Smith Chart & RF Black Magic Lab

Plot any impedance on the Smith Chart by entering ZL, then drag it around to see how Γ, VSWR, and return loss change in real time. Add series/shunt L and C components to build a matching network and watch the impedance trace a path toward the centre — the "magic" of RF design made visible.

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Lesson

The theory — Smith Chart & RF Black Magic Lab

The Smith chart is a picture of a single complex number. Γ = (Z_L − Z_0)/(Z_L + Z_0) takes every load that exists and folds it into a disc of radius one, so the position of the dot is the answer. The five rows underneath are not five measurements. They are that one number, written five ways.

What each symbol means

Γ
the reflection coefficient — the fraction of the incoming wave that comes back, as a magnitude and an angle. Everything else is derived from it.
VSWR
(1+|Γ|)/(1−|Γ|). Runs from 1 at a perfect match to infinity at total reflection, so it magnifies the bad end of the range and compresses the good one.
RL
return loss, −20 log₁₀|Γ|, in decibels. The same quantity turned upside down: bigger is better, and a perfect match is infinite.
y
the normalised admittance 1/z. It exists in the readout because shunt components add to admittance while series components add to impedance.

Where the formula comes from

  1. One measurement starts it all. Feeding 36 − j20 Ω into a 50 Ω system gives Γ = 0.276 ∠ −111.9°. Nothing else on the page is measured; the rest is arithmetic.
  2. The magnitude alone drives four rows. 0.276 gives VSWR 1.76, return loss 11.17 dB, mismatch loss 0.345 dB and 92.4% of the power delivered. The angle −111.9° appears in none of them.
  3. So the angle has exactly one job: it says where on the circle you are, and therefore what to add. The two normalised rows are how you cash it in — z = 0.720 − j0.400 for series parts, which change impedance, and y = 1.061 + j0.590 for shunt parts, which change admittance.
  4. Auto-match solves that geometry. It picks a series inductor of +j42.45 Ω — 7.38 nH at 915 MHz — then a shunt capacitor of +j12.5 mS, or 2.17 pF, and the readout drops to |Γ| = 0.000.

How to read what you see

Put the presets side by side and watch the five names disagree. The 915 MHz antenna sits at VSWR 1.76, which sounds like a problem, and delivers 92.4% of its power, which does not. The 2.4 GHz amplifier input reads VSWR 4.56 and 59.0%. The short circuit reads VSWR 5000 and 0.1%. VSWR ran over three orders of magnitude while the delivered power went from most of it, to some of it, to none — because VSWR has infinity at one end and percentage cannot. Read in decibels the same three loads cost 0.345, 2.293 and 30.971 dB, which is the scale that behaves itself.

Assumes
That the matching elements are lossless, that only one frequency matters, and that a component is small enough to have no length. The nomograph checks the last of those and will tell you when it fails: at 915 MHz a wavelength is 327.6 mm, so a 10 mm trace is 3.05% of it and the lumped model holds. Carry that same 10 mm to 2.4 GHz and it becomes 8.0%, and the λ/10 warning appears. The single-frequency assumption is never checked at all: an L-network is solved at the design frequency and nothing on this page shows how fast it falls apart either side of it.
Breaks when
A mismatch loss of 0.345 dB is, by itself, nothing. No link budget would notice a third of a decibel, and 92.4% delivered is a fine number. Yet the page also warns that a standing-wave ratio like this can damage a transmitter output stage, and that warning is not an exaggeration. The two statements answer different questions. Mismatch loss asks how much power fails to arrive. The hazard asks where the reflected wave goes — back down the line into the amplifier, as heat it was not sized for and as a voltage maximum somewhere along the cable that the insulation was not rated for. A load can waste almost none of your power and still be the thing that destroys the radio. That is why RF parts are specified in VSWR and return loss rather than in percent delivered: the percentage answers the question nobody was worried about.

An infinite half-plane folded into one circle 🖖

The Smith Chart is a conformal map that translates complex load impedance ZL to the complex reflection coefficient Γ on a unit circle: Γ = (z - 1)/(z + 1). Developed by Phillip Smith in 1939, it solves transmission line equations graphically. The chart converts infinite, rectangular impedance coordinates into a circular region. Circles of constant resistance all meet at the rightmost point (open circuit), and arcs of constant reactance trace paths from that same point. In radio frequency engineering, components like capacitors and inductors shift the plotted impedance along these circles and arcs. By adding series or parallel elements, designers can guide any arbitrary load impedance directly to the center of the chart (50 Ω), establishing a perfect match and preventing signal reflections.

Reading the chart like a target 🖖

The whole chart is really a target. Wherever your load lands, the only thing that matters is how far the dot sits from the dead centre. Right at the centre, source and load agree perfectly and no signal bounces back. The further the dot drifts toward the rim, the more power is reflected back at the transmitter instead of reaching the antenna, and that is exactly what VSWR and return loss put into numbers. Takeaway: good design is simply "drag the dot toward the middle."

Invented three times, on three continents 🖖

Phillip Smith usually gets sole credit, yet the same chart built on the map (z − 1)/(z + 1) was devised independently three times in the 1930s. Tōsaku Mizuhashi published his version in Japan in December 1937, and Amiel Volpert unveiled his in the Soviet Union in 1939 — the very same year as Smith. Russian engineers still call it the Volpert–Smith chart, and historians sometimes credit all three as the Mizuhashi–Volpert–Smith chart.

Problem solved in full

  1. A mismatched 915 MHz antenna on a 50 Ω line 6 steps

    A 915 MHz antenna measures 36 − j20 Ω on a 50 Ω line. Work out how badly it is mismatched, then design the two-component network that fixes it — and decide how narrow that fix is.

    1. Normalise. Every rule on the chart is written in units of Z₀, and the admittance is just the reciprocal — you need both because series parts add to impedance and shunt parts add to admittance.

    2. The reflection coefficient is the bilinear map that makes the chart a chart. Multiply through by the conjugate of the denominator; the modulus is what the next line needs.

    3. Three of the panel's rows are the same number wearing different clothes. Only the magnitude of Γ enters any of them — the angle contributes nothing at all, which is exactly why it is free to carry the other half of the information.

    4. Here is where the angle earns its keep. A series reactance changes the imaginary part of the impedance without touching the real part, so pick the value that makes the real part of the resulting admittance equal to 1/50 — one quadratic, two roots, and this is the positive one.

    5. The leftover susceptance is cancelled by a shunt element of the opposite sign. Convert both to components at 915 MHz and the network is done.

    6. Finally, ask how fragile it is. The node Q of an L-network depends only on the two resistances it bridges.

    Answer

    A series inductor of 7.38 nH followed by a shunt capacitor of 2.17 pF, and the reflection goes to zero. The last line is the one worth keeping. The node Q of this match is 0.62, which is small, and a low-Q match is a wide one: hold the load fixed, move the frequency ±10%, and the VSWR only rises from 1.00 to about 1.12. "L-networks are narrowband" is a half-truth — the bandwidth is set by how far the transformation has to move the resistance, not by the topology. Matching 5 Ω into 50 Ω needs Q = 3 and really is fragile; matching 36 into 50 barely notices.

References (3)

Example problems