Why is the cut-in voltage assumed to be 0.7 V for a silicon diode and 0.3V for a germanium diode?

Question

any help is very much appreciated. thanks!

Answer 1

When certain crystals form, the electrons are bound to the atoms in the lattice and cannot serve as conductors through the crystal especially at low temperatures. They can be liberated from the atoms in the lattice and set free to conduct through the crystal by thermal energy or an electric field or few other mechanisms.

When the electron is bound to the atoms it is said to be in the valence band (non conducting); when it is free (conducting) it's in the conduction band. The difference in energy between the valence and the conductin band is called the "band gap". For an electron to jump from the very top of the valence band to the very bottom of the conduction band it must be given the entire band gap's amount of energy. It's a quantum jump of energy and electrons cannot exist in the band gap so they need the entire shot of energy to get to the conduction band.

Without going into the mechanics of crystal doping or Fermi levels, just know that the band gap for germanium is much lower than it is for silicon. It basically takes less energy (proportional to electric field or voltage) to kick the crystal into conduction. This translates to a lower "cut-in" voltage for Ge when compared to Si