Hola @DickCappelsThe ferrite beads I’ve used were around a length of wire and therefore had negligible D.C. losses.
There is not much to calculate. The preferred procedure is to consult the bead manufacturer’s impedance as a function of frequency charts, chose a bead and try it. Repeat if necessary.
Thanks Dick!!There was one user of this site who was having inconsistent performance with an LED oscillator/driver and was using ferrite beads. He had bought some selected values via eBay and sent some to me to test. It turned out that each envelope had a variety of values in it rather than the values marked on the envelope. An inductance meter or network analyzer are your only hope in a case like that.
Painting first is a great idea considering how much trouble it would be to go through a pile of beads and sorting them out!
Thanks Dick!!
capacitor's purpose in bypass mode is to provide CURRENT not voltage. It can provide current faster than the power-supply.Two different but related tasks,
Ferrites are in series with the power
high frequencies , i.e noise is attenuated,
in both directions,
whilst the DC is only mildly attenuated by the resistance of the ferrite.
The capacitance,
is in parallel with the power pin of the chip.
its aim is to provide voltage to the chip when the input voltage dips ,
and to provide the higher current spikes
If the chip did not have the capacitor,
a spike in current, of say 100 ps,
would have to come from the PCB power supply .
the PCB has track impedance as well as the ferrite bead resistance,
so the voltage would drop.
Yes, you are right @BobaMosfet, and as we know , the reason to provide the current is to keep the voltage up,capacitor's purpose in bypass mode is to provide CURRENT not voltage. It can provide current faster than the power-supply.
Yes, we keep the voltage drop across the load at MAX. Without the capacitor, we couldn't deliver enough current, and the voltage would sag. I'm not arguing with you, I'm merely explaining the 'how/why' for people who don't know. People talk about voltage, and frequently forget what they really mean is voltage -drop-.Yes, you are right @BobaMosfet, and as we know , the reason to provide the current is to keep the voltage up,
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