Audited 31 Jul 2026·Last updated 28 Oct 2026·4 citations·Tier 2·0 uses

Impedance Matching Calculator — L-Network Design

Design a two-part L-network between two resistive impedances: network Q, both reactances, inductor and capacitor values, and the VSWR it corrects.

Impedance Matching Calculator

Network arrangement
Purely resistive. A complex source must be resolved into its resistive part plus a conjugate-matching step first — this page does not do that.
Ω
Purely resistive, and it must differ from the source. Equal impedances are already matched, and an L-network between them is degenerate.
Ω
An L-network matches at one frequency. This field sets the component values but has no effect on Q or on either reactance.
Frequency unit
Network Q
1.7321
The network's quality factor, √(R_high ÷ R_low − 1), fixed by the impedance ratio; a higher value means a narrower match.
Series element (µH low-pass · pF high-pass)
0.1378
Shunt element (pF low-pass · µH high-pass)
13.7832
Series reactance |Xs|
86.6025 Ω
Shunt reactance |Xp|
115.4701 Ω
Transformer turns ratio (primary : secondary)
0.5
Reflection coefficient Γ (unmatched)
0.6
VSWR (unmatched)
4
Return loss (unmatched)
4.437 dB
Mismatch loss (unmatched)
1.9382 dB
The network, in words
Matching 50 Ω to 200 Ω at 100 MHz needs a network Q of 1.732051: a 0.137832 µH inductor in series with the 50 Ω side, and a 13.783222 pF capacitor shunting the 200 Ω side. The series element always goes on the lower-impedance side and the shunt element across the higher one. Equivalently, an ideal transformer with a turns ratio of 0.5 : 1 would do the same job over a much wider band. Without any network the reflection coefficient is 0.6, a VSWR of 4 and a mismatch loss of 1.9382 dB — that loss is what the match is worth. Two limits that decide whether these numbers are usable: the source and load must be PURELY RESISTIVE, since a reactive load has to be resolved first and this page does not do that; and an L-network matches at ONE frequency, with a fractional bandwidth that narrows as Q rises, so a Q of 1.7321 here is a comparatively broad match.

Background.

Connect a 50 Ω source straight to a 200 Ω load and 36% of the power arriving at the load is reflected back toward the source. The reflection coefficient is (200 − 50) ÷ (200 + 50) = 0.6, the standing-wave ratio on the line is 4, and the power actually delivered falls by 1.9382 dB. Two well-chosen components between them can make the load look like 50 Ω to the source at the frequency you care about.

Those two components form an L-network: one element in series on the low-impedance side, one in shunt across the high-impedance side. Reactive parts dissipate no power in principle, so the network transforms impedance rather than burning off the excess. You choose only which kind of part goes where. A series inductor with a shunt capacitor makes a low-pass network that also attenuates harmonics, the usual choice after an amplifier. A series capacitor with a shunt inductor makes a high-pass network that blocks DC. Either matches the same pair of impedances with the same Q and the same reactance magnitudes.

The Q is set by the impedance ratio alone: a 4:1 step gives 1.7321 and a 17:1 step exactly 4. The higher it is, the faster the match degrades away from the design frequency and the more it suffers from component tolerance, and two components give you no way to lower it.

Two assumptions sit under every answer. Both impedances are treated as pure resistances, so a real antenna or transistor input with a reactive part needs that reactance dealt with first. And the components are calculated for one frequency only.

What is impedance matching calculator?

Impedance matching means arranging for a load to present, at the source's terminals, the impedance that delivers the most power. For a resistive source that is a load equal to the source resistance, by the maximum power transfer theorem; any difference reflects a share |Γ|² of the incident power, where Γ = (Z_L − Z_S) ÷ (Z_L + Z_S). An L-network is the smallest lossless circuit that does the job: two reactances whose values follow from the two resistances, converted into an inductor and a capacitor once a frequency is named.

How to use this calculator.

  1. Enter the source and load impedances in ohms as resistances. They must differ; equal values are already matched and an L-network between them has no meaning.
  2. Choose low-pass for a series inductor and shunt capacitor, or high-pass for a series capacitor and shunt inductor. Low-pass helps suppress harmonics; high-pass blocks DC and gives the shunt side a DC path to ground.
  3. Enter the design frequency and pick Hz, kHz, MHz or GHz. Only the component values depend on it; Q and both reactances do not.
  4. Build the network with the series element on the lower-impedance side and the shunt element across the higher-impedance side. Reversed, it transforms in the wrong direction.
  5. Check the mismatch loss before building anything. If it is a fraction of a decibel, the network's own losses and tolerances may cost nearly as much as it saves.

The formula.

Q = √(R_high ⁄ R_low − 1) |X_S| = Q·R_low |X_P| = R_high ⁄ Q Γ = (Z_L − Z_S) ⁄ (Z_L + Z_S)

Following Steer's derivation in Microwave and RF Design III, §6.4, the series leg and the shunt leg must have Q values of equal magnitude. That condition leaves no free parameter: Q = √(R_high ÷ R_low − 1), the series reactance is |X_S| = Q × R_low and the shunt reactance is |X_P| = R_high ÷ Q. Multiplying the two shows |X_S| × |X_P| = R_high × R_low, a quick check on any design. Component values follow from ω = 2πf: for low-pass, L = X_S ÷ ω and C = 1 ÷ (ω X_P); for high-pass, C = 1 ÷ (ω X_S) and L = X_P ÷ ω. Inductors are reported in microhenries and capacitors in picofarads. The unmatched figures come from Γ = (Z_L − Z_S) ÷ (Z_L + Z_S): VSWR = (1 + |Γ|) ÷ (1 − |Γ|), return loss = −20 log₁₀|Γ| and mismatch loss = −10 log₁₀(1 − |Γ|²), the last being the share of power lost to reflection. The transformer alternative uses the reflected-resistance relation from OpenStax University Physics, R_primary = (N_P ÷ N_S)² × R_secondary, which inverts to a turns ratio n = √(Z_S ÷ Z_L). Not modelled: losses in real inductors and capacitors, parasitic inductance and capacitance, complex impedances and the full frequency response.

A worked example.

Example

Match a 50 Ω source to a 200 Ω load at 100 MHz with the low-pass arrangement. The impedance ratio is 200 ÷ 50 = 4, so Q = √(4 − 1) = √3 = 1.7321. The series reactance is 1.7321 × 50 = 86.6025 Ω and the shunt reactance is 200 ÷ 1.7321 = 115.4701 Ω; their product is 10,000, which is 200 × 50 as it should be. At 100 MHz, ω = 2π × 10⁸ ≈ 628,318,531 rad/s. The series inductor is 86.6025 ÷ ω = 0.1378 µH, about 138 nH, placed in series with the 50 Ω side. The shunt capacitor is 1 ÷ (ω × 115.4701) = 13.7832 pF, placed across the 200 Ω side. Chosen as high-pass instead, the same reactances become an 18.3776 pF series capacitor and a 0.1838 µH shunt inductor. To confirm the match, combine 200 Ω in parallel with the capacitor's −j115.4701 Ω: the result is 50 − j86.6025 Ω. Adding the inductor's +j86.6025 Ω cancels the reactance and leaves 50 + j0 Ω, exactly what the source wants to see. Before the network, the same pair had a reflection coefficient of 0.6 and lost 1.9382 dB to mismatch.

frequency UnitMHz
source Impedance50
design Frequency100
network Typelowpass-l
load Impedance200

Frequently asked questions.

Can I choose a lower Q for a wider match?
Not with two components; the ratio fixes it. A third component, as in a pi or T network, lets you set Q freely, and cascading L-sections lowers it. Matching 16:1 in one step needs Q = 3.873, but two 4:1 steps through an intermediate 200 Ω between 50 Ω and 800 Ω need only 1.7321 each.
Is a VSWR of 2 worth matching?
Often not for efficiency alone. A 2:1 mismatch reflects 11.1% of the power, a mismatch loss of about 0.5115 dB, which a lossy inductor can eat into. At 4:1 the loss is 1.9382 dB, more than a third of the power. Many transmitters also reduce output to protect themselves into a high VSWR, which is a separate reason to match.
Why is the series element always on the low-impedance side?
Seen from the high side, the low resistance plus the series reactance behaves like a larger resistance, R_low × (1 + Q²) = R_high, in parallel with a reactance, and the shunt element cancels that reactance. Put the shunt element across the low side instead and the same conversion runs the other way, shrinking the resistance further, so no pair of values reaches the target.
Can I use these values for an antenna?
Only if the antenna's feed-point impedance is purely resistive at your frequency, which is rarely exact. A reactive part has to be absorbed into one of the network's elements or cancelled with an extra part before the resistive step is designed. Using these values on a complex load without that step will not give a match.
When is a transformer the better choice?
When you need a match over a wide band. A transformer with turns ratio √(Z_S ÷ Z_L), 1:2 for 50 Ω to 200 Ω, transforms impedance across a broad range of frequencies, limited at the low end by its magnetising inductance and at the high end by leakage inductance and winding capacitance. An L-network is smaller and simpler but matches at a point.
Why do the components change with frequency but Q does not?
Q and both reactances depend only on the two resistances. Frequency enters when a reactance is turned into a part: the same 86.6025 Ω of series reactance needs 0.1378 µH at 100 MHz but 13.7832 µH at 1 MHz. That is why designs for different bands are easiest to compare in ohms of reactance.

References& sources.

  1. [1]Michael Steer, Microwave and RF Design III — Networks, §6.4 'The L Matching Network' (Engineering LibreTexts). Source of the design equations: Eq. (4)/(5) Q_S = |X_S/R_S| and Q_P = |R_L/X_P|; Eq. (6) |Q_S| = |Q_P| = √(R_L/R_S − 1) for R_S < R_L; Eq. (11) √(R_S/R_L − 1) for R_S > R_L; Eq. (12) Q_S = X_S/R_L and Q_P = R_S/X_P; and the topology showing the shunt element on the load side when R_S < R_L. Open access; retrieved 2026-07-29.
  2. [2]Michael Steer, Microwave and RF Design II — Transmission Lines, §2.6 'Reflections at Interfaces' (Engineering LibreTexts). Source of the reflection coefficient Γ = (Z_L − Z_0)/(Z_L + Z_0) (Eq. 2.3.7) and of the reflected-power relation P_R = |Γ|²P_I (Eq. 2.6.13), from which the return-loss and mismatch-loss expressions on this page follow. Open access; retrieved 2026-07-29.
  3. [3]Steven W. Ellingson (Virginia Tech), Electromagnetics I, §3.14 'Standing Wave Ratio' (Engineering LibreTexts). Source of SWR = (1 + |Γ|)/(1 − |Γ|), with SWR defined as 'the ratio of the maximum magnitude of the standing wave to minimum magnitude of the standing wave'. Open access; retrieved 2026-07-29.
  4. [4]OpenStax (Rice University), University Physics Volume 2, §15.6 'Transformers'. Source of the transformer relations V_S/V_P = N_S/N_P (Eq. 15.21) and i_S(t) = (N_P/N_S)i_P(t) (Eq. 15.22), and of the reflected-resistance result R_P = (N_P/N_S)²R_S, which inverts to the impedance-matching turns ratio n = √(Z_s/Z_l) reported here. Open access; retrieved 2026-07-29.

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Q = √(R_high ⁄ R_low − 1) |X_S| = Q·R_low |X_P| = R_high ⁄ Q Γ = (Z_L − Z_S) ⁄ (Z_L + Z_S)
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