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I dont quite understand what importance does having a 2nd or 3rd order RC Lowpass filter do to filtering power rail noises. If the first stage filter only allows frequencies below 15.9 Hz arent the other stages unnecessary? Adding more series resistance to the rail will always not benefit the load since it would be receiving slightly less voltage.

Does adding more stages equate to less noise than the previous one ?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ You have an error in your diagram. The cutoff frequency of the second stage filter is actually 1.59kHz, not 15.9kHz. \$\endgroup\$
    – Mr. Snrub
    Jun 14, 2020 at 23:54
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    \$\begingroup\$ iirc the "order" of a filter essentially dictates how sharply the filter rolls off. Higher order falls off more steeply. The ideal filter would be a square shape. No realizable filter can do that. \$\endgroup\$
    – vicatcu
    Jun 15, 2020 at 0:01
  • \$\begingroup\$ The latter stages load the prior stages. So, you need to solve:$$\begin{align*} \frac{V_O}{R_3} + s\: C_3\: V_O &= \frac{V_Y}{R_3}\tag{$V_O$}\\\\ \frac{V_Y}{R_2} + \frac{V_Y}{R_3} + s\: C_2\:V_Y &= \frac{V_X}{R_2} + \frac{V_O}{R_3}\tag{$V_Y$}\\\\ \frac{V_X}{R_1} + \frac{V_X}{R_2} + s\: C_1\:V_X &=\frac{V_I}{R_1} + \frac{V_Y}{R_2}\tag{$V_X$} \end{align*}$$where \$V_X\$ is the left node and \$V_Y\$ is the middle node, for$$\frac{V_\text{O}}{V_\text{I}}$$From that you can work out the actual behavior. \$\endgroup\$
    – jonk
    Jun 15, 2020 at 2:09
  • \$\begingroup\$ I've never seen a power rail filter like the one in your question - it will be really ineffective. Where did you find the circuit and what made you think it is suitable as a power filter (it isn't). \$\endgroup\$
    – Andy aka
    Jun 15, 2020 at 9:37
  • \$\begingroup\$ I have something like this on my guitar tube amp, coming from transformer through tube fullwave rectifier. \$\endgroup\$ Jun 15, 2020 at 10:44

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I would say the second and third stages are there to deal with self-resonance issues in the capacitors. (So this really isn't a "third-order filter", for any practical purposes.)

If the capacitors were ideal capacitors, then the second and third stages would have no appreciable effect. In an ideal capacitor, the impedance of the capacitor keeps getting lower as the frequency gets higher.

In the real world, all capacitors have some parasitic inductance. At low frequencies, this effect of this parasitic inductance is inconsequential, so the impedance-vs-frequency relationship resembles an ideal capacitor. But at higher frequencies the effect of the parasitic inductance becomes more and more significant (the capacitor actually looks like an inductor-capacitor "LC" circuit). At some point there will be a resonant frequency just like with any other LC circuit, and beyond that frequency the impedance of the capacitor will increase as frequency increases, just like an inductor. At this point the low-pass filter is no longer a low-pass filter!

https://resources.pcb.cadence.com/blog/2019-capacitor-self-resonant-frequency-and-signal-integrity

As you'd probably imagine, larger capacitors have a lower self-resonant frequency and smaller capacitors have a higher self-resonant frequency. So, it is a common practice to put multiple capacitors in parallel -- for example a 0.1µF capacitor in parallel with a 10µF capacitor -- to overcome this effect. Once the frequency gets high enough that the 10µF capacitor is no longer effective, the 0.1µF capacitor is still acting as an effective capacitor so the circuit overall still works.

The extra resistors in this circuit are new to me, but as this article suggests perhaps they are there to dampen out other resonant effects that could happen otherwise.

https://incompliancemag.com/article/using-capacitors-in-parallel-dangerous/

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  • \$\begingroup\$ Spot-on with the answer. I also think the added resistors are teacher-speak to create more math. Parallel capacitors of equal and different values are common, but I have never seen added resistors. Normally specific value inductors would be used, if any at all. \$\endgroup\$
    – user105652
    Jun 15, 2020 at 3:47
  • \$\begingroup\$ Ahh so those resistors are just unnecessary. and the best circuit to use would be resistor paired with parallel capacitor \$\endgroup\$
    – Jake quin
    Jun 15, 2020 at 7:15
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The additional RC sections are useful to attenuate at very high frequencies, where the ESR and ESL of the first capacitor acts to set an attenuation floor.

With 100 ohms in the first section, and 0.1 ohm ESR, you get at most 60dB (1,000:1) attenuation at any frequency, even ignoring the ESL.

If you truly want 100dB attenuation, which would be a mere 5 decades above the F3dB of the first RC section, all these irksome parasitics have to be including in the thinking.

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