讲解 EEC 140AV - WQ 2024 Midterm Exam - I讲解留学生SQL语言

Midterm Exam - I

EEC 140AV - WQ 2024

Total Points: 80

Time Duration: 100 minutes

Question 1 (2 points)

Which of the following three unit cells are primitive cells for the two-dimensional lattice. Circle the correct diagram. (Note:- By definition a primitive cell contains only one lattice point)

Question 2 (2 points)

Which of the following three planes (shown within the first quadrant only) is a (121) plane. Circle the correct diagram.

Question 3 (3 points)

Find the density of Si cell shown below given its lattice constant of 0.5431 nm and atomic mass of 28.085 g/mol.

(a) 2.18 g/cm3 (b) 2.32 g/cm3 (c) 2.03 g/cm3 (d) None of these

Question 4 (4 points)

Considering the following conduction band energy E vs wavevector kx dispersion relation, where the curvature of the band determines the effective mass of charge carriers.

1. Which energy valley has the greater effective mass in the x-direction (circle one): X-valley or Γ-valley.

2. Consider two electrons, one each located at the positions of the heavy crosses. Which electron has the greater velocity magnitude (circle one): The one in the Γ-valley / The one in the X-valley.

Question 5 (3 points)

The equilibrium band diagram for a doped direct gap semiconductor is shown below. What type of semiconductor it is?

(a) n – type (b) p – type (c) Intrinsic (d) Not enough information provided

Question 6 (3 points)

Consider the sketch of electron density vs temperature for an n-type Si below, at T > 500 K, the sample is said to be:

(a) Extrinsic (b) Intrinsic (c) Degenerate (d) In high level injection

Question 7 (6 points)

A hypothetical semiconductor has an intrinsic carrier concentration of 1.0 × 1010 /cm3 at 300 K, it has conduction and valence band effective densities of states Nc and Nv , both equal to 1019 /cm3. (Note: Boltzmann constant in eV is kB = 8.6173303 × 10–5 eV/K)

1. What is the band gap ’Eg ’ of semiconductor in eV?

(a) 1.07 (b) 2.2 (c) 1.4 (d) None of these

2. If the semiconductor is doped with Nd = 1 × 1016 donors/cm3 , what are the equilibrium electron and hole concentrations at 300 K?

(a) no = 104 /cm3 ; po = 1016 /cm3

(b) no = 1016 /cm3 ; po = 104 /cm3

(d) no = 1012 /cm3 ; po = 1016 /cm3

3. If the same piece of semiconductor, already having Nd = 1 × 1016 donors/cm3 , is also doped with Na = 2 × 1016 acceptors/cm3 , what are the new equilibrium electron and hole concentrations at 300 K?

(a) no = 104 /cm3 ; po = 1016 /cm3

(b) no = 1016 /cm3 ; po = 104 /cm3

(d) no = 1012 /cm3 ; po = 1016 /cm3

Question 8 (4 points)

Consider the following equilibrium band diagram for a portion of a semiconductor sample with a built-in electric field.

1. For the band diagram below, the slope of EF is zero. Which of the following is true? Assume moving from left to right is the positive direction (2 points)

(a) The electric field is zero

(b) The total hole current density is zero

(d) The electron drift current density is positive

2. Which direction does the hole diffusion current flow towards in the previous figure? (2 points)

(a) Towards left

(b) Towards right

(d) The same direction as the hole drift current

Question 9 (2 points)

With EF located 0.4 eV above the valence band in a Si sample, what charge state would you expect for most Ga atoms in the sample?

(a) Doubly negative

(b) Singly negative

(d) No charge state (i.e., neutral)

Question 10 (10 points)

1. Ionized impurity scattering dominates for a larger temperature range in a highly doped semiconductor compared to a low doped semiconductor

(a) True (b) False

2. If Si dopes GaAs n-type, then it must occupy As lattice sites

(a) True (b) False

3. Generation occures to restore equilibrium when np > ni(2)

(a) True (b) False

4. The electron mobility of a Si wafer doped with NA = ND = 1017 /cm3 is lower than the electron mobility of an undoped Si wafer

(a) True (b) False

5. All semiconductors become intrinsic at high temperatures

(a) True (b) False

Question 11 (25 points)

Consider the below energy band diagram for a semiconductor at 300 K, and answer the following questions.

1. The kinetic energy of the electron shown on the diagram is found to be: (2 points)

(a) Ec – E peak (b) E peak – Ec (c) 0 (d) EF+ E peak – Ec

2. Equilibrium conditions prevail in this semiconductor (2 points)

(a) True (b) False

3. The semiconductor is degenerate at: (3 points)

(a) –W ≤ x ≤ W (b) Near x = 0 (c) For all x (d) Nowhere

4. Identify the appropriate profiles from below figures for the electrostatic potential and electric field for the energy band diagram shown above. Circle the suitable ones and label separately for the electrostatic potential and electric field cases (6 points)

5. Identify an appropriate semi-log plot electron concentration from the profiles given below for the energy band diagram situation shown above (4 points)

6. The electron drift current flowing at x = xi is: (3 points)

(a) 0

(b) μnni(E peak – Ec )/W

(d) μn ND(E peak – Ec )/W

(e) –μn ND(E peak – Ec )/W

7. The hole diffusion current flowing at x = xi is: (5 points)

(a) 0

(b) μpn i(E peak – Ec )/W

(d) qDp ND/W

(e) –qDp ND/W

Question 12 (16 points)

A laser beam striking a uniformly doped p-type bar of silicon maintained at room temperature causes a steady state excess of Δnp = 1011 /cm3 electrons at x = 0. Note that the laser-induced photo-generation only occurs at x = 0. As pictured below, the bar extends from x = –L to x = L and Δnp (–L) = Δnp(+L) = 0. NA = 1016 /cm3 and the electric field inside the bar is approximately zero.

1. What are the dominant physical processes that determine the steady-state excess electron concentration, Δnp(x), in the regions far from x = 0 (3 points)

(a) Diffusion (b) Drift (c) Recombination (d) Generation

2. What is the equilibrium electron concentration no? (3 points)

(a) 1016 /cm3 (b) 1010 /cm3 (c) 104/cm3 (d) 1011 /cm3

3. Low level injection exists under steady state conditions (2 points)

(a) True (b) False

4. The simplest possible form of equation that must be solved to determine Δnp(x) for 0 < x < L is: (4 points)

(a) 0 = DN –

(b) = –

(d) 0 = DN

(e) = DN – – + GL

5. Suppose L → ∞, the general solution for Δnp(x) for 0 < x < L is: (4 points)

(a) Δnp(x) = A + Bx