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GAT Subject - Electronics & Electrical Engineering Solved Question Bank & Preparation Syllabus (2026) - Apex Rankers

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Electronics Engineering

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Q. 1 Electronics Engineering
Difficulty: Easy (1 Mark)
In an intrinsic (pure) semiconductor at room temperature (300 K), the relationship between electron concentration (n) and hole concentration (p) is:
A
n >> p
B
n = p = ni (intrinsic carrier concentration)
✓ Correct
C
p >> n
D
n = 0 and p = 0
💡 Step-by-Step Explanation & Concept Rationale
In an intrinsic semiconductor, thermal energy excites valence electrons across the bandgap, creating equal numbers of free conduction electrons and holes. Thus, n = p = ni. For Silicon at 300 K, ni ≈ 1.5 * 10^10 cm^-3.
Q. 2 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the forbidden energy bandgap (Eg) at 300 K for Silicon (Si) and Germanium (Ge), respectively?
A
1.12 eV for Si, and 0.66 eV for Ge
✓ Correct
B
0.7 eV for Si, and 0.3 eV for Ge
C
1.42 eV for Si, and 1.12 eV for Ge
D
5.0 eV for Si, and 1.12 eV for Ge
💡 Step-by-Step Explanation & Concept Rationale
At room temperature (300 K), the forbidden energy bandgap of Silicon is approximately 1.12 eV, while for Germanium it is approximately 0.66 eV (0.67 eV). Note that 0.7 V and 0.3 V represent the forward threshold/barrier voltages of their respective PN junctions.
Q. 3 Electronics Engineering
Difficulty: Easy (1 Mark)
Doping intrinsic Silicon with a pentavalent impurity (such as Phosphorus, Arsenic, or Antimony) creates a(n):
A
P-type semiconductor with holes as majority carriers
B
N-type semiconductor with electrons as majority carriers and donor ions
✓ Correct
C
Intrinsic insulator
D
Superconductor
💡 Step-by-Step Explanation & Concept Rationale
Pentavalent atoms have 5 valence electrons. Four form covalent bonds with adjacent Si atoms, while the fifth electron is loosely bound (donor level ~0.05 eV below conduction band) and easily liberated at room temperature, creating an N-type semiconductor where electrons are majority carriers.
Q. 4 Electronics Engineering
Difficulty: Easy (1 Mark)
Doping Silicon with a trivalent impurity (such as Boron, Gallium, or Indium) creates a(n):
A
N-type semiconductor with donor impurities
B
P-type semiconductor with holes as majority carriers and acceptor ions
✓ Correct
C
Intrinsic semiconductor
D
Semimetal
💡 Step-by-Step Explanation & Concept Rationale
Trivalent atoms have 3 valence electrons, leaving one covalent bond vacancy (a hole). They readily accept electrons from nearby bonds, creating mobile holes as majority carriers and fixed negative acceptor ions, forming a P-type semiconductor.
Q. 5 Electronics Engineering
Difficulty: Easy (1 Mark)
The Mass Action Law for a semiconductor under thermal equilibrium states that:
A
n + p = ni
B
n * p = ni^2 (independent of doping level)
✓ Correct
C
n / p = ni
D
n * p = constant * T
💡 Step-by-Step Explanation & Concept Rationale
Under thermal equilibrium, the product of electron concentration n and hole concentration p is constant and equals the square of the intrinsic carrier concentration: n * p = ni^2. If donor doping increases n, hole concentration p decreases proportionately through recombination.
Q. 6 Electronics Engineering
Difficulty: Medium (1 Mark)
The Fermi-Dirac distribution function f(E) gives the probability that an energy state E is occupied by an electron. At energy E = Ef (Fermi energy level), the probability f(Ef) is:
A
1.0 (100%)
B
0.5 (50%) at any temperature T > 0 K
✓ Correct
C
0.0 (0%)
D
0.707
💡 Step-by-Step Explanation & Concept Rationale
f(E) = 1 / [1 + exp((E - Ef) / (k * T))]. When E = Ef, exp(0) = 1, so f(Ef) = 1 / (1 + 1) = 1/2 = 0.5 (or 50%) at all temperatures above absolute zero.
Q. 7 Electronics Engineering
Difficulty: Medium (1 Mark)
In an N-type semiconductor as donor doping concentration Nd increases, the Fermi level (Ef):
A
Shifts downward toward the valence band
B
Shifts upward toward the conduction band (Ec)
✓ Correct
C
Remains precisely at the center of the bandgap
D
Disappears
💡 Step-by-Step Explanation & Concept Rationale
Ef - Ei = k * T * ln(Nd / ni). As donor concentration Nd increases, the density of electrons in the conduction band rises, shifting the Fermi level closer to the conduction band edge Ec. In degenerate semiconductors, Ef enters the conduction band.
Q. 8 Electronics Engineering
Difficulty: Medium (1 Mark)
Einstein's relation relating carrier diffusion coefficient (D) to mobility (mu) in a semiconductor is:
A
D / mu = V_T = k * T / q (thermal voltage)
✓ Correct
B
D * mu = k * T
C
D / mu = q / (k * T)
D
D = mu^2 * V_T
💡 Step-by-Step Explanation & Concept Rationale
Einstein's relation states that Dn / mun = Dp / mup = V_T = k * T / q, where thermal voltage V_T ≈ 25.86 mV (approx. 26 mV) at room temperature (300 K).
Q. 9 Electronics Engineering
Difficulty: Easy (1 Mark)
The depletion region (space-charge layer) formed at an unbiased PN junction consists of:
A
Free mobile electrons and holes
B
Immobile unneutralized ionized donor and acceptor atoms stripped of their free carriers
✓ Correct
C
Neutral silicon atoms only
D
Pure metallic copper
💡 Step-by-Step Explanation & Concept Rationale
When P and N regions meet, electrons diffuse from N to P and holes from P to N. This leaves behind uncovered, fixed positive donor ions on the N-side and fixed negative acceptor ions on the P-side, creating a built-in electric field that opposes further carrier diffusion.
Q. 10 Electronics Engineering
Difficulty: Easy (1 Mark)
The built-in contact potential barrier (V0) of a Silicon PN junction at 300 K is typically around:
A
0.1 V to 0.2 V
B
0.6 V to 0.8 V (nominal 0.7 V)
✓ Correct
C
1.5 V to 2.0 V
D
5.0 V
💡 Step-by-Step Explanation & Concept Rationale
V0 = V_T * ln[(Na * Nd) / ni^2]. For typical doping levels in Silicon at 300 K, the contact barrier potential is approximately 0.7 V. For Germanium, it is approximately 0.3 V.
Q. 11 Electronics Engineering
Difficulty: Easy (1 Mark)
When a PN junction is forward-biased (P-side positive, N-side negative):
A
The depletion layer width widens, and barrier height increases
B
The depletion layer width narrows, the potential barrier decreases (V0 - Vf), and majority carriers cross the junction in large numbers
✓ Correct
C
Current flow drops to zero
D
Breakdown immediately occurs
💡 Step-by-Step Explanation & Concept Rationale
Forward bias opposes the built-in electric field, reducing barrier height to (V0 - Vf) and narrowing the space-charge width. This allows majority carriers to overcome the barrier, resulting in exponential diffusion current: I = Is * [exp(V / (eta*V_T)) - 1].
Q. 12 Electronics Engineering
Difficulty: Easy (1 Mark)
When a PN junction is reverse-biased (P-side negative, N-side positive):
A
Depletion layer narrows
B
Depletion layer widens, barrier height increases (V0 + Vr), and only a tiny reverse saturation current (Is) due to minority carriers flows
✓ Correct
C
Current increases exponentially
D
Resistance drops to zero
💡 Step-by-Step Explanation & Concept Rationale
Reverse bias aids the built-in field, pulling majority carriers away from the junction and widening the depletion layer. Only thermally generated minority carriers swept across by the field constitute the reverse saturation current Is.
Q. 13 Electronics Engineering
Difficulty: Easy (1 Mark)
The reverse saturation current (Is) of a Silicon PN diode approximately doubles for every:
A
1 degree Celsius rise in temperature
B
10 degrees Celsius rise in temperature
✓ Correct
C
50 degrees Celsius rise in temperature
D
100 degrees Celsius rise in temperature
💡 Step-by-Step Explanation & Concept Rationale
Because thermal generation of electron-hole pairs increases exponentially with temperature, the reverse saturation current Is of a diode approximately doubles for every 10 °C rise in operating junction temperature: Is(T2) = Is(T1) * 2^((T2 - T1) / 10).
Q. 14 Electronics Engineering
Difficulty: Medium (1 Mark)
The forward voltage drop (Vf) of a Silicon PN diode at constant current decreases with temperature at a rate of approximately:
A
-2.0 to -2.5 mV per degree Celsius (-2 mV/°C)
✓ Correct
B
+10 mV/°C
C
-1.0 V/°C
D
Zero (temperature independent)
💡 Step-by-Step Explanation & Concept Rationale
Due to the rapid increase in intrinsic carrier concentration ni with temperature, the forward diode voltage at fixed current decreases by roughly -2 mV/°C to -2.5 mV/°C. This negative temperature coefficient is frequently exploited in semiconductor temperature sensors.
Q. 15 Electronics Engineering
Difficulty: Medium (1 Mark)
Transition (or depletion layer) capacitance (CT) in a reverse-biased PN junction dominates over diffusion capacitance because:
A
Charge is stored as immobile ions across the expanding depletion dielectric like a parallel-plate capacitor (CT ∝ 1 / sqrt(V0 + Vr))
✓ Correct
B
Minority carrier diffusion is very high
C
Current is leading voltage by 180 degrees
D
Barrier potential drops to zero
💡 Step-by-Step Explanation & Concept Rationale
Under reverse bias, diffusion current is negligible. The immobile ions on either side of the depletion layer act as charged plates separated by a dielectric of width W, giving transition capacitance CT = epsilon * A / W. Because W ∝ sqrt(V0 + Vr), CT varies inversely with reverse voltage.
Q. 16 Electronics Engineering
Difficulty: Easy (1 Mark)
A Varactor diode (varicap) operates in which region and is primarily used for:
A
Forward bias as a light source
B
Reverse bias, utilizing its voltage-variable transition capacitance for electronic tuning in RF local oscillators and phase-locked loops (PLLs)
✓ Correct
C
Breakdown region as a voltage regulator
D
High-current power rectification
💡 Step-by-Step Explanation & Concept Rationale
A Varactor diode is specifically designed to maximize the voltage-dependent transition capacitance under reverse bias. Varying reverse voltage controls CT smoothly, providing electronic tuning for LC resonant tanks without mechanical variable capacitors.
Q. 17 Electronics Engineering
Difficulty: Medium (1 Mark)
Zener breakdown occurs in PN junctions that are:
A
Lightly doped, with wide depletion layers, breaking down above 6 V via impact ionization
B
Heavily doped, with extremely narrow depletion layers (W < 10 nm), where high electric field (> 10^6 V/cm) pulls valence electrons directly into the conduction band by quantum mechanical tunneling
✓ Correct
C
Forward biased with high current
D
Fabricated exclusively from Germanium
💡 Step-by-Step Explanation & Concept Rationale
Zener breakdown occurs in heavily doped junctions at low reverse voltages (< 5-6 V). The narrow depletion width creates an intense electric field that directly ruptures covalent bonds via quantum tunneling. It exhibits a negative temperature coefficient of breakdown voltage.
Q. 18 Electronics Engineering
Difficulty: Medium (1 Mark)
Avalanche breakdown in a reverse-biased PN junction occurs in:
A
Lightly doped junctions with breakdown voltages greater than 6 V, where thermally generated carriers gain high kinetic energy to ionize lattice atoms by impact ionization (carrier multiplication)
✓ Correct
B
Heavily doped junctions at 2 V
C
Forward-biased diodes
D
Photodiodes in complete darkness only
💡 Step-by-Step Explanation & Concept Rationale
In lightly doped diodes, the depletion layer is relatively wide. High reverse voltage accelerates minority carriers to high velocities, dislodging additional valence electrons upon impact. These secondary carriers cause cumulative carrier multiplication (avalanche), having a positive temperature coefficient of breakdown voltage.
Q. 19 Electronics Engineering
Difficulty: Medium (1 Mark)
A Zener diode with a nominal breakdown voltage of 5.6 V has a temperature coefficient that is:
A
Strongly positive
B
Strongly negative
C
Nearly zero, because the negative temperature coefficient of Zener tunneling cancels the positive temperature coefficient of Avalanche multiplication
✓ Correct
D
Infinite
💡 Step-by-Step Explanation & Concept Rationale
Below 5 V, Zener breakdown (negative temp coeff) dominates. Above 6 V, Avalanche breakdown (positive temp coeff) dominates. Around 5.6 V, both mechanisms contribute equally, cancelling each other's temperature drift to yield superior voltage reference stability.
Q. 20 Electronics Engineering
Difficulty: Medium (1 Mark)
A Tunnel diode (Esaki diode) exhibits negative differential resistance (dV/dI < 0) over a portion of its forward characteristic due to:
A
Extreme degenerate doping (1 in 10^3 atoms) that causes quantum mechanical electron tunneling across an ultra-thin depletion barrier
✓ Correct
B
Thermal heating of the junction
C
Avalanche carrier multiplication
D
External magnetic field
💡 Step-by-Step Explanation & Concept Rationale
Degenerate doping causes the conduction band on the N-side to overlap the valence band on the P-side. Under small forward bias, electrons tunnel directly through the thin barrier (~10 nm). As forward bias increases, band overlap diminishes, causing forward current to drop from peak current (Ip) to valley current (Iv), creating negative resistance used in high-frequency microwave oscillators.
Q. 21 Electronics Engineering
Difficulty: Medium (1 Mark)
A Schottky barrier diode is formed by a metal-semiconductor junction (e.g., gold or platinum on N-type silicon). Its key operational advantages are:
A
Majority-carrier device with virtually zero reverse recovery time (ultra-fast switching) and lower forward voltage drop (approx. 0.2 to 0.3 V)
✓ Correct
B
Extremely high reverse breakdown voltage exceeding 20 kV
C
High diffusion capacitance under reverse bias
D
Negative dynamic resistance
💡 Step-by-Step Explanation & Concept Rationale
Because conduction is by majority carriers (electrons in N-type silicon), there is no minority carrier charge storage in the depletion region. Reverse recovery time is virtually zero (picoseconds), and forward drop is only 0.2-0.3 V, making Schottky diodes ideal for high-efficiency switch-mode power supplies (SMPS) and RF mixers.
Q. 22 Electronics Engineering
Difficulty: Hard (1 Mark)
A PIN diode contains an undoped intrinsic (I) semiconductor layer sandwiched between heavily doped P and N layers. At microwave radio frequencies, it behaves as a:
A
Linear variable RF resistor whose resistance is controlled by DC bias current, useful as an RF switch and attenuator
✓ Correct
B
Zener regulator
C
Digital flip-flop
D
Constant voltage source
💡 Step-by-Step Explanation & Concept Rationale
At low frequencies, a PIN diode acts as a standard rectifier. At RF/microwave frequencies, carriers cannot follow fast AC cycles; the intrinsic layer behaves as an almost pure resistance whose RF value varies continuously from thousands of ohms to under 1 ohm depending on applied DC forward bias current.
Q. 23 Electronics Engineering
Difficulty: Easy (1 Mark)
A Photodiode is always operated in which mode to detect incident optical radiation with linear response and high bandwidth?
A
Forward bias mode
B
Reverse bias mode (photoconductive mode)
✓ Correct
C
Zero bias (photovoltaic mode) only
D
Thermal breakdown mode
💡 Step-by-Step Explanation & Concept Rationale
In reverse bias, dark current is minimal. Incident photons of energy h*nu >= Eg generate electron-hole pairs in the wide depletion layer, which are swept by the strong built-in electric field. Reverse photocurrent is strictly linear with optical power and switching speed is maximal.
Q. 24 Electronics Engineering
Difficulty: Medium (1 Mark)
Light Emitting Diodes (LEDs) are fabricated from direct bandgap compound semiconductors (such as GaAs, GaN, InGaN) rather than indirect bandgap semiconductors like Silicon because:
A
In direct bandgap materials, electrons recombine directly across the bandgap releasing energy predominantly as photons (light) without requiring lattice phonon momentum assistance
✓ Correct
B
Silicon is opaque to electricity
C
Direct bandgap materials have zero thermal resistance
D
Silicon cannot be doped P-type
💡 Step-by-Step Explanation & Concept Rationale
In direct bandgap semiconductors, the conduction band minimum and valence band maximum align at the same crystal momentum (k = 0). Radiative electron-hole recombination is direct and highly efficient, producing photons: lambda = h * c / Eg. In Silicon (indirect bandgap), recombination requires a phonon (lattice vibration), dissipating energy as heat.
Q. 25 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the peak inverse voltage (PIV) rating across each diode in a single-phase center-tapped full-wave transformer rectifier supplying peak secondary voltage Vm across each half?
A
PIV = Vm
B
PIV = 2 * Vm
✓ Correct
C
PIV = Vm / 2
D
PIV = sqrt(2) * Vm
💡 Step-by-Step Explanation & Concept Rationale
In a center-tapped full-wave rectifier, during the non-conducting half cycle, the reverse-biased diode must withstand its own half-winding peak voltage Vm plus the peak voltage delivered by the other conducting half to the load, resulting in PIV = 2 * Vm.
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