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I would like to send ~100W through an RF filter. Can you estimate the current passing through the shunt component without a circuit simulation assuming a reasonably high S21 value?

MLCC's are cheap, even at higher voltages. Inductors are expensive and high-power inductors even more so.

Questions:

  • In the case of a high-pass (pi or tee) filter where the shunt components are inductors, just how much current actually flows through the shunt inductors?

  • I would think the power moving through the inductor is low because the loss is low (nearly lossless S21), but is that a correct assumption?

  • If very little current (<100mA RMS) actually flows through the inductors, then can less expensive and small (like 0402) low-current high-Q RF inductors can be used?

There are 2 votes-to-close this question :( ... but I don't understand why this should be closed because I'm seeking to understand the principle here and not just shove it through a simulator and get the answer, so please read on:

The principle seems to make sense that almost no current passes through a shunt if nearly all the signal goes through the filter. I'm trying to understand why shunt current could be present in this case. Simulations are great, but they don't provide the understanding---just the result. This question is about understanding if (and why) a nontrivial amount of current would pass through a shunt component in the presence of a nearly lossless S21.

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    \$\begingroup\$ Why not just do a simulation to work this out? Getting the answer by calculation would just be doing essentially the same math. \$\endgroup\$
    – The Photon
    Jun 23 at 3:03
  • \$\begingroup\$ If you can define desired input and output transfer function , spectrum, impedance , power output = 100W and choose a loss acceptance criteria that has a cost factor so define a budget for components and heatsink, harmonic suppression in dB etc. then an answer is possible. consider the amplifier losses too as different classes have different efficiencies. All filters have specs for s21, s11 for pass and stop bands. \$\endgroup\$ Jun 23 at 4:05
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    \$\begingroup\$ There is no way to tell this without knowing the exact component properties and routing layout. A simulation with careful attention to parasitic properties of components will ne a useful estimate. \$\endgroup\$
    – tobalt
    Jun 23 at 4:38
  • \$\begingroup\$ @ThePhoton, ok I can simulate it---but if S21 is nearly lossless then does that mean there is almost no current passing through the shunt component, or does it just not work that way? \$\endgroup\$
    – KJ7LNW
    Jun 23 at 22:03
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    \$\begingroup\$ It might help to give us some concrete numbers. What is the operating frequency, and what is your filter's cut-off frequency? What are your L and C values? \$\endgroup\$
    – The Photon
    Jun 23 at 23:31

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You are lacking several terms, and so you aren't asking quite the right question, or providing sufficient information to base an answer on.

There are a few things that can be measured, which can be described that way.

The most likely are the apparent or real power in a given component.

Apparent power is given by, roughly speaking, incident power times system Q of the given component.

Real power is then apparent power divided by Q of the given component.

There are different definitions of Q, because the reactance of the component can be compared with different impedances in the circuit. Namely, "component Q" is the ratio of reactance to resistance of the component itself (for series equivalent, or vice versa for parallel equivalent), while "system Q" is the ratio to the system the filter is embedded in, or the impedance at a local point (when impedance matching is involved), etc. So, for the most common, basic ladder topology filter, that would be relative to 50 ohms or whatever.

Consider the most basic high-pass filter: a capacitor in series between a 50 ohm source and a 50 ohm load. At low frequencies, little current flows, so also little apparent power. At high frequencies, little voltage is dropped, and the source and load voltages are equal to half the open-circuit (Thevenin equivalent) voltage (i.e., s11 = 0). However, exactly at cutoff, Xc = 100R, maximum apparent power transfer (so to speak) occurs; the total series impedance is 100 + j100 ohms, so current is sqrt(2) times lower than in the passband, i.e. -3dB in the load, and apparent power in the capacitor is voltage drop times current flow, or sqrt(2)/2 * sqrt(2)/2 or half the passband power. Again, this is apparent power, in this case the capacitor is ideal so no power is dissipated (all reactive) and the apparent power is recycled through the source and load every cycle. This is an RC network so has a maximum Q of 1/2 (no coincidence!), but you can get higher amounts for LC networks of course.

The system Q of components in a typical filter depends on the transfer function: a sharper filter has higher Q. So, a Chebyshev takes higher Q components, or has higher losses, all things being equal (component type, cutoff, order, etc.)

Finally, the ratio is only true at resonance, or cutoff (transition band), or whatever puts the most apparent power into the network. In general, the amounts will be lower at other frequencies, but as others have said -- the easiest way is to simply simulate it.

Remember to put in realistic parasitics/strays, and use good component models where you can find them. Good luck!

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All the power that goes in will end up going somewhere; either it is reflected back from the input, passed through to the output, or lost as heat.

You have described S21 as nearly lossless. You can look at the squared magnitudes of S11 and S21, the return loss and insertion loss, and estimate a how much power would be lost as heat which then guides component selection.

Keep in mind that choosing lower Q inductors will change your frequency response and increase the amount of loss that ends up as heat.

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  • \$\begingroup\$ Can you elaborate on squared magnitudes of S11 / S21? How does this convert to power dissipation in watts? I did a quick google search didn't find anything obvious there. Also I usually look at S11 / S21 in dB, so do you convert it from dB (to not dB, ie 20*10^x) before squaring? \$\endgroup\$
    – KJ7LNW
    Jun 23 at 22:18
  • \$\begingroup\$ @KJ7LNW 1dB of insertion loss at say 100W input, is 80W output. Where did the 20W go? Absorbed in the filter of course. Assuming s11 is very low of course; if not, subtract that power from incident power first. (Subtraction isn't very straightforward in dB domain, convert to ratios first.) \$\endgroup\$ Jun 24 at 1:20
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If s21 shows expect gain or loss that means impedance was matched but does imply anything about your topology or bandstop currents or impedance matching design.

You have underspecified all the spectral scattering specs.

High power air coils are not expensive.

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