How Do You Calculate the Built-In Potential of a PN Junction at Zero Bias?

[lam58]

The question asks for a pn junction, calculate the total bending through energy at zero bias of the conduction band edge passing from neutral n-type section to the neutral p-type section.

Additional info:

Band gap of silicon Eg = 1.1 eV
Density of states at conduction band (Nc) = 2.8e19 states/cm^3
Density of states at valence band (Nv) = 1e19 states/cm^3
p-type contains 5e16 holes/cm^3 (p)
n-type contains 1e19 electrons/cm^3 (n)
Boltz constant k = 8.614e-5 eV/K
temperature = 300K

My ans:

qV_bi = Eb - ∂1 - ∂2, where ∂1 = (Ef-Ev) and ∂2 = (Ec-Ef)

if holes p = Nv*exp[-(Ef-Ev)/KT] = Nv*exp[-∂1/KT]

and electrons n = Nc*exp[-(Ec-Ef)/KT] = Nc*exp[-∂2/KT]

then ∂1 = KT*ln[Nv/p]
and ∂2 = KT*ln[Nc/n]

therefore qV_bi = EG - {KT*ln[Nv/p]} - {KT*ln[Nc/n]} = 0.62V

Is this correct?

 

[mark726]

Yes, your calculation is correct. The formula you have used is the correct one for calculating the built-in potential (V_bi) of a pn junction at zero bias. The values you have used for the band gap, density of states, and carrier concentrations are also correct. The only thing to note is that the built-in potential is typically given in units of volts (V), not electron volts (eV), so the correct answer would be V_bi = 0.62 V.