GATE EC Electronic Devices Previous Year Questions with Solutions

Full question on page 25 of the official paper (PDF)

GATE 2026, Q.25

Electronic Devices1 markMSQ

A p-n junction is forward biased at 2 V. What is the separation between the quasi-Fermi levels $E_{fn}$ and $E_{fp}$ ?

One or more options are correct.

Option A: 2 eV
Option B: 1 eV
Option C: 2 V
Option D: 1 V

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Key idea: An applied bias $V$ splits the quasi-Fermi levels by $qV$ .

Official answer: A

At a forward bias of 2 V the split is $q \times 2$ V $= 2$ eV, option (A).

Options (C) and (D) give an energy in volts, which isn't an energy unit, and 1 eV doesn't match the bias. The official key lists (A) only.

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Full question on page 29 of the official paper (PDF)

GATE 2026, Q.29

Electronic Devices1 markMSQ

Two BJTs at the same base current: in the active region, BJT 1's collector current rises more steeply with VCE than BJT 2's. Compare their Early voltages |VA1| and |VA2|.

Output characteristics, IC against VCE at constant IB, for two BJTs: after the knee, the BJT 1 curve (solid) rises more steeply than the BJT 2 curve (dashed). — Figure from the official GATE 2026 EC paper, IIT Guwahati

One or more options are correct.

Option A: $|V_{A1}| > |V_{A2}|$
Option B: $|V_{A1}|$ is infinitely large
Option C: $|V_{A1}| < |V_{A2}|$
Option D: $|V_{A2}|$ is finite

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Key idea: The active-region slope is about $I_C/|V_A|$ : the steeper curve has the smaller Early voltage.

Official answer: C and D

Compare the two active-region slopes at the same $I_C$ :

BJT 1 is steeper, so $|V_{A1}| < |V_{A2}|$ : option (C).
BJT 2 still has a nonzero slope, so $|V_{A2}|$ is finite: option (D).
BJT 1 has the steeper slope, so its Early voltage is certainly not infinite, which rules out (B), and (A) is the reverse of (C).

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Full question on page 51 of the official paper (PDF)

GATE 2026, Q.54

Electronic Devices2 marksMSQ

An LED uses a direct-gap semiconductor with $E_g = 1.3$ eV ( $h = 6.63 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s). In which wavelength bands will it not emit?

One or more options are correct.

Option A: 1410 ± 20 nm
Option B: 1090 ± 20 nm
Option C: 950 ± 20 nm
Option D: 510 ± 20 nm

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Key idea: An LED emits in a narrow band just above the gap, at $\lambda \approx hc/E_g$ .

Official answer: A, B and D

The peak wavelength is set by the bandgap:

\lambda = \frac{hc}{E_g} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.3 \times 1.6 \times 10^{-19}}\text{ m} \approx 956\text{ nm}

\begin{aligned} \lambda &= \frac{hc}{E_g} \\ &= \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.3 \times 1.6 \times 10^{-19}}\text{ m} \\ &\approx 956\text{ nm} \end{aligned}

1410 nm (0.88 eV) and 1090 nm (1.14 eV) need photons below the gap. No transition gives them, so no emission: (A) and (B).
950 ± 20 nm contains 956 nm: the LED emits here.
510 nm (2.43 eV) is more than 1 eV above the gap. Carriers relax to the band edges before recombining, so the emission band (a few $kT$ wide) doesn't reach it: (D).

Options (A), (B) and (D).

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Full question on page 58 of the official paper (PDF)

GATE 2026, Q.64

Electronic Devices2 marksNAT

Electron concentration falls linearly from $2 \times 10^{17}$ cm⁻³ at $x = 1$ µm to $10^{16}$ cm⁻³ at $x = 4$ µm. With $\mu_n = 1400$ cm²/V s, $V_T = 25$ mV and $q = 1.6 \times 10^{-19}$ C, find the electron diffusion current density.

Answer format: A/mm², rounded off to two decimal places (the key accepts either sign)

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Key idea: $J_n = qD_n\,dn/dx$ , with $D_n = \mu_n V_T$ from Einstein's relation.

Official answer: -36.50 to -34.50, or 34.50 to 36.50

Einstein's relation gives the diffusion coefficient:

D_n = \mu_n V_T = 1400 \times 0.025 = 35\text{ cm}^2/\text{s}

\begin{aligned} D_n &= \mu_n V_T \\ &= 1400 \times 0.025 \\ &= 35\text{ cm}^2/\text{s} \end{aligned}

The gradient, over $3\ \mu$ m $= 3 \times 10^{-4}$ cm:

\frac{dn}{dx} = \frac{10^{16} - 2 \times 10^{17}}{3 \times 10^{-4}} \approx -6.33 \times 10^{20}\text{ cm}^{-4}

\begin{aligned} \frac{dn}{dx} &= \frac{10^{16} - 2 \times 10^{17}}{3 \times 10^{-4}} \\ &\approx -6.33 \times 10^{20}\text{ cm}^{-4} \end{aligned}

J_n = qD_n\frac{dn}{dx} = 1.6 \times 10^{-19} \times 35 \times (-6.33 \times 10^{20}) \approx -3547\text{ A/cm}^2 = -35.47\text{ A/mm}^2

\begin{aligned} J_n &= qD_n\frac{dn}{dx} \\ &= 1.6 \\ &\quad \times 10^{-19} \\ &\quad \times 35 \\ &\quad \times (-6.33 \times 10^{20}) \\ &\approx -3547\text{ A/cm}^2 \\ &= -35.47\text{ A/mm}^2 \end{aligned}

Electrons diffuse towards $+x$ , so the conventional current flows towards $-x$ . Its magnitude is about 35.47 A/mm².

Note: The official key accepts either sign: -36.50 to -34.50 or 34.50 to 36.50. The question asks for the current density without fixing a positive direction.

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Full question on page 29 of the official paper (PDF)

GATE 2025, Q.31

Electronic Devices1 markMSQ

Which of these can be used as an n-type dopant for silicon?

One or more options are correct.

Option A: Arsenic
Option B: Boron
Option C: Gallium
Option D: Phosphorus

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Key idea: An n-type dopant has one more valence electron than silicon: it comes from group 15.

Official answer: A and D

Silicon has four valence electrons. Arsenic and phosphorus (group 15) have five, so each donates a free electron: they are donors, and make silicon n-type.

Boron and gallium (group 13) have three and accept an electron, making silicon p-type. Options (A) and (D).

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Full question on page 41 of the official paper (PDF)

GATE 2025, Q.46

Electronic Devices2 marksMCQ

A semiconductor has intrinsic carrier concentration $2.5 \times 10^{16}$ m⁻³ at 300 K, and electron and hole mobilities of 0.15 and 0.05 m²/Vs. With $e = 1.6 \times 10^{-19}$ C, what is its intrinsic resistivity at 300 K, in kΩ·m?

Option A: 1.65
Option B: 1.25
Option C: 0.85
Option D: 1.95

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Key idea: In an intrinsic semiconductor both carriers have concentration $n_i$ , so their mobilities add.

Official answer: B

\sigma_i = n_i e (\mu_n + \mu_p) = 2.5 \times 10^{16} \times 1.6 \times 10^{-19} \times 0.2 = 8 \times 10^{-4}\ \text{S/m}

\begin{aligned} \sigma_i &= n_i e (\mu_n + \mu_p) \\ &= 2.5 \\ &\quad \times 10^{16} \\ &\quad \times 1.6 \\ &\quad \times 10^{-19} \\ &\quad \times 0.2 \\ &= 8 \times 10^{-4}\ \text{S/m} \end{aligned}

\rho_i = \frac{1}{\sigma_i} = 1250\ \Omega\,\text{m} = 1.25\ \text{k}\Omega\,\text{m}

\begin{aligned} \rho_i &= \frac{1}{\sigma_i} \\ &= 1250\ \Omega\,\text{m} \\ &= 1.25\ \text{k}\Omega\,\text{m} \end{aligned}

Option (B).

Full question on page 43 of the official paper (PDF)

GATE 2025, Q.48

Electronic Devices2 marksMCQ

The electron mobility in non-degenerate germanium at 300 K is 0.38 m²/Vs. With $k_B = 1.38 \times 10^{-23}$ J/K and $e = 1.6 \times 10^{-19}$ C, what is the electron diffusivity $D_n$ at 300 K, in cm²/s, to the nearest integer?

Option A: 26
Option B: 98
Option C: 38
Option D: 10

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Key idea: The Einstein relation ties diffusivity to mobility through the thermal voltage $kT/e$ .

Official answer: B

\frac{k_B T}{e} = \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} = 0.025875\ \text{V}

\begin{aligned} \frac{k_B T}{e} &= \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \\ &= 0.025875\ \text{V} \end{aligned}

D_n = \mu_n \frac{k_B T}{e} = 0.38 \times 0.025875 = 9.83 \times 10^{-3}\ \text{m}^2/\text{s} = 98.3\ \text{cm}^2/\text{s}

\begin{aligned} D_n &= \mu_n \frac{k_B T}{e} \\ &= 0.38 \times 0.025875 \\ &= 9.83 \\ &\quad \times 10^{-3}\ \text{m}^2/\text{s} \\ &= 98.3\ \text{cm}^2/\text{s} \end{aligned}

To the nearest integer, 98. Option (B). Forgetting that 1 m² is $10^4$ cm² gives 0.0098, and 26 is the thermal voltage in mV.

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Full question on page 52 of the official paper (PDF)

GATE 2025, Q.63

Electronic Devices2 marksNAT

An ideal germanium p-n junction diode has a reverse saturation current of 10 µA at 300 K. With $k_B = 1.38 \times 10^{-23}$ J/K and $e = 1.6 \times 10^{-19}$ C, what forward voltage gives a current of 100 mA at 300 K, in volts?

Answer format: Volts, rounded off to two decimal places

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Key idea: Invert the ideal diode equation: the voltage grows with the logarithm of the current ratio.

Official answer: 0.23 to 0.24

I = I_s\left(e^{V/V_T} - 1\right) \;\Rightarrow\; V = V_T \ln\!\left(1 + \frac{I}{I_s}\right)

\begin{aligned} I &= I_s\left(e^{V/V_T} - 1\right) \\ \Rightarrow\; V &= V_T \ln\!\left(1 + \frac{I}{I_s}\right) \end{aligned}

V_T = \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} = 0.025875\ \text{V}, \qquad V = 0.025875 \times \ln(10001) \approx 0.238\ \text{V}

\begin{aligned} V_T &= \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \\ &= 0.025875\ \text{V}, \\ V &= 0.025875 \\ &\quad \times \ln(10001) \\ &\approx 0.238\ \text{V} \end{aligned}

To two decimal places, 0.24 V, inside the key's range of 0.23 to 0.24.

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Full question on page 19 of the official paper (PDF)

GATE 2024, Q.22

Electronic Devices1 markMCQ

For non-degenerately doped n-type silicon, which plot in the figure shows how the free electron concentration $n$ depends on temperature $T$ ? Each plots $\log n$ against $1/T$ .

Four plots of log(n) against 1/T, labelled A to D. A: steep rise at small 1/T, a long flat plateau, then a fall at large 1/T. B: a horizontal line. C: a plateau then a fall at large 1/T, with no rise at small 1/T. D: a straight line falling as 1/T grows. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Option A: Plot (A)
Option B: Plot (B)
Option C: Plot (C)
Option D: Plot (D)

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Key idea: Read the axis backwards: small $1/T$ is high temperature, so the plot runs from the intrinsic region on the left to freeze-out on the right.

Official answer: A

Going from high to low temperature, left to right along $1/T$ :

High $T$ (small $1/T$ ): intrinsic carriers outnumber the donors, and $n \approx n_i$ rises steeply with $T$ .
Middle range: every donor is ionised, so $n \approx N_D$ , a flat plateau.
Low $T$ (large $1/T$ ): freeze-out, as electrons drop back onto the donors, so $n$ falls.

Only plot (A) has all three regions. Plot (C) misses the intrinsic rise, (B) has only the plateau, and (D) only a single slope. Option (A).

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Full question on page 23 of the official paper (PDF)

GATE 2024, Q.26

Electronic Devices1 markMSQ

The figure shows the equilibrium free-electron concentration $n(x)$ in a doped semiconductor, rising from A to a maximum at B and falling to C. Which statements are true?

A plot of electron concentration n(x) against x, with three marked positions A, B and C: n rises from A to a maximum at B and falls again towards C. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

One or more options are correct.

Option A: Between B and C, the electron diffusion current is directed from C to B.
Option B: Between B and A, the electron drift current is directed from B to A.
Option C: Between B and C, the electric field is directed from B to C.
Option D: Between B and A, the electric field is directed from A to B.

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Key idea: At equilibrium the drift and diffusion currents cancel everywhere, and electron current flows opposite to electron motion.

Official answer: A, B and C

Between B and C, electrons diffuse from B (high $n$ ) towards C. Electrons carry negative charge, so the diffusion current points from C to B: (A) true.
Between A and B, electrons diffuse from B towards A, so the diffusion current points from A to B. At equilibrium drift cancels it, so the drift current points from B to A: (B) true.
Between B and C, the drift current must point from B to C to cancel diffusion. Electron drift current is along the field, so the field points from B to C: (C) true.
The same reasoning between A and B puts the field from B to A, so (D) is false.

Options (A), (B) and (C).

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Full question on page 40 of the official paper (PDF)

GATE 2024, Q.52

Electronic Devices2 marksMSQ

Which statements about a BJT's DC current gain $\beta$ are true?

One or more options are correct.

Option A: Under high-level injection in forward active mode, $\beta$ decreases as the collector current increases.
Option B: Under low-level injection in forward active mode, with the emitter-base current dominated by recombination-generation, $\beta$ decreases as the collector current increases.
Option C: $\beta$ is lower in the saturation region than in the active region.
Option D: A higher $\beta$ leads to a lower collector-to-emitter breakdown voltage.

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Key idea: $\beta$ peaks at moderate current: it is pulled down by recombination at low current and by high injection at high current.

Official answer: A, C and D

(A) True: at high injection the base's majority carriers are no longer negligible, emitter efficiency drops, and $\beta$ falls as $I_C$ rises.
(B) False: when recombination current in the emitter-base junction dominates the base current, it grows more slowly than $I_C$ , so $\beta$ rises with $I_C$ in this region.
(C) True: in saturation the base current exceeds $I_C/\beta_F$ , so the ratio $I_C/I_B$ is below the active-region $\beta$ .
(D) True: avalanche multiplication is amplified by the transistor's gain, so $BV_{CEO} \approx BV_{CBO}/\beta^{1/n}$ falls as $\beta$ rises.

Options (A), (C) and (D).

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Full question on page 45 of the official paper (PDF)

GATE 2024, Q.61

Electronic Devices2 marksNAT

A MOS capacitor on p-type silicon has a 100 nm oxide of relative permittivity 4, a fixed positive oxide charge of $10^{-8}$ C/cm² at the oxide-silicon interface, and a metal work function of 4.6 eV, with $\varepsilon_0 = 8.85 \times 10^{-14}$ F/cm. If the flatband voltage is 0 V, what is the work function of the silicon, in eV?

Answer format: eV, rounded off to two decimal places

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Key idea: The flatband voltage is the work-function difference minus the oxide charge's voltage, $V_{FB} = \phi_{ms} - Q_{ox}/C_{ox}$ .

Official answer: 4.10 to 4.50

C_{ox} = \frac{4 \times 8.85 \times 10^{-14}}{100 \times 10^{-7}} = 3.54 \times 10^{-8}\ \text{F/cm}^2, \qquad \frac{Q_{ox}}{C_{ox}} = \frac{10^{-8}}{3.54 \times 10^{-8}} \approx 0.283\ \text{V}

\begin{aligned} C_{ox} &= \frac{4 \times 8.85 \times 10^{-14}}{100 \times 10^{-7}} \\ &= 3.54 \\ &\quad \times 10^{-8}\ \text{F/cm}^2, \\ \frac{Q_{ox}}{C_{ox}} &= \frac{10^{-8}}{3.54 \times 10^{-8}} \\ &\approx 0.283\ \text{V} \end{aligned}

With $V_{FB} = 0$ , $\phi_m - \phi_s = 0.283$ V, so

\phi_s = 4.6 - 0.283 \approx 4.32\ \text{eV}

\begin{aligned} \phi_s &= 4.6 - 0.283 \\ &\approx 4.32\ \text{eV} \end{aligned}

Within the key's range of 4.10 to 4.50.

Full question on page 46 of the official paper (PDF)

GATE 2024, Q.63

Electronic Devices2 marksNAT

In a non-degenerate n-type semiconductor, 5% of the dopant atoms are neutral. The Fermi level is 0.25 eV below $E_C$ , the donor level has degeneracy 2, and the thermal voltage is 20 mV. What is $E_C - E_D$ , in eV?

Answer format: eV, rounded off to two decimal places

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Key idea: The fraction of donors still holding their electron follows a Fermi function with the degeneracy factor in front of the exponential.

Official answer: 0.17 to 0.19

\frac{N_D^0}{N_D} = \frac{1}{1 + \frac{1}{g}e^{(E_D - E_F)/kT}} = 0.05 \;\Rightarrow\; \frac{1}{2}e^{(E_D - E_F)/kT} = 19

\begin{aligned} \frac{N_D^0}{N_D} &= \frac{1}{1 + \frac{1}{g}e^{(E_D - E_F)/kT}} \\ &= 0.05 \\ &\Rightarrow\; \frac{1}{2}e^{(E_D - E_F)/kT} \\ &\quad = 19 \end{aligned}

E_D - E_F = 0.02 \ln 38 \approx 0.0728\ \text{eV}, \qquad E_C - E_D = 0.25 - 0.0728 \approx 0.18\ \text{eV}

\begin{aligned} E_D - E_F &= 0.02 \ln 38 \\ &\approx 0.0728\ \text{eV}, \\ E_C - E_D &= 0.25 - 0.0728 \\ &\approx 0.18\ \text{eV} \end{aligned}

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Full question on page 46 of the official paper (PDF)

GATE 2024, Q.64

Electronic Devices2 marksNAT

An NMOS in the linear region carries $I_{DS} = 5$ µA at $V_{DS} = 0.1$ V. Keeping $V_{GS}$ fixed, $V_{DS}$ is raised to 1.5 V. With $\mu_n C_{ox}\frac{W}{L} = 50$ µA/V², what is the transconductance at the new operating point, in µA/V?

Answer format: µA/V, rounded off to two decimal places

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Key idea: Use the first point to find the overdrive voltage, then check which region the new $V_{DS}$ puts the device in.

Official answer: 52.40 to 52.60

In the linear region $I_{DS} = k\left[V_{ov}V_{DS} - V_{DS}^2/2\right]$ :

5 = 50\left[0.1\,V_{ov} - 0.005\right] \;\Rightarrow\; V_{ov} = 1.05\ \text{V}

\begin{aligned} 5 &= 50\left[0.1\,V_{ov} - 0.005\right] \\ \Rightarrow\; V_{ov} &= 1.05\ \text{V} \end{aligned}

At $V_{DS} = 1.5$ V $> V_{ov}$ , the transistor is in saturation, where $g_m = k V_{ov}$ :

g_m = 50 \times 1.05 = 52.50\ \mu\text{A/V}

\begin{aligned} g_m &= 50 \times 1.05 \\ &= 52.50\ \mu\text{A/V} \end{aligned}

Using the linear-region expression $g_m = kV_{DS}$ instead would give 75, the usual slip.

Full question on page 46 of the official paper (PDF)

GATE 2024, Q.65

Electronic Devices2 marksNAT

A PN-junction solar cell has a photocurrent of 1 mA, and its maximum power point is at 0.3 V. With a thermal voltage of 30 mV, what is the diode's reverse saturation current, in nA?

Answer format: nA, rounded off to two decimal places

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Key idea: At the maximum power point, the derivative of $P = VI$ with respect to $V$ is zero.

Official answer: 4.00 to 4.26

P = V\left[I_L - I_s\left(e^{V/V_T} - 1\right)\right], \qquad \frac{dP}{dV} = 0 \;\Rightarrow\; I_L = I_s\left[e^{x}(1 + x) - 1\right],\ x = \frac{V}{V_T} = 10

\begin{aligned} P &= V\left[I_L - I_s\left(e^{V/V_T} - 1\right)\right], \\ \frac{dP}{dV} &= 0 \\ \Rightarrow\; I_L &= I_s\left[e^{x}(1 + x) - 1\right],\ x \\ &= \frac{V}{V_T} \\ &= 10 \end{aligned}

I_s = \frac{10^{-3}}{11e^{10} - 1} \approx \frac{10^{-3}}{242290} \approx 4.13\ \text{nA}

\begin{aligned} I_s &= \frac{10^{-3}}{11e^{10} - 1} \\ &\approx \frac{10^{-3}}{242290} \\ &\approx 4.13\ \text{nA} \end{aligned}

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Full question on page 11 of the official paper (PDF)

GATE 2023, Q.16

Electronic Devices1 markMCQ

In a semiconductor, if the Fermi energy level lies in the conduction band, the semiconductor is known as

Option A: degenerate n-type.
Option B: degenerate p-type.
Option C: non-degenerate n-type.
Option D: non-degenerate p-type.

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Key idea: A Fermi level near the conduction band means n-type; one inside a band means degenerate doping.

Official answer: A

Doping with donors pushes $E_F$ up towards $E_C$ . Very heavy donor doping pushes it past $E_C$ into the conduction band, where Boltzmann statistics no longer hold: the semiconductor is degenerate. Option (A).

A degenerate p-type semiconductor has $E_F$ inside the valence band instead.

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Full question on page 12 of the official paper (PDF)

GATE 2023, Q.17

Electronic Devices1 markMCQ

For an intrinsic semiconductor at temperature $T = 0$ K, which of the following statements is true?

Option A: All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.
Option B: All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons.
Option C: All energy states in the valence and conduction band are filled with holes.
Option D: All energy states in the valence and conduction band are filled with electrons.

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Key idea: At 0 K no electron has the thermal energy to cross the band gap.

Official answer: A

At 0 K the Fermi function is a step at $E_F$ , which lies in the gap: every state below it is full and every state above it is empty. So the valence band is full and the conduction band is empty, which is why an intrinsic semiconductor is an insulator at 0 K. Option (A).

Full question on page 13 of the official paper (PDF)

GATE 2023, Q.19

Electronic Devices1 markMCQ

For a MOS capacitor, $V_{fb}$ and $V_t$ are the flat-band and threshold voltages. Which graph in the figure best represents the depletion width $W_{dep}$ as the gate voltage $V_g$ varies?

Four sketches of depletion width against gate voltage across accumulation, depletion and inversion. A: zero until V_fb, rising steeply and flattening at a value reached before V_t. B: zero until V_fb, rising through depletion and levelling off at V_t, flat in inversion. C: constant in accumulation and depletion, falling to zero at V_t. D: rising from V_fb to a peak and falling back to zero at V_t. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: Graph (A)
Option B: Graph (B)
Option C: Graph (C)
Option D: Graph (D)

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Key idea: There is no depletion layer in accumulation; it grows through depletion and stops growing once the inversion layer forms.

Official answer: B

Below $V_{fb}$ (accumulation) majority carriers pile up at the surface and $W_{dep} = 0$ . Between $V_{fb}$ and $V_t$ the depletion layer widens as $V_g$ rises. At $V_t$ the inversion layer forms and screens any further charge, so $W_{dep}$ stays at its maximum, $W_{max}$ , through inversion.

Graph (B) shows exactly this: zero, a smooth rise through depletion, and a flat line beyond $V_t$ . (A) saturates before $V_t$ , and (C) and (D) fall to zero at $V_t$ . Option (B).

Full question on page 41 of the official paper (PDF)

GATE 2023, Q.56

Electronic Devices2 marksNAT

In an extrinsic semiconductor, the hole concentration is $1.5n_i$ , where $n_i = 1 \times 10^{10}$ cm⁻³ is the intrinsic carrier concentration. What ratio of electron to hole mobility makes the hole and electron drift currents equal?

Answer format: rounded off to two decimal places

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Key idea: Equal drift currents in the same field need $n\mu_n = p\mu_p$ , and $np = n_i^2$ .

Official answer: 2.20 to 2.30

n = \frac{n_i^2}{p} = \frac{n_i}{1.5}, \qquad \frac{\mu_n}{\mu_p} = \frac{p}{n} = \frac{1.5 n_i}{n_i/1.5} = 2.25

\begin{aligned} n &= \frac{n_i^2}{p} \\ &= \frac{n_i}{1.5}, \\ \frac{\mu_n}{\mu_p} &= \frac{p}{n} \\ &= \frac{1.5 n_i}{n_i/1.5} \\ &= 2.25 \end{aligned}

The value of $n_i$ cancels. Answer: 2.25.

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Full question on page 42 of the official paper (PDF)

GATE 2023, Q.61

Electronic Devices2 marksNAT

In a semiconductor device, the Fermi level is 0.35 eV above the valence band edge. The effective density of states in the valence band at $T = 300$ K is $1 \times 10^{19}$ cm⁻³. What is the thermal equilibrium hole concentration in silicon at 400 K, in units of $10^{13}$ cm⁻³? Given $kT$ at 300 K is 0.026 eV.

Answer format: × 10¹³ cm⁻³, rounded off to two decimal places

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Key idea: $p = N_V e^{-(E_F - E_V)/kT}$ , and both $N_V$ (as $T^{3/2}$ ) and $kT$ change with temperature.

Official answer: 60.00 to 70.00

Scale both to 400 K, keeping $E_F - E_V = 0.35$ eV as the question gives it:

N_V(400) = 10^{19}\left(\frac{400}{300}\right)^{3/2} = 1.540 \times 10^{19}\ \text{cm}^{-3}, \qquad kT = 0.026 \times \frac{400}{300} = 0.03467\ \text{eV}

\begin{aligned} N_V(400) &= 10^{19}\left(\frac{400}{300}\right)^{3/2} \\ &= 1.540 \\ &\quad \times 10^{19}\ \text{cm}^{-3}, \\ kT &= 0.026 \times \frac{400}{300} \\ &= 0.03467\ \text{eV} \end{aligned}

p = 1.540 \times 10^{19}\, e^{-0.35/0.03467} = 1.540 \times 10^{19} \times 4.12 \times 10^{-5} = 6.34 \times 10^{14}\ \text{cm}^{-3}

\begin{aligned} p &= 1.540 \\ &\quad \times 10^{19}\, e^{-0.35/0.03467} \\ &= 1.540 \\ &\quad \times 10^{19} \\ &\quad \times 4.12 \\ &\quad \times 10^{-5} \\ &= 6.34 \times 10^{14}\ \text{cm}^{-3} \end{aligned}

That is $63.4 \times 10^{13}$ cm⁻³. Answer: about 63.4. Leaving out the $T^{3/2}$ factor gives 41.2, outside the key's range.

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