Calculating Donor Concentration in Si at 300K | Semiconductor Homework Q1 & Q2

[th3bl0b]

Homework Statement

Q1
A Si sample is doped with 10^16 per cm^-3 boron atoms and a certain
number of shallow donors. The fermi level (Ef) is 0.36 eV above Ei
(intrinsic energy level) at 300K. What is the donor concentration Nd?

For Si at 300K ni(intrinsic carrier concentration) = 1.5 x 10^10 per
cm cube

Q2
A Si sample contains 10^16 per cm^-3 In(indium) acceptor atoms and
a certain number of shallow donors. The In (indium) acceptor level is
0.16 eV above Ev(Valence band edge), and Ef is 0.26eV above Ev at
300K. How many in atoms in cm per cube are unionized (i.e. neutral)?

For Si at 300K ni(intrinsic carrier concentration) = 1.5 x 10^10 per
cm cube

Homework Equations

n(o) x p(o) = ni^2

n(o) = ni x e((Ef-Ei))/KT)

p(o) = ni x e((Ei-Ef)/KT)

The Attempt at a Solution

I haven't been able to make a good decision on how to start...

If you have any suggestions, let me know. I'm open to everything!

Thanks

 

[KataKoniK]

for sharing this interesting problem! I would approach these questions by using the equations given and the known values for the intrinsic carrier concentration (ni) of silicon at 300K.

For Q1, we can use the equation n(o) x p(o) = ni^2, where n(o) is the concentration of free electrons and p(o) is the concentration of holes. Since the sample is doped with boron atoms, the concentration of acceptors (Na) is equal to the concentration of holes (p). We can then rearrange the equation to solve for the donor concentration (Nd).

n(o) = ni x e((Ef-Ei))/KT)

p(o) = Nd

Substituting in the given values, we get:

Ni x Nd = ni^2

10^16 x Nd = (1.5 x 10^10)^2

Nd = (1.5 x 10^10)^2 / 10^16

Nd = 2.25 x 10^-3 per cm^-3

Therefore, the donor concentration Nd is 2.25 x 10^-3 per cm^-3.

For Q2, we can use the same equation n(o) x p(o) = ni^2, but this time we are given the concentration of acceptors (Na) and we need to find the concentration of unionized donors (n). We can use the equation for p(o) and substitute in the given values to solve for n.

p(o) = ni x e((Ei-Ef)/KT)

Na = p(o)

Substituting in the given values, we get:

Ni x Na = ni x e((Ei-Ef)/KT)

10^16 x Na = (1.5 x 10^10) x e((0.16-0.26)/KT)

Na = (1.5 x 10^10) x e(-0.1/KT)

Since we don't have the value for temperature (T), we cannot solve for Na. However, we can use the equation n(o) = ni x e((Ef-Ei))/KT) to find the concentration of unionized donors (n).

n(o) = ni x e((Ef-Ei))/KT)

n(o) = (1.5 x 10^10) x e((0.26-0.36)/KT)

n(o)