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When not using power planes, in order to prevent >100 MHz noise from travelling excessively around the board via the power distribution network, one can place filters just outside the decoupling caps nearest to the consumer, inline with the power supply connections. The classical topologies are T-Filters and Pi-Filters, both of which are available in small integrated form factors.

schematic

simulate this circuit – Schematic created using CircuitLab

But when to use which filter topology ?

When simulating this naively with Spice, it looks like T-Filters would be almost always better to prevent noise entering the power distribution, using the same values for the L and C. At the same time, they are better to prevent noise from reaching the consumer. However, Pi-Filters seem to be much more popular.

So what am I missing ? And assuming my sims are wrong, what would be a scenario for T-filters ?

To simulate this, I set either V1 or the IC as a voltage source and observed the rejection on the other end, respectively. Typical values for the L and C of the filter are 100 nH and 100 pF, respectively. For both parts I included some typical parasitics, e.g. 0.1 nH for the C and 0.3 pF for the L, aswell as some ESR. I don't model the Network itself as a transmission line, but a inductor with parallel capacitance and parallel resistor of 10 Ohm. It is not an extended plane, but large polys over GND plane, so the impedance will be rather low. However, I am not sure what Rser to assume for the voltage source (when the IC is the voltage source). Putting a high Rser of e.g. 100 Ohms there makes the Pi Filter better, but I am not sure what Rser would be realistic.

As can be seen below, the only advantage of the Pi (red curves) in my sims, is less noise reaching the consumer between 1 and 10 GHz.

Noise reaching the Network:

enter image description here

Noise reaching the consumer:

enter image description here

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  • \$\begingroup\$ Maybe your simulation is wrong or mis-applied the components? Maybe your simulation didn't do something that real life does? No data sheets for comparable parts? \$\endgroup\$
    – Andy aka
    Jul 6 '21 at 14:32
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    \$\begingroup\$ I think you need to put the circuit up there. There'll be a reason and it's likely to be surrounding source output impedance being zero thus favouring T filters rather than Pi filters. Ditto load impedances and not taking into account power distribution being a transmission line etc.. \$\endgroup\$
    – Andy aka
    Jul 6 '21 at 14:48
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    \$\begingroup\$ Pi filters give you two bites at the parallel cherry, T filters give you series impedance at the ports which can be essential depending on what you connect. I had a very difficult isolation problem which was caused by the length of a power line being resonant at some frequency. It had a cap to ground at one end and a kohm impedance ferrite bead at the other, so it went off at lambda/4. Changing the bead for one with a few 10s of ohms impedance terminated the line and killed the resonance. Look at the entire environment of your filter to determine what's needed where and when. \$\endgroup\$
    – Neil_UK
    Jul 6 '21 at 15:52
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    \$\begingroup\$ @tobalt look up 'a second bite at the cherry'. Generally a parallel element has an unimpressive attenuation due to the inductive reactance of its physical length. A pi filter gives you two of those in series. The series element between them allows their attenuation to be multiplicative. \$\endgroup\$
    – Neil_UK
    Jul 6 '21 at 18:53
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    \$\begingroup\$ @tobalt both pi and T are nominally 3rd order. However the presence of C2 in your original diagram will work better with a T, as it has a series port impedance, as you say, making it 4th overall. I worked at 6GHz for a while, and one of the engineers spent a few weeks characterising various purchased filters, and filters constructed from chip Ls, Cs and ferrite beads, and found poor correlation between even the best models we had of the components and the actual deep attenuation they gave in a practical circuit, though the passband was usually right. If it matters, build it. \$\endgroup\$
    – Neil_UK
    Jul 6 '21 at 20:06
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T types are preferred when the I/O impedances are low, while Π filters are preferred for high I/O impedances. It's not a rule, but a bit of common sense, because if the I/O impedances are low and you're using a Π lowpass, the capacitors will be of little help, whereas the series inductors will contribute more. Of course, if the values for the LC elements are comparable with the I/O impedances, it won't matter much if it's a T or a Π.

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  • \$\begingroup\$ i have heard that 50 Ohm would be regarded as neither high nor low. but since everything in the power distribution network is probably far below 50 Ohm, i wonder why Pi filters are still so popular. \$\endgroup\$
    – tobalt
    Jul 6 '21 at 18:42
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    \$\begingroup\$ @tobalt "50 Ohm would be regarded as neither high nor low" -- compared to what? If it's compared with the reactive elements of the filter and they are the same then the results will be pretty much the same, even if you have parasitics: shunt C will have PCB inductances (capacitances will add up), while series L will have PCB capacitances (inductances will add up). If the I/O impedances are reactive (such as I see in your pictures) then shunt or series may matter. Ultimately, testing will be the decisive answer, since a simulation is only as good as the model. \$\endgroup\$ Jul 6 '21 at 19:16
  • \$\begingroup\$ I thought that 50 Ohm would be average because that's what all the filter makers use for their datasheet plots, but power distribution is usually much wider/lower impedance in my case. The LTspice guru advocating testing means something :-) I will definitely provide footprints for both options. sounds like T filters can be better indeed sometimes. \$\endgroup\$
    – tobalt
    Jul 6 '21 at 19:28
  • \$\begingroup\$ The T section might prove more useful, in that the series L will be larger than a PCB L, and for power rails you're dealing with currents; shunt C will hardly matter for those small PCB L. OTOH in a T with a high output impedance the 2nd L will hardly matter (if not properly calculated). Also, @Neil_UK has good points. I will never put simulation above practice, or claim to be master in some domain, but thank you for the vote of confidence. :-) \$\endgroup\$ Jul 6 '21 at 19:37

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