Correct answer Carries: 4.
Wrong Answer Carries: -1.
In reverse bias, the width of the depletion region:
In reverse bias, the applied voltage increases the barrier height and electric field, widening the depletion region as more immobile charges are exposed due to reduced carrier diffusion.
In an intrinsic semiconductor, the equilibrium condition implies:
At equilibrium, the rate of generation of electron-hole pairs equals the rate of recombination, maintaining a constant number of carriers (\( n_e = n_h = n_i \)).
Which material has the smallest energy gap among C, Si, and Ge?
The energy gaps are: C (diamond) = 5.4 eV, Si = 1.1 eV, Ge = 0.7 eV. Ge has the smallest energy gap.
In an n-type semiconductor, the number of electrons contributed by donors is:
In an n-type semiconductor, each pentavalent donor atom contributes one extra electron, making the electron concentration dependent on doping level, not just intrinsic generation.
In an intrinsic semiconductor, the number of charge carriers increases with:
In an intrinsic semiconductor, thermal energy excites electrons from valence to conduction band, creating electron-hole pairs, and this process increases with rising temperature.
Which of the following is an elemental semiconductor?
Elemental semiconductors are pure elements like Si and Ge, while compound semiconductors (e.g., GaAs, CdS) consist of multiple elements. Among the options, Ge is elemental.
In a rectifier circuit, the diode conducts when it is:
A diode conducts in forward bias (p-side positive, n-side negative), allowing current to flow through the load, which is the basis of rectification in both half-wave and full-wave circuits.
Which of the following is a trivalent dopant used in p-type semiconductors?
Trivalent dopants (valency 3) like Boron (B), Aluminium (Al), and Indium (In) are used in p-type semiconductors to create holes. Phosphorus (P) is pentavalent.
In a full-wave rectifier with centre-tap transformer, the output frequency for a 50 Hz input is:
A full-wave rectifier doubles the input frequency by using both half-cycles, so for a 50 Hz input, the output frequency is \( 50 \times 2 = 100 \, \text{Hz} \).
The barrier potential in a p-n junction opposes the flow of:
The barrier potential arises due to the space-charge region and opposes the diffusion of majority carriers (electrons from n-side to p-side and holes from p-side to n-side) at equilibrium.
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