In an intrinsic semiconductor at room temperature, if the bandgap energy is Eg, what is the relationship between electron and hole concentrations?
Answer: C
In an intrinsic semiconductor, the number of electrons equals the number of holes as they are generated in pairs. Both equal the intrinsic carrier concentration ni.
Q.22Easy
When a p-type semiconductor is doped with donor atoms instead of acceptors, what happens to the Fermi level position?
Answer: B
Adding donor atoms to a p-type semiconductor introduces electrons, shifting the material towards n-type behavior and moving the Fermi level toward the conduction band.
Q.23Medium
The conductivity of a semiconductor is given by σ = e(neμe + nhμh). If temperature increases from 300K to 400K, which factor primarily determines the change in conductivity?
Answer: A
Though mobility decreases with temperature (T^-23), the exponential increase in carrier concentration (proportional to exp(-Eg/2kT)) dominates, resulting in net increase in conductivity.
Q.24Medium
In a reverse-biased p-n junction, the depletion width increases when:
Answer: B
Increasing reverse bias voltage creates a stronger electric field, pushing charge carriers away from the junction and widening the depletion region according to W ∝ √V.
Q.25Medium
A silicon semiconductor has ni = 1.5 × 10^10 cm^-3 at 300K. If doped with 10^16 cm^-3 donor atoms, calculate the hole concentration at thermal equilibrium.
Answer: B
Using mass action law: ne·nh = ni². Since ne ≈ Nd = 10^16 cm^-3, then nh = ni²/ne = (1.5×10^10)²/10^16 = 2.25×10^4 cm^-3
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Q.26Medium
The reverse saturation current (I₀) in a p-n junction increases exponentially with temperature. This is primarily because:
Answer: A
I₀ ∝ ni² which increases exponentially with temperature. This thermal generation of minority carriers in the neutral regions near the junction constitutes the reverse saturation current.
Q.27Hard
In a forward-biased p-n junction diode at room temperature, if the applied voltage is increased from 0.3V to 0.6V, by approximately what factor does the forward current increase? (assume kT/e ≈ 26mV)
Answer: D
Forward current I = I₀exp(eV/kT). Ratio = exp(e×0.3/kT) = exp(0.03.026) ≈ exp(11.5) ≈ 10^5. For ΔV = 0.3V more, increase is exp(11.5) ≈ 1000 times.
Q.28Easy
Which of the following statements about the bandgap of semiconductors is correct?
Answer: B
The bandgap energy decreases with increasing temperature at a rate of approximately -2 to -4 meV/K, described by the Varshni equation: Eg(T) = Eg(0) - αT²/(T+β)
Q.29Medium
In a Zener diode, the Zener breakdown occurs due to:
Answer: B
Zener breakdown (sharp, reversible breakdown at lower voltages) occurs when a strong electric field in the depletion region enables direct tunneling of electrons across the narrow bandgap.
Q.30Medium
A compound semiconductor like GaAs has a direct bandgap while Si has an indirect bandgap. Which statement about light emission is correct?
Answer: B
Direct bandgap semiconductors like GaAs allow radiative recombination (photon emission) without phonon assistance, making them ideal for LEDs and lasers.
Q.31Medium
The minority carrier diffusion length Ln in a semiconductor depends on which parameters?
Answer: B
Diffusion length Ln = √(Dn·τn) where Dn is diffusion coefficient and τn is minority carrier lifetime. It represents the average distance a carrier travels before recombination.
Q.32Medium
In an n-channel JFET (Junction Field Effect Transistor), when the gate-source voltage (Vgs) becomes more negative, what happens to the channel conductance?
Answer: B
More negative Vgs increases the reverse bias on the gate junction, expanding the depletion region and narrowing the conducting channel, thus decreasing conductance.
Q.33Easy
The pinch-off voltage Vp in a JFET is the gate voltage at which:
Answer: A
At pinch-off voltage, the depletion region extends completely across the channel width, cutting off the flow of carriers and reducing drain current to nearly zero (IDSS becomes zero).
Q.34Medium
A photodiode operates in reverse bias to:
Answer: B
Reverse bias widens the depletion region, allowing photogenerated carriers throughout this region to be swept out by the electric field, improving collection efficiency and response time.
Q.35Medium
In a solar cell, the maximum power output occurs when the operating point satisfies which condition?
Answer: C
Maximum power point (MPP) occurs at the knee of the I-V characteristic where the product V×I is maximum, typically at ~80% of Voc and ~90% of Isc.
Q.36Hard
The fill factor of a solar cell is defined as:
Answer: A
Fill Factor (FF) = Pmax/(Voc × Isc) = (Vm × Im)/(Voc × Isc). It indicates how close the I-V curve is to a rectangle, typically 0.7-0.85 for practical solar cells.
Q.37Hard
In an avalanche photodiode (APD), the internal gain (multiplication factor M) increases significantly due to:
Answer: B
APD operates in high reverse bias where impact ionization (collision ionization) produces secondary electron-hole pairs, creating avalanche multiplication and signal amplification internally.
Q.38Hard
The specific contact resistance of a metal-semiconductor junction is proportional to:
Answer: B
Contact resistance ρc ∝ exp(φB/kT)/Nc where φB is barrier height. Higher barrier leads to exponentially higher resistance following Thermionic emission theory.
Q.39Medium
In the Einstein relation for semiconductors, the ratio of diffusion coefficient to mobility (D/μ) equals:
Answer: B
Einstein relation: D/μ = kT/e. At room temperature (300K), this thermal voltage ≈ 26 mV. This fundamental relation connects drift and diffusion processes in semiconductors.
Q.40Hard
Which combination of semiconductor parameters would result in the highest intrinsic carrier concentration at 300K?
Answer: B
ni = √(Nc·Nv)·exp(-Eg/2kT) where Nc, Nv depend on effective masses. Small Eg and high density of states both exponentially increase ni. GaAs has higher ni than Si due to smaller Eg.
