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Ferrite Beads V 10Ω Or 22Ω Resistor - Best Practice?
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author Ferrite Beads V 10Ω Or 22Ω Resistor - Best Practice?
Reality Checkpoint
Now that I am trying to understand what I am building, rather than just complete a project in a 'paint by numbers' fashion can some kind soul explain to this dullard which is best practice and why?

I understand that the role of the bead is to attenuate high frequency noise, and when run in series with the power rail and capacitors it can act as a low pass filter.

So this all sounds exactly what you need. So why do I read that the use of 10Ω or 22Ω resistors in place is also recommended? Is it just for protection of the circuit?

And if I am building solely analogue circuits isn't there less noise compared to digital circuits presuming that the power network in to my house and power supply is fairly clean? Wouldn't it therefore be better to have a power conditioner in between the network and power supply and use the resistors? If so, would you use 1/4W or 1/2W?

So many questions, so little time! help
Ferrite beads are only useful at attenuating RF, so if you're building a digital board and want to prevent it from radiating radio interference, that may help. At audio frequencies they will look like a jumper.

With a small value resistor (followed by filtering capacitors), you have a single-pole RC lowpass filter with a cutoff frequency of 1/(2*pi*R*C) where R is in ohms and C in farads.

The only major downside is that the voltage on the power rail after the resistor is reduced by the average current (in amps) times the resistance.
I just started a thread about this, so let's see if I understand it correctly. Hopefully someone will jump in if I get something wrong. I'm thinking through this problem by writing it out, so please excuse the long-windedness smile.

Answering your questions in reverse order:

First, don't assume that things are clean unless you know they are. Your house power has all kinds of high-frequency signals because of everything else electric and all the rf happening in a given house with long wires routing between rooms. Ideally these are small. Ideally they should be filtered before the power even gets to the transformer. But unless you know that you have that ideal situation, it's very possible that you don't. And that's only for a linear supply; for a switching supply, high-frequency noise is inherent, although again ideally it should be filtered before it makes it from the power supply to the rails.

Second, analogue circuits don't necessarily produce only low-frequency noise. Discontinuities in waveforms---the jump in the sawtooth, the direction switching of the triangle, the rise and fall of a square wave--plus discontinuities in cvs---the transition time of a sample and hold, a sequencer, a very fast envelope generator, or a quantizer---all have high-frequency content, and depending on how they are designed, may draw this high-frequency content as high-frequency current spikes from one rail to another, or from the power rails to ground.

Now, you might know that even in an ideal sawtooth or square wave the amplitude decreases at a rate of 1/f, and so even if there is high-frequency content, there's more low-frequency content. This is true. But the high-frequency content is more of a problem.

What you have to understand is how different current draws in one module end up appearing as voltage fluctuations in the power inlet of another module in the first place. Basically, current turns into voltage through Ohm's law: V=IR, or when we take inductive and capacitive reactance into account, V=IZ, where Z is total impedance. The current from the module must flow through the wiring harness to the distribution system, usually a board with headers, but possibly busbars or a flying bus, and then through that distribution system to the wires which lead back to the power supply, and then through whatever output impedance (hopefully very low) the power supply provides. These things all have a low, but finite, resistance. But any conductor also has a low, but finite inductance. As the conductor gets thicker, the resistance drops at a faster rate than the inductance, and so it's harder to get rid of. What this amounts to is that the voltage drop for a given current draw increases considerably with frequency.

So lastly, coming to your actual question, when a voltage fluctuation appears in the power bus, that voltage fluctuation then goes through the wiring to the module, plus whatever impedances it encounters, and then hits the decoupling capacitor, which depending on where the voltage was before may be at a higher or lower voltage than the voltage that it sees now. The capacitor will then charge or discharge through those impedances back to the power bus, thus maintaining a more steady voltage than the bus. A higher impedance means a slower discharge, and hence a steadier voltage. Since a ferrite bead has a higher impedance at higher frequencies, this adds to the frequency-dependent effect of the capacitor, and hence more effectively filters higher frequencies, and less effectively filters lower frequencies. This hopefully makes up for the way the inductance in the distribution system creates greater voltage drops at higher frequencies than at lower frequencies. A resistor has more or less the same impedance at all frequencies, but it does essentially the same thing: keeps the decoupling capacitor from discharging quickly into the bus when the bus voltage fluctuates. Some people prefer the resistor, because the crossover frequency of the RC filter is lower than for the ferrite bead, and so it does more for keeping out the bus noise which results from the bus resistance, rather than just the (probably higher voltage) noise resulting from the bus inductance.

As a bonus, the resistor will go up in smoke if you pump too much current through it, possibly saving your module when you connect it wrongly. If you want this to happen, use the lowest wattage resistor which won't get damaged by your module's normal current intake. Since there's not much voltage across it, that's probably no bigger than 1/4W, possibly 1/8W, but it depends on your module.

Most analyses I've seen stop there. But there is another factor as well: the module's current draw. When the capacitor is higher voltage than the bus, the module pulls current from the capacitor and the capacitor discharges to both the module and the bus. This goes on until the capacitor has the same voltage as the bus. Then the module draws current from both the capacitor and the bus, until the voltage of the capacitor reaches such a level that there is zero current flow and everything comes from the bus. When the capacitor begins at a lower voltage than this, the module pulls current from the bus, and the capacitor pulls current from the bus, until the current flow into the capacitor is zero.

Because of this, higher frequency demands in current requirements can put the burden on the capacitor instead of the bus, and thereby keep these high-frequency currents entirely off the bus. This is good for other modules, as they won't have to work to filter noise which your module doesn't put on the bus in the first place. On the other hand, lower frequency changes in current demands will exhaust the capacitor's current reservoir, and draw current directly from the bus. When this happens, the module sees a voltage drop of V = IZ, where Z is the impedance of the wires to the bus, plus whatever input impedance is present at this frequency. If you use an input resistor, this is V = IR, at any frequency where the capacitor isn't supplying appreciable current. On the other hand, if you use a ferrite bead, at low frequencies the voltage drop will be much smaller, since the impedance of the bead is small at low frequencies.

Making the input impedance higher will cause the capacitor to supply more of the current requirements for the module, but will also cause the module to see bigger voltage drops. At any given frequency of current fluctuation, then, there is a tradeoff between letting these fluctuations spread to the rails and keeping the voltage drop low for our own module. Which one is better depends on the circuit, but since the voltage drops for our own module tend to correspond to the signals themselves, the tradeoff ends up being between distortion and crosstalk---and minimizing crosstalk is probably usually the better option. This is certainly the case where triggers, gates, and digital signals are concerned, since these circuits tend to have a very high rejection of not only power supply noise, but also signal noise/distortion itself.

This is all complicated, but I would say as a rule of thumb, if your module doesn't create much high-frequency current draw (including sharp-edged "low-frequency" draw), or if you need ultra-low distortion, use a ferrite. E.g. a mixer, amp, filter or some oscillators. If it does create high-frequency noise, use an inductor, with the value somewhat proportional to the noise created. If your module really doesn't care about a tiny amount of distortion and/or you want the bonus protection, then maybe use a resistor. Others will disagree, however.
Reality Checkpoint
555x555 and mmagin

Thank you so much for your replies.

You have given me a lot of info 555x555 we're not worthy that I need to read properly rather than skim on the bus on my way to work! I will get back to you.

Thank you both again.
My opinion on ferrite beads being somewhat pointless is because they're most useful well over 50 MHz, and they're often not even specified below 1 MHz, for example: 38-10R-10_Dwg.pdf

Also, if you're fine with incurring voltage drop, you could always use, for example a 10 volt LDO regulator on each of your module's power rails, provided you follow all the datasheet recommendations with regard to stability.
smudge wiggle
I recently started as a trial using 100uH axial inductors from Tayda for filtering the power rails and then another one to the supply of the microcontroller should I be using one.
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