Solid State Devices Practice Quiz (22 questions)

1. How is a semiconductor classified by its electrical conductivity?

A. Its conductivity matches that of an ideal insulator.

B. Its conductivity is lower than that of all insulators.

C. Its conductivity lies between that of conductors and insulators.

D. Its conductivity is greater than that of conductors.

Answer: (C)

Explanation

A semiconductor has electrical conductivity intermediate between conductors and insulators. A material more conductive than conductors would not fit this classification.

2. Which energy band is the highest range filled with electrons at 0 K?

A. The forbidden gap, which contains allowed electron states.

B. The donor band, created by added impurities.

C. The valence band, the highest electron-filled range.

D. The conduction band, where electrons can move freely.

Answer: (C)

Explanation

At 0 K, the valence band is the highest energy range filled with electrons. The conduction band is associated with electrons able to move and conduct current, rather than being the highest filled band at 0 K.

3. Which description correctly compares the band gaps of conductors, semiconductors, and insulators?

A. Conductors and semiconductors have large gaps, while insulators have overlapping bands.

B. Conductors have a large gap, semiconductors have a gap near zero, and insulators have a small gap.

C. Conductors have a gap near zero, semiconductors have a small gap, and insulators have a large gap.

D. Conductors have a small gap, semiconductors have a large gap, and insulators have a gap near zero.

Answer: (C)

Explanation

Conductors have overlapping valence and conduction bands and approximately zero gap; semiconductors have small gaps, while insulators have large gaps. The description assigning a large gap to conductors reverses the conductor and insulator behavior.

4. What distinguishes an intrinsic semiconductor from an extrinsic semiconductor?

A. An intrinsic semiconductor is doped, while an extrinsic one remains pure.

B. An intrinsic semiconductor has donor impurities, while an extrinsic one has acceptors.

C. An intrinsic semiconductor is pure, while an extrinsic one is doped with impurities.

D. An intrinsic semiconductor has majority holes, while an extrinsic one has majority electrons.

Answer: (C)

Explanation

Intrinsic material is pure, whereas extrinsic material has impurities added to control its conductivity. Donor versus acceptor impurities describes doping types, not the distinction between intrinsic and extrinsic material.

5. A semiconductor is doped with group V donor atoms; which carriers are expected to be in the majority?

A. Electrons and holes in equal numbers as majority carriers.

B. Donor ions, with holes as the minority carriers.

C. Holes, with electrons as the minority carriers.

D. Electrons, with holes as the minority carriers.

Answer: (D)

Explanation

Donor doping produces N-type material, in which electrons are the majority carriers and holes are the minority carriers. Holes as majority carriers characterize P-type rather than N-type material.

6. Which expression gives the Fermi–Dirac probability that a state of energy $E$ is occupied?

A. $f(E)=\frac{1}{1+\exp((E-E_F)kT)}$

B. $f(E)=\frac{1}{1+\exp((E-E_F)/kT)}$

C. $f(E)=\frac{1}{1+\exp(E_F/kT)}$

D. $f(E)=\frac{1}{1+\exp((E_F-E)/kT)}$

Answer: (B)

Explanation

The Fermi–Dirac expression uses the energy difference $E-E_F$ divided by $kT$ in the exponential. Reversing the energy difference changes how occupancy varies with energy and does not match the stated relation.

7. At finite temperature, what is the occupancy probability for a state whose energy equals the Fermi level?

A. It is zero, with occupancy changing gradually below that energy.

B. It is one-half, with occupancy changing gradually around that energy.

C. It is one, with occupancy changing abruptly above that energy.

D. It depends on whether the material is intrinsic or extrinsic.

Answer: (B)

Explanation

At finite temperature, the occupancy probability at $E=E_F$ is one-half, and occupancy changes gradually around the Fermi level. The sharp filled-versus-empty transition applies at 0 K, not at finite temperature.

8. How does the Fermi-level position differ among intrinsic, N-type, and P-type semiconductors?

A. It is near midgap intrinsically, nearer the valence band in N-type, and nearer the conduction band in P-type material.

B. It is near the valence band intrinsically, nearer the conduction band in N-type, and nearer midgap in P-type material.

C. It is near the conduction band intrinsically, nearer midgap in N-type, and nearer the valence band in P-type material.

D. It is near midgap intrinsically, nearer the conduction band in N-type, and nearer the valence band in P-type material.

Answer: (D)

Explanation

The Fermi level is near the band-gap midpoint in intrinsic material, shifts toward the conduction band for N-type material, and toward the valence band for P-type material. The reversed N-type and P-type positions conflict with their respective carrier types.

9. What relationship holds between electron and hole concentrations in an intrinsic semiconductor?

A. They are equal, with each concentration given by $\sqrt{N_C N_V}\exp\left(-\frac{E_g}{2kT}\right)$.

B. Their product equals the square of the intrinsic carrier concentration, but their values may differ.

C. The electron concentration is approximately the donor concentration, while holes are minority carriers.

D. The hole concentration is set by the acceptor density, while electrons are minority carriers.

Answer: (A)

Explanation

An intrinsic semiconductor has equal electron and hole concentrations, each determined by the intrinsic concentration expression. The mass action law applies at equilibrium to both intrinsic and doped material, but does not imply unequal concentrations in an intrinsic semiconductor.

10. At thermal equilibrium, how are electron and hole concentrations related to the intrinsic carrier concentration?

A. Their product equals the intrinsic carrier concentration: $np=n_i$.

B. Their ratio equals the intrinsic carrier concentration: $n/p=n_i$.

C. Their sum equals the intrinsic carrier concentration: $n+p=n_i$.

D. Their product equals the square of the intrinsic carrier concentration: $np=n_i^2$.

Answer: (D)

Explanation

The equilibrium mass action law states that the product of electron and hole concentrations is $n_i^2$. Their sum is not fixed by this law, so adding the concentrations does not give the intrinsic concentration.

11. For N-type material with $N_D\gg n_i$, what are the approximate electron and hole concentrations?

A. Electrons are approximately $n_i^2/N_D$, and holes are approximately $N_D$.

B. Electrons are approximately $N_D$, and holes are approximately $n_i^2/N_D$.

C. Electrons are approximately $n_i$, and holes are approximately $N_D^2/n_i$.

D. Electrons and holes are both approximately $N_D$.

Answer: (B)

Explanation

In strongly N-type material, donor atoms set the majority electron concentration near $N_D$, while the mass action law gives the minority hole concentration as $n_i^2/N_D$. Reversing these concentrations confuses which carrier type is the majority in N-type material.

12. Why does the intrinsic carrier concentration increase strongly as a semiconductor is heated?

A. The exponential factor $\exp(-E_g/2kT)$ increases as temperature rises.

B. The bandgap increases with temperature, raising the number of thermally generated carriers.

C. The exponential factor $\exp(-E_g/2kT)$ decreases as temperature rises.

D. The donor concentration rises as thermal energy creates additional dopant atoms.

Answer: (A)

Explanation

As temperature rises, the exponential factor in the intrinsic concentration expression increases, leading to a strong increase in intrinsic carriers. Dopant-controlled concentration can dominate in the extrinsic region, but that does not explain the temperature-driven rise in intrinsic concentration.

13. Which sequence describes the carrier regimes of a semiconductor as temperature increases?

A. Intrinsic region, freeze-out, then extrinsic region, as dopants progressively lose ionization.

B. Freeze-out, intrinsic region, then extrinsic region, as dopant carriers replace thermal carriers.

C. Freeze-out, extrinsic region, then intrinsic region, as dopants ionize and thermal generation later dominates.

D. Extrinsic region, intrinsic region, then freeze-out, as thermal generation declines with heating.

Answer: (C)

Explanation

The semiconductor progresses from freeze-out, where dopants are insufficiently ionized, through the extrinsic region, where they are mostly ionized, and finally to the intrinsic region, where thermal generation dominates. The reversed ordering incorrectly places the intrinsic regime before the extrinsic regime.

14. Which expression gives semiconductor conductivity in terms of electron and hole concentrations and mobilities?

A. $\sigma=q(n+p)(\mu_n-\mu_p)$

B. $\sigma=q(n\mu_n+p\mu_p)$

C. $\sigma=q(n\mu_n-p\mu_p)$

D. $\sigma=q(n\mu_p+p\mu_n)$

Answer: (B)

Explanation

Conductivity includes additive contributions from electrons and holes, weighted by their respective mobilities and multiplied by the charge magnitude. Subtracting the hole contribution would incorrectly treat it as reducing the conductivity.

15. A carrier moves because it is in a region of high concentration and travels toward a region of lower concentration, without an applied electric field being the driving cause. What transport mechanism is this?

A. Drift, driven by an electric field acting on the carrier.

B. Drift, driven by a concentration gradient.

C. Diffusion, driven by an electric field acting on the carrier.

D. Diffusion, driven by a concentration gradient.

Answer: (D)

Explanation

Diffusion is carrier motion from high to low concentration, driven by a concentration gradient. Drift is the plausible alternative mechanism, but it is driven by an electric field rather than by a concentration difference.

16. What happens when a magnetic field perpendicular to the current is applied across a conducting sample?

A. The sample becomes electrically neutral because the field cancels the motion of its carriers.

B. Charge carriers are deflected sideways, creating charge separation and a transverse Hall voltage.

C. Charge carriers accelerate along the current, increasing the voltage measured in the same direction.

D. The magnetic field reduces carrier density uniformly, producing a voltage drop along the sample.

Answer: (B)

Explanation

The perpendicular magnetic field deflects moving charge carriers sideways, separating charge and producing a transverse Hall voltage. A voltage drop along the current direction does not describe the Hall effect's transverse response.

17. A Hall sample has thickness $d$ and carries current $I$ in magnetic field $B$; which relation gives its Hall voltage?

A. $V_H=\frac{IBd}{nq}$

B. $V_H=\frac{nqd}{IB}$

C. $V_H=\frac{IB}{nqd}$

D. $V_H=\frac{nqB}{Id}$

Answer: (C)

Explanation

The Hall voltage is proportional to current and magnetic field and inversely proportional to carrier concentration, charge, and sample thickness. The expression with thickness in the numerator reverses the thickness dependence.

18. A material has a negative Hall coefficient; what carrier type does this indicate?

A. N-type material, with holes as the majority carriers.

B. N-type material, with electrons as the majority carriers.

C. Intrinsic material, with equal electron and hole populations.

D. P-type material, with holes as the majority carriers.

Answer: (B)

Explanation

A negative Hall coefficient indicates N-type material, while a positive coefficient indicates P-type material. A positive coefficient, rather than a negative one, is associated with P-type material.

19. Which equation describes the Shockley current-voltage relation for a diode?

A. $I_D=I_S\left(e^{V_D/(nV_T)}-1\right)$

B. $I_D=I_S\left(e^{nV_D/V_T}-1\right)$

C. $I_D=I_S\left(e^{-V_D/(nV_T)}-1\right)$

D. $I_D=nI_S\left(e^{V_D/V_T}-1\right)$

Answer: (A)

Explanation

The Shockley equation relates diode current exponentially to diode voltage, with the exponent scaled by the ideality factor and thermal voltage. Placing the ideality factor in the numerator changes the specified voltage scaling.

20. How does an ideal diode behave when forward-biased and when reverse-biased?

A. It blocks current forward-biased and acts as a short circuit reverse-biased.

B. It acts as a resistor forward-biased and limits current reverse-biased.

C. It maintains a fixed voltage drop forward-biased and conducts freely reverse-biased.

D. It acts as a short circuit forward-biased and blocks current reverse-biased.

Answer: (D)

Explanation

The ideal diode conducts without voltage drop when forward-biased, behaving like a short circuit, and blocks current when reverse-biased. A fixed forward voltage drop belongs to a non-ideal diode model, not the ideal model.

21. A diode is in series with supply voltage $V_S$ and resistor $R$; which equation and intercepts describe its load line?

A. $V_D=I_DR-V_S$, with intercepts $(V_D=-V_S,I_D=0)$ and $(V_D=0,I_D=V_S/R)$

B. $V_D=V_S-I_DR$, with intercepts $(V_D=0,I_D=0)$ and $(V_D=V_S,I_D=V_S/R)$

C. $V_D=V_S+I_DR$, with intercepts $(V_D=V_S,I_D=0)$ and $(V_D=0,I_D=V_S/R)$

D. $V_D=V_S-I_DR$, with intercepts $(V_D=V_S,I_D=0)$ and $(V_D=0,I_D=V_S/R)$

Answer: (D)

Explanation

Kirchhoff's voltage law gives $V_D=V_S-I_DR$, yielding the stated voltage-axis and current-axis intercepts. Using a plus sign for the resistor term gives the wrong slope for the series circuit load line.

22. What does the Q-point represent in a diode circuit?

A. The point where the diode reaches breakdown, specifying its maximum reverse voltage and current.

B. The intercept of the load line with the voltage axis, specifying the supply voltage and zero current.

C. The point where the diode current becomes zero, specifying its threshold voltage and resistance.

D. The intersection of the diode I–V curve and load line, specifying steady-state voltage and current.

Answer: (D)

Explanation

The Q-point is the operating point where the diode's I–V characteristic meets the circuit load line, establishing steady-state diode voltage and current. The voltage-axis intercept is a load-line endpoint, not generally the circuit's operating point.