# Semiconductor diode barrier potential

I have little confusion regarding barrier potential (or built-in voltage) and cut-in voltage of diode. Are these are same or what is the difference between built-in voltage and cut-in voltage in diode?

Built in voltage is a physical parameter that depends on how the diode is built, for a p-n junction it holds: $$V_{bi}=V_T\ln{\left(\frac{N_AN_D}{n_i^2}\right)}$$ where $V_T=\frac{kT}{q}$ is the "thermal voltage", I don't know how it's called in English, $N_A$ and $N_D$ are the carriers concentrations and $n_i^2=p_0n_0$ is the intrinsic concentration.
Cut in voltage is not a physical parameter at all. It can not be computed in any way, it does not (strictly) depend on the diode$^1$. It's just a voltage over with a great bunch of engineers would say "ok now that diode is conducting". Cut in voltage, symbol $V_\gamma$, is assumed to be equal to 0.7V in most cases, for schottky diodes it can be much lower, something around 0.3V or whatever.
• $V_{bi}$ is a diode physical parameter that can be derived from the theory of operation
• $V_\gamma$ is just the product of a gentlemen's agreement
$^1$Obviously it does depend on how the diode is built but there's no such a relationship as there is for $V_{bi}$.