Space charge width pn junction

[MorrowUoN]

Homework Statement

Calculate the width of the space charge region in a pn junction when a reverse biased voltage is applied. Consider a silicon pn junction at T=300K with doping concentrations of N_a=10^16 cm^-3 and N_d=10^15 cm-3. Assume that N_i=1.5*10^10 cm-3 and V_R=5V

Homework Equations

W=((2E_f(V_bi+V_r)/e)((N_a+N_d)/((N_a*N_d))))^1/2

The Attempt at a Solution

I correctly calculated the in built potential to be 0.635V however, I do not know how to calculate the fermi energy

 

[phyzguy]

Check that formula again. It does not involve the Fermi energy.

 

[MorrowUoN]

Check that formula again. It does not involve the Fermi energy.

What is the value of epsilon in the formula attached?

 

[marcusl]

It's the permittivity of the material.

 

[paranormal]

As a scientist, it is important to have a solid understanding of the concepts and equations involved in your calculations. The Fermi energy, denoted by Ef, is a measure of the highest energy state occupied by electrons at absolute zero temperature. It is typically used to describe the energy level at which the probability of finding an electron is 50%. In the context of a pn junction, the Fermi energy is used to determine the width of the space charge region.

To calculate the Fermi energy, you will need to use the equation: Ef = kTln(N/Ni), where k is the Boltzmann constant, T is the temperature in Kelvin, N is the number of dopant atoms, and Ni is the intrinsic carrier concentration. In this case, N = Na + Nd, so you will need to substitute the given values for Na, Nd, and Ni into the equation to solve for Ef.

Once you have calculated the Fermi energy, you can then use the equation W = ((2Ef(Vbi+Vr)/e)((Na+Nd)/((Na*Nd))))^1/2 to determine the width of the space charge region. Make sure to double check your calculations and units to ensure accuracy.