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Can anyone recommend a technique for compensating for inductor self-resonant frequency (SRF) in power inductors used in damped low pass filters? Example circuit below:
How should I compensate for the self-resonant frequency of inductor Z1? Let's say the cutoff frequency of this filter was 100 kHz and the inductor SRF is 5 MHz. Naturally I'd want the filter to attenuate frequencies above 5 MHz substantially, however the inductor may not allow that and may even make matters worse.
As @jp314 said, series combinations can be used.
Note that the smaller inductor will be series-resonant with Cp of the larger, so the common node between them may need damping (R+C across one, or to GND). And so on down the line, if you use an extended chain of them.
A practical example is this Picotronics bias tee: a bias tee is a low-frequency diplexing filter, splitting "DC" and "AC" from a combined node. As the name suggests, extremely wide bandwidth is demanded -- well beyond 10GHz. Resonances in the bias inductors would cause dips in the response (reduced transmission, and increased reflection). So a very well-behaved inductor is needed.
(Photo courtesy Highland Technology)
In this case, they added a shunt resistor to most inductors, and notice the (mostly) geometric series of values: (large), 10uH, 1uH, 0.33uH, (puny-ass coil on ferrite rod). (If we assume a proper geometric series, the ring is probably 30-100uH, and the rod ~100nH.) There's also an R+C to GND just before the last one, and at the DC port (under the ring I think).
Here's the schematic. I suspect the "100uH" are mislabeled ("100J" = 10 x 10^0, J = 5% tolerance), but otherwise you can see what's up.
What's really going on, is:
If you build a stacked model like this -- a chain of nonideal inductors (meaning, having a representative RLC equivalent circuit each) with tapered values and optional damping elements -- what you will find is, it acts as a further multiplexing filter: the total (end to end) impedance remains fairly high, but bounded -- resistive on average -- and if you track power dissipation vs. frequency, you will see that most of the power loss into the network goes into just one of the resistors at a time.
The dispersion points (dip and peak) may vary in magnitude and width (depends on how well damped the RLC network is at that point); they're exaggerated here for sake of explanation. The flat regions between dispersion points, can be at different levels -- corresponding to the respective resistor that is exposed in each frequency band.
Now, this is all well and good for a bias tee, but what about a filter?
Now that I've sort of explained a small corner network theory, let's consider that.
Ideally, we'd want the filter to remain 2nd order (or whatever), up to an arbitrarily high frequency. The effect of parallel resistance and capacitance (in the inductor) is to add a real zero, or zero pair (may be real or complex), to the transfer function -- which is to say, the circuit approximates an R+C (1st order) or C+C (0th order, capacitor divider) at high frequencies.
If we're looking for some minimum stop-band attenuation, we can stop once we have a good enough inductor (or relaxed enough spec!): the response may rebound, but it stays low enough we don't care.
If we want to keep the asymptote going, we must cancel out these zeroes with additional poles -- more series inductors.
But we cannot avoid the relations above. The best we can do, is trade off between inductors, one after another, riding up the slope of each one at a time (the straight regions between dispersion points may still be sloped, not flat) -- the constraint is that no flat region can be above the asymptote of the region below it. So, if the first flat region were sloped, it must be below the blue dash line; and so the next, etc. The average slope will be sub-proportional (i.e., approximating \$Z \sim F^{1-\alpha}\$, say for \$\alpha \in [0.1, 0.6]\$), which is equivalent to saying the average Q of this composite inductor, over this frequency range, is on the order 1/α. (Or something like that, I don't know the exact relation. For α = 0.5, Z ∼ √F and R = X, i.e., Q = 1 -- an ideal diffusion (Warburg) element.)
In any case, the point is, impedance can only go up so fast, and in particular, can only go so high, especially at very high frequencies. It's practicable to have an impedance of 100s of kohms at 1MHz (though you'll need a fairly big, well-made inductor for it), but at 1GHz say, you're basically stuck with chip components (lumped elements), or chunks of transmission line (which are never very far from Zo in practical cases).
So, for general purposes, the better solution is to save your effort trying to make a better inductor -- instead, cascade the filter to more stages. Additional stages don't need to be equal Fo, mind: you can knock off the "peaks" of a previous stage, using subsequent stages. Remember to choose the common port type appropriately, i.e. a 2nd-order LC, L-input, Fo = 10kHz, has a shunt C at its output; this is Z ~ 0 input (input-shorted mode) for the next stage, so choose the appropriate prototype.
Beware, most online filter calculators only know the double-terminated case; but a filter only needs one port terminated, and that will the final output port in your case. (Which, if not provided by the load, can be supplied by an R+C -- as in the OP example, which is also supplied from a zero-impedance ideal voltage source!) The calculator at https://rf-tools.com/lc-filter/ does properly account for one port shorted; it has no way to enter infinite impedance (open) however!
In this way, arbitrarily high attenuations can be achieved, using finite components, even when the impedance ratios they offer are quite narrow. (For example, a transmission line low-pass might simply use dozens of elements, making up for its poor sharpness per stage, with just a ton of stages.)
Put a 2nd lower value (and higher SRF) inductor in series with the 1st.
If the filter is set to deliver a cut-off frequency of 100 kHz and the SRF of the inductor is 5 MHz, the ratio of the two frequencies is 50. This means that the parasitic capacitor associated with the inductor is 50² lower in value than the capacitor that makes the filter have a cut-off at 100 kHz. Therefore, at high frequencies in excess of the SRF, the attenuation due to the parasitic capacitor and filter capacitor is, as near as dammit, \$20\times \log_{10} (2500)\$ = 68 dB.
That's still quite a decent amount of attenuation so, if you really need more, then add a small value inductor in series with the first inductor that has an SRF at (say) 50 MHz. Continue until you have enough attenuation.
However, as you approach frequencies in excess of 100 MHz, you may find that the filter capacitor turns inductive and so, you should put a smaller value (with a naturally higher SRF) in parallel. Continue until you have enough attenuation but, watch out for interactive resonances between the capacitors; it's usually small and unimportant but, this answer is partially generalizing at this point so, I have to give that advice.
There is no “compensation”. You need to add another filter in series that will operate at and above the first filter’s SRF. This can be as simple as having several inductors in series. Inductor parasitic capacitances should be shunted by capacitors that have SRFs in the same ballpark as the inductors they shunt. Shielding between the stages may be needed as well. The inductors will radiate and can magnetically couple. Adjacent stages should have inductors at right angles to each other.
