How to Design an L-Network and Choose a Solution

An L-network is one series reactance and one shunt susceptance. For a load of $25 + j50\,\Omega$ and a real $50\,\Omega$ source at $1\,\mathrm{GHz}$, the solver returns four networks, not one.

The low-pass solution is a series $9.746\,\mathrm{nH}$ at the source and a shunt $4.106\,\mathrm{pF}$ at the load. It passes DC, its nodal Q is $1.22$, and on a series $R$–$L$ continuation of this load it is $16.7\,\mathrm{dB}$ down at $2\,\mathrm{GHz}$. The high-pass solution is only $4.4\,\mathrm{dB}$ down at $2\,\mathrm{GHz}$, and it does not pass DC. The widest $\ge 10\,\mathrm{dB}$ band is a two-capacitor network, $42\%$ fractional, and it blocks DC too. The right one is the one that matches the supply, the harmonic, and the bandwidth you measured. Nodal Q does not make that choice for you.

How does an L-section move the load?

Series elements stay on a constant-resistance circle. Shunt elements stay on a constant-conductance circle. The match for a real source equal to the chart reference is the center of the Smith chart. You need one arc of each kind, in one order or the other, and each order has two directions.

On that chart the load is $25 + j50\,\Omega$. The first move is the shunt capacitor, along a conductance circle. The series inductor then rides a resistance circle into $50\,\Omega$. The path is drawn from the load back toward the source, which is the order the chart wants. The parts list is the other way around: source, then load. Both orders are labeled that way on the L-network calculator.

If the load resistance is already on the match circle, one of those elements shrinks to nothing and you have a single-element match. A purely reactive load has no L-section at all. A lossless network cannot manufacture the real part a conjugate match requires.

Why are there two or four solutions?

Write $Z_L = R_L + jX_L$ and $Z_s = R_s + jX_s$. The target admittance is $1/Z_s^* = G_t + jB_t$.

Topology A puts the series element next to the load and the shunt next to the source. It has a real solution when $R_L \le 1/G_t$. Topology B puts the shunt next to the load. It has a real solution when the load conductance $G_L \le 1/R_s$. A load can satisfy one test, the other, or both. Each allowed topology has two signs. Four is the maximum. The formulas and the ABCD check are on the L-network page. Every ideal solution here sits under $|\Gamma| = 10^{-9}$ at the design frequency.

$25 + j50\,\Omega$ into $50\,\Omega$ passes both tests, because $R_L = 25 < 50$ and $G_L = 0.008\,\mathrm{S} < 0.02\,\mathrm{S}$. So there are four.

The $2\,\mathrm{GHz}$ column is not a second design. The $25 + j50\,\Omega$ point is continued as a series resistor and inductor, $L = X/\omega$ at $1\,\mathrm{GHz}$, and the same parts are re-evaluated. A real antenna will not follow that extrapolation. It is still the right comparison between these four networks, because the load model is held still while the topology changes.

Which solution do you build?

Three questions, in this order.

Does the node need a DC path? “Passes DC” means there is no series capacitor and no shunt inductor. Only the low-pass row, B+, passes. A series capacitor blocks the supply. A shunt inductor is a DC short to ground. The two mixed rows are both capacitors, so they block, and they are neither a classical low-pass nor a classical high-pass. The flavor name on the row is the honest one.

What should happen at the second harmonic? At $2\,\mathrm{GHz}$ the low-pass network delivers $16.7\,\mathrm{dB}$ less transducer gain than a perfect match. The high-pass network delivers $4.4\,\mathrm{dB}$ less. If the stage before this match is a power amplifier, the low-pass row is the one that loads the harmonic. If you need the match to pass a higher band, it is the wrong row. This is not a filter specification. There is no layout, and the load model is the series $R$–$L$ stated above.

How wide is the match you will actually measure? The $\ge 10\,\mathrm{dB}$ spans run from $777\,\mathrm{MHz}$–$1.198\,\mathrm{GHz}$ on the widest row down to $887\,\mathrm{MHz}$–$1.154\,\mathrm{GHz}$ on the high-pass. Fractional bandwidth is $42\%$ versus $27\%$. The widest row blocks DC. If the radio is a narrow channel at $1\,\mathrm{GHz}$ and the supply has to reach the pin, take the narrower low-pass and keep the DC path. If the load is AC-coupled anyway, the extra bandwidth of A+ is free.

Part values in the table are ideal. The calculator can snap them to E24 or to the RF grid ($0.1\,\mathrm{nH}$ and $0.1\,\mathrm{pF}$ through $9.9$, then E24). Snapping moves the perfect null. Do that after you have picked the topology, not before.

When is the target Zs* instead of 50 Ω?

A conjugate match sets the impedance seen by the source equal to $Z_s^*$. A $Z_0$ match sets it equal to the chart reference. They coincide when $Z_s$ is real and equal to $Z_{\mathrm{ref}}$.

They do not coincide here. Source $40 + j30\,\Omega$, load $20 + j40\,\Omega$, chart still normalized to $50\,\Omega$, design frequency $1\,\mathrm{GHz}$. The target is $40 - j30\,\Omega$.

The open marker is $50\,\Omega$. The path does not end there. Ending on the center would be a match to $50\,\Omega$, and this source is not $50\,\Omega$. Available power is absorbed at $Z_s^*$, which is the Kurokawa condition the solver enforces: $\Gamma_{\mathrm{in}} = (Z_{\mathrm{in}} - Z_s^*)/(Z_{\mathrm{in}} + Z_s)$.

Again only one row passes DC: series $3.022\,\mathrm{nH}$ and shunt $5.132\,\mathrm{pF}$, nodal Q $1.22$, $\ge 10\,\mathrm{dB}$ from $833\,\mathrm{MHz}$ to $1.128\,\mathrm{GHz}$. The widest row is the two capacitors, $35\%$ fractional. Same choice as before. The new fact is where the path stops.

A conjugate match of a transistor is not a stability analysis. This solver does not draw stability circles and does not know $K$.

What nodal Q does not tell you

Nodal Q is $|X|/R$ on the series node after the series element, or $|B|/G$ on the shunt node. For a real-to-real step between $R_{\mathrm{high}}$ and $R_{\mathrm{low}}$ it collapses to $\sqrt{R_{\mathrm{high}}/R_{\mathrm{low}} - 1}$, and that value is the lowest Q two lossless elements can have. A $5\,\Omega$ load into $50\,\Omega$ has nodal Q $3$. A wider real-to-real match needs more than one L-section. Pi and T networks, which are two L-sections through a virtual resistance, are not calculated here.

$f_0/Q_n$ is an estimate of bandwidth, not the $\ge 10\,\mathrm{dB}$ span in the table. The $Q_n = 1$ solution has $f_0/Q_n = 1\,\mathrm{GHz}$ and a $10\,\mathrm{dB}$ band $421\,\mathrm{MHz}$ wide. Use the sweep. The estimate is useful when you are comparing two real-to-real steps before you have drawn either one.

Component Q is a different number. It is the loss in the parts, applied after the topology is chosen. It does not pick the sign. What it costs, on this kind of network and on a higher-$Q_n$ step, is in why an L-match is narrow.

A quarter-wave transformer and a single-stub tuner solve the same landing-on-the-match-circle problem with transmission lines. They are not in this calculator yet. A microstrip width, if you already know the impedance you want, is on the microstrip calculator. On 1.6 mm FR-4 that width is 50 Ω microstrip on FR-4.

References

• D. M. Pozar, Microwave Engineering, the chapter on impedance matching and tuning. The L-network page computes a check in the shape of Pozar’s Example 5.1 and labels it as computed, not as a transcribed textbook line.
• C. Bowick, RF Circuit Design, the chapter on L-network matching.
• K. Kurokawa, “Power Waves and the Scattering Matrix,” IEEE Transactions on Microwave Theory and Techniques, 1965.
• R. M. Fano, “Theoretical limitations on the broadband matching of arbitrary impedances,” Journal of the Franklin Institute, 1950. A finite lossless ladder cannot match an arbitrary load over an arbitrary band. The $Q_n$ floor for a real step is the two-element version of that limit.

Related

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

FAQ

How many L-network solutions are there?

Up to four. Topology A puts the series element at the load and is allowed when the load resistance is no larger than 1/Gt. Topology B puts the shunt at the load and is allowed when the load conductance is no larger than 1/Rs. Each allowed topology has two signs. A load already on the match circle can collapse to one element.

Which L-network should I build?

For 25 + j50 Ω into 50 Ω at 1 GHz, the low-pass solution passes DC and is 16.7 dB down at 2 GHz on a series R–L continuation of the load. The high-pass solution is only 4.4 dB down at 2 GHz and does not pass DC. The widest ≥10 dB band is a two-capacitor network that also blocks DC. Choose from the DC path, the harmonic, and the measured band.

What is the difference between a conjugate match and a Z0 match?

A Z0 match makes the input impedance equal to the chart reference, usually 50 Ω. A conjugate match makes it equal to Zs*. Those are the same only when the source is real and equal to Zref. With Zs = 40 + j30 Ω the target is 40 − j30 Ω, off the center of a 50 Ω chart.

Does nodal Q equal the matched bandwidth?

No. Nodal Q is |X|/R on the series node or |B|/G on the shunt node, and f0/Qn is an estimate. The bandwidth reported with these solutions is the span where return loss stays at or above 10 dB. On the 25 + j50 Ω example the Qn = 1 solution has a 42% fractional 10 dB band, not a 1 GHz 10 dB band.