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?


1 Answer 1


Built in voltage and cut in voltage are not the same thing at all.

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.

Just keep in mind the huge difference:

  • \$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}\$.


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