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?
Forward Bias p-n Junction Fermi Level Energy Difference
For a p-n junction diode under forward bias, the energy difference between the quasi-Fermi level for electrons ($E_{fn}$) and the quasi-Fermi level for holes ($E_{fp}$) is directly proportional to the applied bias voltage ($V$).
Quasi-Fermi Level Energy Difference Calculation
The relationship is defined by the equation:
$E_{fn} - E_{fp} = qV$Where:
- $E_{fn}$ is the quasi-Fermi level for electrons.
- $E_{fp}$ is the quasi-Fermi level for holes.
- $q$ is the elementary charge (magnitude of electron charge, $1e$).
- $V$ is the applied forward bias voltage.
Applying Given Voltage
Given the forward bias voltage $V = 2$ V. Substituting this value into the equation:
$E_{fn} - E_{fp} = (1e) \times (2 \text{ V})$The result is expressed in electron volts (eV) because the charge $q$ is $1e$.
$E_{fn} - E_{fp} = 2 \text{ eV}$Conclusion
Therefore, the energy difference between the quasi-Fermi levels $E_{fn}$ and $E_{fp}$ is 2 eV when the p-n junction diode is forward biased with 2 V. This matches option A.
Important Questions from PN Junction
In a semiconductor diode, the cut-in voltage is the voltage:
The leakage current in a pn junction is of the order of:
Diode junction breakdowns above 5 V are caused by:
The width of the depletion layer in a P-N junction diode
ln a p-n junction diode the forward current
