When I put in resistor R14 (100 Ω) I get this:
But when I remove R14 I get this:
Nothing happens to the MixerOut1 wave. Why not?
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2\$\begingroup\$ The inductor of a LC filter should usually be on the lower impedance side of the capacitor. Removing R14 increased the impedance on that side. Try running it again with L3 "before" the capacitor and R14 removed. \$\endgroup\$– virCommented Oct 27, 2022 at 22:22
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\$\begingroup\$ @vir Perhaps I got that filter in this website : leleivre.com/rf_butterworth_LPF.html Perhaps it must be addapted both to input and output? Not sure if it's because of that, makes little sense since we are in Band base. I did what you told and it didnt work \$\endgroup\$– ScipioCommented Oct 27, 2022 at 22:27
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1 Answer
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The results you are getting are reasonable. Refer to the simulation below. The circuit on the left is your circuit, except I don't know what is connected to the + input on U6. Thus, R1 could be a very high value. This circuit is a L-R filter. The 3 uF capacitor on the left of the inductor in your circuit doesn't do much since a low impedance source is driving it. In fact, it will probably cause instability.
If you change the resistor to 100 ohms (R2), then this starts to behave closer to what you want, but it's still a L-R filter.
As @vir suggested, move the 3 uF capacitor to the right of the inductor and now you have a L-R-C filter. With R being high impedance, you will get severe peaking, not what you want. You can reduce peaking by decreasing R as shown in the right-most circuit.
The resource you cite is fine, but you failed to use a series resistor on the input and a resistor to ground on the output of the filter. Zo is the input and output resistance values. Either use a L-section filter (L-RC) or a pi-section filter (R-C-L-CR). You attempted to use a backwards L-section filter which isn't very effective. You may want to choose a higher value of Zo like 1k or 10k to get reasonable component values.
