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Impedance frequency curves for ferrite beads

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AAnand928

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I was going through the application note AN-1368 by Analog devices regarding Ferrite bead selection and design.

Attached picture is the impedance vs frequency graph of Ferrite bead part number: BMB2A1000LN2.

I understand the inductive region lies on the left side of the Z= XL point in the Impedance vs frequency graph.

But how is the capacitive region determined? I mean how to determine the frequency range of capacitive region shown in the attached graph

ZvsF.PNG

Thanks in advance
 

FvM

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But how is the capacitive region determined? I mean how to determine the frequency range of capacitive region shown in the attached graph
Not sure what you are exactly asking? The capacitive region starts right of X=0 respectively Z=max and goes up to infinite frequency.
 

crutschow

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The region where the capacitance starts to dominate is to the right, as the impedance starts to decrease.
 

AAnand928

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To be more clear, my query is ..
Inductive region is on the left side of the X=R point.. Likewise capacitive region is on the right side of the frequency where absolute value of the capacitance equals R. From graph it is evident about Inductive region. But i am not clear how this particular area is obtained as the capacitive region.
ZvsF1.png

Thanks
 

AAnand928

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But. If you see clearly the capacitive region is not marked from where impedance is decreasing.. Instead a different are is marked as the capacitive region.

ZvsF1.png

Thanks
 

FvM

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I suggest this view

ZvsF.PNG

The impedance is at no frequency purely inductive or purely capacitive, there's always a resistive (lossy) component.
The diagram marks regions of mostly inductive and mostly capacitive impedance, how far they extend is a matter taste, I think.

- - - Updated - - -

In the diagram, "mostly" inductive/capacitive can be roughly translated to |phi(Z)| > 70°
 

Easy peasy

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they are choosing where XL is greater than the resistance to be the inductive part, and similarly where Cparallel Xc is greater than the resistive part ( at that freq) to be the capacitive part.
Note they are assuming a parallel LC ckt where the impedance is maximum at resonance - as you go very high in freq you get just Rac, and very low in freq you get Rdc

The peak in XL is shown - and is interesting - the peak in Xc (curve) is, disappointingly, not shown - but could be calculated.
 

FvM

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Here's the full impedance curve of Murata blm03ag601sn1 derived from published S-parameters.

blm03ag601sn1.PNG
 

andre_teprom

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But i am not clear how this particular area is obtained as the capacitive region
The inductive effect comes from the ferrite bead being nothing else than an inductor, but designed to operate under dissipative region of the core (saturated) for unwanted components of signal, so inductance is primarily obtained straight by its geometry and material with which is manufactured; the capacitive component is there due to parasitic effects, being therefore estimated indirectly by subtracting Z (measured) from XL (estimated). Since the ferrite is placed in series with signal path, and capacitive and inductive reactance are opposite to themselves (-90o and +90o), both are allways subtracted, and beyond a certain frequency one of them starts to dominate, the point where XL=Xc, the peak of the figure above.
 

Easy peasy

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the dissipative region is not when the core is saturated - saturation is generally to be avoided if you want best bead performance - a lot of engineers overlook this when they put DC thru a bead ...
 

FvM

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Hello FvM, how does R go down with ever increasing freq .... ?
This happens because the complex Z plotted in the graph doesn't correspond to the physical circuit. Complex Z = R + jX would be represented by the series circuit of parallel LC with R. A physically meaningful equivalent circuit looks a bit different. A parallel R represents most of the losses in the mid frequency range which is bypassed by a parallel C, thus the effective ESR is decreasing with increasing Cpar effect.

I derived a good matching equivalent circuit for the above shown 600 ohms ferrite bead

blm03ag601sn1.PNG

- - - Updated - - -

the dissipative region is not when the core is saturated - saturation is generally to be avoided if you want best bead performance - a lot of engineers overlook this when they put DC thru a bead ...
Below the saturation curve of a 600 ohms 0603 ferrite bead illustrating this statement

Saturation 2506036017Y0.PNG

Fair-Rite did a good job in documenting ferrite bead saturation behavior. You are often searching in vain in other vendor's data sheets.

Referring to the above equivalent circuit, saturation reduces the inductance but almost keeps the lossy components. The inductance may drop to 50% of the initial value at only 10-20% of rated current. Consider that chip bead current rating is only thermal related.
 

c_mitra

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Hello FvM, how does R go down with ever increasing freq .... ? ...
Of course these are models made with RLC elements. At every point the three graphs satisfy Z^2=R^2+X^2. Therefore the three graphs are not independent. Given any two you can compute the others (within a +/- sign).

I believe (pardon me) that R is a part of X (LC component) and as the dissipation is frequency dependent, R too becomes frequency dependent.

Also note that X changes sign (LC transition point) is not exactly when the Z peaks. And that is also due to dissipation.

R decreases at still higher frequency because Z deceases; R graph cannot go above the Z graph.

And there are impedance models that cannot be explained in terms of LCR (see e.g., https://en.wikipedia.org/wiki/Warburg_element).

Clearly R is frequency dependent (and in a complex way) because both L and C are (but the graphs remind me of absorption and dispersion processes).

Only Z is measured directly and R and X are the two quadrature components.

I may be wrong but that is how I understand these experimental graphs.
 

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