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I have a board with this differential PI filter:

schematic

simulate this circuit – Schematic created using CircuitLab

Say it's designed to pass signals with frequency X and bandwidth B1, and now I want to redesign the filter so that it stays with the same configuration but change it's values so that it will pass signals with frequency Y and bandwidth B2.

Luckily, I have a program that allows me to input the desired filter type (lowpass, highpass, bandpass), configuration (PI, T), frequency, and impedance. It provides me the values of the inductor and capacitors. Unfortunately, it provides me the filter as a single ended PI filter, like this:

schematic

simulate this circuit

My questions:

  1. How do I change the values of the inductors and Capacitors that are given to me on a single ended basis (if at all), to transform the filter to a differential one?
  2. How do I make sure my new differential filter has the same impedance I want it to have (the single ended filter has that impedance, not the new differential filter).
  3. If I am limited in my choice of inductors (say I don't have the value I need L1, but I have L1+150 or L1-200), is there a way to compensate for it, for example by changing the values of the capacitors? (I can't put inductors in parallel or series).

Bonus question - how do I measure the impedance of the filter, using either ADS (advanced system design) or LTspice?

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  • \$\begingroup\$ The PI filter you have shown has a frequency response that depends on the driving source impedance and load impedance for (1) and, (2) they are low pass hence B = Y. You can simulate the above of course and get answers. \$\endgroup\$ – Andy aka Jul 18 '20 at 7:53
  • \$\begingroup\$ @Andyaka Alright, So if I know the driving source impedance and the load impedance I'm supposed to be matched to, how do I fit my filter to have the desired frequency response? Also, are you saying the bandwidth = pass frequency? \$\endgroup\$ – nettek Jul 18 '20 at 8:11
  • \$\begingroup\$ Additionally, I did not mention it in my question (didn't think it's relevant, tell me if it is), right after the source signal there is a transformer (I know my load impedance changes according to the transformer impedance ratio). I mentioned I inputted the impedance in the filter designer program - this is the load impedance I calculated. \$\endgroup\$ – nettek Jul 18 '20 at 8:13
  • \$\begingroup\$ What is the filter meant to do i.e. why have you got a need for it? All circuit details in the signal chain are relevant especially if drawn as a schematic. \$\endgroup\$ – Andy aka Jul 18 '20 at 9:00
  • \$\begingroup\$ I receive a RF signal with a certain frequency. The purpose of the filter is to make sure only the signal passes (within the allowed bandwidth) and not any images of the signal, or something. It's also used for impedance matching. \$\endgroup\$ – nettek Jul 18 '20 at 9:48
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The resulting asymmetrical Π will have C/2 and 2L as values:

result

This is also one way to simulate it. Note that the source now has twice the impedance and twice the amplitude of the signal in order to match the outputs (so you should account for that when you're recalculating the filter).


In response to the comments, I have re-uploaded a different image (the outputs are slightly modified to be all visible; the phases show they're identical by overlapping, though). The values for the components are completely bogus (values ranging from 1 to 5, at a whim).

For the symmetrical case, the load is better viewed as two series resistances. Transforming this to an asymmetrical filter can be done in two ways:

  1. Keep the values of the filter the same (LC elements), but halven the source's and load's resistances. This makes V(c) be the "half" equivalent of the symmetrical case, since you can visually draw a horizontal line separating the two sides. To have the same magnitude at the output, the source needs to be twice as high (ac 2), but the power will also doubled.

  2. Keep the overall values for the source and the load, but halven the LC elements (V(d)). This is done to preserve the equivalent source/load of the symmetrical case. The source also has twice the magnitude due to the two input sources being in series (V1 and V2), but now the power is preserved.

In both cases, the source and the load are considered together, i.e. they're either both full valued, or both half-valued. If, during the transformation process, the source and the load cannot end up proportionally the same (e.g. discarding one winding from a center tapped trafo, but keeping the load), then the whole filter needs to be recalculated, since the equations account for both the source and the load (they cannot be separated).

One last comment: in general, when the source's impedance is zero (voltage source), or less than or equal to the load, a T section is used; when the source is greater than the load, or infinite (current source), then a Π section is used. I have made no comments about choosing one or the other since I consider you to know what you're doing.

At any rate, the same steps for transformation apply for the T as for the Π section. For the LC elements, the shunt elements are considered in series (C1 and C4 result in C7) and the series elements are in parallel (L1 and L2 result in L4).

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  • \$\begingroup\$ Thank you! about the source doubling it's impedance, two questions: 1) Is this still true if there is a transformer between the source and the filter? 2) My source has an impedance of 50 Ohm, I can't change it, so what do I do? \$\endgroup\$ – nettek Jul 18 '20 at 17:35
  • \$\begingroup\$ 1) If you mean a center-tapped trafo, then yes, it behaves just like any other differential source. 2) The load in the 1st case is actually half in the 2nd, halvened the rest so you can better see the effect. You can either double Rin and halven the values for the LC components, or halven Rout and keep the LC values. If you can't change Rin and Rout, then you need to recalculate the values of the LC. Also, as a minor reminder, in general, high Rin or low Rout (or both) means Π section; low Rin/high Rout/both means T. \$\endgroup\$ – a concerned citizen Jul 18 '20 at 22:03
  • \$\begingroup\$ Additional questions, if you don't mind: 1) You said something about the load being half (of half), but I don't understand since in both cases Rload = 5. Can you elaborate? 2) So if my Rout = 400Ohm, I need to input (in my program) an Rout of 200 Ohm. But the LC values given to me are the same ones used in a differential filter? 3) Do C5 and C6 have to be in a 1:2 ratio? Can they be of the same value? \$\endgroup\$ – nettek Jul 19 '20 at 10:59
  • \$\begingroup\$ @Eran I have updated my answer. \$\endgroup\$ – a concerned citizen Jul 19 '20 at 12:27

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