4 ms·

A PN junction (ie, a diode) only passes current one way, so a sine wave applied across a PN junction gets turned into only half a sine wave, with either the pos

by tlb 8y ago

A PN junction (ie, a diode) only passes current one way, so a sine wave applied across a PN junction gets turned into only half a sine wave, with either the positive or negative half clipped depending on the polarity. The fourier series of a rectified sine wave is mostly the second harmonic. A natural nonlinear junction usually consists of a thin layer of insulating oxide between two pieces of metal. It only conducts when the voltage exceeds some threshold where the electrons tunnel through the insulator. When a sine wave is applied, both the tops and bottoms of the waveform are clipped. Because it's symmetrical around zero, the energy it re-radiates is mostly at the third harmonic.

na85 8y ago
Makes sense, thanks.
DoctorOetker 8y ago
does this in only detect junctions connected to an [explicit or implicit] antenna or does it also detect any junction without the junction needing to be connected to an antenna? my intuition is this method would require the detected junction to be connected to at least an implicit antenna to which the detector can couple, but I could be wrong?
- CamperBob2 8y ago
  Anything that's more or less conductive can be an antenna if the frequency is high enough. That's one reason to design these instruments to operate in the 2.4 GHz ISM band, where the wavelength is around 12 centimeters. Junctions associated with metallic objects over 1 cm in size should be readily detectable.
hatsunearu 8y ago
Another way to think about it: If the voltage-to-current transfer characteristic of whatever you're working with is symmetric about the y axis, then taking the maclaurin series gives you a polynomial with only odd terms. If you apply a sinusoid to that, you get odd multiple harmonics. Working through the same reasoning with an asymmetric transfer characteristic means you get even polynomials as well, which means you get both even and odd harmonics.

by tlb 8y ago

na85 8y ago

Makes sense, thanks.

DoctorOetker 8y ago

does this in only detect junctions connected to an [explicit or implicit] antenna or does it also detect any junction without the junction needing to be connected to an antenna? my intuition is this method would require the detected junction to be connected to at least an implicit antenna to which the detector can couple, but I could be wrong?

CamperBob2 8y ago
Anything that's more or less conductive can be an antenna if the frequency is high enough. That's one reason to design these instruments to operate in the 2.4 GHz ISM band, where the wavelength is around 12 centimeters. Junctions associated with metallic objects over 1 cm in size should be readily detectable.

hatsunearu 8y ago

Another way to think about it: If the voltage-to-current transfer characteristic of whatever you're working with is symmetric about the y axis, then taking the maclaurin series gives you a polynomial with only odd terms. If you apply a sinusoid to that, you get odd multiple harmonics. Working through the same reasoning with an asymmetric transfer characteristic means you get even polynomials as well, which means you get both even and odd harmonics.