I'm working on a 144 MHz antenna using a gamma match. The best result (resonance 50+j0 at 144.400 MHz) I reached is a 10 mm diameter gamma rod, 15 cm long and a capacity of 10 pF.

I thought to obtain the 10 pF with a 4 mm rod inside the gamma rod (internal diameter 8 mm). The dielectric is air.

Since I will use 1 kW I wonder if this configuration (dielectric air, spacing 2 mm) is sufficient for the power I'll use or I have to find a different configuration (having more spacing at the dielectric).

More in general, how can I esteem the maximum power of a gamma match with air dielectric? Is there a rule to decide the minimum better dielectric spacing in function of the power to use?


1 Answer 1


So at 1 kW for a truly resistive impedance of 50 Ω, the voltage would be

$$ V = \sqrt{PR} = \sqrt{1000 \cdot 50} = 223.6\,\mathrm{V} $$

So now you need to establish the minimum gap to avoid arcing for 223.6 volts. An example calculator online would be here.

But at 223.6 volts, I believe your 4 mm rod inside an internal-dia-8-mm would be an air gap of 2 mm, which should be sufficient. Then again, maybe not — in case of humidity in the air, which may change the dielectric constant of the air.

  • $\begingroup$ dielectric constant is a parameter NOT having directly connection to high voltage endurance. Agree, in wet condition air has lower endurance, but is is not dielectric constant. Sorry, i had learn physics not in English ;) $\endgroup$
    – Jacek Cz
    Jul 5, 2016 at 14:18

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