Match a 2.4 GHz Chip Antenna to 50 Ω

At $2.45\,\mathrm{GHz}$ the demo chip antenna is $10.64 + j5.56\,\Omega$. Into $50\,\Omega$ that is $|\Gamma| = 0.653$, return loss $3.70\,\mathrm{dB}$, VSWR $4.76$, and $2.41\,\mathrm{dB}$ of mismatch loss. The ideal low-pass L-section is a shunt $2.499\,\mathrm{pF}$ at the source and a series $968.4\,\mathrm{pH}$ at the load. Only topology A exists, because the resistance is below $50\,\Omega$ and the conductance is not low enough for topology B.

That impedance is only meaningful at a stated reference plane. A NanoVNA .s1p is the plane where you connected the calibration standards. Fifty picoseconds of line still in the fixture, $15.0\,\mathrm{mm}$ at velocity $c$, moves this same antenna to $24.6 + j54.8\,\Omega$. The low-pass parts for that shifted impedance are a series $4.521\,\mathrm{nH}$ and a shunt $1.602\,\mathrm{pF}$. Recalculate after you move the plane. Do not tweak the first network and hope.

The demo is synthetic. It is not a vendor antenna and not a NanoVNA capture.

What impedance are you matching?

The model is a $10\,\Omega$ resistor, $3.2\,\mathrm{nH}$, and $1.5\,\mathrm{pF}$ in series, with $0.35\,\mathrm{pF}$ across the combination. At $2.45\,\mathrm{GHz}$ the series arm is still inductive, the pad capacitor has started to matter, and the resistance you see is $10.64\,\Omega$ rather than the $10\,\Omega$ in the series arm. A short chip antenna often looks like this: a small real part and a reactance that swings through resonance. The small real part is matchable. A purely reactive feed is not.

Unmatched, the $2.41\,\mathrm{dB}$ mismatch loss is power that never enters the antenna. A lossless conjugate match gives it back, at one frequency. The $\ge 10\,\mathrm{dB}$ span of this ideal low-pass section on the antenna model is $2.311$–$2.584\,\mathrm{GHz}$, about $273\,\mathrm{MHz}$. Nodal Q is $1.92$. Why that band is much narrower than $f_0/Q_n$ is the load moving, which is worked in why an L-match is narrow.

Where is the reference plane in an S1P file?

Save a one-port Touchstone file on the NanoVNA and drop it on the Smith chart calculator. The tool reads $\Gamma$ from the file and converts it with the file’s reference impedance, interpolated in $\Gamma$, not in $Z$. It will not invent samples past the ends of the file. If the phase steps more than $20^\circ$ between points, the band edges are not trustworthy.

The number you get is the impedance at the plane in the file. A typical calibration is at the end of the VNA cable, on the connector you screwed the standards onto. Everything past that connector is inside the answer: a pigtail, a fixture, a length of microstrip, the pad you have not soldered yet. Port extension and electrical delay are the VNA’s way of sliding that plane. This calculator does not apply them. If you want the feed, extend the port on the instrument until the plane is the feed, then export the file. If you forget, the chart will happily match the connector.

The arc is $50\,\mathrm{ps}$ of lossless $50\,\Omega$ line in front of the feed. Clockwise, toward the generator, as a Smith chart rotates. The new point, $24.6 + j54.8\,\Omega$, is a different design. Its low-pass solution is topology B, series $4.521\,\mathrm{nH}$ then shunt $1.602\,\mathrm{pF}$, not a small edit of the $968.4\,\mathrm{pH}$ coil. Fifty picoseconds is not a large fixture. It is enough.

Two checks before you trust the file. The real part should be positive and in a range a chip antenna can have. A real part of a few tenths of an ohm, or a huge one, is usually the plane or a bad cal, not a new law of antennas. And the resonance you see should sit near the frequency printed on the antenna only after the fixture is removed. Matching the fixture’s resonance is a common way to get a beautiful $S_{11}$ on the bench and a poor one on the board.

Plotting the file without designing a network is the S-parameter viewer. The match tool is the one that proposes parts.

Which network, and which real values?

Topology A is the only one available at the feed, so the choice is the sign. Low-pass is shunt C and series L. It passes DC. High-pass is shunt L and series C. The series capacitor blocks DC, and the shunt inductor is a DC short if it lands across a bias pin. For a plain antenna feed with no DC on the line, either sign works. The low-pass row is the one plotted above. On this model its $\ge 10\,\mathrm{dB}$ band is slightly wider than the high-pass band ($2.311$–$2.584\,\mathrm{GHz}$ versus $2.331$–$2.590\,\mathrm{GHz}$). Neither will suppress a harmonic the way a low-pass does on a fixed $R$–$L$ load. The antenna has already left the matched point by $3\,\mathrm{GHz}$.

Ideal parts are not a reel. The RF grid used by the calculator is $0.1\,\mathrm{nH}$ and $0.1\,\mathrm{pF}$ through $9.9$, then E24. The combined search may leave the nearest step when a neighbor wins on the sweep. On this antenna the low-pass pair snaps to $1\,\mathrm{nH}$ and $2.3\,\mathrm{pF}$, and the return loss at $2.45\,\mathrm{GHz}$ falls from a numerical match to $23.5\,\mathrm{dB}$. That is still a match. The worked table on the Smith chart page is the same engine, including the snapped row.

Put the shunt capacitor on the source side of the series inductor. Swapping them is the other topology, and this load does not have that topology.

What inductor Q costs here

Q is applied after the topology is chosen. Series resistance of an inductor is $\omega L/Q$. Capacitor ESR is $1/(\omega C Q)$. With the ideal $2.499\,\mathrm{pF}$ and $968.4\,\mathrm{pH}$ held still, $Q_L = 15$ and $Q_C = 300$ give $-0.42\,\mathrm{dB}$ of transducer gain at $2.45\,\mathrm{GHz}$. $Q_L = 50$ and the same capacitor give $-0.15\,\mathrm{dB}$. The difference is $0.28\,\mathrm{dB}$.

Those Q values are the starting points typed into the form, not a datasheet for this $0.97\,\mathrm{nH}$ coil. Snapping to $1\,\mathrm{nH}$ and $2.3\,\mathrm{pF}$ and then setting $Q_L = 30$, $Q_C = 300$ — the row the calculator shows — gives $-0.24\,\mathrm{dB}$ and $27.0\,\mathrm{dB}$ return loss. That mixes the snap with the loss. The comparison that isolates Q is the table in why an L-match is narrow. A $100\,\mathrm{MHz}$ minimum Q on a multilayer coil is not the Q at $2.45\,\mathrm{GHz}$.

What this example is not

No vendor antenna, no measured .s1p, no pad, no via, no ground inductance, no de-embedding. The $50\,\mathrm{ps}$ line is lossless and exactly $50\,\Omega$. A real pigtail is none of those. The reference plane is whatever plane you export. If the file is wrong, the parts are a precise match to the wrong impedance.

Share links for these inputs are free. A BOM or a network Touchstone file asks you to sign in. The measured samples, if you add them, stay in the URL hash.

References

• D. M. Pozar, Microwave Engineering, the chapter on impedance matching and tuning.
• P. H. Smith, “Transmission Line Calculator,” Electronics, vol. 12, January 1939. The rotation with line length is the original use of the chart.
• The synthetic element values are the demo in IndieRF Match, documented with the calculator, not a manufacturer’s measured antenna.

Related

• How to read a Smith chart
• Why an L-match is narrow
• Smith chart calculator

FAQ

What impedance does a NanoVNA S1P file contain?

The reflection coefficient at the calibration plane, in the file’s reference impedance. IndieRF Match does not remove fixture delay or port extension. A 50 ps electrical delay on the synthetic 2.45 GHz antenna moves 10.64 + j5.56 Ω to 24.6 + j54.8 Ω, and the low-pass parts change with it.

How do you match this 2.4 GHz chip antenna to 50 ohms?

At 2.45 GHz the synthetic demo antenna is 10.64 + j5.56 Ω. The ideal low-pass L-section is a 2.499 pF shunt at the source and a 968.4 pH series inductor at the load. On the RF value grid those snap to 2.3 pF and 1 nH. The demo is not a vendor measurement.

Why is the real part only about 10 ohms?

The model is a 10 Ω resistor in series with 3.2 nH and 1.5 pF, with 0.35 pF across that combination. A small real part is what a short chip antenna often looks like at the feed. It is still matchable. A purely reactive load is not, because a lossless network cannot create the missing resistance.

Does a higher component Q fix a narrow antenna match?

No. Component Q sets insertion loss. On this low-pass section, inductor Q of 15 instead of 50, with capacitor Q held at 300, costs 0.28 dB of transducer gain at 2.45 GHz. The 273 MHz width of the 10 dB band comes from the antenna impedance moving with frequency.