L-Network Impedance Matching Calculator
L-Network Impedance Matching Calculator
Match a source resistance to a load with two reactive parts: both L-network solutions (low-pass and high-pass), their inductor and capacitor values, where the shunt part goes, the network’s Q — and a sweep of how much power reaches the load either side of the design frequency.
L-network match
Pozar’s Example 5.1: a 200 − j100 Ω load matched to 100 Ω at 500 MHz
L-network design equations
- Rlarge, Rsmall
- the larger and smaller of Rs and RL; the shunt part goes across Rlarge
- Xseries
- reactance of the series part: an inductor (low-pass) or a capacitor (high-pass)
- Xshunt
- reactance of the shunt part: a capacitor (low-pass) or an inductor (high-pass)
- RL + jXL
- a complex load: Pozar’s general solution absorbs XL into the two parts
Worked example
Pozar's Example 5.1: a 200 − j100 Ω load matched to 100 Ω at 500 MHz
RL = 200 Ω is larger than Rs = 100 Ω, so the shunt part goes across the load; Q = √(200 ÷ 100 − 1) = 1.00
Solution 1: B = 0.0029 S → shunt C = 0.9228 pF; X = 122.5 Ω → series L = 38.98 nH
Solution 2: B = −0.0069 S → shunt L = 46.14 nH; X = −122.5 Ω → series C = 2.599 pF
Either network turns the load into exactly 100 Ω at 500 MHz; the sweep shows how they differ off frequency
Resistive matches at 10 MHz
| Match | Q | Shunt across | Low-pass: series L + shunt C | High-pass: series C + shunt L |
|---|---|---|---|---|
| 50 Ω → 12.5 Ω | 1.732 | source | 344.6 nH + 551.3 pF | 735.1 pF + 459.4 nH |
| 50 Ω → 25 Ω | 1.000 | source | 397.9 nH + 318.3 pF | 636.6 pF + 795.8 nH |
| 50 Ω → 100 Ω | 1.000 | load | 795.8 nH + 159.2 pF | 318.3 pF + 1.592 µH |
| 50 Ω → 200 Ω | 1.732 | load | 1.378 µH + 137.8 pF | 183.8 pF + 1.838 µH |
| 50 Ω → 300 Ω | 2.236 | load | 1.779 µH + 118.6 pF | 142.4 pF + 2.135 µH |
| 50 Ω → 1000 Ω | 4.359 | load | 3.469 µH + 69.37 pF | 73.03 pF + 3.651 µH |
| 75 Ω → 300 Ω | 1.732 | load | 2.067 µH + 91.89 pF | 122.5 pF + 2.757 µH |
How an L-network matches impedances
A source delivers its maximum power when the load it sees equals its own resistance. When the load is different — a 200 Ω antenna feed on a 50 Ω transmitter, a transistor’s few-ohm input on a 50 Ω line — an L-network of one inductor and one capacitor transforms it. The shunt part goes across the larger resistance and turns it into a smaller resistance in series with a reactance; the series part cancels that reactance. Because neither part dissipates power, all of the available power reaches the load at the design frequency.
Two solutions. For a resistive load there are always two: a low-pass form with the inductor in series and the capacitor in shunt, which also attenuates harmonics and passes DC (useful for biasing a transistor through the match), and a high-pass form with a series capacitor and shunt inductor, which blocks DC. Matching 50 Ω to 200 Ω at 14.2 MHz, for instance, gives Q = 1.732 and either 970.6 nH with 97.06 pF, or 129.4 pF with 1.294 µH.
Q is not a free choice. The resistance ratio alone sets Q = √(Rlarge ÷ Rsmall − 1), and with it the bandwidth, roughly the design frequency divided by Q for a match of this kind. A large ratio means a high Q, a narrow match and more sensitivity to part tolerances and losses; Bowick’s RF Circuit Design shows how two L-sections in cascade, or a pi or T network, trade that off. The chart shows the power reaching the load as the frequency moves from half to twice the design value, holding the load impedance fixed (exact for a resistive load; a real reactive load changes with frequency as well).
Complex loads. A load with reactance, such as an antenna off resonance or a device’s input, is handled with Pozar’s general closed form, which absorbs the reactance into the two parts rather than cancelling it with a third. The default example is Pozar’s own: 200 − j100 Ω matched to 100 Ω at 500 MHz gives 0.9228 pF and 38.98 nH, or 2.599 pF and 46.14 nH. Enter the load reactance as negative for a capacitive load.
Building it. At VHF and above the values get small: under a picofarad or a few nanohenries, comparable to the parasitics of the parts and the board, so choose parts with high self-resonant frequency, keep leads short, and tune on the bench with a network analyser. Real inductors have losses (finite Q) that the ideal model leaves out. To convert the power levels involved, see the dBm to watts calculator.
Frequently asked questions
How do I design an L-matching network?
Compute Q = √(R_large ÷ R_small − 1). The series part has reactance Q × R_small and the shunt part R_large ÷ Q, with the shunt across the larger resistance. Convert with L = X ÷ 2πf and C = 1 ÷ (2πf X).
Which side does the shunt element go on?
Across the larger of the two resistances. If the load is larger than the source, the shunt part is across the load; if smaller, across the source.
What is the difference between the two solutions?
For a resistive load, one is low-pass (series L, shunt C) and the other high-pass (series C, shunt L). Choose low-pass to reduce harmonics or pass DC bias, high-pass to block DC.
Can an L-network match a complex load?
Yes. Enter the load’s reactance and the page uses the general solution (Pozar, §5.1), which absorbs the reactance into the two parts. 200 − j100 Ω to 100 Ω at 500 MHz needs 0.92 pF and 38.98 nH, or 2.60 pF and 46.14 nH.
Why is my matching bandwidth so narrow?
Because the resistance ratio fixes Q, and bandwidth is roughly frequency ÷ Q. A 50 Ω to 1,000 Ω match has Q ≈ 4.36. For a wider match use two L-sections through an intermediate resistance.
Related calculators
References
- Pozar DM. Microwave Engineering, 4th ed. Wiley, 2012. Section 5.1, Matching with lumped elements (L networks), Example 5.1.
- Bowick C. RF Circuit Design, 2nd ed. Newnes, 2008. Chapter 4, Impedance matching: the L network, Q and bandwidth.
- scikit-rf documentation. Impedance Matching example reproducing Pozar’s Example 5.1 (0.92 pF / 38.98 nH; 2.60 pF / 46.14 nH).