Solving Quasi Fermi Levels: EHP Generation & Conductivity Change

[EE_or_Bust]

Can someone please help with this problem. I've been looking at it for a week!
I can't figure out what equation to use.

A si sample with 10^15/cm^3 donors is uniformly optically excited at room temperature such that 10^19 EHPs (Electron Hole Pairs) are generated per second. Find the separation of the quasi Fermi level and the change of conductivity upon shinig the light. Electron and hole lifetimes are both 10μs. =12. {Hint: to find Fn-Fp , use the relation relating (np) and nd (ni, Fn, Fp). μn=1300[cm^2/ V-s].

I know that Fn and Fp are the quasi fermi electron and hole levels. And I know the common quasi formulas are
n=(ni)(e^-(ei-Fn)/KT)
p=(ni)(e^-(-Fp-Ei)/KT)

How do I use these equations to find the separation of the quasi fermi level?

 

[Valkarie]

First, use the given information to solve for ni. nd = 10^15/cm^3ni = (nd)^(1/2) = 10^7.5/cm^3Next, solve for the quasi-Fermi levels: Fn = Ei - KT ln (nd/ni) Fp = -Ei - KT ln (nd/ni) Substitute in the given information to solve for Fn and Fp: Fn = Ei -kt ln (10^15/10^7.5) = 0.026V Fp = -Ei -kt ln (10^15/10^7.5) = -0.026V The separation of the quasi Fermi level is then the difference between Fn and Fp, which is 0.052V. To find the change in conductivity upon shining the light, use the relation Δσ=μn(nd)(Fn-Fp)Substitute in the given information to solve for Δσ: Δσ = μn (10^15/cm^3) (0.052V) = 6.6 x 10^-13 (Ω-cm)^-1 Therefore, the separation of the quasi Fermi level is 0.052V and the change in conductivity upon shining the light is 6.6 x 10^-13 (Ω-cm)^-1.

 

[piercas]

The separation of the quasi Fermi level can be found by using the relation between the electron and hole concentrations (n and p) and the intrinsic carrier concentration (ni). This relation is given by the equation:

np = ni^2 * e^( (Fn-Fp)/kT)

Where k is the Boltzmann constant and T is the temperature. This equation is known as the law of mass action and it describes the balance between the generation and recombination of electron-hole pairs in a semiconductor material.

To solve for the separation of the quasi Fermi level, we can rearrange this equation to:

Fn-Fp = kT * ln(np/ni^2)

Now, we can plug in the given values for n, p, and ni:

Fn-Fp = (8.62 x 10^-5 eV/K) * ln(10^19/10^30)

= (8.62 x 10^-5 eV/K) * ln(10^-11)

= (8.62 x 10^-5 eV/K) * (-25.8)

= -0.00222 eV

Therefore, the separation of the quasi Fermi level is -0.00222 eV.

To find the change in conductivity upon shining light, we can use the equation:

σ = q * μ * (n + p)

Where q is the elementary charge, μ is the carrier mobility, and n and p are the electron and hole concentrations.

We can find the change in conductivity by subtracting the initial conductivity (before shining light) from the final conductivity (after shining light). The initial conductivity can be calculated using the given donor concentration:

σ_initial = q * μ * (n + p)

= (1.6 x 10^-19 C) * (1300 cm^2/V-s) * (10^15/cm^3 + 0/cm^3)

= 2.08 x 10^-4 (Ω*m)^-1

The final conductivity can be calculated by adding the additional electron-hole pairs generated by the light:

σ_final = q * μ * (n + p + Δn + Δp)

= (1.6 x 10^-19 C) * (1300 cm^2/V-s) * (10^15/cm^3 + 10^19/cm^3)

= 2.08 x 10^-4 (Ω*m)^-1