Constant-G Circle
Reading Conductance on the Admittance Smith Chart
The standard Smith chart is an impedance chart whose grid is built from constant-resistance and constant-reactance arcs. When a network adds elements in parallel rather than in series, working in impedance becomes awkward because parallel components add in admittance, not impedance. The fix is to rotate the chart 180° about its center, converting it into an admittance chart. On that admittance chart the resistance circles reinterpret as constant-conductance circles and the reactance arcs reinterpret as constant-susceptance arcs. A constant-G circle therefore traces every load that presents the same real part of admittance, regardless of how inductive or capacitive its imaginary part may be.
Every constant-G circle is tangent to the left edge of the chart at the short-circuit point (−1, 0), mirroring the way constant-R circles all pass through the open-circuit point on the right. The outermost circle, g = 0, is the unit boundary representing a purely reactive (lossless) load; as conductance rises, the circles shrink and slide leftward, collapsing to a point at the short-circuit when g is infinite. The single most important member of the family is the g = 1 circle, the only conductance value that can be matched to a 50 Ω source using nothing but a shunt reactive element.
In practice an engineer measures a load with a vector network analyzer, normalizes the admittance to the line characteristic admittance Y0 (0.02 S for a 50 Ω system), and plots the point. Sliding toward the generator along the constant-VSWR circle moves the point until it crosses the g = 1 circle at two candidate locations, each corresponding to a different stub position and length. This geometric procedure is the backbone of stub matching and is automated inside most RF design tools.
Geometry and Matching Equations
y = Y / Y0 = g + jb, Y0 = 1 / Z0 = 0.02 S (50 Ω)
Constant-G circle center and radius (unit chart):
Center = ( −g / (1 + g), 0 ), Radius = 1 / (1 + g)
Single-stub match condition:
At the stub plane, g = 1; shunt stub supplies bstub = −bload ⇒ y = 1 + j0
Where g = normalized conductance, b = normalized susceptance, Y0 = characteristic admittance. Example: the g = 1 circle is centered at (−0.5, 0) with radius 0.5; the g = 3 circle is centered at (−0.75, 0) with radius 0.25.
Constant-G Circle Family Reference
| Normalized g | Center (x, 0) | Radius | Passes through center? | Typical use |
|---|---|---|---|---|
| g = 0 | (0, 0) | 1.0 | No (outer boundary) | Lossless / purely reactive load |
| g = 0.5 | (−0.33, 0) | 0.67 | No | 100 Ω-equivalent shunt load |
| g = 1 | (−0.5, 0) | 0.5 | Yes | Single-stub match target |
| g = 2 | (−0.67, 0) | 0.33 | No | 25 Ω-equivalent shunt load |
| g → ∞ | (−1, 0) | 0 | Collapses to short | Short-circuit termination |
Frequently Asked Questions
How does a constant-G circle differ from a constant-R circle on the Smith chart?
They are the same circle family seen in two domains. Constant-R circles are fixed normalized resistance on the impedance chart and all pass through the open-circuit point on the right; constant-G circles are fixed normalized conductance on the admittance chart and all pass through the short-circuit point on the left. The admittance chart is the impedance chart rotated 180° about the center, so reflecting a constant-R circle through the origin yields a constant-G circle. Series elements move along constant-R circles; shunt elements move along constant-G circles.
Why does single-stub matching rely on the unity constant-G circle?
A shunt stub adds pure susceptance in parallel, so it changes only the imaginary part of the admittance, never the conductance. The normalized conductance must therefore already equal 1 (the g = 1 circle) before the stub is added, because g = 1 is the only value where canceling the susceptance lands you exactly at the chart center, y = 1 + j0. The series line between load and stub rotates the admittance along a constant-VSWR circle until it crosses g = 1, and the stub then supplies bstub = −b to finish the match.
How do you compute the conductance of a constant-G circle from its geometry?
On a unit-radius chart the constant-G circle for conductance g is centered at (−g/(1+g), 0) with radius 1/(1+g). The g = 1 circle is centered at (−0.5, 0) with radius 0.5, the g = 0 circle is the outer unit boundary, and as g approaches infinity the circle shrinks to the short-circuit point at (−1, 0). To recover g from a measured admittance, use g = Re(Y/Y0) with Y0 = 0.02 S for a 50 Ω system.