Do these impedance matching components interfere with each other? - Page 1
Forum section: electronics / beginners

author: tehtehteh
author_rank: Regular Contributor
timestamp: September 09, 2021, 11:21:28 pm
content: In the red box is an impedance matching network that gives 50 ohms at the C9 L2 junction. In the green box are trimmer capacitors for tuning the antenna. There is a 50 ohm trace between the two.The capacitors C9, C12 and C13 are in parallel. So would tuning the antenna also be changing the value of C9 and therefore taking the impedance away from 50 ohms before the transmission line?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 10, 2021, 01:45:03 am
content: A complete answer requires knowing the layout, including antenna geometry, trace lengths, transmission line geometry or impedance, and so on up to whatever the relevant port / reference datum is (I'm assuming, something like an rx/tx module pin or whatever). Then the impedance of every element (transmission lines -- traces -- are elements too) can be calculated and the impedance, bandwidth, and other stuff like radiation resistance, pattern, gain, etc. as well (give or take what scope of simulation you're doing).So, that said, note that any analysis of the schematic assumes the schematic is correct -- representative of the elements in the circuit.Note that C13 is NOT in parallel with C12, any more than C9 and C7 are in parallel, or all of them together. Which is true enough at low frequencies, but that's not where things are interesting. Presumably, the antenna (being drawn as a somewhat open loop) manifests as significant inductance (an odd shaped trace off the ground plane, presumably -- hence also giving better coupling to radiating fields), so serves much like L2, and consequently C13 and the antenna are series resonant. Then C12 would likely serve to match the (presumably) low radiation resistance to something near 50 ohms; in which case the trace between C12 and C9 can be arbitrary length, at least at the center frequency; but who knows what happens at other frequencies (likely this arrangement creates resonances at multiples of the electrical length of that trace). Likewise, C9-L2-C7-C6 serve as some matching network up to the, well it's probably a TX output because of the RFC pullup, isn't it. Could still be internally muxed to an RX I suppose. Also provides some lowpass filtering -- it helps to remove harmonics, that would otherwise be prone to radiation from the electrically-small antenna (if it is, I mean -- which seems likely, else why bother loading it with capacitors?). That is, at higher frequencies (multiples of the driven frequency), the antenna doesn't look so electrically small, so can radiate relatively well in comparison.Also I call it an RFC, but 27nH isn't really that much higher than everything else in the circuit, probably it's part of the matching too.Which, for capacitors around 12pF and inductors around 20nH, implies Zo = sqrt(L/C) = 35Ω -- probably close enough to 50Ω and there's some fudge in there to implement the desired filter response or matching. And, Fo = 1 / (2 pi sqrt(LC)) = 325MHz, again give or take some margin, so, whatever bands are around there. LORA something or other?Tim

author: tehtehteh
author_rank: Regular Contributor
timestamp: September 10, 2021, 10:13:22 am
content: I think your assumptions are mostly correct. The feed is a 125 ohm power amplifier output of a 433.92MHz transmitter IC. I was rigorously following the data sheet and application notes because I know very little about RF. According to these L1 is indeed a choke, C6 is to block DC, and the pi network is to filter out harmonics, and obviously the impedance matching aspect of some or all of these too.If I put these into a smith chart app (assuming I've done it right) the impedance it gives is about 32 ohms, so perhaps these particular example values they gave were specific to the layout of the dev kit board to achieve 50 ohms, or perhaps the accuracy does not matter, I don't have the knowledge to answer this.

author: tehtehteh
author_rank: Regular Contributor
timestamp: September 10, 2021, 10:40:58 am
content: Thinking about it some more, what's the point of me using that smith chart if I'm guessing the IC's impedance as only resistive. I'd need to know the imaginary part too if I wanted it to be accurate right?

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: September 10, 2021, 11:49:24 am
content: Here is a way of looking at it:C12 "tunes" your Pi matching network as it is in parallel with C9The antenna is a magnetic loop consisting of the loop and C13. The loop is an inductor and C13 tunes it into resonance.The Pi matching network including C12, C9 L2 and C7 "passes" the desired frequency and filters out undesirable freq. actually looks like the "plate and load" tuning in old tube radiosThe magnetic loop antenna is tuned to resonance of the desired freq with C13, it is a simple LC circuit, like all mag loops.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 10, 2021, 01:10:05 pm
content: In fact the IC is most likely capacitive, an open drain output (hence why you have to provide bias voltage externally). You're looking at it a bit backwards -- you've calculated what the antenna sees (on the assumption that the transmitter is resistive -- which I doubt, but it could still be the case), but the important figure is what the transmitter sees. Which is looking into the network, to the right from DP1. For which, there's no termination specified (your model is missing the antenna), so you can expect nonsense results for that (as shown, at least).This is all a bit confusing to start, but once you grasp which nodes, and directions, you're measuring for impedance (or transmission or whatever other N-port properties you're after), it makes a lot of sense. RF looks like black magic at first, butallsufficiently advanced science looks like magic; raise your science to match.In particular, this is a big reason why we use ground planes, shields, transmission lines and (usually) very small components: these design practices allow us to make dramatic simplifications. Planes and shields block fields between different parts of the circuit, isolating them and confining signals locally, allowing us to analyze parts of the circuit separately. Small components can be placed very close together, i.e. within small fractions of a wavelength, so a lumped-equivalent approximation applies.Hence, we can reduce a problem -- if not an intractable one, then one that at least might demand full 3D field simulation -- to something simple enough that we can draw it on a napkin. Notice that we may not even need to know the speed of light: a lumped-element equivalent circuit (pure R, L, C) is instantaneous, there's no propagation delay between nodes! Such a circuit can be said to be point-like or zero-dimensional.Or, if we can't remove some propagation delays from the circuit (while still getting an acceptable approximation), then if we can approximate those delays as transmission lines, and transmission lines only have one dimensional property (length), so the circuit is one-dimensional. (The C9-C12 trace could be a case of this.)And when we can separate parts (because they're isolated from each other, except for few, direct connections: ports), we can work with them in isolation. But how? Say you wanted to look at the CLC filter part in isolation, how would that work? You could set up a test between a 50 ohm source and load, but what about its response for other impedances? You can only work with it, as some kind of transformation -- the impedance on one side, is dependent on the impedance applied to the other.The Smith chart -- with the use of circular and cartesian geometry on a cleverly chosen grid -- expresses that transformation, as a geometrical path, starting from one arbitrary point, transforming it to another. It's a relative path, so you can slide it around and see what happens for any arbitrary "input" impedance; and it's also reciprocal (it works the same in reverse). Which, actually we can apply reciprocity right now: since you've measured the right end of the network shown, you can terminate that end with the conjugate impedance (same R, opposite X), and you'll get exactly 125 ohms looking in to the network from DP1.(Uh, I think that's right about geometry, anyway? It's honestly been a long time since I used Smith charts, I normally just crank the numbers in a sim.)So, although you've perhaps approached this a bit backwards as I said at the top -- it's not necessarily wrong or unusable.Knowing how to use it, is obviously the big challenge.Another way to look at the impedance transformation is, since the impedance at one port depends on the other's, and vice versa: we have a pair of numbers, and transform them to a pair. We can stack them up in vectors, respectively; and how do you transform vectors, with a matrix of course. And this generalizes for any number of ports, not just 2, so we get an N-port coupling matrix in the general case. So if we wanted to solve for the CLC filter by itself (a 2-port network), this is how we would do it -- fill in a 2x2 matrix. And to use it, just multiply it by a vector representing the impedances thus attached, and out rolls the impedances on the other sides; or the gain and phase of the network (transmission matrix), or equivalently, any other combination of properties we might wish to express.And, a port is simply a point in the circuit where signals flow both ways, the voltage and current of which is known (say because you can cut and divert it to an impedance analyzer). You can arbitrarily declare a port anywhere on a transmission line, in which case the two ports (left and right sides) are simply joined to each other, of course.More generally, for modeling purposes: a port is a 2-terminal component with equal and opposite currents (i.e., no common mode, it's just floating like any other 2-terminal component), some voltage drop, and some dependent relationship to 0 or more other ports. (You can of course have a 1-port: an inductor has self impedance, etc. Nice and trivial!) An ideal 2-port "pass-thru" network is a, well, not quite a transmission line, because a transmission line has common mode current flow as well -- more like an ideal 1:1 transformer. Although, we can approximate those by winding a transmission line on a magnetic core!In practice, we might simplify our use of ports, by simply grounding one side of them all. Like a bunch of BNC connectors on a metal box, or the various points in a circuit like this one, laid out over ground plane.Tim

author: tehtehteh
author_rank: Regular Contributor
timestamp: September 10, 2021, 02:05:08 pm
content: Yeah I noticed the page says "Below is your system, note impedance is looking towards the BLACK BOX" but to somebody that's new to smith charts that sentence means nothing to me.I assumed if I know the IC is 125 ohms and I want to that to become 50 that I would start with what I know and work towards 50.

author: tehtehteh
author_rank: Regular Contributor
timestamp: September 11, 2021, 07:47:43 am
content: I switched it all around but it still doesn't get anywhere near 125 ohms.