# Re: magnetrons as diodes

```Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Jim,

```
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??
```
Gerry R.

```
```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.)
```

```
```

```