# What does the impedance look like around a small resonant loop?

Say I have a hoop of conductor, broken at the top, with a capacitor across the gap. The inductance of the hoop and the capacitance form a series resonant LC circuit: Now, say this loop is excited by a distant transmitter. What will the currents and the voltages look like in this loop? At what point are current and voltage in phase? How does the phase angle change as I move around the loop? Is there a point at which $|Z| = 50\Omega$ ? Can I predict where it will be? Can I predict what the phase angle will be at this point?

(The objective, you might guess, is to match an electrically small, resonant loop to a transmission line. I've read many ways to do this, and I can get it to work through trial and error, but I desire a better understanding of why it works, and what's happening inside the loop.)

• If circumference is << wavelength, basically the current is the same everywhere throughout the thing, and so is the voltage across any equal-length section. – Kaz Jun 3 '13 at 2:45
• @Kaz if that's true, then why do antennas of this type seem to universally place the feedpoint opposite the capacitor? It can't be the same all around, or it wouldn't matter. Maybe I'm asking the wrong question? – Phil Frost Jun 3 '13 at 15:07
• I agree with @Kaz and your reply about placing the feed-point opposite the cap might need some clarification. The way I see it, if the feed is inserted anywhere in the loop (given that it is a small loop) the 50ohm impedance of feed derails your question a bit - the loop isn't very resonant any more and the dominant impedance will be the 50 ohms. Forgive me if I'm missing the point. – Andy aka Jun 25 '13 at 10:21
• @Andyaka What's a "dominant impedance"? – Phil Frost Jun 25 '13 at 11:38
• The series tuned circuit has 50 ohms inserted. If it were resonant at 1MHz an inductance of 1uH and capacitance of 25nF would have impedances of only 6.3 ohms thus the 50ohm is dominant. A single turn coil is what you are talking about. If it were multi-turn (say 3 turns) then the 50ohm would be less dominant and the Q would still be decent. Single turn coils and caps up to any frequency will be the same providing loop is small compared to wavelength I reckon. I am also assuming you mean small coils about the size of a coin. – Andy aka Jun 25 '13 at 11:49

## 3 Answers

This kind of stuff is notoriously difficult to predict. Analisys will only get you so far. Then you have to experiment and tweak.

Basically you have a parallel L-C resonant circuit. If this is a small loop relative to the wavelength, then you have mostly a lumped system. That means that the current thru the whole loop is pretty much the same at any one time. The impedance between any two feed points goes up linearly with the angle between them. Think of the loop as a auto-transformer. Don't let the fact that it only has one turn get in the way of that. Tapping off some section of the loop is like two taps in the auto-transformer winding. Given that the whole winding has some fixed impedance at the desired frequency, the impedance between the tap points goes down proportional with the fraction of the winding being tapped.

I used a loop like that in a product to receive 434 MHz. The loop was about 1 inch in diameter, and the tap points to get 50 Ω were about 1/4 turn apart. I don't remember the cap value, but somewhere around a few pF.

In the prototype, I put two pads for the resonating cap in series. That allowed easier experimental tuning of the loop to the desired resonant frequency. I also put a whole bunch of tap points in the loop, then we tried different combinations experimentally to pick the best impedance match to the circuit.

## Added:

I dug the board I mentioned above. Here is the layout of the resonant loop antenna: C24 and C25 in series are the resonating cap. Being in series made it easier to tweak the overall capacitance a small amount. The total ended up being a few pF. R19 is only there because someone wanted to experiment with damping. It was never used. The two thru-hole pads to the left of the loop by C23 and C27 are the RF feed points into the circuit. Wires were installed between them and the right holes in the loop, which were determined by experimentation. Now that I see this again, I think the feed points were more like 45-55° apart. The diameter of the loop was .9 inches, and the RF frequency was 434 MHz.

• The loop you have is about one-eighth the wavelength but I'm not sure if Phil is referring to something whose size is electrically significant to the wavelength. He mentions "electrically small" and this becomes a trivial exercise so maybe your example is useful? – Andy aka Jun 25 '13 at 13:18
• @Andyaka "electrically small" is usually considered as $< \lambda/10$, but $\lambda/8$ is pretty close, and I think it still applies. – Phil Frost Jun 25 '13 at 14:11
• @Andy: As Phil said, 1/8 wavelength is still mostly lumped. In any case, the final details were found experimentally. Analisys to the detail level is difficult exactly because of these boundary cases and other difficult to know or quantize effects. – Olin Lathrop Jun 25 '13 at 14:14
• It's going to be a non-exact answer then – Andy aka Jun 25 '13 at 14:20
• Did you take a look at analyzing this sort of thing using a 3D RF modeling tool (Such as CST, Genesys, etc...)? I suspect you could get fairly close to real-world tests using a electromagnetic FEA-based modeling tool. Of course, it may take more time to model then actually do the testing, but it's interesting to think about. – Connor Wolf Jun 27 '13 at 16:35

The loop and capacitance for a parallel (LC tank) circuit not a series circuit. A 'small' loop couples mainly to the magnetic field component of the electromagnetic wave which is the opposite to the Hertzian dipole (couple to the electric field).

A 'large' loop - self resonant loop (diameter >= wavelength) can be treated as a folded dipole.

For high frequencies the loop may be physically small but 'large' in terms of wavelength.

At resonance the impedance (A-B) is undefined (depending on the Q value) but it will be a large number. The currents in the capacitor and inductor are antiphase.

There are two ways to couple into the loop. (think of it as a transformer)

(1) construct a smaller (feed) loop which acts as a secondary coil (2) Tap into the 'inductor' to form an 'autotransformer'.

For some midnight reading giving chapter and several verses on small loop aerials try http://www.aa5tb.com/loop.html#info

• Tapping into it will provide a more stable matching point. It is also easier to predict the values using ADS or some other EM solver because 1/8 wavelength can't really be treated as lumped. – user6972 Jun 27 '13 at 17:07

It seems you really want an impedance match to a LC network and use it to inductively couple to a signal. Assuming your circumference total (c) is small $<0.1\lambda$ so that it's not an EM structure.

You can put the feed line and tap points wherever you want since you aren't worried about making an EM resonate structure. I would just match the impedance rather than try to pick weird tap points to make it work. You can pick different tap points, but those traces are really hard to predict what their differential inductance will be and it changes as the feed point changes.

Compute the inductance expected from your trace/pcb structure and pick a cap that will roughly resonate it at your desired frequency.

Measure S11 to check resonance center and adjust if needed.

You can then attach a network analyzer and measure S11 as your reference point. From there you can just use a smith chart and move the impedance to 50ohms with a matching network.

Once you've determined your matching network values, remeasure the S11 to see if there is any unexpected matching shifts (which usually happens). Then tweak the match.

You will most likely need a step-up match topology, so just put in a standard 3 component step up match topology in your layout.

EDIT (added background): If wavelength of operation is < 0.1 for the total loop length and the length-to-diameter ratio for the solenoidal coil is greater than about 3 (Lc/2b ≥ 3.0) where Lc is the conductor length and b is the average radius, then you can treat this as a radiating inductor rather than an EM structure. This lets you assume uniform current and you can tap in where ever you like for your feed points.

The field pattern is going to look just like that for a magnetic monopole (as Jim explained it's magnetically coupled). The lumped capacitor that you see often used is to compensate for the distributed capacitance of the sides of the traces. To add to Jim's model I suggest it might be better like this: Were C is your distributed cap from the traces, Le is your external inductance based on the geometry of the coupling and Li is the internal inductance of the trace. Likewise Rr is the radiation resistance and Ri is the internal or ohmic loss. These parameters are more important with more than one loop.

Li is calculated just based on the inductance of a straight piece of transmission line the same length as your loop. If you are using multiple lines, then you need to include the mutual coupling.

For a single loop Le can be based on a coil inductance formula:

$L_e=\mu_o*b*[ln(\frac{8*b}{a})-2]$ (where $\mu_o$ is the permittivity of free space, b is the radius, a is the radius of the trace). See online calculator

Rr is computed using

$R_r=\frac{\sqrt{\mu_o*\epsilon_o}}{6*\pi}*\beta^4*(N*A)^2$

Where N is the loops=1, A is the area, $\beta$ is $\frac{2*\pi}{\lambda}$

Commonly this is matched with one of the following techniques: I would choose the first one on the left because it is more predictable then varying S2 and L2 and it has a little wider bandwidth. I would also add two more elements as placeholders in your layout in an L shape (series, shunt) for additional matching options if needed.

If you gave us your operating frequency/bandwidth/size constraints we might be able to help some more. A simple ferrite loaded antenna might make your receiver work much better.

Other References: Calculating inductance (you can set Y=0 for your case)