The forward voltage of a Si diode will drop by about 2.1 mV/°C (negative temperature coefficient, NTC). For a Ge diode this NTC is a bit higher at 2.5 mV/°C.
The effect is almost perfectly linear, so if you want you can make the diode a temperature sensor by supplying it with a constant current.
Reverse current increases exponentially with rising temperature. For a 1N4148 it increases tenfold for every 20°C temperature rise.
Please fasten your seatbelts - our flight today will fly over a bit of the math to get from power (dBm) to field strength (V/m).
The most pragmatic way to think about the "strength" of any RF signal is in terms of how much power it can deliver to the terminals of a practical receiving antenna.
Ultimately the laws of physics dictate that for a given receiving antenna, at a given frequency, there is a simple linear relationship between the square root of the power at the receive antenna terminals and the strength (in Volts/meter) of the RF field:
\$Field\ Strength\ = k\cdot\sqrt {P_{rx}}\ \ \ \ \cdots (1)\$
You can think of the factor \$k\$ as the "conversion factor" of that antenna at that specific frequency.
That's it - once you know your antenna conversion factor \$k\$ and the received power (in Watt), you simply calculate the field strength using the above formula.
So, to answer your questions: 1. How does it work? Primarily due to the inverse square law of radiators and aperture. 2. Yes, it's possible. 3. Once you know k for your antenna, use Equation (1) above to calculate RF field strength - see example below.
Example:
The \$k\$ factor for an ideal half-wave dipole, which has a gain of 1.643 (2.14dB) can be found in literature - \$k\$ is essentially the square root of (\$120\pi\ \div \$Antenna Effective Area).
\$k = \sqrt{120\pi} \cdot \sqrt{\frac{4\pi }{1.643\cdot\lambda^2}} \approx\frac{53.7}{\lambda}\$
Let's say we are receiving 0dBm (.001W) at the terminals of our half-wave dipole at 100 MHz. (\$\lambda\$=3m) Using Eq. (1) we see
Field Strength = \$\frac{53.7}{\lambda} \cdot \sqrt{.001\ }\$ = 0.566 V/m
TLDR: Convert dBm to Watt, get square root, multiply by k.
Best Answer
Germanium diodes are not obsolete, they are better than silicon ones and have better temperature stability and parameters. The function that you are looking for is a datasheet parameter. You can easily measure it, its a function of current or voltage change based on temperature. Its called a diode temperature characteristic.
The characteristic will change in quadrants 1 and 3 based on the temperature, meaning the voltage and current through the diode will change with the temperature in a border.
The characteristic and the original image.
Formulas and explanations.