How Coaxial Cable Diameter Ratios Set 50-Ohm Impedance

The characteristic impedance of a coaxial cable in radio frequency (RF) electronics is fundamentally determined by its physical geometry and the insulating material inside it. Specifically, the ratio between the outer shield's inner diameter and the center conductor's outer diameter dictates the balance of capacitance and inductance along the transmission line. To achieve the industry-standard 50-ohm impedance, engineers calculate a precise diameter ratio combined with the dielectric constant of the insulator, ensuring optimal power handling and minimal signal Attenuation. This article explains the physics, mathematical formulas, and practical engineering choices behind setting the 50-ohm impedance using coaxial diameter ratios.

The Physics of Characteristic Impedance

Characteristic impedance (\(Z_0\)) is not a measure of DC resistance, which changes with cable length. Instead, it is the opposition a transmission line presents to an alternating current (RF signal) as it propagates down the line. This impedance is determined by the ratio of the cable's inductance per unit length (\(L\)) to its capacitance per unit length (\(C\)), expressed as:

\(Z_0 = \sqrt{\frac{L}{C}}\)

Both inductance and capacitance are directly controlled by the physical dimensions of the conductors:

Capacitance (\(C\)): The inner conductor and outer shield act like the plates of a cylindrical capacitor. Bringing them closer together (decreasing the diameter ratio) increases the capacitance, which lowers the impedance.
Inductance (\(L\)): The magnetic field generated by the current exists in the space between the inner conductor and the outer shield. Moving the conductors further apart (increasing the diameter ratio) gives the magnetic field more space to form, increasing inductance and raising the impedance.

Because the physical spacing affects both \(L\) and \(C\) simultaneously, the characteristic impedance becomes strictly dependent on the ratio of the diameters rather than the absolute size of the cable.

The Coaxial Impedance Formula

To calculate the exact impedance of a coaxial cable, RF engineers use the following logarithmic formula:

\(Z_0 = \frac{138}{\sqrt{\varepsilon_r}} \times \log_{10}\left(\frac{D}{d}\right)\)

Where:

\(Z_0\) is the characteristic impedance in ohms (\(\Omega\)).
\(D\) is the inside diameter of the outer shield.
\(d\) is the outside diameter of the inner conductor.
\(\varepsilon_r\) is the relative permittivity (dielectric constant) of the insulating material between the conductors.

This formula proves that as long as the insulator material (\(\varepsilon_r\)) remains the same, a massive rigid coaxial line used in a cell tower and a microscopic coax used inside a smartphone will both have a 50-ohm impedance if their \(D/d\) ratio is identical.

Calculating the Ratio for 50 Ohms

The exact ratio required to hit 50 ohms changes depending on the dielectric material used to separate the inner conductor from the outer shield. Different materials alter the speed of the RF wave and the capacitance between the conductors.

1. Air Dielectric (\(\varepsilon_r \approx 1.0\))

In high-power transmission lines where air is the primary insulator, the dielectric constant is 1.0. To find the required ratio for 50 ohms:

\(50 = \frac{138}{\sqrt{1}} \times \log_{10}\left(\frac{D}{d}\right)\)

\(\frac{50}{138} = 0.3623 = \log_{10}\left(\frac{D}{d}\right)\)

\(\frac{D}{d} = 10^{0.3623} \approx 2.3\)

For an air-dielectric coaxial cable, the inner surface of the outer shield must be 2.3 times larger than the outer surface of the center conductor to create a 50-ohm environment.

2. Solid Polyethylene (PE) Dielectric (\(\varepsilon_r \approx 2.25\))

Standard cables like RG-58 or RG-213 use solid polyethylene, which has a dielectric constant of about 2.25. Because the plastic increases capacitance, the conductors must be spaced further apart to raise the impedance back up to 50 ohms.

\(50 = \frac{138}{\sqrt{2.25}} \times \log_{10}\left(\frac{D}{d}\right)\)

\(50 = \frac{138}{1.5} \times \log_{10}\left(\frac{D}{d}\right)\)

\(50 = 92 \times \log_{10}\left(\frac{D}{d}\right)\)

\(\frac{50}{92} = 0.5435 = \log_{10}\left(\frac{D}{d}\right)\)

\(\frac{D}{d} = 10^{0.5435} \approx 3.5\)

For solid PE cables, the outer shield diameter must be 3.5 times larger than the inner conductor.

3. Solid PTFE (Teflon) Dielectric (\(\varepsilon_r \approx 2.1\))

Using the same math for PTFE, a common material in high-frequency, low-loss cables, the required \(D/d\) ratio comes out to approximately 3.35.

Why 50 Ohms is the RF Standard

The 50-ohm value is not arbitrary; it is a calculated engineering compromise derived directly from testing different \(D/d\) ratios in air-filled coaxial lines during the 1930s:

Maximum Power Handling (~30 Ohms): If the center conductor is made thicker (a \(D/d\) ratio of 1.65 in air), the cable can handle maximum voltage and power without arcing. This occurs at 30 ohms.
Lowest Signal Loss (~77 Ohms): If the center conductor is made thinner (a \(D/d\) ratio of 3.59 in air), the cable experiences the lowest possible signal attenuation over distance. This occurs at 77 ohms (the reason 75-ohm cable is used for low-power, long-distance video and cable TV).

For general RF electronics, which require both transmitting high power and receiving weak signals with low loss, the arithmetic mean between 30 ohms and 77 ohms is 53.5 ohms. Early engineers rounded this to 50 ohms, locking in the standard \(D/d\) ratios used in RF manufacturing today.