Semiconductor Electronics

Answer: (a)

1. In a semiconductor the band gap is small (about $1\text{ eV}$).
2. At room temperature some electrons get thermal energy to jump from the valence band to the conduction band.
3. They leave holes behind, so the valence band is partially empty and the conduction band is partially filled.
4. So the answer is (a).

Answer: (c)

1. Insulators have a very large band gap ($E_g > 3\text{ eV}$), so almost no electrons reach the conduction band.
2. A very few electrons can still be excited, so conductivity is not exactly zero but extremely small.
3. So the answer is (c) extremely small.

Answer: (a)

1. In an insulator the band gap is very large, $E_g > 3\text{ eV}$.
2. Thermal energy cannot lift electrons across it, so the conduction band is empty.
3. So the answer is (a) very high, empty.

Answer: (b)

1. A hole forms only when an electron leaves an otherwise filled valence band across a gap.
2. In good conductors the valence and conduction bands overlap, so there is no gap and no holes are created.
3. So the answer is (b).

Answer: (d)

1. Small band gap ($\sim 1\text{ eV}$) lets some valence electrons jump to the conduction band at room temperature.
2. This leaves holes in the valence band (partially empty) and electrons in the conduction band (partially filled).
3. So the answer is (d).

Answer: (c)

1. At room temperature thermal energy excites some electrons across the small gap of a semiconductor.
2. The conduction band becomes partially filled and the valence band partially empty (holes).
3. So the answer is (c).

Answer: (c)

1. At room temperature some valence electrons cross the small band gap into the conduction band.
2. So the conduction band is partially filled and the valence band partially empty.
3. So the answer is (c).

Answer: (c)

1. Insulators have a very large band gap ($E_g > 3\text{ eV}$).
2. No electrons can cross it, so the conduction band is empty.
3. So the answer is (c) very high and empty.

Answer: (b)

1. In an insulator all valence electrons are tightly bound, so the valence band is completely filled.
2. The large gap keeps electrons from reaching the conduction band, so it stays empty.
3. So the answer is (b).

Answer: (b)

1. Resistivity is smallest for a good conductor.
2. Silver is a metal (best conductor); Ge and Si are semiconductors; glass is an insulator.
3. So the answer is (b) Silver.

Answer: (a)

1. In conductors (metals) the valence and conduction bands overlap; there is no band gap.
2. So a large number of free electrons is always available for conduction.
3. So the answer is (a).

Answer: (b)

1. Semiconductors have a small band gap, about $1\text{ eV}$ (Si $1.1\text{ eV}$, Ge $0.7\text{ eV}$).
2. Large gap means insulator; overlapping bands means conductor.
3. So the answer is (b) separated by a small energy gap.

Answer: (a)

1. Insulators: band gap very large ($> 3\text{ eV}$).
2. Electrons cannot cross it, so the conduction band is empty.
3. So the answer is (a) very high, empty.

Answer: (b)

1. (a) True: in conductors the bands generally overlap.
2. (b) False: resistivity of a semiconductor decreases as temperature rises (more electron-hole pairs).
3. (c) True: conductivity increases with temperature.
4. (d) True: $E_g \sim 10\text{ eV}$ means an insulator.
5. So the false statement is (b).

Answer: (c)

1. C ($1s^2 2s^2 2p^2$) has its four bonding electrons in the 2nd orbit; Si ($\ldots 3s^2 3p^2$) has them in the 3rd orbit.
2. Electrons in the 2nd orbit are closer to the nucleus and more tightly bound, so the band gap of C (about $5.4\text{ eV}$) is much larger than that of Si ($1.1\text{ eV}$).
3. Hence C is an insulator and Si a semiconductor.
4. So the answer is (c).

Answer: (a)

1. Band gaps: C $\approx 5.4\text{ eV}$, Si $1.1\text{ eV}$, Ge $0.7\text{ eV}$; $kT \approx 0.025\text{ eV}$ at room temperature.
2. Small gaps of Si and Ge allow a significant number of electrons to be thermally excited; the large gap of C allows almost none.
3. So the answer is (a).

Answer: (c)

1. In the figure $E_c$ and $E_v$ are the widths of the conduction and valence bands, $E_g$ is the gap between them.
2. Smaller lattice constant means atoms are closer, so their wave functions overlap more.
3. Greater overlap spreads each level over a wider range: $E_c$ and $E_v$ increase.
4. Wider bands come closer to each other, so the gap $E_g$ decreases.
5. This gives (b); the answer key gives (c).

Answer: (c)

1. (a) True: in conductors the bands overlap.
2. (b) True: $E_g \sim 10\text{ eV}$ is far above $3\text{ eV}$, so such substances are insulators.
3. (c) False: as $T$ rises more electron-hole pairs form, so resistivity of a semiconductor decreases.
4. (d) True: conductivity increases with temperature.
5. So the false statement is (c).

Answer: (a)

1. Band gaps: carbon (diamond) $E_g \approx 5.4\text{ eV}$, Si $1.1\text{ eV}$, Ge $0.7\text{ eV}$.
2. So $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$.
3. So the answer is (a) $(E_g)_C > (E_g)_{Si}$.

Answer: (d)

1. Conductivity rises only when photon energy $\ge E_g$, so $E_g = \dfrac{hc}{\lambda_{max}}$.
2. $E_g(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})} = \dfrac{1240}{2480}$
3. $E_g = 0.5\text{ eV}$
4. So the answer is (d) $0.5\text{ eV}$.

Answer: (b)

1. When atoms come together in a solid, their identical energy levels cannot hold the same electrons in the same state.
2. Pauli's exclusion principle forces each level to split into many closely spaced levels, which form a band.
3. So the answer is (b) Pauli's exclusion principle.

Answer: (a)

1. At room temperature some electrons cross the small gap of a semiconductor.
2. Valence band becomes partially empty (holes) and conduction band partially filled.
3. So the answer is (a).

Answer: (c)

1. Metals and superconductors: bands overlap, $E_g = 0$.
2. Semiconductors: $E_g \approx 1\text{ eV}$; insulators: $E_g > 3\text{ eV}$.
3. So the answer is (c) insulator.

Answer: (a)

1. In an intrinsic semiconductor every broken covalent bond gives one free electron and one hole.
2. So $n_e = n_h = n_i$ (non-zero at ordinary temperature).
3. So the answer is (a) $n_e = n_h$.

Answer: (c)

1. Each electron that leaves a bond in a pure semiconductor leaves one hole behind.
2. So $n_h = n_e$.
3. So the answer is (c).

Answer: (b)

1. Conduction only from breaking of covalent bonds means no impurity carriers are present.
2. A pure (undoped) semiconductor is called intrinsic.
3. So the answer is (b) intrinsic.

Answer: (d)

1. A trivalent acceptor atom needs one extra electron to complete its bonds.
2. Its empty level lies in the gap, just above the valence band (about $0.01$ to $0.05\text{ eV}$), so valence electrons easily jump into it, leaving holes.
3. So the answer is (d) just above the valence band.

Answer: (a)

1. A p-type semiconductor is doped with a trivalent impurity that creates holes.
2. So holes are majority carriers and electrons are minority carriers.
3. So the answer is (a) large number of holes and few electrons.

Answer: (c)

1. The fifth electron of a pentavalent donor is loosely bound.
2. Its energy level (donor level) lies in the band gap just below the conduction band, so it is easily excited into the conduction band.
3. So the answer is (c).

Answer: (b)

1. n-type: doped with a pentavalent impurity (P, As, Sb), which donates extra electrons.
2. So electrons are majority carriers and holes minority carriers; only statement C fits.
3. So the answer is (b) C.

Answer: (b)

1. Doping adds a large number of extra carriers (electrons or holes).
2. Conductivity $\sigma = e(n_e\mu_e + n_h\mu_h)$ increases, so resistivity decreases.
3. So the answer is (b) decreases.

Answer: (b)

1. Mass action law: $n_e n_h = n_i^2$
2. $n_e = \dfrac{(2\times10^{16})^2}{4\times10^{22}} = \dfrac{4\times10^{32}}{4\times10^{22}}$
3. $n_e = 10^{10}\ \text{m}^{-3}$
4. So the answer is (b) $10^{10}/\text{m}^3$.

Answer: (d)

1. Each P atom gives one electron, so $n_e \approx 10^{21}\ \text{m}^{-3}$ (much larger than $n_i$).
2. Mass action law: $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(10^{16})^2}{10^{21}}$
3. $n_h = 10^{11}\ \text{m}^{-3}$
4. So the answer is (d) $10^{11}$ per m$^3$.

Answer: (b)

1. Intrinsic: $E_F$ lies midway between C and V. n-type: $E_F$ near C. p-type: $E_F$ near V.
2. Figure (2): $E_F$ just above V, so p-type.
3. Figure (1): $E_F$ in the middle, so intrinsic. Figure (3): $E_F$ just below C, so n-type.
4. So (2), (1), (3) are p, intrinsic and n: the answer is (b).

Answer: (c)

1. n-type: pentavalent dopant gives extra electrons, so electrons are majority carriers.
2. Holes are then minority carriers.
3. So the answer is (c).

Answer: (c)

1. The donor electron of a pentavalent impurity is loosely bound.
2. Its level lies in the band gap, just below the conduction band.
3. So the answer is (c).

Answer: (b)

1. Phosphorus is a donor, so $n_e \approx 10^{21}\ \text{m}^{-3}$.
2. $n_h = \dfrac{n_i^2}{n_e} = \dfrac{10^{32}}{10^{21}} = 10^{11}\ \text{m}^{-3}$
3. So the answer is (b) $10^{11}$.

Answer: (c)

1. Raising $T$ breaks more covalent bonds, creating more electron-hole pairs.
2. More carriers lowers the resistance of the semiconductor, so $I = V/R$ rises.
3. So the answer is (c) increase.

Answer: (c)

1. Conductor: carrier number is fixed, but more lattice vibrations increase collisions, so $\rho$ increases.
2. Semiconductor: carrier number rises sharply with $T$, so $\rho$ decreases.
3. So the answer is (c) increases, decreases.

Answer: (a)

1. For a semiconductor, carrier density rises roughly exponentially with $T$, so $\rho$ falls steeply and then levels off.
2. Graph C shows this falling, curved (decaying) shape; A and B rise (metal-like) and D is a straight line.
3. So the answer is (a) C.

Answer: (a)

1. (A) In n-type, holes are minority carriers: true. (B) The dopant is pentavalent: true.
2. (C) Mass action law gives $n_e n_h = n_i^2$, so $n_e n_h \neq n_i^2$ is false; (D) is therefore not the intended relation either.
3. (E) Donors release only electrons; holes come from thermal pair generation: true.
4. So the answer is (a) (A), (B), (E) only.

Answer: (c)

1. Doping greatly increases the number of charge carriers.
2. So conductivity rises and resistivity decreases.
3. So the answer is (c) decreases.

Answer: (c)

1. Donor electrons of a pentavalent impurity occupy the donor level.
2. The donor level lies in the band gap, close to the conduction band (about $0.01$ to $0.05\text{ eV}$ below it).
3. So the answer is (c).

Answer: (d)

1. p-type: trivalent dopant (B, Al, Ga, In) creates a hole for each atom.
2. So holes are majority carriers and electrons minority carriers.
3. So the answer is (d).

Answer: (a)

1. Phosphorus is pentavalent, so heavy doping gives a strongly n-type material.
2. Since $n_e n_h = n_i^2$, a very large $n_e$ makes $n_h$ very small.
3. So the answer is (a) $n_e \gg n_h$.

Answer: (d)

1. To create a hole, a photon must lift a valence electron to the acceptor level: $E = \dfrac{hc}{\lambda} \ge 6$ eV.
2. $\lambda_{max} = \dfrac{hc}{E} = \dfrac{1242}{6}$ nm
3. $\lambda_{max} = 207$ nm
4. So the answer is (d) 207 nm.

Answer: (a)

1. At balance (no deflection): $\dfrac{R_{AB}}{R_{BC}} = \dfrac{R_{AD}}{R_{DC}} \Rightarrow \dfrac{0.8}{R_{BC}} = \dfrac{1}{3}$, so $R_{BC} = 2.4\ \text{m}\Omega$.
2. Temperature change in 10 s: $\Delta T = -2 \times 10 = -20\ ^\circ\text{C}$.
3. $R = R_0(1 + \alpha\,\Delta T) \Rightarrow 2.4 = 3(1 - 20\alpha) \Rightarrow |\alpha| = \dfrac{0.6}{3 \times 20} = 1 \times 10^{-2}\ ^\circ\text{C}^{-1}$
4. A semiconductor has a negative temperature coefficient.
5. So the answer is (a) $-1 \times 10^{-2}\ ^\circ\text{C}^{-1}$.

Answer: (b)

1. Carriers produced only by breaking covalent bonds means a pure semiconductor with no dopant.
2. This is an intrinsic semiconductor.
3. So the answer is (b) intrinsic.

Answer: (b)

1. In an intrinsic semiconductor electrons and holes are created in pairs.
2. So $n_h = n_e$.
3. So the answer is (b).

Answer: (a)

1. Doping adds many extra charge carriers to the semiconductor.
2. More carriers mean higher conductivity, so resistivity decreases.
3. So the answer is (a) Decreases.

Answer: (d)

1. A p-type semiconductor needs a trivalent (acceptor) impurity.
2. Sb, P and As are pentavalent; boron is trivalent.
3. So the answer is (d) Boron.

Answer: (d)

1. Semiconductor resistivity falls with temperature because the number of carriers rises roughly as $e^{-E_g/2kT}$.
2. So $\rho$ is very large at low $T$ and decreases along a decaying curve that flattens out.
3. Graphs (a) and (c) rise with $T$, (b) is a straight line; only (d) shows the decaying curve.
4. So the answer is (d).

Answer: (c)

1. Higher $T$ gives more electrons enough energy to jump across the band gap, so $n_e$ in the conduction band increases.
2. More carriers means lower resistance.
3. So the answer is (c) $n_e$ increases, resistance decreases.

Answer: (d)

1. Intrinsic: equal $n_e$ and $n_h$, so $E_F$ is midway between the bands: A-II.
2. n-type: excess electrons push $E_F$ near the conduction band: B-III.
3. p-type: excess holes pull $E_F$ near the valence band: C-I. Metals: $E_F$ lies inside the conduction band: D-IV.
4. So the answer is (d) A-II, B-III, C-I, D-IV.

Answer: (c)

1. Each Sb (pentavalent) atom donates one free electron, so $n_e \approx N_D$.
2. 1 ppm means 1 dopant atom per $10^{6}$ Si atoms: $N_D = \dfrac{4 \times 10^{28}}{10^{6}} = 4 \times 10^{22}\ \text{m}^{-3}$
3. So the answer is (c) $4 \times 10^{22}\ \text{m}^{-3}$.

Answer: (c)

1. Conductors: more lattice vibration at higher $T$ increases collisions, so $R$ increases.
2. Semiconductors: more carriers are generated at higher $T$, so $R$ decreases.
3. So the answer is (c).

Answer: (a)

1. Statement I: pentavalent donors add free electrons, so electron density increases: true.
2. Statement II: each donor that gives an electron becomes a fixed positive ion, so the crystal stays electrically neutral: false.
3. So the answer is (a).

Answer: (b)

1. Mass action law: $n_e = \dfrac{n_i^2}{n_h}$
2. $n_e = \dfrac{(1.6\times10^{16})^2}{4\times10^{22}} = \dfrac{2.56\times10^{32}}{4\times10^{22}}$
3. $n_e = 6.4\times10^{9}\ \text{m}^{-3}$
4. So the answer is (b).

Answer: 5

1. Mass action law: $n_e n_h = n_i^2$
2. $n_e = \dfrac{(1.5\times10^{16})^2}{4.5\times10^{22}} = \dfrac{2.25\times10^{32}}{4.5\times10^{22}}$
3. $n_e = 5\times10^{9}\ /\text{m}^3$
4. So the answer is 5.

Answer: (d)

1. Higher $T$ creates more electron-hole pairs in a semiconductor.
2. So resistance decreases and conductivity $\sigma = e(n_e\mu_e + n_h\mu_h)$ increases.
3. So the answer is (d) decreases and increases.

Answer: (d)

1. More donor doping in n-type adds electrons, so $E_F$ moves up towards the conduction band.
2. More acceptor doping in p-type adds holes, so $E_F$ moves down towards the valence band.
3. So the answer is (d).

Answer: (d)

1. Current $I = neAv_d = neA\mu E$, so for equal carrier concentration and equal field, $I \propto \mu$.
2. Electrons are more mobile than holes: $\mu_e > \mu_h$.
3. So the answer is (d) current in n-type > current in p-type.

Answer: (c)

1. Copper (metal): on cooling, lattice vibrations reduce, collisions fall, so resistivity decreases.
2. Silicon (semiconductor): on cooling, fewer electrons cross the band gap, carrier number falls, so resistivity increases.
3. So the answer is (c) decreases in copper and increases in silicon.

Answer: (a)

1. Thermal breaking of each covalent bond creates one electron and one hole.
2. So $n_e = n_h$, both non-zero at ordinary temperature.
3. So the answer is (a) $n_e = n_h$.

Answer: (c)

1. Doping adds extra charge carriers (electrons from donors or holes from acceptors).
2. More carriers means higher conductivity $\sigma = e(n_e\mu_e + n_h\mu_h)$, so resistivity $\rho = 1/\sigma$ falls.
3. So the answer is (c) decreases.

Answer: (b)

1. An n-type semiconductor needs a pentavalent (donor) impurity.
2. Phosphorus has 5 valence electrons; Al is trivalent, Ge is tetravalent, Ag is a metal.
3. So the answer is (b) Phosphorous.

Answer: (b)

1. At $0$ K no electron has enough energy to cross the band gap ($\approx 1.1$ eV for Si).
2. The conduction band is empty and the valence band full, so there are no free carriers.
3. So the answer is (b) insulator.

Answer: (a)

1. p-type is made by adding trivalent (acceptor) atoms such as B, Al, In.
2. Each acceptor creates a hole, so holes are the majority carriers.
3. So the answer is (a).

Answer: (c)

1. For an n-type semiconductor, $\sigma = n e \mu_e$ and $\rho = \dfrac{1}{n e \mu_e}$.
2. $\rho = \dfrac{1}{10^{19} \times 1.6 \times 10^{-19} \times 1.6} = \dfrac{1}{2.56}$
3. $\rho \approx 0.39\ \Omega\,\text{m}$
4. So the answer is (c) $0.4\ \Omega\,\text{m}$.

Answer: (a)

1. $\sigma = e(n_e \mu_e + n_h \mu_h)$
2. $n_e \mu_e = 5 \times 10^{18} \times 2.0 = 1.0 \times 10^{19}$, $n_h \mu_h = 5 \times 10^{19} \times 0.01 = 0.05 \times 10^{19}$
3. $\sigma = 1.6 \times 10^{-19} \times 1.05 \times 10^{19} = 1.68\ (\Omega\text{-m})^{-1}$
4. So the answer is (a) $1.68\ (\Omega\text{-m})^{-1}$.

Answer: (a)

1. Cu (metal): $R = R_0(1 + \alpha \Delta T)$, a linear increase with temperature.
2. Undoped Si: carrier density $\propto e^{-E_g/2kT}$, so resistance decreases exponentially with temperature.
3. So the answer is (a) linear increase for Cu, exponential decrease for Si.

Answer: (a)

1. n-type is made by adding pentavalent (donor) atoms such as P, As, Sb.
2. Donors give free electrons, so electrons are majority and holes are minority carriers.
3. So the answer is (a).

Answer: (c)

1. Antimony is pentavalent, so it acts as a donor in Ge.
2. Ge becomes n-type: free electrons outnumber holes, and resistance decreases.
3. So the answer is (c).

Answer: (a)

1. Indium is trivalent, so the doped Si is p-type.
2. $n_e = \dfrac{n_i^2}{n_h} = \dfrac{(1.5\times10^{16})^2}{4.5\times10^{22}} = \dfrac{2.25\times10^{32}}{4.5\times10^{22}}$
3. $n_e = 5\times10^{9}\ \text{m}^{-3}$
4. So the answer is (a) p-type with $n_e = 5\times10^9\ \text{m}^{-3}$.

Answer: (b)

1. (a) Trivalent doping gives p-type: true.
2. (b) In n-type the majority carriers are electrons, not holes: false.
3. (c) Minority carriers in p-type are electrons: true. (d) Intrinsic resistance falls as $T$ rises: true.
4. So the false statement is (b).

Answer: (d)

1. The diagram shows a band gap $E_g$ between $E_C$ and $E_V$, so it is a semiconductor (not a metal).
2. Holes (open circles) in the valence band outnumber electrons (filled circles) in the conduction band.
3. Holes are the majority carriers, so the material is p-type.
4. So the answer is (d) a p-type semiconductor.

Answer: (c)

1. $I = neAv_d \Rightarrow v_d \propto \dfrac{I}{n}$ for the same sample.
2. $\dfrac{v_e}{v_h} = \dfrac{I_e}{I_h} \times \dfrac{n_h}{n_e} = \dfrac{7}{4} \times \dfrac{5}{7} = \dfrac{5}{4}$
3. So the answer is (c) $\dfrac{5}{4}$.

Answer: (d)

1. Copper (metal): fewer lattice vibrations on cooling, so resistance decreases.
2. Germanium (semiconductor): fewer free carriers on cooling, so resistance increases.
3. So the answer is (d).

Answer: (c)

1. Copper (metal): cooling reduces collisions with the vibrating lattice, so resistance decreases.
2. Germanium (semiconductor): cooling reduces carrier generation, so resistance increases.
3. So the answer is (c).

Answer: (b)

1. In a metal the number of free electrons is fixed; rising temperature only increases scattering, so $R$ rises.
2. In a semiconductor the number of carriers rises sharply with temperature, so $R$ falls.
3. So the answer is (b) variation of the number of charge carriers with temperature.

Answer: (c)

1. At absolute zero no electron in Si can cross the band gap.
2. Conduction band empty, valence band full: no free charge carriers.
3. So the answer is (c) insulator.

Answer: (c)

1. Conductor: more lattice vibration at higher $T$, so specific resistance increases.
2. Semiconductor: more electron-hole pairs at higher $T$, so specific resistance decreases.
3. So the answer is (c) increases, decreases.

Answer: (a)

1. n-type silicon needs a pentavalent donor.
2. P (group 15) is pentavalent; Al and B are trivalent, Mg is divalent.
3. So the answer is (a) P.

Answer: (d)

1. A hole is a vacancy left in a covalent bond when an electron is absent.
2. It behaves like a positive charge $+e$.
3. So the answer is (d) missing electrons.

Answer: (a)

1. p-type is doped with trivalent acceptors, each creating a hole.
2. So holes far outnumber free electrons.
3. So the answer is (a) holes.

Answer: (a)

1. p-type Ge needs a trivalent acceptor.
2. Indium is trivalent; P, As and Sb are pentavalent.
3. So the answer is (a) indium.

Answer: (b)

1. Arsenic is pentavalent, so it donates a free electron in Si.
2. Electrons become majority carriers, but the material is still a semiconductor.
3. So the answer is (b) n-type semiconductor.

Answer: (d)

1. Copper is a metal: cooling reduces lattice vibrations, so its resistance decreases.
2. Germanium is a semiconductor: cooling reduces the number of free carriers, so its resistance increases.
3. So the answer is (d).

Answer: (d)

1. Heating breaks covalent bonds and creates electron-hole pairs.
2. Each broken bond adds one free electron and one hole, so both increase by the same number.
3. So the answer is (d) number of electrons and holes increases equally.

Answer: (c)

1. At $0$ K the valence band of Si is full and the conduction band is empty.
2. No free carriers exist, so Si cannot conduct.
3. So the answer is (c) insulator.

Answer: (c)

1. The depletion region has fixed positive ions on the n-side and negative ions on the p-side.
2. The field points from n to p and is strongest at the junction (middle of the depletion region); reverse bias makes it even larger.
3. So the answer is (c) electric field is maximum.