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I have been struggling to obtain the Bode plot of common-mode response of a filter. I first simulated it using LTspice, then I used a USB network analyzer. Both do not match each other and to the given plot of the filter.

Here is the general datasheet of the filter. The specific model I have is called FN2060-10-06. Below is the components parameters for this model:

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And below shows the circuit diagram of this filter and the frequency response plots:

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First I want to show how I simulated this circuit in LTspice and what Bode plots I have obtained:

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Secondly, I used a USB network analyzer where it has the following pin-outs:

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Above 1+ and 1- are differential input terminals for channel 1; 2+ and 2- are differential input terminals for channel 2. W1 is one of the signal generator output.

So I set Ch1 as the reference channel using this software and made the wiring between the filter and this USB network analyzer as below where 220 Ohm is to limit the current:

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And the network analyzer plotted the following between 1Hz to 30 MegHz:

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So something I'm doing is not correct because neither the simulation nor the network analyzer results do not match the datasheet at all. All I wanted was to find a simple method to obtain the common mode response of this filter. Does anybody have experience with such setup? If wrong, how can I correct the simulation and/or the network analyzer setup?

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  • \$\begingroup\$ L1 and L2 are coupled inductors as are L3 and L4 so at a minimum you will need to add two K lines (coupling coefficients) \$\endgroup\$ – Warren Hill Dec 20 '18 at 15:26
  • \$\begingroup\$ Are my parameters wrong on my circuit like K1 L1 L2 1 ? \$\endgroup\$ – panic attack Dec 20 '18 at 16:27
  • \$\begingroup\$ I can't see "K1 L1 L2 1" on your circuit but that's the idea try "K1 L1 L2 0.998" as the windings will not couple perfectly and there will be some leakage inductance. \$\endgroup\$ – Warren Hill Dec 21 '18 at 7:41
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Your simulation model has not taken into account the full testing regime listed just above the graphs: -

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That testing regime details such things as source resistance and measurement node resistances. With these resistances added you will not get the response shown in your current simulation.

So, if you don't model the CISPR test precisely you won't get the same response (or nearly the same response).

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  • \$\begingroup\$ Does that mean 50 Ohm between Vcm and N; and 50 Ohm between P” and N” needed to be added to the current simulation? \$\endgroup\$ – panic attack Dec 20 '18 at 16:31
  • \$\begingroup\$ The comment to my answer also mentions inductor coupling issue in my sim. \$\endgroup\$ – panic attack Dec 20 '18 at 16:32
  • \$\begingroup\$ Sorry, but you'll have to look into the test properly. I don't have the details floating around in my head for this. But certainly a 50 ohm source and measurement load is highly likely. Regards the K, I think you have assumed 100% coupling but it doesn't matter for the common-mode simulation you are doing because both inductors are in-phase magnetically and see the same input voltage. \$\endgroup\$ – Andy aka Dec 20 '18 at 16:41
  • \$\begingroup\$ I see thanks; one last question could you also drop a short comment my network analyzer setup? I mean is that correct wiring for common mode. Because this differs even much more than the datasheet plot. Maybe mostly again due to setup parameters but I was also curious if it were correct implementation at least wiring. \$\endgroup\$ – panic attack Dec 20 '18 at 16:43
  • \$\begingroup\$ I've never used a network analyser so I think I'm out of my depth on that one but it looks to me like it might have a 50 ohm source or measurement port because the resonant peak is not as prolific as seen in the sim graph. \$\endgroup\$ – Andy aka Dec 20 '18 at 16:52

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