Question

Consider a GaAs p-n junction diode. The p-type acceptor is Zn, with a dopant concentration of 2 x 10^17 cm^-3 and the n-tpye donor is Si, with a dopant concentration of 5 x 10^16 cm^-3. Working at temp= 300 K.

a) Calculate the Fermi Level in the p-type material. Assume that the hole density is equal to the density of the p-dopant atoms.

b) Performing the same calculation for the n-type material gives a fermi level of 1.363 eV. What is the zero-bias built- in potential at the junction?

c) Calculate the width of the depletion zone for this case

Answer 1

PN Junction Diode

P-Type dopant/Acceptor: Zn: No. of electrons: 30. Accepts 2 electrons per atom.

No. of acceptor atoms: N_A=2*10¹⁷ cm^-3

N-Type Dopant/Donor: Si: No of electrons 14. Donates 2 electrons per atom.

No. of Donor atoms: N_D=2*10¹⁶ cm^-3

Intrinsic charge carrier density at T=300K is approximately taken to be:

n_i = 1.5*10¹⁰ cm^-3

T=300K

1. Fermi Level on P-side: (E_ip is the intrinsic fermi level on p side before poining of n and p sides to form a junction.)

and,

Fermi-level on n side: (E_in is the intrinsic fermi level on p side before poining of n and p sides to form a junction.)

2. Zero bias Built-in Potential V₀: (Potential od the pn junction when no external bias is applied):

Alternatively, you can find this by just adding above Fermi level values found:

or,

(I don't get it when you say Fermi level to be 1.363 eV when cals are done for n-side, You don't need it as it can already be calculated from given quantities.)

3. Width of the depletion zone is given by the formula:

Substitute the given values for all :

Relative Permittivity of GaAs

Thus permittivity:

Other values are already mentioned above.

Upon calculating, you will get: