Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
In a MOSFET, the threshold voltage (Vth) increases when:
Answer: B
Vth = Vfb + 2φF + (√(2εsqNa(2φF))/Cox). Threshold voltage increases with oxide thickness (Cox decreases) and increases with substrate doping. Temperature has weak effect.
Q.682Easy
An LED emits light because:
Answer: A
In LEDs (forward biased p-n junctions in direct bandgap semiconductors like GaAs), electron-hole recombination releases energy as photons. Indirect bandgap materials (Si, Ge) produce mainly heat.
Q.683Medium
The photo-generated current in a solar cell is proportional to:
Answer: B
Photo-generated current IL is directly proportional to incident light intensity (photon flux). At open circuit, the current is independent of applied voltage; voltage dependence matters for load characteristics.
Q.684Hard
Impurity scattering dominates in semiconductors when:
Answer: A
At low temperatures, phonon scattering is suppressed, but impurity scattering (interaction with ionized donors/acceptors) remains, limiting mobility. At high T, phonon scattering dominates.
Q.685Hard
A depletion mode MOSFET differs from enhancement mode in that:
Answer: A
Depletion mode MOSFETs have a conducting channel at Vgs = 0 and require gate voltage to turn OFF. Enhancement mode requires positive gate voltage to create a channel. This is a key structural difference.
Q.686Hard
The temperature coefficient of bandgap for silicon is approximately:
Answer: B
Bandgap of semiconductors decreases with increasing temperature. For Si: dEg/dT ≈ -2.3 meV/K near 300 K. This is derived from Varshni equation and is crucial for device design.
Q.687Hard
In a heterojunction like AlGaAs/GaAs, the main advantage is:
Answer: B
Heterojunctions use materials with different bandgaps to create band offsets that confine carriers (electrons and holes) to specific regions, improving device efficiency. AlGaAs has larger Eg than GaAs, confining carriers in GaAs.
Q.688Easy
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.689Easy
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.690Medium
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.691Medium
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.692Medium
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
Q.693Medium
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.694Hard
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.695Easy
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.696Medium
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.697Medium
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.698Medium
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.699Medium
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.700Easy
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).
