Radial ferrite bead has gone obselete leaving no possible alternative parts...

[treez]

Hi,
This 7A Ferrite bead, (its through hole, radial) has gone obsolete…..we need to find a like-for-like replacement

Murata BL02RN2R1M2B ferrite bead
https://www.murata.com/en-us/products/productdata/8796738420766/ENFG0004.pdf

Its used on multiple SMPS’s in a company’s stable of products. Sometimes its used as a series L for an output filter for offline Flyback or forward converters. Also, sometimes its being used in series with the FET drain in various offline DCM flybacks or forward converters.

The maximum current it sees in any product is 5A RMS.

We need to find a replacement, but there is nothing available anywhere. Does anybody sell the bare, Bent wires, that we could insert a cylinder ferrite over to make one ourselves?

The following Ferrite bead, B-01-R, by Kemet, is only rated for 5A, which is odd because its made of a piece of 0.65mm wire, just the same as the BL02RN2R1M2B, which says its rated to 7A, even though its wire is longer than the B-01-R.
Also, its difficult to compare these beads because the Kemet one gives its impedance as 2 Ohms, whereas the BL02RN2R1M2B gives its impedance as 1.1uH at 1MHz.

Kemet B-01-R ferrite bead
file:///I:/Temp/AM/OBSELECENCE/BL02RN2R1M2B%20_ferrite%20bead/KEM_LF0070_BEAD-L.pdf

We want to find a replacement which has less inductance, because it is used in the switching node of various SMPS's (eg in series with the flyback FET drain). ..But not that much less inductance, because then it won't have as good filtering effect in other products where its used as part of an SMPS output LC filter.

 

[FvM]

Also, its difficult to compare these beads because the Kemet one gives its impedance as 2 Ohms, whereas the BL02RN2R1M2B gives its impedance as 1.1uH at 1MHz.

Both products have impedance curves and can be well compared.

 

[treez]

Thanks,
Also, the B-01-RS is rated 5A, but the BL02RN2R1M2B is rated 7A....but they are both comprised of a piece of wire which is approx 25mm long and 0.65mm thick.
In both cases, even with 7A in this wire, the wire would only dissipate 82mW, so surely both parts could conduct well over 7A? So why are they rated at 7A (and 5A for the B-01-RS)?
Presumably the current rating is to do with saturation of the ferrite bead bit?

 

[FvM]

Quite obviously, the different current rating with same wire diameter reflects different temperature limits assumed by the manufacturer. At the end of the day, you have to set and verify current ratings for your application conditions yourself.

It should be noted, that ferrite bead current ratings are purely thermally defined. Reduction of impedance will occur at considerably lower currents. See a Fair-Rite ferrite bead curve for reference. Unfortunately Fair-Rite is the only vendor who gives detailed current dependency for leaded and chip ferrite beads.

 

[treez]

Also, i am not sure with these soft ferrite beads if you can glean info about the actual "inductance" comparison between the two (we must have same or less inductance in the replacement part, otherwise it will potentially cause overvoltage ringing at the FET drain).

...Because each part only gives "impedance" vs freq charts, instead of "inductance" vs frequency.

....these soft ferrites actually end up looking to an extent resistive.........and since each chart only shows "impedance", we dont know what proportion of that is inductance and what proportion is resistance?......so we cant make the comparison......we need the replacement to be less inductive than the BL02RN2R1M2B

--- Updated Mar 17, 2021 ---

Quite obviously, the different current rating with same wire diameter reflects different temperature limits assumed by the manufacturer.

Thanks, and also, i am sure you would agree that a 25mm piece of 0.65mm thick copper wire would not break at 10A, let alone 7A ?(it would only be dissipating 0.175W at 10A). This even at say 80degC ambient.

 

[c_mitra]

Thanks,
Also, the B-01-RS is rated 5A, but the BL02RN2R1M2B is rated 7A....but they are both comprised of a piece of wire which is approx 25mm long and 0.65mm thick.
Presumably the current rating is to do with saturation of the ferrite bead bit?

Right; the smaller the bead (size does matter) the lower is the saturation current. If you have a large DC current as bias, you are basically shifting the magnetization curve on one side and the effective inductance and impedance falls. The solution is simple: use two beads!

In post #5, you can see the effective impedance as a function of frequency under different bias current. As the bias current increases, the effective impedance falls. The behavior is most pronounced at around 100MHz and is practically absent at 1MHz.

You can safely assume that all soft ferrites will show a similar pattern. The most common model is a resistor in series with an inductor but this is not sufficient to explain the dispersion effects properly. I do not have a ready reference however.

 

[treez]

Thanks, with these Z curves, we need to know L , and not too interested in R.
Now Z = SQRT(R^2+ XL^2).
So how may we work out what is the actual inductance when we are given a Z figure.?.....or can we not?

 

[c_mitra]

So how may we work out what is the actual inductance when we are given a Z figure.?.....or can we not?

I am afraid, the latter.

It depends on the model.

Remember that Z is an observable quantity; it can be measured directly. So are the in-phase and out-of-phase components. They can be determined by simple quadrature.

Honestly, R should be a constant. Z at the lowest freq should be equal to R. It should be around 1 Ohm or less. But I plead my ignorance.

But then the inductance should also be independent of the frequency. So I take the 0A curve (vide post #4) and estimate the L at 10^7 and at 10^8 Hz and take the mean. That is the best that can be done.