B= n . I . ue / le in teslas amps and meters.

For gapped structures, out of saturation:

B = n . I . uo / lg

l in amps uo = 4 . pi . 10E-7 lg = gap length meters

AC flux, as always, is

deltaB = E . t / ( n . a )

volts, seconds, meters and Teslas, Half the delta is arithmetically additive/subtractive to the DC flux value.

RL

In this case, just using the nearly DC low ripple current waveform through the inductor, I could simply use the relation:

B=(n.I.ue)/le ue is that the permeability? le is the the magnetic path length?

I guess this would be useful too if I ever needed to make an electromagnet powered by a DC current source.

D from BC

Oops my OP is a little fuzzy...

There's no current bias through the inductor. The inductor current waveform is the result of being in a continuous mode converter. The converter doesn't let the inductor current drop to zero. Using the inductor voltage waveform in a B calculation and then adding on B from the inductor current waveform doesn't seem right. But ok for dc bias as described.

Bdc=N*Idc*u Is this a lean equation?.. No other inductor parameters like Across and lmag path length?? But then, I'm ok with B error up to 20%.

Note: I'm watching the u and aware it can be current dependent. D from BC

yes - the permeability of the medium used to store the energy.

In gapped media, the gap will tend to dominate, hence dominance of the gap length.

In homogenous media the full path length carries the permeability of the material used. This may degrade with direct current bias.

An electromagnet develops a useful external field in the (varying) gap.The field does not have to be DC; is often more effective with AC due to the potential reduction in residual magnetism.

RL

I wish I knew. Just what *IS* the purpose of quoting Bmax for ferrite, anyway? That doesn't help me a damn when I want to measure the f***er. Geometry is fixed, give me amp-turns, vs. gap if need be!

Tim

-- Deep Fryer: A very philosophical monk. Website @

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I'm thinking of the electromagnet used by the auto wreckers to pick up cars.. AC or DC? So an electromagnets core might stay a little magnetized after a shot of DC. Wouldn't an AC powered electromagnet just be kinda jiggy when attracting say some iron.. Reminds me of those buzzing solenoids for doors. D from BC

Not exactly, I was thinking of the case where the manu plots B vs H at DC for you.

The function is the same for a shaped field, and the average forces applied average out.

There are many ways to configure electronic noise makers, not all of them intentional. No doubt German-made models would be configured to produce a more 'noble' sound, if noise was actually considered to be neccessary to do the job of unlatching an electronic door.

RL

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