Question:medium
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?
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?
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?
Show Hint
The split between quasi-Fermi levels equals the applied voltage converted to energy: \(E_{fn}-E_{fp}=qV\).
Updated On: Jul 20, 2026
- \(2\) eV
- \(1\) eV
- \(2\) V
- \(1\) V
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Picture what happens under forward bias.
At zero bias, electrons and holes share one Fermi level all through the diode. Push the diode into forward bias and the picture splits into two levels: $E_{fn}$ describing the electrons and $E_{fp}$ describing the holes.
Step 2: Recall the shortcut result for a forward biased diode.
A well known consequence of forward biasing a p-n junction is that the product $np$ inside the depletion layer rises above its equilibrium value $n_i^2$ by a factor $\exp(qV/kT)$:
\[ np = n_i^2\,e^{qV/kT} \]
Step 3: Read this in terms of the two Fermi levels.
Writing $n$ and $p$ with their own quasi-Fermi levels always gives $np = n_i^2\exp\big((E_{fn}-E_{fp})/kT\big)$. Matching the exponents of the two expressions for $np$ tells us directly that
\[ E_{fn}-E_{fp} = qV \]
Step 4: Plug in numbers.
The bias is $V = 2$ V, so the split is $q(2\text{ V}) = 2$ eV. This is an energy, so eV is the natural unit here, not V.
Step 5: Rule out the rest.
$1$ eV is off by a factor of two. $2$ V and $1$ V both quote a voltage where an energy is asked for, so both are wrong even before checking the number.
\[ \boxed{2\text{ eV}} \]
At zero bias, electrons and holes share one Fermi level all through the diode. Push the diode into forward bias and the picture splits into two levels: $E_{fn}$ describing the electrons and $E_{fp}$ describing the holes.
Step 2: Recall the shortcut result for a forward biased diode.
A well known consequence of forward biasing a p-n junction is that the product $np$ inside the depletion layer rises above its equilibrium value $n_i^2$ by a factor $\exp(qV/kT)$:
\[ np = n_i^2\,e^{qV/kT} \]
Step 3: Read this in terms of the two Fermi levels.
Writing $n$ and $p$ with their own quasi-Fermi levels always gives $np = n_i^2\exp\big((E_{fn}-E_{fp})/kT\big)$. Matching the exponents of the two expressions for $np$ tells us directly that
\[ E_{fn}-E_{fp} = qV \]
Step 4: Plug in numbers.
The bias is $V = 2$ V, so the split is $q(2\text{ V}) = 2$ eV. This is an energy, so eV is the natural unit here, not V.
Step 5: Rule out the rest.
$1$ eV is off by a factor of two. $2$ V and $1$ V both quote a voltage where an energy is asked for, so both are wrong even before checking the number.
\[ \boxed{2\text{ eV}} \]
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