Current Maximum
How Current Antinodes Form on a Mismatched Line
When a transmission line is terminated in anything other than its characteristic impedance, part of the incident wave reflects back toward the source. The forward and reflected waves superimpose to create a stationary interference pattern: the standing wave. Voltage and current each form their own standing-wave pattern, but the two are spatially shifted by a quarter wavelength because the reflected current wave inverts in sign relative to the reflected voltage wave at the load. The result is that wherever the voltage envelope reaches a peak, the current envelope reaches a trough, and vice versa. The current maximum is simply the antinode of the current envelope, the location where the magnitude of the line current is largest.
The amplitude of the current at the antinode depends on how badly the line is mismatched. With a reflection coefficient magnitude of |Γ|, the standing-wave ratio is S = (1 + |Γ|) / (1 - |Γ|). The current at the antinode is (1 + |Γ|) times the forward-wave current, while at the current minimum it falls to (1 - |Γ|) times that value. A perfectly matched line has |Γ| = 0, no standing wave, and a uniform current with no distinct maximum; a total reflection (open, short, or pure reactance) drives |Γ| to 1 and produces a current antinode twice the forward amplitude sitting directly on top of a true current null.
Because the impedance seen looking into the line equals the local voltage divided by the local current, the current maximum is also a point of minimum impedance. On a lossless line that impedance is purely real and equal to Z0 divided by the SWR. Designers exploit this: a quarter-wave section transforms the low impedance at a current antinode into a high impedance a quarter wavelength away, the basis of stub matching and quarter-wave transformers.
Position and Spacing Equations
|I(d)| = |I+| × |1 − Γe−j2βd|
Antinode amplitude and minimum:
Imax = |I+|(1 + |Γ|) , Imin = |I+|(1 − |Γ|)
Spacing & offset:
Δd (max to max) = λ/2 , offset from Vmax = λ/4
Impedance at a current maximum:
Z(Imax) = Z0 / S (minimum, purely real on a lossless line)
Where I+ = forward-wave current, Γ = reflection coefficient, β = 2π/λ phase constant, d = distance from load, S = VSWR ≈ (1 + |Γ|)/(1 − |Γ|). Example: S = 3 on a 50 Ω line → Z(Imax) ≈ 16.7 Ω and Imax/Imin = 3.
Current Maximum vs. Other Standing-Wave Points
| Standing-wave point | Current | Voltage | Local impedance | Dominant stress | Found at (load) |
|---|---|---|---|---|---|
| Current maximum | Peak (antinode) | Minimum (node) | Z0/S (low, real) | I2R conductor heating | Short circuit |
| Current minimum | Minimum (node) | Peak (antinode) | Z0×S (high, real) | Dielectric breakdown | Open circuit |
| Voltage maximum | Minimum (node) | Peak (antinode) | Z0×S (high, real) | Arcing / E-field | Open circuit |
| Voltage minimum | Peak (antinode) | Minimum (node) | Z0/S (low, real) | Conductor heating | Short circuit |
| Matched line (no SWR) | Uniform | Uniform | Z0 everywhere | Evenly distributed | ZL = Z0 |
Frequently Asked Questions
How far apart are successive current maxima on a transmission line?
Adjacent current maxima are spaced one half wavelength apart. At 1 GHz in air, λ/2 is 150 mm, so antinodes repeat every 150 mm; on coax with a 0.66 velocity factor that physical spacing shrinks to about 99 mm. Each current maximum sits λ/4 (75 mm in air at 1 GHz) from the neighboring voltage maxima because the current and voltage standing-wave patterns are offset by 90°.
Where does a current maximum form on a shorted versus open stub?
A short-circuited stub leaves the load current unconstrained, so a current maximum (and voltage minimum) forms right at the short and repeats every λ/2 toward the source. An open-circuited stub forces the end current to zero, placing a current minimum at the open and the first current maximum a quarter wavelength back. This is why a λ/4 shorted stub looks like a parallel-resonant open and a λ/4 open stub looks like a series-resonant short.
Why does the current maximum point heat up the most on a mismatched feedline?
Conductor loss is proportional to I2R, so the spot carrying the largest RMS current dissipates the most heat. At VSWR 5 the reflection magnitude is |Γ| = 4/6 = 0.67, so the antinode current is (1 + |Γ|) = 1.67 times the forward-wave current, while the current null drops to 0.33 times it; the antinode-to-null current ratio equals the VSWR of 5. For the same forward power, the antinode I2R loss runs about 1.672 ≈ 2.8 times the matched-line value. The current maxima are therefore the hot spots that cap power handling, while the voltage maxima are the points most prone to arcing; both must be checked when rating a feedline for forward power and mismatch.