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Electronic Devices and Circuits Practice Hub

Practice electronic devices and circuits with question sets on diodes, BJTs, MOSFETs, and GATE-level problems on device physics for ECE students.

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BeginnerIntrinsic vs dopedIntermediateMOSFET IDS calcAdvancedGATE body effect

Hub intro

Electronic devices connects semiconductor physics to practical circuit behaviour. This hub covers the p-n junction, BJT operation regions, MOSFET characteristics, and basic device modelling. Beginner questions address band theory and device definitions. Intermediate questions require applying device equations and identifying operating regions. Advanced questions mirror GATE problems on threshold voltage, carrier transport, and small-signal modelling traps.

Difficulty levels

BeginnerIntermediateAdvanced

Practice sets

Beginner: Semiconductor Fundamentals
Beginner
What is the difference between intrinsic and extrinsic semiconductors?
Answer: An intrinsic semiconductor is a pure semiconductor with equal concentrations of electrons and holes generated by thermal excitation. An extrinsic semiconductor is doped with impurities to create an excess of either electrons (n-type, donor doping) or holes (p-type, acceptor doping).
In silicon at room temperature, ni ≈ 1.5×10^10 cm^-3. In n-type silicon doped with ND donors, the electron concentration n ≈ ND (for ND >> ni) and the hole concentration p = ni²/ND (minority carriers). Doping allows the conductivity to be controlled over many orders of magnitude, which is the basis for all semiconductor devices.
Define the built-in potential (contact potential) of a p-n junction and write its expression.
Answer: Vbi = (kT/q) × ln(NA × ND / ni²), where NA is acceptor concentration, ND is donor concentration, ni is intrinsic carrier concentration, k is Boltzmann's constant, T is temperature in kelvin, and q is electron charge.
Vbi is the equilibrium electrostatic potential difference across the depletion region. It arises because diffusion of majority carriers leaves behind fixed ionised impurities, creating a space-charge region. The electric field from this space charge opposes further diffusion, establishing equilibrium. Vbi cannot be directly measured as a terminal voltage because contact potentials at the metal-semiconductor interfaces exactly compensate it.
In a BJT operating in the active region, what are the bias conditions for the base-emitter and base-collector junctions?
Answer: In the active region: the base-emitter junction (BEJ) is forward biased and the base-collector junction (BCJ) is reverse biased.
Forward biasing the BEJ injects minority carriers (electrons for n-p-n) into the base. The narrow base allows most of these electrons to diffuse to the collector without recombining. The reverse-biased BCJ sweeps these electrons into the collector, producing the collector current. In saturation, both junctions are forward biased. In cutoff, both are reverse biased. In reverse active mode (rarely used), BCJ is forward and BEJ is reverse.
What is the depletion approximation used in p-n junction analysis?
Answer: The depletion approximation assumes that the depletion (space-charge) region is completely depleted of free carriers and the neutral regions outside are completely neutral. The transition between depletion and neutral regions is assumed to be abrupt.
Under this approximation, the charge density within the depletion region is +qND on the n-side and -qNA on the p-side (from ionised donors and acceptors). This allows straightforward integration of Poisson's equation to find the electric field and potential. The approximation is accurate for typical doping levels because the transition region is very narrow compared to the depletion width.
State the condition for an n-channel MOSFET to operate in the triode (linear) region versus the saturation region.
Answer: Triode: VDS < VGS - VT. Saturation: VDS ≥ VGS - VT (pinch-off condition).
When VDS < VGS - VT, the channel exists from source to drain and acts like a voltage-controlled resistor. When VDS = VGS - VT, the channel is pinched off at the drain end. For VDS > VGS - VT, the pinch-off point moves toward the source and the drain current saturates at ID(sat) = (µn Cox W)/(2L) × (VGS - VT)². In saturation, drain current is approximately independent of VDS (ideal MOSFET).
Intermediate: Device Equations and Circuit Analysis
Intermediate
An n-channel MOSFET has µnCox = 200 µA/V², W/L = 20, VT = 1 V. Calculate IDS for VGS = 3 V and VDS = 1 V. State the operating region.
Answer: VGS - VT = 2 V. Since VDS = 1 V < 2 V, the device is in the triode region. IDS = µnCox(W/L)[(VGS-VT)VDS - VDS²/2] = 200×10^-6 × 20 × [2×1 - 0.5] = 4×10^-3 × 1.5 = 6 mA.
Step 1: Check region: VGS-VT = 3-1 = 2 V. VDS = 1 V < 2 V, so triode. Step 2: Triode current: IDS = µnCox(W/L)[(VGS-VT)VDS - VDS²/2] = 200µ × 20 × [2×1 - 1²/2] = 4000µ × [2 - 0.5] = 4000µ × 1.5 = 6000 µA = 6 mA. If we had mistakenly used the saturation formula: ID(sat) = (200µ×20/2)×(2)² = 2000µ×4 = 8 mA, which overestimates the current. Always check VDS vs VGS-VT first.
Derive the expression for the small-signal transconductance gm of a MOSFET biased in saturation.
Answer: gm = dID/dVGS|VDS=const = µnCox(W/L)(VGS - VT) = √(2µnCox(W/L)ID).
In saturation: ID = (µnCox W)/(2L) × (VGS - VT)². Differentiating with respect to VGS: gm = dID/dVGS = µnCox(W/L)(VGS-VT). This can also be written as gm = √(2µnCox(W/L)ID) by substituting (VGS-VT) = √(2ID/(µnCox(W/L))). Physically, gm is the efficiency with which the gate voltage controls the drain current. A larger W/L or a larger bias current ID both increase gm.
Explain Early effect in a BJT and how it modifies the output characteristics.
Answer: The Early effect (base-width modulation) refers to the narrowing of the effective base width as the reverse bias on the BCJ increases. This causes the collector current to increase slightly with increasing VCE, manifested as a non-zero slope in the IC vs VCE characteristics in the active region.
As VCE increases, the BCJ reverse bias increases, widening the depletion region into the base. The effective base width decreases, reducing recombination and increasing the gradient of minority carrier concentration, thus increasing IC. Extrapolating the IC vs VCE lines back to the VCE axis, they intersect at a single point -VA, the Early voltage (typically 50 to 200 V for silicon BJTs). The modified IC expression is: IC = IS × exp(VBE/VT) × (1 + VCE/VA).
A diode has saturation current IS = 10 nA. Calculate the forward current at VD = 0.6 V and T = 300 K. Use VT = kT/q = 26 mV.
Answer: ID = IS × (e^(VD/VT) - 1) = 10×10^-9 × (e^(0.6/0.026) - 1) ≈ 10 nA × e^23.08 ≈ 10 nA × 10^10 ≈ 100 mA.
VD/VT = 0.6/0.026 ≈ 23.08. e^23.08 ≈ e^23 × e^0.08 ≈ 9.74×10^9 × 1.083 ≈ 1.055×10^10. ID = 10×10^-9 × 1.055×10^10 ≈ 105.5 mA ≈ 100 mA. The -1 term is negligible for forward bias. A change of 60 mV in forward voltage (one decade in e^(VD/VT)) changes the current by a factor of 10. This exponential sensitivity is the key property of the p-n junction.
What is the difference between drift current and diffusion current in a semiconductor? Which dominates in the depletion region?
Answer: Drift current is driven by an electric field: J_drift = q(µn × n + µp × p) × E. Diffusion current is driven by a carrier concentration gradient: J_diff = q(Dn × dn/dx - Dp × dp/dx). In the depletion region of a p-n junction at equilibrium, both currents are present but cancel exactly, giving zero net current.
At thermal equilibrium, no net current flows. The built-in electric field drives drift of majority carriers toward the junction (electrons toward n-side, holes toward p-side, i.e., both drift away from the depletion region). Simultaneously, the concentration gradients drive diffusion of majority carriers toward the other side. These two currents balance for each carrier type separately, as required by the principle of detailed balance. Under forward bias, the balance is upset and diffusion dominates; under reverse bias, drift (of minority carriers) dominates.
Advanced: GATE-Style Problems
Advanced
An abrupt p-n junction has NA = 10^17 cm^-3 and ND = 10^15 cm^-3. Given ni = 1.5×10^10 cm^-3 and VT = 26 mV, calculate the built-in voltage Vbi.
Answer: Vbi = VT × ln(NA × ND / ni²) = 0.026 × ln(10^17 × 10^15 / (1.5×10^10)²) = 0.026 × ln(10^32 / 2.25×10^20) = 0.026 × ln(4.44×10^11) ≈ 0.026 × 26.82 ≈ 0.697 V.
ln(4.44×10^11) = ln(4.44) + 11×ln(10) = 1.49 + 11×2.303 = 1.49 + 25.33 = 26.82. Vbi = 0.026 × 26.82 ≈ 0.697 V ≈ 0.7 V. The GATE trap is to confuse ln (natural log) with log10. Using log10: ln(x) = 2.303 × log10(x). log10(4.44×10^11) = log10(4.44) + 11 = 0.647 + 11 = 11.647. ln = 2.303 × 11.647 ≈ 26.82 (consistent). Always use natural logarithm in the Vbi formula.
An n-channel MOSFET is in saturation with ID = 1 mA and VGS - VT = 0.5 V. Another identical MOSFET has VGS - VT = 1 V. What is the drain current of the second transistor, assuming both are in saturation?
Answer: ID ∝ (VGS - VT)². For VGS-VT doubling from 0.5 V to 1 V: ID scales by (1/0.5)² = 4. ID2 = 4 × 1 mA = 4 mA.
In saturation, ID = (µnCox W)/(2L) × (VGS - VT)². The ratio ID2/ID1 = (VGS2-VT)²/(VGS1-VT)² = (1)²/(0.5)² = 1/0.25 = 4. ID2 = 4 × 1 mA = 4 mA. GATE trap: some students use a linear relationship (doubling VGS-VT doubles ID). The square-law relationship is a defining feature of MOSFET operation in saturation.
In a BJT, the common-emitter current gain β = 100 and the base transport factor αT = 0.99. Find the emitter injection efficiency γ.
Answer: α = β/(β+1) = 100/101 ≈ 0.990. α = γ × αT. γ = α/αT = 0.9901/0.99 ≈ 1.001. This is greater than 1, which is physically impossible, indicating the given values are inconsistent.
The common-base current gain α = γ × αT, where γ is the emitter injection efficiency and αT is the base transport factor. α = β/(β+1) = 100/101 = 0.9901. If αT = 0.99, then γ = 0.9901/0.99 = 1.0001 > 1, which is impossible since γ ≤ 1 by definition. GATE examiners sometimes provide inconsistent device parameters to test whether you can identify the violation. A correct answer here is to note the inconsistency and point out that either β or αT is misstated.
A MOSFET has threshold voltage VT0 = 0.5 V at VSB = 0. With body bias VSB = 2 V, the threshold voltage increases to 0.9 V. Identify the body effect coefficient γ, given 2φF = 0.6 V.
Answer: VT(VSB) = VT0 + γ(√(2φF + VSB) - √(2φF)). 0.9 = 0.5 + γ(√(0.6+2) - √0.6) = 0.5 + γ(√2.6 - √0.6) = 0.5 + γ(1.612 - 0.775) = 0.5 + 0.837γ. 0.4 = 0.837γ. γ ≈ 0.478 V^(1/2).
The body effect (substrate bias effect) increases the threshold voltage when the source-body junction is reverse biased. The body effect coefficient γ has units of V^(1/2). γ = (0.9-0.5)/(√2.6-√0.6) = 0.4/(1.6125-0.7746) = 0.4/0.8379 ≈ 0.477 V^0.5. GATE problems often give VT0, VSB, and 2φF and ask for γ, or give γ and ask for VT. Know the formula and be careful with the square root terms.
A long-channel MOSFET has W = 100 µm, L = 1 µm, µnCox = 50 µA/V², VT = 1 V, λ = 0.1 V^-1. For VGS = 2 V and VDS = 5 V, calculate IDS including channel-length modulation.
Answer: Device is in saturation (VDS = 5 > VGS-VT = 1). IDS = (µnCox W)/(2L) × (VGS-VT)² × (1+λVDS) = (50µ×100)/(2×1) × (1)² × (1+0.5) = 2500µ × 1.5 = 3.75 mA.
Step 1: Check region: VGS-VT = 1 V, VDS = 5 V > 1 V, so saturation. Step 2: Basic saturation current: (50µ×100/2×1)×1² = (5000µ/2) = 2500 µA = 2.5 mA. Step 3: With channel-length modulation: IDS = 2.5 mA × (1 + λ×VDS) = 2.5 mA × (1 + 0.1×5) = 2.5 × 1.5 = 3.75 mA. The GATE trap is forgetting the (1+λVDS) factor when it is explicitly given, or applying it in the triode region where it does not belong.

Lab exercises

  • Measure the I-V characteristics of a 1N4148 diode and extract the ideality factor n and saturation current IS: /labs/diode-iv-characteristic-lab
  • Plot the drain characteristics (ID vs VDS) of an n-channel MOSFET for multiple VGS values and identify triode and saturation regions: /labs/mosfet-drain-characteristics-lab
  • Measure the BJT common-emitter output characteristics and extract β and the Early voltage VA: /labs/bjt-output-characteristics-lab

Revision checklist

  • Can you state the bias conditions for all four BJT operating regions (active, saturation, cutoff, reverse active)?
  • Can you derive the MOSFET triode and saturation current expressions from the channel charge model?
  • Do you know the body effect formula and can you calculate the new threshold voltage for a given VSB?
  • Can you use the Shockley diode equation to estimate forward current for a given forward voltage?
  • Can you explain why the drift and diffusion currents cancel at thermal equilibrium in a p-n junction?