I will show how to do this in an upcoming video. It is actually very easy by looking at an impedance curve for a ferrite bead and the impedance equations for an RLC parallel circuit. Give it a try and see if you can do it, or you can tune into the video once it comes out in the next couple of weeks.
Hello Zach, thank you for all the awesome content. Would you recommend putting a 100Ω@100MHz ferrite bead on a VBUS line near the USB connector for a general high frequency noise attenuation? I mean it's recommended by many IC manufacturers and it's even mentioned in the USB spec.
Hello. I'm also a beginner pcb layouter and I'm watching your videos. Can you make a video where you can explain practically where and how the impedance is used? Maybe a few examples.The problem is that you deal with many topics with only theory and I don't know where to apply all these rules and tips.Can you show us some practical placement tips? Power planes routing?
Hi Adrian, we have quite a few videos where we show examples of the concepts inside a PCB layout, and I'm making sure to do more of these that are only focused on layout examples inside of Altium Designer. Thanks for watching!
Sure. The ferrite can be simply modeled as a parallel RLC circuit. This is not an exact model, it is only an approximation and it will overestimate the reactance at mid-range frequencies. The impedance value at the peak in the impedance curve gives the value of R. The bandwidth (Q value) is used to determine L using the value of R, and then the natural frequency (known from the peak in the impedance curve) and L can be used to determine the product LC. From that you solve for C, and now you know all three circuit quantities. This first thing to note is that the ferrite's behavior can be approximated as being a parallel RLC circuit. This is because, at resonance, the impedance curve of the ferrite is maximum and purely resistive. So at the peak in the impedance curve, the impedance will be equal to the resistance. To determine the capacitance and inductance in this model, you have to use the FWHM of the impedance curve (the Q-factor) and the resonant frequency. This gives you 2 equations with 2 unknowns, so you can solve these two equations using your input values of resistance, Q-factor, and resonant frequency. Just note that this model is only an approximation. A better way to incorporate the ferrite is to get a SPICE model for your specific component. You might need to create your own SPICE model. This is something I'm still learning to do, but once I get better at it I can make a video on it.
@@Zachariah-Peterson I was doing the calculations on my own and I never get the values you have. The resistance is just a read off value. But how do you get the Q-factor. I assume the FWHM is at 75Ohm so starts more or less @30MHz and ends at more or less @4000MHz. After I would find my Q factor, am i right to assume that the formulas needed to solve to L and C are : Q = R*sqrt(C/L)
Thank you again, Dr. Peterson.
Thanks for listening
I use ferrites and caps for a 1Hz 300ps rise time signal going in a 1 to 8 output buffer. The power supplies are now happy.
hi dr. Peterson, how did you calculate R,L and C in the ferrite model? for example to calculate C did you use C=1/2*pi*f*Xc? Thank you
I will show how to do this in an upcoming video. It is actually very easy by looking at an impedance curve for a ferrite bead and the impedance equations for an RLC parallel circuit. Give it a try and see if you can do it, or you can tune into the video once it comes out in the next couple of weeks.
Interesting video, waiting for part two
Coming soon!
Hello Zach, thank you for all the awesome content.
Would you recommend putting a 100Ω@100MHz ferrite bead on a VBUS line near the USB connector for a general high frequency noise attenuation? I mean it's recommended by many IC manufacturers and it's even mentioned in the USB spec.
Hello. I'm also a beginner pcb layouter and I'm watching your videos. Can you make a video where you can explain practically where and how the impedance is used? Maybe a few examples.The problem is that you deal with many topics with only theory and I don't know where to apply all these rules and tips.Can you show us some practical placement tips? Power planes routing?
Hi Adrian, we have quite a few videos where we show examples of the concepts inside a PCB layout, and I'm making sure to do more of these that are only focused on layout examples inside of Altium Designer. Thanks for watching!
Great as always
Thank you! Cheers!
Hello! Can you explain how you calculated inductance and capacitance for the model of ferrite?
Sure. The ferrite can be simply modeled as a parallel RLC circuit. This is not an exact model, it is only an approximation and it will overestimate the reactance at mid-range frequencies. The impedance value at the peak in the impedance curve gives the value of R. The bandwidth (Q value) is used to determine L using the value of R, and then the natural frequency (known from the peak in the impedance curve) and L can be used to determine the product LC. From that you solve for C, and now you know all three circuit quantities.
This first thing to note is that the ferrite's behavior can be approximated as being a parallel RLC circuit. This is because, at resonance, the impedance curve of the ferrite is maximum and purely resistive. So at the peak in the impedance curve, the impedance will be equal to the resistance. To determine the capacitance and inductance in this model, you have to use the FWHM of the impedance curve (the Q-factor) and the resonant frequency. This gives you 2 equations with 2 unknowns, so you can solve these two equations using your input values of resistance, Q-factor, and resonant frequency.
Just note that this model is only an approximation. A better way to incorporate the ferrite is to get a SPICE model for your specific component. You might need to create your own SPICE model. This is something I'm still learning to do, but once I get better at it I can make a video on it.
@@Zachariah-Peterson I was doing the calculations on my own and I never get the values you have. The resistance is just a read off value. But how do you get the Q-factor. I assume the FWHM is at 75Ohm so starts more or less @30MHz and ends at more or less @4000MHz. After I would find my Q factor, am i right to assume that the formulas needed to solve to L and C are : Q = R*sqrt(C/L)
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