No need for ferrite bead.... just use resistor?

[cupoftea]

The following ferrite bead has an impedance of 100 Ohms at 100MHz....

BLM18EG101TN1D

https://uk.farnell.com/murata/blm18eg101tn1d/ferrite-bead-0-045ohm-2a-0603/dp/1515714.

At 100MHz, the skin depth of copper is 6um.
So we imagine that an actual 100R,0603 resistor has an impedance of much more than 100 Ohms at 100MHz?
..And so therefore, if you have a signal wire leaving a PCB, and it happens to have a series 100R resistor on the PCB....then put the resistor right by the connector (where the signal wire leaves the board) and use the resistor as your "high frequency bead". Would you agree?

 

[tyassin]

Hi

A 100R, 0603 resister is 100R at 100MHz. If you use FR4 then the wavelength is 1.41m at 100MHz and the resister is physically much smaller. There might be some small parasitic inductance of a few nH.

I don't see the skindepth as an issue as the metal in the resister is properly thicker anyway.

 

[cupoftea]

Thanks so you agree....just use resistor instead of ferrite bead? (in the case i mentioned)

 

[andre_luis]

A ferrite bead is intended to act as a high losly device above its cutoff frequency.
Why would you replace by a resistor which would add an intrinsic insertion attenuation at the whole spectrum ?

 

[FvM]

Ferrite beads are used when you want the DC resistance to be much lower than the high frequency impedance. If this is no requirement, a resistor can work as filter. Your thread started with a wrong assumption about frequency depedency of chip resistors but doesn't give an actual filter requirement. Thus we can hardly guess which solution serves your application.

 

[cupoftea]

Thanks, we want a 100R resistor in there, and we also want the same or better attenuation at 100MHz and below.
I wonder what each would be like at 200MHz?