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I'm trying to understand a pi filter, which typically looks like this:-

typical pi filter

I've seen lots of these especially with the valve guys, and C1 = C2 always. So I made this in LTSpice which is for a 100mA load:-

spice model

Notice the parameterised value of C1. The AC analysis produces this however:-

spice analysis

There is only one curve! If C1 = C2 and C1 = {X}, you'd expect four curves. To check my model, I interchanged the parameterised capacitor so that C2 = {X} and got four widely spaced curves as expected. The analysis is suggesting that C1 has no effect, and I'm confused (again). Have I misunderstood the nature of a pi filter, or is my LTSpice model incorrect?

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  • \$\begingroup\$ The transfer function of the filter - as seen from left to right - does not depend on C1, since V1 is applied between one lead of L1 and ground. \$\endgroup\$ Commented Apr 28, 2017 at 14:27

2 Answers 2

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You are feeding the filter from a zero ohm source.

Ask yourself what the ratio of C1's impedance to the source impedance is, and you'll see why it appears to have no effect.

Introduce a 150 ohm series resistor (to match R1) between V1 and C1 and it'll start to make more sense.

Any time you see a Pi filter, work out what the source and load impedances - terminations - are. The filter is designed with these terminations in mind, and if mis-terminated, it won't have the expected frequency response. Usually source and load terminations are the same, but sometimes the filter is designed to work from a low source impedance into a high load impedance, to avoid the 6dB loss (half the voltage) of a normally terminated filter.

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  • \$\begingroup\$ Yes. I've done as you suggest and it makes all the difference. 4 curves appear. But do you mean source impedance or source internal resistance? If it's impedance, at what frequency as isn't it frequency dependant? \$\endgroup\$
    – Paul Uszak
    Commented Apr 28, 2017 at 23:16
  • \$\begingroup\$ Internal resistance (which is impedance with zero imaginary). In SPICE world, voltage sources have (machine) zero resistance. \$\endgroup\$ Commented Apr 29, 2017 at 6:05
  • \$\begingroup\$ Resistance is the real component of impedance. Most termination schemes (like your 150 ohm R1) have no imaginary component, in which case impedance and resistance are the same. \$\endgroup\$
    – user16324
    Commented Apr 29, 2017 at 10:02
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One advantage of a pi filter is that it's symmetric: signals travelling from right to left are affected the same way as signals travelling from left to right. Additionally, you'd ordinarily have an impedance on your feedline as well, if you are wanting to model high-frequency systems where transmission line effects may become important--in this case, the value of C1 does matter.

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  • \$\begingroup\$ I don't feel this answer is nearly complete enough, so feel free to modify it as you see fit! I could even be wrong. \$\endgroup\$
    – Hearth
    Commented Apr 28, 2017 at 13:53
  • \$\begingroup\$ Sorry, I don't understand what left to right signals are. Isn't that the quintessential definition of AC? \$\endgroup\$
    – Paul Uszak
    Commented Apr 28, 2017 at 13:53
  • \$\begingroup\$ What I'm saying is, if you put your AC source on the right side of this filter, and the load resistor on the left side, you would have exactly the same result. \$\endgroup\$
    – Hearth
    Commented Apr 28, 2017 at 13:54

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