Built-in Voltage

In pn-junction there is a high concentration of holes on the p-side and a high concentration of electrons on the n-side. The mobile charge carriers diffuse from high concentration to low concentration so the holes diffuse to the n-side and the electrons diffuse to the p-side. This establishes a charge imbalance with positive charge on the n-side and negative charge on the p-side resulting in an electric field that points from n to p. This electric field pushes the holes back to the p-side and the electrons back to the n-side. At zero applied voltage, the diffusion current is cancelled by the drift current caused by the electric field. There is a voltage drop across the junction in this case which can be calculated by integrating the electric field across the junction. This voltage is called the built-in voltage, $V_{bi}$. In the band diagram there is an offset of $eV_{bi}$ between the valence band on the two sides of the junction. If both sides of the junction are heavily doped, the Fermi energy is close to the valence band on the p-side and close to the conduction band on the n-side. This makes $eV_{bi}$ nearly equal to the band gap energy $E_g$.

$N_A$ = 

 1/cm³ 

$N_D$ = 

 1/cm³  $E_g$ =  eV

$N_v(300)$ = 

 1/cm³

$N_c(300)$ = 

 1/cm³

$\epsilon_r$ = 

$T$ = 

K
  

$E_g=$  eV  $W=$  μm  $x_p=$  μm  $x_n=$  μm  $V_{bi}=$  V  $n_i=$  1/cm³

Band diagram

 [eV]

x [μm]

The energy associated with the built-in voltage $eV_{bi}$ (orange) is the band gap energy $E_g$ (green) minus the distance between the conduction band and the Fermi energy on the n-side (pink) minus the difference between the Fermi energy and the valence band on the p-side (blue).

$$eV_{bi} = E_g - (E_c - E_f)|_n - (E_f-E_v)|_p.$$

Far from the junction on the p-side, the hole concentration is $p= N_v\exp\left(\frac{E_v-E_f}{k_BT}\right)\approx N_A$. This can be solved for the energy difference between the valence band and the Fermi energy,

$$(E_f-E_v)|_p = k_BT\ln\left(\frac{N_v}{N_A}\right).$$

Similarly, far from the junction on the n-side, the hole concentration is $n= N_c\exp\left(\frac{E_f-E_c}{k_BT}\right)\approx N_D$. This can be solved for the energy difference between the conduction band and the Fermi energy,

$$(E_c-E_f)|_n = k_BT\ln\left(\frac{N_c}{N_D}\right).$$

The built-in voltage can then be expresed as,

$$eV_{bi} = E_g - k_BT\ln\left(\frac{N_c}{N_D}\right) - k_BT\ln\left(\frac{N_v}{N_A}\right).$$

The expression for the intrinsic carrier density $n_i$ is,

$$n_i^2=N_cN_v\exp\left(\frac{-E_g}{k_BT}\right).$$

Solving this for $E_g$ yields,

$$E_g = k_BT\ln\left(\frac{N_cN_v}{n_i^2}\right).$$

The built-in voltage can then be written as,

$$eV_{bi} = k_BT\ln\left(\frac{N_cN_v}{n_i^2}\right) - k_BT\ln\left(\frac{N_c}{N_D}\right) - k_BT\ln\left(\frac{N_v}{N_A}\right).$$

Using the properties of the logarithm, this can be simplied to,

$$\boxed{\quad eV_{bi} = k_BT\ln\left(\frac{N_DN_A}{n_i^2}\right).\quad}$$