Why is 50 ohms the standard impedance for coaxial cables?

It is a compromise between power handling and signal loss. In the 1920s, Bell Labs engineers calculated that for an air-dielectric coaxial cable, maximum power handling (highest breakdown voltage) occurs at an impedance of 30 ohms. Minimum attenuation (lowest signal loss) occurs at 77 ohms. They chose 50 ohms as the exact mathematical compromise between those two extremes. For television systems where power handling doesn't matter but signal loss over long distances is critical, the industry uses 75-ohm cable.

Why do cables lose power (attenuation) at high frequencies?

Two reasons: Conductor loss and Dielectric loss. Because of the 'skin effect,' high-frequency RF currents are forced to the extreme outer edge of the center conductor. This drastically reduces the usable cross-section of the copper, increasing its resistance and generating heat. Second, as the high-frequency electric field rapidly flips back and forth, it causes friction within the molecules of the Teflon or foam dielectric, generating even more heat. Higher frequencies mean more friction and less usable copper, resulting in higher attenuation.

What happens if I push a frequency that is too high into a cable?

It will trigger higher-order waveguide modes (usually the TE11 mode). Coaxial cable is designed to propagate energy in the TEM (Transverse Electromagnetic) mode, where the electric and magnetic fields are perfectly perpendicular to the direction of travel. If the wavelength of the signal becomes smaller than the circumference of the cable's dielectric space, the wave starts bouncing diagonally down the cable like it would in a waveguide. This ruins the signal's phase and completely destroys the 50-ohm characteristic impedance.

Passive Components

Coaxial Cable

Before coaxial cable, telegraph and early radio signals were sent over "open wire" lines. While cheap, open wires act like giant antennas, radiating their energy away and absorbing massive amounts of interference from lightning and other radios. Coaxial cable solved this via geometric shielding. By wrapping a solid center conductor in a dielectric, and then wrapping that dielectric in a braided metal outer shield, the electromagnetic wave is entirely trapped inside the cable. The wave propagates in the Transverse Electromagnetic (TEM) mode, meaning its electric field reaches radially from the center pin to the shield, and its magnetic field circles the pin. Because the outer shield is tied to ground, no electric fields can escape, and no external noise can get in. However, the exact ratio of the inner pin diameter to the outer shield diameter strictly dictates the cable's characteristic impedance.

Category: Passive Components

Propagation Mode: Transverse Electromagnetic (TEM)

Primary Advantage: Perfect field containment / shielding

Common Coaxial Cable Types

Cable Type	Impedance	Typical Dielectric	Cut-off Frequency	Primary Application
RG-58	50 Ω	Solid PE	~1 GHz	Low-frequency lab testing, Ham radio
RG-6	75 Ω	Foam PE	~3 GHz	Cable television, satellite receivers
LMR-400	50 Ω	Gas-Injected Foam	~6 GHz	Cell tower base station antennas
Semi-Rigid (.085")	50 Ω	Solid PTFE (Teflon)	~60 GHz	Internal RF module routing, high-frequency

Characteristic Impedance (Z₀):
Z₀ = [ 138 / √ε_r ] · log₁₀( D / d )
Where ε_r is the dielectric constant of the insulator, D is the inner diameter of the outer shield, and d is the outer diameter of the center conductor. To change a cable from 50Ω to 75Ω, you simply make the center pin thinner (decreasing d).

Cut-off Frequency Limit (f_c):
f_c ≈ [ 2 · c ] / [ π · √ε_r · (D + d) ]
If the RF frequency exceeds f_c, the cable stops supporting the TEM mode and begins acting like a multimode waveguide, causing massive signal distortion. To push higher frequencies (like 60 GHz radar), the physical diameter of the cable must become microscopically small.

Common Questions

Frequently Asked Questions

Can I use 75-ohm TV cable for my 50-ohm transmitter?

Technically yes, but it will cause a severe impedance mismatch. Connecting a 50-ohm radio to a 75-ohm cable generates a VSWR of 1.5:1. This means roughly 4% of your transmit power will instantly reflect off the cable connection and travel back into your amplifier, generating heat. For low-power systems it might survive, but for high-power transmitters, the reflections can destroy the final amplifier stage.

Why are microwave cables so stiff?

At high microwave frequencies (above 10 GHz), the standard braided wire used for the outer shield of cheap cables becomes "leaky." The microscopic gaps in the woven braid allow high-frequency RF to escape. To prevent this, precision microwave cables use "semi-rigid" construction, replacing the flexible braid with a solid, continuous copper or aluminum tube. This provides 100% shielding effectiveness but makes the cable stiff and difficult to bend.

Why do long cables need equalizers?

Coaxial cable attenuation is not flat; it increases with frequency. If you send a wideband signal (containing both 100 MHz and 1 GHz components) down a 100-foot cable, the 1 GHz portion will lose 10 dB of power, while the 100 MHz portion will only lose 2 dB. The signal arrives severely tilted. Engineers install an "equalizer" at the receiver end, which artificially attenuates the low frequencies to match the degraded high frequencies, flattening the response back out.

Related Terms

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Passive Design

Coaxial Impedance Calculator

Input your physical cable dimensions and the dielectric constant of your insulator. Instantly calculate the characteristic impedance (Z0), the theoretical attenuation per 100 feet, and the absolute cut-off frequency where TEM mode fails.

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