I have always had the idea that the forward voltage drop in a semiconductor diode was related in a simple way to the bandgap energies in the semiconductor. However this is apparently not the case:

• germanium has a bandgap of 0.66 eV, but germanium diodes have a typical forward drop of 0.2 V

• silicon has a bandgap of 1.12 eV, but silicon diodes have a typical drop of 0.6 V

I'm aware of the Shockley equation describing the current in a diode as a function of diode voltage drop $V_D$ and temperature $T$, $$ I(V_D) =I_0 \left( e^{eV_D/kT} -1 \right) \approx I_0 e^{eV_D/kT} $$ where the scale current goes something like $$ I_0 = A \,T^3 e^{ -{E_\text{gap}}/{kT} } $$ and the constant $A$ depends on the geometry of the diode, the degree of doping, the charge mobility, and probably some other stuff, too.

I recognize that there's some arbitrariness to the approximation of a "turn-on voltage": the exponential grows so fast that if your choice for the current threshold differs from mine by a factor of a thousand, we'll only disagree about the turn-on voltage by about a couple tenths of a volt. However, I've had for years the impression that there's something fundamental about silicon that gives silicon diodes a forward drop of roughly 0.6V. Is that the case? Or is there some constellation of design choices that conspire to give the same drop both across most p-n diodes and across the p-n junctions of bipolar transistors?

I was motivated to ask this question by a similar question about forward voltage drops in LEDs. I was expecting to answer that question with some data from a student using LED turn-on voltage and the wavelength of light to measure the Planck constant. However those data are a lot more complicated than I expected: in fact, most of my LEDs apparently emit multiple wavelength components with comparable strength, and there doesn't seem to be much correlation the turn-on voltage between and the most prominent color in the LED spectrum. I don't seem to be able to say much more than "LEDs have turn-on voltages between two and three volts."

I have read a little bit about band-bending diagrams on Wikipedia, which suggest the potential barrier $\phi_B$ across an interface is different from the band gap, but I can't figure out why.

I have always had the idea that the forward voltage drop in a semiconductor diode was related in a simple way to the bandgap energies in the semiconductor. However this is apparently not the case:

• germanium has a bandgap of 0.66 eV, but germanium diodes have a typical forward drop of 0.2 V

• silicon has a bandgap of 1.12 eV, but silicon diodes have a typical drop of 0.6 V

I'm aware of the Shockley equation describing the current in a diode as a function of diode voltage drop $V_D$ and temperature $T$, $$ I(V_D) =I_0 \left( e^{eV_D/kT} -1 \right) \approx I_0 e^{eV_D/kT} $$ where the scale current goes something like $$ I_0 = A \,T^3 e^{ -{E_\text{gap}}/{kT} } $$ and the constant $A$ depends on the geometry of the diode, the degree of doping, the charge mobility, and probably some other stuff, too.

I recognize that there's some arbitrariness to the approximation of a "turn-on voltage": the exponential grows so fast that if your choice for the current threshold differs from mine by a factor of a thousand, we'll only disagree about the turn-on voltage by about a couple tenths of a volt. However, I've had for years the impression that there's something fundamental about silicon that gives silicon diodes a forward drop of roughly 0.6V. Is that the case? Or is there some constellation of design choices that conspire to give the same drop both across most p-n diodes and across the p-n junctions of bipolar transistors?

I was motivated to ask this question by a similar question about forward voltage drops in LEDs. I was expecting to answer that question with some data from a student using LED turn-on voltage and the wavelength of light to measure the Planck constant. However those data are a lot more complicated than I expected: in fact, most of my LEDs apparently emit multiple wavelength components with comparable strength, and there doesn't seem to be much correlation the turn-on voltage between and the most prominent color in the LED spectrum. I don't seem to be able to say much more than "LEDs have turn-on voltages between two and three volts."

I have read a little bit about band-bending diagrams on Wikipedia, which suggest the potential barrier $\phi_B$ across an interface is different from the band gap, but I can't figure out why.