The built-in potential of a p-n junction _______.

Concept:

Energy Band diagram of P-N diode:

E_Fi = intrinsic Fermi level

E_FP = Fermi level of P-type

E_FN = fermi level of n-types

\({E_1} = {E_{FN}} - {E_{Fi}} = KT\ln \left( {\frac{{{N_D}}}{{{n_i}}}} \right)\)

\({E_2} = {E_{Fi}} - {E_{FP}} = KT\ln \left( {\frac{{{N_A}}}{{{n_i}}}} \right)\)

Potential Barrier or Built in potential E₀ = E₁ + E₂

\({E_0} = KT\ln \left( {\frac{{{N_D}}}{{{n_i}}}} \right) + KT\ln \left( {\frac{{{N_A}}}{{{n_i}}}} \right) \Rightarrow \)

\({E_0} = KT\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)eV\)

\({V_{bi}} = \frac{{KT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)V\;\;\;\;\;\;\;\;\;\frac{{KT}}{q} = {V_T}\)   ----(1)

So we can say Built-in potential is equal to the difference in fermi level of two sides (E₁ + E₂) and on increasing doping, V_bi will increase

Since,

 \(n_i^2 = {A_0}{T^3}\;{e^{ - E_G/KT}}\) 

So if we, \(T \uparrow \;\;\;{V_T} \uparrow \;\;\;\;\;\;\;\)
So, the built-in potential of a p-n junction depends on both temperature and doping concentration.