I am wondering if its possible to buffer LC filters? I am also wondering about the I/O impedance of the LC filters if you can buffer them. I do know the component values you pick for the LC filters is based on impedance and the frequency range you need.

I know when you calculate LC filters usually you need to need to know the input and output impedances of the circuit to get the correct component values but if you buffer with op amps how do you set the impedances into the calculation? I am also guessing the input and output impedance calculations of a LC filter would probably chance with whatever op amp you use?

buffered LC filter example:

Buffered LC filter example

  • \$\begingroup\$ What kind of voltage does the non inverting pin of the first opamp(from left to right) see?AC or DC? \$\endgroup\$
    – Miss Mulan
    Jul 29 at 21:29
  • \$\begingroup\$ You have correctly terminated both ends of the filter here (the two 50 ohm resistors), with the buffers outside that system. That is absolutely OK fine. \$\endgroup\$ Jul 29 at 21:47
  • \$\begingroup\$ @user_1818839 Because virtually no current is being pulled from LC filter section due to the buffering wouldn't the impedance be technically 0 or maybe infinite? op-amps do have a input and output impedance , so would you calculate using those values? \$\endgroup\$ Jul 29 at 21:50
  • \$\begingroup\$ OK if the source opamp has a 1 ohm Zout (closed loop) you should reduce the source resistor by 1 ohm ... here, to 49 ohms. But practically Zout will be much less. The output buffer has practically infinite Zin, so the load is the 50R resistor across the filter output. (If it didn't, the load would be Zin and 50R in parallel) \$\endgroup\$ Jul 29 at 21:55
  • 1
    \$\begingroup\$ Excuse me, but if you use opamps to buffer LC filters to make them less sensitive to load and source impedances you are in the usable frequency range of opamps. Then you could probably also design the circuit as an active RC filter without those often difficult to get inductors. \$\endgroup\$
    – user287001
    Jul 29 at 22:58

1 Answer 1


Three-ish things:

  1. An LC filter needs at least one termination. With buffering, you have the freedom to choose both port impedances (within reason, see below). This can avoid the 6dB voltage gain penalty of a doubly-terminated filter, for example. (Different component values are required, you can't just change/remove a termination arbitrarily; https://rf-tools.com/lc-filter/ for example allows unequal matching, or one port shorted, though it does not support an open-circuit port! Most reasonably-comprehensive filter tables include open/short, e.g. Zverev.)

  2. Op-amps are not ideal, and if we're considering LC filters, presumably it's at frequencies where op-amps are especially far from ideal (else we'd probably prefer to save the cost/space and use active RC filters instead!). We need to know the closed-loop output impedance of the left amp, and the input impedance of the right amp, and subtract them from the respective termination element and first reactance as appropriate.

    The source impedance, will generally be complex (output typically has an inductive characteristic, |Z(f)| rising, because loop gain is falling; but it also falls at even higher frequencies, as internal and pin capacitances take over), so we can't assume the source termination is only Rs + Re(Zout). We may need to introduce another reactive component (matching network) to balance this out, or approximate it as L or C and subtract the amount from the filter's first element (choosing series or shunt type on that port, respectively).

    Op-amp inputs should be mostly capacitance in practice (so, subtract from a final shunt capacitor, for the lowpass case) -- for common voltage-mode op-amps. But beware this is not true of current-mode op-amps! Or if we use for example a common-base amplifier, the input impedance can be extremely low indeed (negative even, if there's some base/gate capacitance on it), and the terminator may be series rather than shunt.

    Note also, since these are buffers, primarily for isolation or current gain (but they could be made with voltage gain as well, with little impact on the filter), they could be implemented with amplifiers other than operational types. We might consider the case for RF amps, used to increase isolation (reduce s12) and maybe also some gain (s21) as needed; these might be discrete circuits (like the aforementioned grounded-base/gate amp), MMICs, or inline modules. The impedance of which might be whatever; well-behaved RF amps at least tend to have resistive inputs, but in any case, what it is, matters.

  3. Note also, this all assumes the amps are well behaved, not oscillating due to unwelcome impedances (a matching capacitor, shunting the output of a non-C-driving type amp, may cause it to oscillate!), and able to drive the load impedance at the full required signal level.

    Which, impedance is a fairly free variable here, but mind that it can't be so low that the output is unable to drive it, and it can't be so high that the required inductors are impossibly large, or capacitors impossibly small. Do not neglect the effect of stray capacitance (shunting across the element, or to GND), especially on sharp and narrow filters (the ladder topology for a narrow bandpass prototype can be especially troublesome in this regard; consider a coupled-resonators topology instead).

If you have accurate SPICE models available, simulating the filter is a good idea. Also include component models where possible (some manufacturers provide data for much of their product line, e.g. Coilcraft), and estimations of strays (component, pad and trace capacitance, etc.).

Finally, actually testing the filter (preferably with a spec+TG or VNA, but it can be done with a scope and sweep, too), and adjusting values between reality and simulation until they match reasonably well, can be both quite illuminating (find the strays, see the effect on response, matching, etc.) and helpful to converge on desired response faster.


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