Context: I am a physics grad student with basic knowledge of circuits, but impedance matching and RF domain are beyond my normal scope. My PCB is actually copper bonded to a ceramic substrate that is going in my vacuum chamber, so knowing the details of what this circuit is actually doing is important for calculating heat dissipation requirements as well as because I won't be able to modify it after I install it in the chamber. There is no resistor element, all resistance is from the wire traces themselves.

So my goal is to have ~1 Amp current peak going through my PCB traces at ~5 MHz in order to have the right magnetic fields for the set of boards to act as a trap. The only components I have are a tuning capacitor (high voltage compatible) and then whatever I choose to use for my impedance matching. By using different tuning capacitors to make the circuit resonate at different frequencies, I believe my parasitic capacitance is approximately 4 pF, my inductance is about 42 uH, and to get the circuit to resonate at 5 MHz I use a tuning capacitor of 20pF.

My "transmission" port is a wound pick up coil that is on the board near the larger drawn coils. Using a network analyzer, I have a couple of different scenarios which is where my confusion lies. If I attach the circuit to a 1:1 toroidal transformer, my power coupling is quite bad, my dip in reflection is only -4 dB. however, the width of the resonance is .1 MHz at 5 MHz, so my Q is 50. Using \$Q = \omega L/R \$ that means my resistance is 28 ohms. Even if I change my turns ratio though, I still cannot get better power transfer than an ~4 dB dip in reflection. Why can't I get better power coupling than this? Is the inductance of the toroidal transformer the issue? However, if I use an L matching circuit, (capacitor across input port, inductor in series with the load, (values work out to 470 nH and 560 pF) I can get -30 dB in reflection, but my Q drops down to 25. What is real here? I would think that all the power coupled would have to be dissipated in my coil, but why is the apparent resistance doubled by using the L match impedance matching? Or is the drop in Q just a function of the source now seeing an apparent 50 ohms, but not really driving that?

Any help or references would be much appreciated, I've been fighting with this for a while and all my labmates are also physicists, not EEs, so we don't have much experience with this. Thanks!

Edit: here is the schematic for scenario one: toroidal transformer and 2: L-match circuit circuit drawing

Another edit: the actual coils in question: coils on boards

  • 2
    \$\begingroup\$ Consider adding some paragraph breaks to make your question easier to read. A schematic of your circuit might also make it easier for us to follow what you are doing. \$\endgroup\$
    – The Photon
    Aug 27, 2014 at 16:26
  • \$\begingroup\$ Thanks, I'm new here. I cleaned up the post and added diagrams. \$\endgroup\$ Aug 27, 2014 at 16:43

3 Answers 3


Concerning the fact that Q dropped when you used an impedance matching network:

You used a simple L matching network consisting of just two components (a capacitor and an inductance) which both are completely determined by the input and output impedance of your situation.
This leaves you no parameter to control Q. You get whatever Q is a result of the component values you need for matching.

So I suppose what you need is a more sophisticated matching network containing at least three components (e.g. a T- or a Pi-network) that has another degree of freedom which allows you to control not only input and output impedance but also Q.

Since you are also asking for references I recommend very much the chapter about impedance matching and Smith charts in Chris Bowick, "RF Circuit Design". It explains and contains an example problem for impedance matching while also caring about Q.

  • \$\begingroup\$ Does the Q dropping mean that power is necessarily being dropped in the matching circuit? Or is the power still all being dissipated in the main circuit and just appearing different due to the matching. Strictly speaking, I don't care about Q, I only need to know where the power is going to. Thanks for the reference, I'll try and get my hands on it \$\endgroup\$ Aug 27, 2014 at 23:53
  • \$\begingroup\$ If you assume perfect inductance and capacitance (no R) no additional power is dissipated in the matching network. Of course real components will also have some resistance but if you choose the right ones it shouldn't be a problem. \$\endgroup\$
    – Curd
    Aug 28, 2014 at 6:29
  • \$\begingroup\$ I thought having high Q is important for you. \$\endgroup\$
    – Curd
    Aug 28, 2014 at 6:32
  • \$\begingroup\$ I had been approaching the Q measurement as being indicative of the losses in the circuit, I know that for a simple LC tank lower values are due to more losses. I want to keep losses low if possible because heat dissipation in vacuum is tricky, in addition to needing to acquire a larger amplifier if the matching circuit also had loss. From the reference you gave though, it appears that no matter how I do the matching with other elements I will get a lower Q, even if no more resistance is added. Does this mean that Q in general no longer directly tells you about loss in a more complex circuit? \$\endgroup\$ Aug 28, 2014 at 18:41
  • \$\begingroup\$ I am not totally sure but I think here using Q as a measure of how much power is wasted doesn't make sense because Qs of networks with different impedances are compared (Q=X/R; it can be influenced by R but also by X) \$\endgroup\$
    – Curd
    Aug 28, 2014 at 20:36

Oh no, I'm a physicist too. (I hope I can still help.) So I'm going to guess that both your numbers are correct.
The Q of a resonant circuit can certainly change as you load it down with more power. (Is there something in those coils?) There will also be some resistive loss in the matching network.
When imputing power at RF you need to know something about the source impedance. (What's driving it?) And then maybe something about the connection (transmission line) between the source and the load. 5 MHz, is fairly low freq, RF-wise, how long is the connection?

  • \$\begingroup\$ The RF amplifier driving the circuit has a 50 ohm source impedance. The transmission line is currently a ~4 foot BNC and then a twisted wire pair of about 4 inches, though removing the BNC (plugging into the network analyzer bare) does not change the Q of the circuit appreciably \$\endgroup\$ Aug 27, 2014 at 19:07
  • \$\begingroup\$ @AlejandraCollopy, OK the 50 ohm source impedance is going to make 1 amp hard. (I think.) And driving it with 50 ohms means the cable length doesn't matter. Does the 28 ohms of load come close to the DC resistance of your coil? (well, or is there a lot of copper conductor nearby?) \$\endgroup\$ Aug 27, 2014 at 19:21
  • \$\begingroup\$ It's a 25 Watt RF amplifier, so I think it can supply the necessary voltage for an amp at ~25 ohms. The DC resistance is only ~2 ohms, I am currently working on making another set of boards with a material that has thicker copper cladding to try to drive the resistance down. However, I am still puzzled about the matching of impedances and what that means with regards to power transfer to the circuit \$\endgroup\$ Aug 27, 2014 at 19:24
  • \$\begingroup\$ Do RF amplifiers work as current or voltage sources? I had been thinking that they are voltage sources. Also, the above measurements were all done at low power, with the network analyzer itself driving the circuit. \$\endgroup\$ Aug 27, 2014 at 19:36
  • \$\begingroup\$ Ahh, the picture helps. So if the DC resistance is 2 ohms where does the 28 ohms of loss come from? Is that a big pcb copper layer underneath the coils? Re: RF amplifiers, I guess it's assumed that they are voltage sources. (almost certainly if it's a 50 ohm source impedance.) (I'm not really sure where your loss is, I have some guesses. But consider this, if it's going in vacuum and dissipating heat, where is that heat going to go? What kind of caps? \$\endgroup\$ Aug 27, 2014 at 20:49

At resonance the capacitance and inductance of a series tuned circuit cancel out, so it appears to be a pure resistor (mostly the resistance of the coil). To get maximum power transfer you must match the source and load impedances. With the generator at 50 Ohms and the load at 28 Ohms (or less) they are not even close to matched, so a large amount of the power will be reflected back into the generator.

The Q of a series resonant circuit is equal to the inductive reactance (or capacitive reactance - they are the same) at resonance, divided by the total resistance in the series circuit. This includes the resistance of the source. If the source and load are properly matched and tuned to resonance then they will have equal resistances, so if the tuned circuit coil and capacitor add up to 28 Ohms then the total resistance in the series circuit should be 56 Ohms, and the Q is (42uH*5Mhz*2pi)/56 Ohms = 24. This low Q is not a concern unless you want to filter out off-resonant frequencies.

But if the DC resistance of the coil is only 2 Ohms, why is the calculated resistance over 10 times higher? At 5MHz, skin effect causes most of the current to flow just under the surface of the coil traces, which dramatically increases their effective resistance. Thicker cladding won't help much. To get lower AC resistance you need to increase the surface area.

You say that changing the turns on the toroidal transformer made no difference, but with the correct turns ratio it should match the generator to the load. If power coupling is still low then it's because the transformer is not designed for efficient operation at 5MHz and 25W (not enough turns, wire too thin, high core loss?).

  • \$\begingroup\$ My cladding is currently 35 um, which is almost exactly the expected skin depth of copper at 5 MHz, which makes me think that increasing the thickness will help somewhat. Additionally, The substrate may have some dielectric coefficient, removing material from the space between the traces improved the Q quite a bit. The core I am wrapping the transformer around is designed for use from 2-10 MHz, but the wire thickness and number of turns may still be my issue. I will try changing both of those. Thanks! \$\endgroup\$ Aug 28, 2014 at 18:47

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