GATE EC

Energy bands and carrier concentration

Energy bands and carrier concentration covers how the discrete levels of isolated atoms broaden into a valence band and a conduction band separated by a forbidden gap $E_{g}$ , and how temperature and doping set the equilibrium electron and hole concentrations. The core results are Fermi–Dirac occupation and its Boltzmann approximation, $n_{0}$ and $p_{0}$ in terms of $N_{c}$ , $N_{v}$ and $E_{F}$ , the intrinsic concentration $n_{i}$ and its temperature dependence, the mass-action law $n_{0} p_{0} = n_{i}^{2}$ , charge neutrality with compensated doping, the position of the Fermi level, and the direct/indirect band-gap distinction that governs light emission. Conventions: concentrations in cm $^{- 3}$ , energies in eV, $q = 1.6 \times 1 0^{- 19} C$ , $k T \approx 0.0259 eV$ at 300 K.

9 previous-year questions · 9 solved · 6 practice questions

1. Formation of energy bands in solids

When atoms are brought together to form a crystal, the electron wavefunctions of neighbouring atoms overlap and each discrete atomic level splits — as the Pauli exclusion principle requires — into as many closely spaced levels as there are atoms. With of order $1 0^{22}$ – $1 0^{23}$ atoms per cm $^{3}$ (about $5 \times 1 0^{22} cm^{- 3}$ in silicon) these levels are so close together that they form effectively continuous energy bands. In a semiconductor, the highest band that is completely filled at $T = 0 K$ is the valence band (top edge $E_{v}$ ), and the next band above it, empty at 0 K, is the conduction band (bottom edge $E_{c}$ ). Between them lies the forbidden gap $E_{g} = E_{c} - E_{v}$ : a perfect crystal has no allowed electron states in this energy range.

Whether a material is a conductor, insulator or semiconductor is decided by (a) whether the highest occupied band is completely full or only partly full at $T = 0 K$ , and (b) the size of the gap $E_{g}$ separating a full valence band from the empty conduction band.

A hole is an empty state in the otherwise full valence band. It behaves as a carrier of charge $+ q$ with its own effective mass, so electrons in the conduction band (concentration $n$ ) and holes in the valence band (concentration $p$ ) are treated as two separate carrier populations.

EXAM TIP

Band diagrams plot electron energy upward. $E_{c} - E_{F}$ is how far the Fermi level sits below the conduction-band edge, and $E_{F} - E_{v}$ is how far it sits above the valence-band edge. Keep this sign convention fixed whenever you solve for $E_{F}$ .

2. Conductors, insulators and semiconductors

The size of the band gap $E_{g}$ (in eV) separates the three classes. A larger gap means far fewer carriers can be thermally excited across it, so intrinsic conduction falls sharply as $E_{g}$ increases.

Material	Band structure at 0 K	Typical $E_{g}$ (eV)	Conduction
Conductor (metal)	Partially filled band, or overlapping valence and conduction bands	No gap	High; the carriers are already present, and conductivity falls as $T$ rises (more lattice scattering)
Semiconductor	Valence band full, conduction band empty, moderate gap	Roughly $0.1$ – $3.5$ (at 300 K: Ge $\approx 0.66$ , Si $\approx 1.12$ , GaAs $\approx 1.42$ )	Zero at 0 K (behaves as an insulator); rises steeply with $T$ as carriers are thermally generated
Insulator	Valence band full, conduction band empty, large gap	Typically above about 4 (e.g. diamond $\approx 5.5$ , SiO $_{2}$ $\approx 9$ )	Negligible at ordinary temperatures

EXAM TIP

In this simple band model a semiconductor and an insulator differ only in the size of $E_{g}$ , and the boundary between them is a convention, not a sharp line: wide-gap materials such as GaN ( $E_{g} \approx 3.4 eV$ ) are used as semiconductors. Consequence: a semiconductor's resistance falls as temperature rises (negative temperature coefficient), while a metal's rises.

3. Intrinsic semiconductors, Fermi–Dirac statistics and effective density of states

A pure (intrinsic) semiconductor has no impurities: every electron thermally excited into the conduction band leaves exactly one hole in the valence band, so the electron and hole concentrations are equal, $n = p = n_{i}$ . The probability that a state of energy $E$ is occupied by an electron at temperature $T$ is given by the Fermi–Dirac distribution, and the energy at which this probability equals $\frac{1}{2}$ is the Fermi level $E_{F}$ .

FERMI–DIRAC DISTRIBUTION

f (E) = \frac{1}{1 + exp ( \frac{E - E _{F}}{k T} )}

Probability that an allowed state at energy $E$ is occupied by an electron, in thermal equilibrium. $k = 8.62 \times 1 0^{- 5} eV/K$ . For any $T > 0$ , $f (E_{F}) = \frac{1}{2}$ . At $T = 0 K$ , $f = 1$ for $E < E_{F}$ and $f = 0$ for $E > E_{F}$ .

PROBABILITY THAT A STATE IS EMPTY (HOLE OCCUPANCY)

1 - f (E) = \frac{1}{1 + exp ( \frac{E _{F} - E}{k T} )}

Used for holes in the valence band; follows directly from $f (E)$ .

When $E - E_{F} \geq 3 k T$ , the Fermi–Dirac function reduces to the Maxwell–Boltzmann tail $f (E) \approx exp [- (E - E_{F}) / k T]$ . Likewise, for $E_{F} - E \geq 3 k T$ , $1 - f (E) \approx exp [- (E_{F} - E) / k T]$ . A semiconductor whose $E_{F}$ lies at least about $3 k T$ below $E_{c}$ and at least about $3 k T$ above $E_{v}$ is called non-degenerate, and the simple exponential expressions for $n_{0}$ and $p_{0}$ below hold only in that case. Very heavy doping can push $E_{F}$ close to or into a band (the degenerate case), where they fail.

DENSITY OF STATES NEAR THE CONDUCTION-BAND EDGE

g_{c} (E) = \frac{4 π ( 2 m _{n}^{*} ) ^{3/2}}{h ^{3}} E - E_{c}, E \geq E_{c}

Number of states per unit volume per unit energy for a parabolic band. $g_{v} (E)$ has the same form with $m_{p}^{*}$ and $E_{v} - E$ for $E \leq E_{v}$ . The electron concentration is $n_{0} = \int g_{c} (E) f (E) d E$ taken over the conduction band, which gives the $N_{c}$ expression below.

EQUILIBRIUM ELECTRON AND HOLE CONCENTRATIONS

n_{0} = N_{c} exp (- \frac{E _{c} - E _{F}}{k T}), p_{0} = N_{v} exp (- \frac{E _{F} - E _{v}}{k T})

Valid in thermal equilibrium for a non-degenerate semiconductor ( $E_{F}$ at least about $3 k T$ from $E_{c}$ and from $E_{v}$ ), whatever the doping. $N_{c}$ and $N_{v}$ are the effective densities of states of the conduction and valence bands.

EFFECTIVE DENSITIES OF STATES

N_{c} = 2 (\frac{2 π m _{n}^{*} k T}{h ^{2}})^{3/2}, N_{v} = 2 (\frac{2 π m _{p}^{*} k T}{h ^{2}})^{3/2}

In SI units these come out in m $^{- 3}$ ; divide by $1 0^{6}$ for cm $^{- 3}$ . Both scale as $(m^{*})^{3/2} T^{3/2}$ , so they depend on the carrier effective masses and rise with temperature. They are material properties, not constants: in a numerical problem, use the values given.

INTRINSIC CARRIER CONCENTRATION

n_{i} = N_{c} N_{v} exp (- \frac{E _{g}}{2 k T})

Holds in thermal equilibrium, non-degenerate; $E_{g} = E_{c} - E_{v}$ . For silicon at $T = 300 K$ the commonly used value is $n_{i} \approx 1.5 \times 1 0^{10} cm^{- 3}$ (some texts use $1.0 \times 1 0^{10} cm^{- 3}$ ). $n_{i}$ depends on the material and the temperature, so always use the value the problem states.

TEMPERATURE DEPENDENCE OF $N_I$

n_{i}^{2} = N_{c} N_{v} exp (- \frac{E _{g}}{k T}) \propto T^{3} exp (- \frac{E _{g}}{k T})

This follows from $N_{c} N_{v} \propto T^{3}$ , and the exponential factor dominates. Taking $E_{g}$ as constant, two temperatures are related by $\frac{n _{i} ( T _{2} )}{n _{i} ( T _{1} )} = (\frac{T _{2}}{T _{1}})^{3/2} exp [\frac{E _{g}}{2 k} (\frac{1}{T _{1}} - \frac{1}{T _{2}})]$ .

In an intrinsic sample $E_{F}$ sits at the intrinsic level $E_{i}$ . It is exactly at mid-gap only if $N_{c} = N_{v}$ (that is, $m_{n}^{*} = m_{p}^{*}$ ) or at $T = 0 K$ . Otherwise it is displaced slightly toward the band with the smaller effective mass (smaller density of states).

INTRINSIC FERMI LEVEL

E_{i} = \frac{E _{c} + E _{v}}{2} + \frac{k T}{2} ln (\frac{N _{v}}{N _{c}}) = \frac{E _{c} + E _{v}}{2} + \frac{3}{4} k T ln (\frac{m _{p}^{*}}{m _{n}^{*}})

Obtained by setting $n_{0} = p_{0}$ in the Boltzmann expressions. If $m_{p}^{*} > m_{n}^{*}$ , $E_{i}$ lies slightly above mid-gap. The shift is of order $k T$ , small compared with $E_{g}$ .

CARRIER CONCENTRATIONS REFERRED TO $E_I$

n_{0} = n_{i} exp (\frac{E _{F} - E _{i}}{k T}), p_{0} = n_{i} exp (\frac{E _{i} - E _{F}}{k T})

Equivalent to the $N_{c}$ , $N_{v}$ forms and valid under the same conditions (equilibrium, non-degenerate, any doping). Multiplying the two gives $n_{0} p_{0} = n_{i}^{2}$ directly.

4. Extrinsic (doped) semiconductors and the Fermi level

Adding controlled impurities (doping) changes the carrier concentrations by orders of magnitude without changing $E_{g}$ . A donor (a Group V atom such as P or As in Si) introduces a level just below $E_{c}$ , a few tens of meV below it in Si. Its extra electron is easily ionised into the conduction band, making the material n-type. An acceptor (a Group III atom such as B or Ga in Si) introduces a level just above $E_{v}$ . It captures an electron from the valence band and leaves a hole, making the material p-type. Electrons are the majority carriers in n-type material and holes are the majority in p-type; the other carrier in each case is the minority carrier.

Identify the dopant concentrations $N_{d}$ (donors/cm $^{3}$ ) and $N_{a}$ (acceptors/cm $^{3}$ ). Take both as fully ionised at room temperature unless the problem says otherwise.
Find the majority-carrier concentration from charge neutrality (Section 5). If both dopants are present (compensation), use the net doping $N_{d} - N_{a}$ , not $N_{d}$ alone.
The approximation $n_{0} \approx N_{d} - N_{a}$ (or $p_{0} \approx N_{a} - N_{d}$ ) is valid only when $∣ N_{d} - N_{a} ∣ ≫ n_{i}$ . If they are comparable, for example at high temperature, use the exact quadratic solution in Section 5.
Find the minority-carrier concentration from the mass-action law, $p_{0} = n_{i}^{2} / n_{0}$ or $n_{0} = n_{i}^{2} / p_{0}$ .
Locate $E_{F}$ from the majority-carrier concentration using the formulas below.

FERMI LEVEL POSITION, N-TYPE

E_{F} - E_{i} = k T ln (\frac{n _{0}}{n _{i}}), E_{c} - E_{F} = k T ln (\frac{N _{c}}{n _{0}})

From the Boltzmann expressions; non-degenerate, equilibrium. $E_{F}$ lies above $E_{i}$ when $n_{0} > n_{i}$ .

FERMI LEVEL POSITION, P-TYPE

E_{i} - E_{F} = k T ln (\frac{p _{0}}{n _{i}}), E_{F} - E_{v} = k T ln (\frac{N _{v}}{p _{0}})

From the Boltzmann expressions; non-degenerate, equilibrium. $E_{F}$ lies below $E_{i}$ when $p_{0} > n_{i}$ .

WORKED EXAMPLE

A silicon sample at 300 K is doped with $N_{d} = 1 \times 1 0^{16} cm^{- 3}$ donors (fully ionised, no acceptors). Take $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ and $k T = 0.0259 eV$ . Where is $E_{F}$ relative to $E_{i}$ ?

Solution

$N_{d} ≫ n_{i}$ , so $n_{0} \approx N_{d} = 1 \times 1 0^{16} cm^{- 3}$ .
$n_{0} / n_{i} = 1 0^{16} / (1.5 \times 1 0^{10}) = 6.67 \times 1 0^{5}$ .
$ln (6.67 \times 1 0^{5}) \approx 13.41$ .
$E_{F} - E_{i} = 0.0259 \times 13.41 \approx 0.347 eV$ .

Answer: $E_{F}$ lies about $0.347 eV$ above $E_{i}$ (toward $E_{c}$ ).

Because more donors raise $n_{0}$ , the formulas above show that a higher $N_{d}$ moves $E_{F}$ closer to $E_{c}$ (further above $E_{i}$ ). Symmetrically, a higher $N_{a}$ moves $E_{F}$ closer to $E_{v}$ . The change is only logarithmic: each tenfold increase in doping moves $E_{F}$ by $k T ln 10 \approx 0.06 eV$ at 300 K. In a non-degenerate semiconductor $E_{F}$ stays inside the gap, at least about $3 k T$ from either band edge.

5. Mass-action law, charge neutrality and compensated doping

MASS-ACTION LAW

n_{0} p_{0} = n_{i}^{2}

Holds in thermal equilibrium in a non-degenerate semiconductor, doped or undoped, with $n_{i}$ evaluated at the same temperature. It does not hold under illumination or carrier injection (non-equilibrium).

CHARGE NEUTRALITY (GENERAL)

n_{0} + N_{a}^{-} = p_{0} + N_{d}^{+}

Total negative charge equals total positive charge in the bulk, where $N_{a}^{-}$ and $N_{d}^{+}$ are the ionised acceptor and donor concentrations. With full ionisation, $N_{a}^{-} = N_{a}$ and $N_{d}^{+} = N_{d}$ .

COMPENSATED DOPING (FULL IONISATION)

n_{0} - p_{0} = N_{d} - N_{a}

Combine with $n_{0} p_{0} = n_{i}^{2}$ to solve for $n_{0}$ and $p_{0}$ . If $∣ N_{d} - N_{a} ∣ ≫ n_{i}$ , the material is n-type when $N_{d} > N_{a}$ (then $n_{0} \approx N_{d} - N_{a}$ ) and p-type when $N_{a} > N_{d}$ (then $p_{0} \approx N_{a} - N_{d}$ ); the minority carrier follows from mass action.

EXACT EQUILIBRIUM SOLUTION ($N_D > N_A$)

n_{0} = \frac{N _{d} - N _{a}}{2} + (\frac{N _{d} - N _{a}}{2})^{2} + n_{i}^{2}, p_{0} = \frac{n _{i}^{2}}{n _{0}}

This is the positive root of $n_{0}^{2} - (N_{d} - N_{a}) n_{0} - n_{i}^{2} = 0$ (full ionisation, non-degenerate). For $N_{a} > N_{d}$ swap the roles: $p_{0} = \frac{N _{a} - N _{d}}{2} + (\frac{N _{a} - N _{d}}{2})^{2} + n_{i}^{2}$ . Needed whenever $∣ N_{d} - N_{a} ∣$ is not much larger than $n_{i}$ . With $N_{d} = N_{a}$ it gives $n_{0} = p_{0} = n_{i}$ .

WORKED EXAMPLE

A silicon sample is doped with $N_{d} = 2 \times 1 0^{15} cm^{- 3}$ phosphorus atoms and no acceptors, fully ionised, at 300 K where $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Find $n_{0}$ and $p_{0}$ .

Solution

$N_{a} = 0$ , so neutrality gives $n_{0} - p_{0} = N_{d} = 2 \times 1 0^{15} cm^{- 3}$ .
Since $N_{d} ≫ n_{i}$ , $p_{0} ≪ n_{0}$ , so $n_{0} \approx N_{d} = 2 \times 1 0^{15} cm^{- 3}$ (majority carrier).
Mass action gives the minority carrier: $p_{0} = n_{i}^{2} / n_{0} = (1.5 \times 1 0^{10})^{2} / (2 \times 1 0^{15})$ .
$p_{0} = 2.25 \times 1 0^{20} / (2 \times 1 0^{15}) = 1.125 \times 1 0^{5} cm^{- 3}$ .

Answer: $n_{0} \approx 2 \times 1 0^{15} cm^{- 3}$ , $p_{0} \approx 1.125 \times 1 0^{5} cm^{- 3}$ (n-type).

WORKED EXAMPLE

Another silicon sample is doped with both $N_{d} = 8 \times 1 0^{15} cm^{- 3}$ donors and $N_{a} = 3 \times 1 0^{15} cm^{- 3}$ acceptors (compensated doping), both fully ionised, at 300 K ( $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ ). Find the majority- and minority-carrier concentrations.

Solution

Net doping: $N_{d} - N_{a} = 8 \times 1 0^{15} - 3 \times 1 0^{15} = 5 \times 1 0^{15} cm^{- 3} > 0$ , so the material is n-type.
Since $N_{d} - N_{a} ≫ n_{i}$ : $n_{0} \approx N_{d} - N_{a} = 5 \times 1 0^{15} cm^{- 3}$ .
Mass action gives the minority holes: $p_{0} = n_{i}^{2} / n_{0} = (1.5 \times 1 0^{10})^{2} / (5 \times 1 0^{15}) = 2.25 \times 1 0^{20} / (5 \times 1 0^{15})$ .
$p_{0} = 4.5 \times 1 0^{4} cm^{- 3}$ .

Answer: $n_{0} \approx 5 \times 1 0^{15} cm^{- 3}$ (majority, electrons), $p_{0} \approx 4.5 \times 1 0^{4} cm^{- 3}$ (minority, holes).

EXAM TIP

In compensated material, use the net doping $N_{d} - N_{a}$ for the majority carrier, never $N_{d}$ or $N_{a}$ alone. Compensation cancels only the ionised charge in the neutrality equation. All $N_{d} + N_{a}$ ionised impurities still scatter carriers, so a compensated sample does not regain the mobility of a lightly doped one.

6. Temperature dependence of $n_i$ and $E_F$

$n_{i} \propto T^{3/2} exp (- E_{g} /2 k T)$ rises steeply with $T$ . The exponential factor dominates the weaker $T^{3/2}$ factor, and $E_{g}$ itself also decreases slightly as $T$ rises, which adds to the increase. A doped sample passes through three ranges as $T$ rises. At low $T$ it is in freeze-out: not all dopants are ionised, so $n_{0} < N_{d} - N_{a}$ . Over a wide middle range it is in the extrinsic (saturation) range: $n_{0} \approx N_{d} - N_{a}$ , essentially constant with $T$ , which is the normal operating range for devices. At high enough $T$ , $n_{i}$ grows until it is comparable to $∣ N_{d} - N_{a} ∣$ and the sample enters the intrinsic range, where doping no longer controls the carrier concentration.

THERMAL VOLTAGE

V_{T} = \frac{k T}{q}

$V_{T} \approx 25.9 mV$ at $T = 300 K$ , with $k = 1.38 \times 1 0^{- 23} J/K$ and $q = 1.6 \times 1 0^{- 19} C$ . Equivalently $k T \approx 0.0259 eV$ at 300 K, so an energy difference in eV divided by $k T$ in eV gives the dimensionless exponent in the Boltzmann expressions directly.

As $T$ rises from the extrinsic range toward the intrinsic range, the Fermi level of a doped sample moves away from $E_{c}$ (n-type) or $E_{v}$ (p-type) and approaches $E_{i}$ . The reason: $n_{i}$ grows rapidly while the dopant concentration stays fixed, so $n_{0} / n_{i}$ (or $p_{0} / n_{i}$ ) falls toward 1, and $E_{F} - E_{i} = k T ln (n_{0} / n_{i})$ falls toward zero.

7. Direct and indirect band-gap semiconductors

Band structure is drawn on an energy versus crystal-momentum ( $E$ – $k$ ) diagram. In a direct band-gap material the conduction-band minimum and the valence-band maximum occur at the same $k$ ; in an indirect band-gap material they occur at different $k$ . The curvature of the band near an edge sets the carrier effective mass.

EFFECTIVE MASS FROM THE $E$–$K$ DIAGRAM

m^{*} = \frac{ℏ ^{2}}{( d ^{2} E / d k ^{2} )}

Evaluated at the band edge, with $ℏ = h /2 π$ . A sharply curved band gives a small effective mass. For a parabolic conduction band, $E = E_{c} + \frac{ℏ ^{2} k ^{2}}{2 m _{n}^{*}}$ .

Property	Direct gap (e.g. GaAs, InP)	Indirect gap (e.g. Si, Ge, GaP)
Band extrema in $k$ -space	Same $k$	Different $k$
Band-to-band recombination	Energy $\approx E_{g}$ carried off by a photon; the photon's momentum is negligible and no momentum change is needed	A phonon (lattice vibration) must also be emitted or absorbed to supply the difference in crystal momentum
Radiative recombination efficiency	High	Low; most recombination is non-radiative, through defect (trap) levels in the gap
Optical absorption just above $E_{g}$	Strong	Weak (phonon-assisted); becomes strong only at higher photon energies
Typical use	LEDs, laser diodes	Transistors, ICs, and also photodetectors and solar cells; not used as efficient light emitters

BAND-GAP WAVELENGTH

λ = \frac{h c}{E _{g}} \Rightarrow λ (μ m) \approx \frac{1.24}{E _{g} ( eV )}

The wavelength of a photon of energy $E_{g}$ . It gives the approximate emission wavelength of band-to-band recombination and the longest wavelength (cut-off) the material absorbs band-to-band. Uses $h c \approx 1.24 eV μ m$ .

WORKED EXAMPLE

Estimate the emission wavelength of a GaAs LED, taking $E_{g} = 1.42 eV$ .

Solution

$λ \approx 1.24/ E_{g}$ with $E_{g}$ in eV and $λ$ in $μ$ m.
$λ \approx 1.24/1.42 \approx 0.873 μ m$ .

Answer: $λ \approx 0.873 μ m$ (about 873 nm, near-infrared).

EXAM TIP

The causal chain to know: indirect gap → band-to-band recombination needs a phonon → radiative recombination is improbable → poor light emitter. Silicon is a poor LED material because its gap is indirect, not because of the size of its gap.

PREVIOUS-YEAR QUESTIONS

9 questions from the EC papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2022 · GATE ECE 2020 · GATE ECE 2018

GATE 2025 EC Q.31

GATE 2025 · ECQ.31 · 1 mark · MSQ

Which of the following can be used as an n-type dopant for silicon? Select the correct option(s).

(A)
Arsenic
(B)
Boron
(C)
Gallium
(D)
Phosphorous

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GATE 2024 EC Q.22

GATE 2024 · ECQ.22 · 1 mark · MCQ

For non-degenerately doped n-type silicon, which one of the following plots represents the temperature ( $T$ ) dependence of free electron concentration ( $n$ )?

GATE 2024 EC question 22: Four small plots, each with axes log (n) (cm^-3) on the vertical axis and 1/T (K^-1) on the horizontal axis (increasing… — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
Figure — plot showing (left to right): a decreasing curve from high $lo g (n)$ , then a flat plateau, then another decreasing curve near the right end (three-region S-shaped curve)
(B)
Figure — plot showing a single flat horizontal line (constant $lo g (n)$ across the whole range)
(C)
Figure — plot showing a flat plateau on the left that bends downward only near the right end (two-region curve, no intrinsic rise on the left)
(D)
Figure — plot showing a single straight line with constant negative slope from upper-left to lower-right (simple monotonic decrease)

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GATE 2024 EC Q.22: answer and solution

GATE 2024 EC Q.63

GATE 2024 · ECQ.63 · 2 marks · NAT

A non-degenerate n-type semiconductor has 5% neutral dopant atoms. Its Fermi level is located at 0.25 eV below the conduction band ( $E_{C}$ ) and the donor energy level ( $E_{D}$ ) has a degeneracy of 2. Assuming the thermal voltage to be 20 mV, the difference between $E_{C}$ and $E_{D}$ (in eV, rounded off to two decimal places) is ____.

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GATE 2023 EC Q.16

GATE 2023 · ECQ.16 · 1 mark · MCQ

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

(A)
degenerate n-type.
(B)
degenerate p-type.
(C)
non-degenerate n-type.
(D)
non-degenerate p-type.

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GATE 2023 EC Q.17

GATE 2023 · ECQ.17 · 1 mark · MCQ

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

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

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GATE 2023 EC Q.61

GATE 2023 · ECQ.61 · 2 marks · NAT

In a semiconductor device, the Fermi-energy level is 0.35 eV above the valence band energy. The effective density of states in the valence band at $T = 300 K$ is $1 \times 1 0^{19} cm^{- 3}$ . The thermal equilibrium hole concentration in silicon at 400 K is _______ $\times 1 0^{13} cm^{- 3}$ (rounded off to two decimal places). Given kT at 300 K is 0.026 eV.

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GATE 2022 EC Q.17

GATE 2022 · ECQ.17 · 1 mark · MCQ

In a non-degenerate bulk semiconductor with electron density $n = 1 0^{16} cm^{- 3}$ , the value of $E_{C} - E_{F n} = 200 meV$ , where $E_{C}$ and $E_{F n}$ denote the bottom of the conduction band energy and electron Fermi level energy, respectively. Assume thermal voltage as 26 meV and the intrinsic carrier concentration is $1 0^{10} cm^{- 3}$ . For $n = 0.5 \times 1 0^{16} cm^{- 3}$ , the closest approximation of the value of $(E_{C} - E_{F n})$ , among the given options, is ____.

(A)
226 meV
(B)
174 meV
(C)
218 meV
(D)
182 meV

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GATE 2020 EC Q.6

GATE 2020 · ECQ.6 · 1 mark · MCQ

A single crystal intrinsic semiconductor is at a temperature of 300 K with effective density of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic Fermi level is shifted from mid-bandgap energy level by

(A)
18.02 meV.
(B)
9.01 meV.
(C)
13.45 meV.
(D)
26.90 meV.

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GATE 2018 EC Q.45

GATE 2018 · ECQ.45 · 2 marks · NAT

A junction is made between $p^{-}$ Si with doping density $N_{A 1} = 1 0^{15} cm^{- 3}$ and $p$ Si with doping density $N_{A 2} = 1 0^{17} cm^{- 3}$ .

Given: Boltzmann constant $k = 1.38 \times 1 0^{- 23} J\cdotK^{- 1}$ , electronic charge $q = 1.6 \times 1 0^{- 19} C$ . Assume 100% acceptor ionization.

At room temperature ( $T = 300 K$ ), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be ____.

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PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

In an indirect band-gap semiconductor, radiative (photon-emitting) recombination of an electron–hole pair across the band gap necessarily also requires:

(A)
Nothing beyond the photon; energy conservation alone suffices
(B)
A phonon, to conserve crystal momentum in addition to the photon carrying away the energy
(C)
A second photon of equal and opposite momentum
(D)
The electron and hole to have equal effective mass

Answer and solution

Answer: (B)

In an indirect-gap material the conduction-band minimum and the valence-band maximum lie at different crystal momentum $k$ .
A photon of energy near $E_{g}$ carries negligible momentum compared with this difference, so the photon alone cannot conserve momentum.
A phonon supplies or absorbs the momentum difference. The photon carries away the energy, which is $E_{g}$ minus or plus the small phonon energy. Both must take part in the same event.
Hence B. Because this process involves three particles, it is much less probable than direct-gap recombination.

PRACTICE 21 mark · MCQ

An n-type silicon sample is in the extrinsic (saturation) range at moderate temperature. Its temperature is then raised further toward the intrinsic range, with the doping unchanged. Over this rise, the Fermi level $E_{F}$ , measured relative to the intrinsic level $E_{i}$ :

(A)
Moves further away from $E_{i}$ , toward $E_{c}$
(B)
Moves toward $E_{i}$
(C)
Stays exactly fixed at its extrinsic-range position
(D)
Jumps discontinuously to $E_{v}$

Answer and solution

Answer: (B)

In equilibrium, $E_{F} - E_{i} = k T ln (n_{0} / n_{i})$ .
With full ionisation, $n_{0} = \frac{N _{d}}{2} + (\frac{N _{d}}{2})^{2} + n_{i}^{2}$ . In the extrinsic range $n_{i} ≪ N_{d}$ , so $n_{0} \approx N_{d} ≫ n_{i}$ and $E_{F}$ lies well above $E_{i}$ .
As $T$ rises, $n_{i}$ grows rapidly (the $exp (- E_{g} /2 k T)$ factor) while $N_{d}$ is fixed, so $n_{0} / n_{i}$ falls toward 1 and $ln (n_{0} / n_{i})$ falls toward 0.
So $E_{F} - E_{i}$ decreases and $E_{F}$ moves toward $E_{i}$ : option B.

PRACTICE 31 mark · MSQ

Which of the following statements about the effective densities of states $N_{c}$ and $N_{v}$ are correct?

(A)
$N_{c}$ depends on the conduction-band electron effective mass $m_{n}^{*}$ , and $N_{v}$ on the valence-band hole effective mass $m_{p}^{*}$
(B)
$N_{c}$ and $N_{v}$ are independent of temperature
(C)
$N_{c}$ and $N_{v}$ increase with temperature approximately as $T^{3/2}$
(D)
$N_{c}$ and $N_{v}$ depend on the applied bias across the sample, not on the material

Answer and solution

Answer: (A), (C)

$N_{c} = 2 (2 π m_{n}^{*} k T / h^{2})^{3/2}$ and $N_{v} = 2 (2 π m_{p}^{*} k T / h^{2})^{3/2}$ , so A and C are correct.
B is false: both scale as $T^{3/2}$ .
D is false: $N_{c}$ and $N_{v}$ are material properties (effective mass) that also depend on temperature. They do not depend on an applied bias.

PRACTICE 42 marks · NAT

A silicon sample is doped with $N_{d} = 1 \times 1 0^{16} cm^{- 3}$ donors and no acceptors, fully ionised, at $300 K$ where $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Calculate the equilibrium minority-carrier (hole) concentration $p_{0}$ in cm $^{- 3}$ .

Answer and solution

Answer: $22500$ (accepted range $\pm 500$ )

Since $N_{d} ≫ n_{i}$ , the sample is n-type and $n_{0} \approx N_{d} = 1 \times 1 0^{16} cm^{- 3}$ .
Apply the mass-action law: $p_{0} = n_{i}^{2} / n_{0}$ .
$p_{0} = (1.5 \times 1 0^{10})^{2} / (1 \times 1 0^{16}) = 2.25 \times 1 0^{20} / (1 \times 1 0^{16})$ .
$p_{0} = 2.25 \times 1 0^{4} cm^{- 3} = 22500 cm^{- 3}$ .

PRACTICE 52 marks · NAT

A silicon sample at $300 K$ is compensation-doped with $N_{d} = 5 \times 1 0^{16} cm^{- 3}$ donors and $N_{a} = 2 \times 1 0^{16} cm^{- 3}$ acceptors, both fully ionised. Take $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ and $k T = 0.0259 eV$ . Calculate $E_{F} - E_{i}$ in eV (positive if $E_{F}$ lies above $E_{i}$ ).

Answer and solution

Answer: $0.376$ (accepted range $\pm 0.005$ )

Net doping: $N_{d} - N_{a} = 5 \times 1 0^{16} - 2 \times 1 0^{16} = 3 \times 1 0^{16} cm^{- 3}$ , so the material is n-type.
Since $N_{d} - N_{a} ≫ n_{i}$ , $n_{0} \approx 3 \times 1 0^{16} cm^{- 3}$ . (Check: $p_{0} = n_{i}^{2} / n_{0} = 2.25 \times 1 0^{20} / (3 \times 1 0^{16}) = 7.5 \times 1 0^{3} cm^{- 3}$ , negligible.)
$E_{F} - E_{i} = k T ln (n_{0} / n_{i}) = 0.0259 ln (\frac{3 \times 1 0 ^{16}}{1.5 \times 1 0 ^{10}}) = 0.0259 ln (2 \times 1 0^{6})$ .
$ln (2 \times 1 0^{6}) = 0.693 + 13.816 = 14.509$ , so $E_{F} - E_{i} = 0.0259 \times 14.509 \approx 0.376 eV$ .

PRACTICE 61 mark · NAT

Silicon has a band gap $E_{g} = 1.12 eV$ at 300 K. Using $h c = 1.24 eV μ m$ , calculate the longest wavelength (in $μ$ m) of light that can generate electron–hole pairs by band-to-band absorption in silicon.

Answer and solution

Answer: $1.107$ (accepted range $\pm 0.01$ )

Band-to-band absorption needs a photon energy of at least $E_{g}$ , so the cut-off wavelength is $λ_{c} = h c / E_{g}$ .
$λ_{c} = 1.24/1.12 \approx 1.107 μ m$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

Formation of energy bands in solids, energy bands in intrinsic and extrinsic semiconductors, equilibrium carrier concentration, direct and indirect band-gap semiconductors

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