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Re: magnetrons as diodes
Original poster: "Gerry Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
lamda of 2.45G is 12.2cm so 6cm is 1/2 lamba. Wouldnt this squared
be 1/4 square wavelength? or what am I missing?
Also, 6cm squared is 36 sq cm so there is a factor of 2 that I dont
understand. Maybe this is an rms value??
Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>
Just to get anyone interested in the ballpark: 2.45 GHz is 12.2
cm. The effective aperture of a 6 cm long dipole is about 1/8th
square wavelength, or roughly 18 square centimeters. So a dipole
will intercept about 40 mW at the ANSI exposure limit. Into 72 ohms
(the impedance of the dipole), this works out to about 1.6-2.7 volts.
A smaller probe intercepts less power. Make it 1 cm, and you
intercept 1/36th the power (about 1 mW) and the voltage is
correspondingly smaller (a bit less than a volt).
A variety of diodes will work. What you want is a zero bias planar
schottky or an old germanium (1n34, for instance). But, almost
anything will work (1n914, 1n5711, etc.)