Parallel Wire Impedance Calculator
Calculate the characteristic impedance of a parallel wire (two-wire) transmission line from the wire diameter and center-to-center spacing. Useful for designing twin-lead, ladder line, and balanced RF feeder systems.
Impedance Estimator for Parallel Wire Transmission Lines
Parallel Wire Impedance Formula
Where:
- \(Z_0\) — characteristic impedance of the line (Ω)
- \(D\) — center-to-center distance between the two wires (mm)
- \(d\) — diameter of each wire (mm)
This formula gives the impedance of a balanced two-wire line in air, assuming thin, parallel, cylindrical conductors. Equivalently, \(Z_0 = 120\ln(2D/d)\) Ω, since \(276\log_{10}(x) = 120\ln(x)\).
Parallel Wire Impedance – Calculation Example
Example: a ladder line built from two wires of 2 mm diameter, spaced 20 mm apart (center to center), running through air.
\[ \frac{2D}{d} = \frac{2 \times 20}{2} = 20 \]
\[ Z_0 = 276\,\log_{10}(20) = 276 \times 1.30103 \approx 359.08\ \Omega \]
Result: the characteristic impedance is approximately 359.08 Ω.
How to Use the Calculator
- Enter the Wire diameter in millimeters — the diameter of each of the two identical conductors.
- Enter the Distance between wires in millimeters — measured from the center of one wire to the center of the other.
- Read the Characteristic Impedance in ohms (Ω) in the result field.
The calculator requires \(D > d/2\); otherwise the wires would overlap, and an error message is shown. Any consistent units work, because only the ratio \(D/d\) matters — the millimeter labels are just a convenience.
What Is Parallel Wire Impedance?
A parallel wire line (also called a two-wire line, twin-lead, or ladder line) is a balanced transmission line made of two parallel conductors held at a fixed spacing. It is one of the oldest and simplest transmission line geometries, still widely used in amateur radio, shortwave receiving antennas, and balanced feeders where low loss matters.
The characteristic impedance \(Z_0\) describes how voltage and current relate to each other as a wave travels along the line. When the line is terminated in a load equal to \(Z_0\), all the transmitted power is absorbed by the load and no reflection occurs. If the load impedance differs from \(Z_0\), part of the signal reflects back toward the source, creating standing waves, higher losses, and potential damage to transmitters. Matching the line, the antenna, and the equipment to the same impedance is therefore essential for maximum power transfer.
Notice what the formula tells you physically: the impedance depends only on the geometry — the ratio of spacing to wire diameter. Widening the spacing increases the impedance; using thicker wires decreases it. A typical 300 Ω twin-lead, for example, achieves its impedance purely through a particular choice of \(D/d\) (plus a modest reduction from its plastic dielectric). Because the conductors must be small compared to the wavelength, the formula is accurate when the spacing is well under about 1/10 of the wavelength of the signal being carried.
Assumptions, Limitations and Practical Tips
- Air dielectric: the calculator assumes \(\varepsilon_r = 1\). For a line embedded in a homogeneous dielectric (e.g. ribbon cable or insulated twin-lead), divide the result by \(\sqrt{\varepsilon_r}\). Real twin-lead has only partial dielectric filling, so the true impedance lies between the air value and the fully filled value.
- Thin-wire approximation: the formula is derived for \(D \gg d\). It remains reasonably accurate down to \(D \approx 2d\), but the error grows when the wires nearly touch.
- Negligible skin-effect corrections: conductor loss and internal inductance are ignored; \(Z_0\) is treated as purely resistive (real), which is an excellent approximation at radio frequencies.
- Uniform geometry: the line must be straight and maintain constant spacing. Bends, twists, spacers, or nearby objects disturb the impedance and can unbalance the line.
- Balanced operation: a two-wire line should be fed symmetrically. Connecting it directly to unbalanced coaxial cable requires a balun.
If you are working with other transmission line geometries, try the Microstrip PCB Impedance or Stripline Impedance calculators, or the Stripline PCB Differential Impedance calculator for paired traces on a circuit board. To match the line to a specific load, see the PI Network Impedance Matching calculator. For related wire properties, the Straight Wire Inductance and Wire Gauge calculators may also be useful.
Frequently Asked Questions
How do I calculate parallel wire impedance by hand?
Take the center-to-center spacing \(D\) and the wire diameter \(d\) in any consistent units, then apply \(Z_0 = 276\,\log_{10}(2D/d)\) ohms. For example, with 1 mm wires spaced 10 mm apart: \(2D/d = 20\), so \(Z_0 = 276 \times 1.301 \approx 359\) Ω. A scientific calculator with a log button is all you need; the ratio has no units.
What is a typical or normal impedance for a two-wire line?
Common values range from about 70 Ω (very close spacing with thick wires) up to 600 Ω (widely spaced ladder line). Commercial 300 Ω twin-lead and 450 Ω window line are the most familiar examples. Amateur radio open-wire feeders are often built in the 450–600 Ω range because higher impedance lines have lower loss per unit length.
Why does 276 appear in the formula?
The constant comes from \(\sqrt{\mu_0 / \varepsilon_0} \cdot \pi^{-1}\), which evaluates to about 120 Ω for air. Since \(\log_{10}(x) = \ln(x)/\ln(10)\) and \(\ln(10) \approx 2.303\), you get \(120 \times \ln(2D/d) \approx 276\,\log_{10}(2D/d)\). Both forms are identical; the 276 version is traditional in radio engineering.
How do I account for wire insulation or a plastic dielectric?
The formula assumes air between the wires. If the space is completely filled by an insulator with relative permittivity \(\varepsilon_r\), divide the result by \(\sqrt{\varepsilon_r}\). Real twin-lead is only partly filled with plastic, so the actual impedance falls somewhere between the air value and the fully filled value — typically 10–20% below the air calculation.
Can I use inches or other units instead of millimeters?
Yes. The result depends only on the ratio \(D/d\), so any consistent units give the same answer — inches, millimeters, or even both converted the same way. Just never mix units between the two inputs (e.g., spacing in inches with diameter in mm), or the ratio and the impedance will be wrong.
What spacing do I need for a specific impedance, like 450 Ω?
Rearrange the formula: \(D = (d/2) \cdot 10^{Z_0/276}\). For 450 Ω with 2 mm wire, \(10^{450/276} \approx 42.3\), so \(D \approx 42\) mm center to center. If the wires are too far apart for your supports, thicker wire brings the spacing down — doubling the diameter halves the required spacing.
Does wire spacing affect SWR or how much power the line can carry?
Spacing sets the characteristic impedance, which determines matching and therefore SWR — but only if the load doesn't match. Closer spacing lowers impedance and also lowers the voltage between wires for a given power, allowing more power before arc-over. Very close spacing raises capacitance and loss. Keep spacing under roughly 1/10 of the wavelength for the formula to stay accurate.