Question

Consider a p-n junction diode when it is forward biased with \(2\) V.

Which of the following is/are the correct magnitude(s) of the energy difference between quasi Fermi-levels, \(E_{fn}\) in the n-side and \(E_{fp}\) in the p-side?

• A. \(2\) eV
• B. \(1\) eV
• C. \(2\) V
• D. \(1\) V

Hint:

The split between quasi-Fermi levels equals the applied voltage converted to energy: \(E_{fn}-E_{fp}=qV\).

Answer 1

The Correct Option is A

Step 1: Recall what a quasi-Fermi level is.
In a p-n junction at equilibrium there is one common Fermi level, \(E_F\), on both sides. Once the diode is forward biased, the carrier populations move out of equilibrium and we need two separate levels: \(E_{fn}\) for electrons and \(E_{fp}\) for holes.

Step 2: Write the carrier concentrations using quasi-Fermi levels.
The electron and hole densities anywhere in the device can be written as
\[ n=n_i\exp\left(\frac{E_{fn}-E_i}{kT}\right),\qquad p=n_i\exp\left(\frac{E_i-E_{fp}}{kT}\right) \]
Multiplying these two expressions,
\[ np=n_i^2\exp\left(\frac{E_{fn}-E_{fp}}{kT}\right) \]

Step 3: Connect this product to the applied bias.
For a forward biased junction with negligible series resistance, the standard result for the junction (the law of the junction) gives, through the depletion region,
\[ np=n_i^2\exp\left(\frac{qV}{kT}\right) \]
where \(V\) is the applied forward voltage.

Step 4: Equate the two expressions for \(np\).
Comparing the exponents,
\[ \frac{E_{fn}-E_{fp}}{kT}=\frac{qV}{kT} \]
so
\[ E_{fn}-E_{fp}=qV \]
The split between the two quasi-Fermi levels, measured in energy, equals the applied voltage measured in electron-volts.

Step 5: Put in the given voltage.
Here \(V=2\) V, so
\[ |E_{fn}-E_{fp}|=q(2\text{ V})=2\text{ eV} \]

Step 6: Check the other options.

(B) 1 eV: This would need \(V=1\) V, but the diode is biased at \(2\) V, so this is wrong.

(C) 2 V: This mixes up units. The Fermi level split is an energy, so it must be expressed in eV, not V.

(D) 1 V: Wrong for the same unit reason as (C), and also has the wrong number.

Final Answer:
The magnitude of the energy split between the quasi-Fermi levels is
\[ \boxed{2\text{ eV}} \]