Why an L-Match Is Narrow, and What Q Costs

The ideal low-pass match for the synthetic $2.45\,\mathrm{GHz}$ chip antenna is $\ge 10\,\mathrm{dB}$ return loss from $2.311$ to $2.584\,\mathrm{GHz}$. That is $273\,\mathrm{MHz}$, an $11.2\%$ fractional bandwidth. Nodal Q of that same network is $1.92$, and $f_0/Q_n$ is $1.27\,\mathrm{GHz}$. Holding the $2.45\,\mathrm{GHz}$ impedance constant, so the load cannot move, widens the $\ge 10\,\mathrm{dB}$ span to $1.833$–$2.958\,\mathrm{GHz}$, which is $1.12\,\mathrm{GHz}$. The antenna moved. The L-section did not suddenly become high-Q.

Component Q is the other disappointment, and it is loss rather than bandwidth. On this antenna, typing inductor Q of $15$ instead of $50$, with capacitor Q held at $300$, costs $0.28\,\mathrm{dB}$ of transducer gain at $2.45\,\mathrm{GHz}$. The same inductor-Q step on a real $5\,\Omega$ load matched to $50\,\Omega$, where nodal Q is $3$, costs $0.57\,\mathrm{dB}$. A low-Q coil hurts more when the network is working harder.

Why is the matched band only 273 MHz?

The demo antenna is a series $10\,\Omega$, $3.2\,\mathrm{nH}$, and $1.5\,\mathrm{pF}$, with $0.35\,\mathrm{pF}$ across them. At $2.45\,\mathrm{GHz}$ that is $10.64 + j5.56\,\Omega$. A few hundred megahertz either way, the series $L$ and $C$ have changed the reactance by more than the L-section was asked to cancel. The match was a cancellation at one frequency. It does not travel with the antenna.

• Unmatched
• Antenna model
• Z held at 2.45 GHz

Nodal Q is $1.92$ on every row. The parts are the same ideal low-pass section, shunt $2.499\,\mathrm{pF}$ and series $968.4\,\mathrm{pH}$. The only change is what the load does after you leave $2.45\,\mathrm{GHz}$.

The constant-impedance row is the thought experiment the $f_0/Q$ estimate is aiming at. The $\ge 10\,\mathrm{dB}$ width, $1.12\,\mathrm{GHz}$, is the same order as $f_0/Q_n = 1.27\,\mathrm{GHz}$ and is not the same definition. $f_0/Q_n$ is not a $10\,\mathrm{dB}$ bandwidth. It is a scale. Using it as the width you will measure on an antenna over-predicts this one by about four times.

The middle row continues the single point as a series $R$–$L$ or $R$–$C$. It is what the calculator does when you type an impedance and leave the sweep model on its default. It is wider than the antenna and narrower than a frozen $Z$, and it is still not a measurement. A typed $10.6 + j5.6\,\Omega$ will look more forgiving on the sweep plot than the antenna that produced that number at one marker.

The teal peak on the plot is the honest width for this model. The amber curve is the same parts, lied to about the load. If your measured match is narrower than the calculator’s constant-$Z$ sweep, the usual reason is the one in the table: the load left.

A real-to-real step has a floor. Matching $5\,\Omega$ to $50\,\Omega$ with two lossless elements cannot have nodal Q below $\sqrt{50/5 - 1} = 3$. At $2.4\,\mathrm{GHz}$ that network is a shunt $3.979\,\mathrm{pF}$ and a series $994.7\,\mathrm{pH}$, and $f_0/Q_n = 800\,\mathrm{MHz}$. Want it wider, with a resistive load that really is resistive, and you need more than one L-section. Pi and T networks are that construction. They are not calculated here. Fano’s 1950 limit is the general statement: a finite lossless ladder does not match an arbitrary impedance over an arbitrary band.

Why does the match walk off the marker?

Three different mechanisms get called “drift.”

The load moved, which is the table above. Temperature, a hand on the antenna, and a plastic cover all change a chip antenna’s resonance. The parts are still the parts you solved at the old impedance.

The reference plane moved. A NanoVNA file is the calibration plane. Fifty picoseconds of leftover line takes this antenna from $10.64 + j5.56\,\Omega$ to $24.6 + j54.8\,\Omega$ and replaces the $968.4\,\mathrm{pH}$ coil with a different topology. That case is drawn in matching the chip antenna. Port extension belongs on the instrument, before the file is exported. This solver will not invent it.

The parts are lossy, so the cancellation that was exact with infinite Q is no longer exact. Return loss stays finite. On this antenna, $Q_L = 15$ and $Q_C = 300$ still leave $26.7\,\mathrm{dB}$ of return loss at $2.45\,\mathrm{GHz}$. The marker moves a little. It does not walk to the band edge. If the peak has left the channel, look at the load and the plane before you look at Q.

Solving at the wrong frequency does the same thing as a moved plane. Match the frequency you will use. A network designed at $2.45\,\mathrm{GHz}$ for an antenna whose resonance is at $2.40\,\mathrm{GHz}$ is a match to an impedance the antenna does not have on channel.

How does component Q show up as loss?

Q is $|\mathrm{reactance}|$ divided by series resistance. The solver does not use it to pick a topology. It uses it when the network is evaluated. An inductor becomes $R = \omega L/Q$ in series with $L$. A capacitor becomes $\mathrm{ESR} = 1/(\omega C Q)$. Constant Q, the default, keeps the number you typed at every frequency. The skin-effect option scales only the inductor, $Q(f) = Q_0\sqrt{f/f_0}$. Capacitors stay at the typed Q either way.

• Ideal
• QL = 50
• QL = 15

Read the antenna columns first. Ideal parts are a match, transducer gain $0\,\mathrm{dB}$. $Q_L = 50$, $Q_C = 300$ costs $0.15\,\mathrm{dB}$. $Q_L = 15$, $Q_C = 300$ costs $0.42\,\mathrm{dB}$. Dropping the capacitor from $300$ to $100$ while the inductor stays at $15$ costs another $0.05\,\mathrm{dB}$. On this network the coil is the loss. The capacitor is not free, but it is the smaller term, because a $Q$ of $100$ on a picofarad capacitor is still a small ESR next to $\omega L/15$ on this inductor.

The $5\,\Omega$ columns are the same typed Q on a network with nodal Q $3$ instead of $1.92$. $Q_L = 50$ costs $0.30\,\mathrm{dB}$. $Q_L = 15$ costs $0.87\,\mathrm{dB}$. The step from $50$ to $15$ is $0.57\,\mathrm{dB}$ there and $0.28\,\mathrm{dB}$ on the antenna. Insertion loss tracks how much reactance the part is carrying, which is what nodal Q is measuring. A match you already know is narrow will also run hotter in the coil, for the same catalog Q.

The Q values in the table are inputs. They are the form’s starting points, labeled as such on the calculator. They are not the Q of a $0.97\,\mathrm{nH}$ or a $1.0\,\mathrm{nH}$ part at $2.45\,\mathrm{GHz}$.

Published figures, so you can see what a datasheet actually promises:

• Murata’s loss-reduction note compares $7.5\,\mathrm{nH}$ parts at $800\,\mathrm{MHz}$ and lists typical Q of $59$ for LQW15AN7N5H00, $33$ for LQG15HN7N5J02, and $27$ for LQP03TN7N5H02. Their FAQ attributes the higher Q of the LQW series to the wire-wound construction. Those three numbers are at $800\,\mathrm{MHz}$, at $7.5\,\mathrm{nH}$.
• Murata reference specification JELF243B-0010, for the multilayer LQG15HS series, lists a minimum Q of $8$ for the $10\,\mathrm{nH}$ part LQG15HS10NJ02, measured at $100\,\mathrm{MHz}$. A $100\,\mathrm{MHz}$ floor is not a $2.45\,\mathrm{GHz}$ Q. The PDF is JELF243B-0010.
• Coilcraft’s 0402HP-10N page lists a typical Q of $62$ at $900\,\mathrm{MHz}$ and $90$ at $1.7\,\mathrm{GHz}$ for that $10\,\mathrm{nH}$ part, measured on an Agilent/HP 4287A with a 16197 fixture. It is a $10\,\mathrm{nH}$ wirewound chip. It is not a Q for every value in the series, and it is not a Q at $2.45\,\mathrm{GHz}$.

C0G and high-Q RF capacitors are why the table can hold $Q_C$ at $100$ or $300$ without the capacitor dominating. The Q that matters is the curve at your frequency and your capacitance. A $1\,\mathrm{MHz}$ catalog Q is the same kind of trap as the $100\,\mathrm{MHz}$ inductor minimum. The component-Q section of the calculator points at those capacitor curves and tells you the $1\,\mathrm{MHz}$ figure is not the RF figure.

Self-resonance does what a typed Q does not. Past the inductor’s self-resonant frequency the part is a capacitor. The calculator can add a parallel $C = 1/(\omega_{\mathrm{SRF}}^2 L)$ if you type an SRF. Leave it empty and that pole is absent. A $0.97\,\mathrm{nH}$ coil with an SRF of a few gigahertz is a different circuit from the ideal $L$ in the table.

What this model leaves out

No pad, no via, no ground return, no trace between the parts, no component tolerance. Two $1\,\mathrm{nH}$ coils from the same reel can sit on opposite sides of the $968.4\,\mathrm{pH}$ you wanted, and the peak moves. Measure the board at the plane you matched.

The antenna numbers are the synthetic demo, not a measured .s1p. A quarter-wave transformer would trade a different frequency sensitivity for the lumped nodal Q. It is not solved here yet.

References

• R. M. Fano, “Theoretical limitations on the broadband matching of arbitrary impedances,” Journal of the Franklin Institute, 1950.
• D. M. Pozar, Microwave Engineering, the chapter on impedance matching and tuning. Nodal Q for the real step is the standard two-element result.
• Murata, “Proposal of the Loss Reduction in the RF Circuit,” and the LQG15HS reference specification JELF243B-0010. Coilcraft 0402HP-10N product page. Figures are quoted from those pages in the section above, at the frequency each page states.

Related

• Match a 2.4 GHz chip antenna
• Which L-network to build
• Smith chart calculator

FAQ

Why is my matched bandwidth narrower than f0/Q?

f0/Q is an estimate for a load that does not move. On the synthetic 2.45 GHz antenna the nodal Q is 1.92, and f0/Qn is 1.27 GHz. Holding the 2.45 GHz impedance constant, the ≥10 dB span is 1.12 GHz. On the antenna model the same L-section is only 273 MHz wide, because the impedance leaves the point you matched.

Why does a match drift off frequency?

The parts were solved at one impedance. A chip antenna’s reactance swings through resonance, a port extension rotates Γ, and a constant-Q loss shifts the perfect cancellation. Fifty picoseconds of leftover line moves this antenna from 10.64 + j5.56 Ω to 24.6 + j54.8 Ω. Recalculate at the plane you will solder to.

How does inductor Q change insertion loss?

Q is |reactance| / series resistance. The solver applies R = ωL/Q and ESR = 1/(ωCQ) when it evaluates the network. On the antenna’s low-pass section, QL = 15 and QC = 300 give −0.42 dB transducer gain, against −0.15 dB at QL = 50. The same QL step on a 5 Ω to 50 Ω match, nodal Q 3, costs 0.57 dB instead of 0.28 dB.

Are multilayer and wirewound Q values interchangeable?

No. Use the Q versus frequency curve for the inductance you will buy. Murata’s comparison of 7.5 nH parts at 800 MHz lists typical Q of 59, 33, and 27 for three constructions. Coilcraft’s 0402HP-10N page lists typical Q of 90 at 1.7 GHz for that 10 nH part. A 100 MHz minimum Q is not the Q at 2.45 GHz, and none of those figures is the Q of the 0.97 nH section in the antenna example.