Thermal equilibrium requires what concentration product in a nondegenerate semiconductor?
A n₀p₀ = nᵢ²
B n₀ − p₀ = nᵢ
C n₀/p₀ = nᵢ
D n₀ + p₀ = nᵢ
Answer: A. n₀p₀ = nᵢ² The mass-action law states that the equilibrium electron and hole concentrations satisfy n₀p₀ = nᵢ².
Given nᵢ = 10¹⁰ cm⁻³ and n₀ = 10¹⁶ cm⁻³, what is p₀ at equilibrium?
A 10⁶ cm⁻³
B 10³ cm⁻³
C 10⁵ cm⁻³
D 10⁴ cm⁻³
Answer: D. 10⁴ cm⁻³ Using n₀p₀ = nᵢ² gives p₀ = (10¹⁰)²/10¹⁶ = 10⁴ cm⁻³.
Which expression gives semiconductor conductivity for electrons and holes?
A qμₙμₚ/(n + p)
B q(n + p)/(μₙ + μₚ)
C q(nμₙ + pμₚ)
D nμₙ − pμₚ
Answer: C. q(nμₙ + pμₚ) Conductivity is the sum of electron and hole contributions: q(nμₙ + pμₚ).
Carrier mobility μ under electric field E gives what drift-speed magnitude?
A μ + E
B μ/E
C μE
D E/μ
Answer: C. μE The drift-speed magnitude is mobility multiplied by electric-field magnitude: v_d = μE.
Which relation links diffusion coefficient D and mobility μ at thermal equilibrium?
A D/μ = kT/q
B Dμ = kT/q
C D + μ = kT/q
D D − μ = kT/q
Answer: A. D/μ = kT/q The Einstein relation gives D/μ = kT/q for nondegenerate carriers at thermal equilibrium.
In equilibrium, which current balances diffusion current at a p–n junction?
A Drift current
B Photocurrent
C Displacement current
D Avalanche current
Answer: A. Drift current The built-in field drives drift current opposite to diffusion current, giving zero net current at equilibrium.
Raising reverse bias causes what change in an abrupt junction’s depletion width?
A It increases
B It vanishes
C It stays fixed
D It decreases
Answer: A. It increases Greater reverse bias increases the electric field and widens the depletion region.
How does reverse bias affect a junction’s depletion capacitance?
A It becomes infinite
B It decreases
C It is unchanged
D It increases sharply
Answer: B. It decreases Higher reverse bias widens the depletion region, increasing charge separation and reducing junction capacitance.
Near room temperature, which forward-voltage rise gives roughly ten times more ideal-diode current?
A 60 mV
B 0.6 mV
C 6 mV
D 600 mV
Answer: A. 60 mV At room temperature, an ideality-one diode’s current increases by about a decade per 60 mV increase in forward voltage.
With an ideal diode carrying 1 mA at room temperature, what is rd when n = 1?
A 26 Ω
B 26 kΩ
C 2.6 Ω
D 260 Ω
Answer: A. 26 Ω The small-signal resistance is approximately V_T/I_D; 25.9 mV divided by 1 mA is about 26 Ω.
Under forward bias, which carriers are injected across a p–n junction?
A No carriers
B Forward-injected carriers
C Only lattice ions
D Only photons
Answer: B. Forward-injected carriers Forward bias injects majority carriers across the junction, where they become minority carriers in the opposite region.
In an abrupt junction, where does most depletion width extend?
A Metal contact
B Equally on both sides
C More heavily doped side
D More lightly doped side
Answer: D. More lightly doped side Charge neutrality requires depletion to extend farther into the more lightly doped side.
Short-circuiting a solar cell sets its terminal voltage to what value?
A Breakdown voltage
B Zero volts
C Open-circuit voltage
D Threshold voltage
Answer: B. Zero volts Short circuit means the cell terminals are connected with negligible external resistance, so terminal voltage is approximately zero.
Open-circuit operation leaves what solar-cell terminal current?
A Approximately zero
B Breakdown current
C Short-circuit current
D Photocurrent only
Answer: A. Approximately zero With no external path, an open-circuit solar cell delivers zero terminal current while developing its open-circuit voltage.
Applying a small positive gate bias to a p-type MOS capacitor produces what surface condition?
A The oxide conducts freely
B Holes are depleted
C The substrate becomes metal
D Electrons accumulate first
Answer: B. Holes are depleted A modest positive gate bias repels majority holes from the surface, creating a depletion region.
Long-channel nMOS saturation begins when which VDS condition holds?
A VGS < VT
B VDS < 0
C VDS = VGS + VT
D VDS ≥ VGS − VT
Answer: D. VDS ≥ VGS − VT The long-channel saturation condition is V_DS ≥ V_GS − V_T, assuming V_GS exceeds threshold.
What causes drain current to rise slightly with VDS in MOSFET saturation?
A Gate leakage
B Impact ionization only
C Channel-length modulation
D Hole accumulation
Answer: C. Channel-length modulation Channel-length modulation shortens the effective channel as V_DS rises, increasing saturation current.
Increasing reverse VSB in a long-channel nMOS does what to threshold voltage?
A Makes mobility infinite
B Removes the gate oxide
C Lowers threshold voltage
D Raises threshold voltage
Answer: D. Raises threshold voltage The body effect increases nMOS threshold voltage as reverse source-to-body bias increases.
Drain-induced barrier lowering is most associated with which MOSFET condition?
A Zero drain voltage
B Very long channels
C Short-channel high drain bias
D Strong optical injection
Answer: C. Short-channel high drain bias In a short-channel MOSFET, high drain bias can lower the source-channel barrier and reduce threshold voltage.
For an NPN BJT in forward-active mode, which junction-bias pair is correct?
A Both forward
B BE reverse, BC forward
C Both reverse
D BE forward, BC reverse
Answer: D. BE forward, BC reverse Forward-active operation injects carriers through a forward-biased base–emitter junction while the base–collector junction is reverse biased.
What does the Early effect describe in a BJT?
A Avalanche in the collector
B Emitter oxidation
C Gate depletion
D Base-width modulation
Answer: B. Emitter oxidation Increasing collector-base reverse bias narrows the effective base width, causing collector current to vary with V_CE.
Given beta = 99, what is alpha approximately for that BJT?
A 1.99
B 0.90
C 0.99
D 0.50
Answer: C. 0.99 Since α = β/(β + 1), a beta of 99 gives α = 99/100 = 0.99.
Which current relation is correct for a BJT’s terminal currents?
A IB = IC + IE
B IE = IC − IB
C IE = IC + IB
D IC = IE + IB
Answer: C. IE = IC + IB By current conservation at the transistor terminals, emitter current equals collector current plus base current.
Why is the base region of a typical BJT made thin?
A To increase emitter resistance
B To reduce carrier recombination
C To eliminate collector current
D To block all injection
Answer: B. To reduce carrier recombination A thin base lets most injected carriers cross into the collector before recombining.
What physical change produces diffusion capacitance in a forward-biased diode?
A Depletion width
B Oxide charge
C Metal resistance
D Changing stored charge
Answer: D. Changing stored charge Forward bias stores injected minority-carrier charge; the voltage dependence of that charge appears as diffusion capacitance.