⚛️ Unit 5 – Part B (Descriptive Q&A)

1. Obtain an expression for carrier concentration in an intrinsic semiconductor...

In an intrinsic (pure) semiconductor, the number of electrons in the conduction band (n) is equal to the number of holes in the valence band (p). This is the intrinsic carrier concentration (n_i).

1. Electron Concentration (n):

The concentration of electrons in the conduction band is found by integrating the product of the density of states g(E) and the probability of occupation F(E) from the bottom of the conduction band (E_c) to infinity.

n = ∫_Ec^∞ g(E) F(E) dE

Using the Maxwell-Boltzmann approximation (as E_F is far from E_c), this integral simplifies to:

n = N_c e^{-(E_c - E_F) / k_BT}

Where N_c is the effective density of states in the conduction band: N_c = 2 (2πm_e*k_BT / h²)^3/2

2. Hole Concentration (p):

Similarly, the concentration of holes in the valence band is:

p = ∫_-∞^E_v g(E) [1 - F(E)] dE

This simplifies to:

p = N_v e^{-(E_F - E_v) / k_BT}

Where N_v is the effective density of states in the valence band: N_v = 2 (2πm_h*k_BT / h²)^3/2

3. Intrinsic Carrier Concentration (n_i):

In an intrinsic semiconductor, n = p = n_i. We can multiply the expressions for n and p:

n · p = n_i² = [N_c e^{-(E_c - E_F) / k_BT}] · [N_v e^{-(E_F - E_v) / k_BT}]

n_i² = N_cN_v e^{-(E_c - E_F + E_F - E_v) / k_BT}

Since E_c - E_v = E_g (the bandgap energy):

n_i² = N_cN_v e^{-E_g / k_BT}

Taking the square root gives the expression for intrinsic carrier concentration:

n_i = (N_cN_v)^1/2 e^{-E_g / 2k_BT}

2. Derive an expression for carrier concentration of an Extrinsic semiconductor (n – type & P-Type) and explain the variation of Fermi level with temperature.

n-type Semiconductor:

• Carrier Concentration: Doped with N_d donor atoms. At room temperature, almost all donors are ionized. Since the number of thermally generated holes (p) is negligible, the charge neutrality equation (n = N_d + p) simplifies to: n ≈ N_d.
• Fermi Level: We use the expression for n: n = N_c e^{-(E_c - E_F) / k_BT}.
  Setting n = N_d: N_d = N_c e^{-(E_c - E_F) / k_BT}
  Solving for E_F: (N_d / N_c) = e^{-(E_c - E_F) / k_BT}
  ln(N_d / N_c) = -(E_c - E_F) / k_BT
  E_F = E_c - k_BT ln(N_c / N_d)
  This shows the Fermi level is located just below the conduction band.
• Variation with Temperature: At T=0K, E_F is halfway between the donor level (E_d) and E_c. As temperature increases, E_F moves downwards, approaching the intrinsic level (E_i) at very high temperatures when the semiconductor starts to behave intrinsically.

p-type Semiconductor:

• Carrier Concentration: Doped with N_a acceptor atoms. At room temperature, p ≈ N_a.
• Fermi Level: We use the expression for p: p = N_v e^{-(E_F - E_v) / k_BT}.
  Setting p = N_a: N_a = N_v e^{-(E_F - E_v) / k_BT}
  Solving for E_F: (N_a / N_v) = e^{-(E_F - E_v) / k_BT}
  ln(N_a / N_v) = -(E_F - E_v) / k_BT
  E_F = E_v + k_BT ln(N_v / N_a)
  This shows the Fermi level is located just above the valence band.
• Variation with Temperature: At T=0K, E_F is halfway between the acceptor level (E_a) and E_v. As temperature increases, E_F moves upwards, approaching the intrinsic level (E_i) at very high temperatures.

3. Describe the construction and working principle of Schottky diode. Give its advantages, disadvantages and applications

Construction:

A Schottky diode is a metal-semiconductor junction. It is formed by bringing a metal (like gold, silver, platinum, or tungsten) into direct contact with a lightly doped n-type semiconductor (e.g., silicon). This creates a rectifying junction known as a Schottky barrier.

Working Principle:

1. Barrier Formation: When the metal and n-type semiconductor are joined, electrons from the semiconductor (which have a higher Fermi level) diffuse into the metal until their Fermi levels align. This leaves behind a region of positive ionized donor atoms in the semiconductor, creating a depletion region and a built-in potential barrier.
2. Forward Bias: When a positive voltage is applied to the metal (anode) and negative to the n-type (cathode), the barrier height is reduced. Electrons (majority carriers) are easily injected from the semiconductor into the metal, resulting in a large forward current.
3. Reverse Bias: When a negative voltage is applied to the metal, the barrier height increases, significantly blocking the flow of electrons. Only a very small reverse leakage current flows.

Unlike a p-n diode which relies on minority carrier injection, the Schottky diode is a majority carrier device (conduction is by electrons only).

Advantages:

• Fast Switching Speed: No minority carrier storage, so it has a near-zero reverse recovery time. This makes it ideal for high-frequency applications.
• Low Forward Voltage Drop: Typically 0.2V to 0.4V, compared to 0.6V to 0.7V for a silicon p-n diode. This means less power loss and higher efficiency.
• Low Noise: Generates less random noise than p-n diodes.

Disadvantages:

• High Reverse Leakage Current: The reverse current is significantly higher than in a p-n diode.
• Low Reverse Voltage Rating: They typically have lower reverse breakdown voltages (e.g., < 100V).

Applications:

• Switching Mode Power Supplies (SMPS): Used as rectifiers due to their high speed and efficiency.
• RF Applications: Used as detectors and mixers in high-frequency radio circuits.
• Reverse Polarity Protection: To protect circuits from being connected backwards.
• Solar Cells: To prevent charge from leaking back out of the battery at night.

4. What is Hall Effect? Derive an expression for the Hall Voltage. Explain an experimental method used to measure the Hall Coefficient of a specimen. What are the uses of Hall effect?

Hall Effect:

When a current-carrying conductor (or semiconductor) is placed in a transverse (perpendicular) magnetic field, a voltage is developed across the specimen in a direction perpendicular to both the current and the magnetic field. This phenomenon is called the Hall effect.

Derivation of Hall Voltage (V_H):

1. Consider a p-type semiconductor slab with current I_x (in x-dir) and magnetic field B_z (in z-dir).
2. The holes (charge +q) move with drift velocity v_x. They experience a magnetic Lorentz force: F_m = q(v_x B_z) in the -y direction.
3. Holes accumulate on the bottom face, creating an electric field E_H (Hall field) in the +y direction.
4. This field exerts an electric force: F_e = qE_H in the +y direction.
5. At equilibrium, the forces balance: F_e = F_m → qE_H = qv_xB_z → E_H = v_xB_z.
6. The Hall Voltage V_H across the width 'w' is V_H = E_H · w = v_xB_zw.
7. We know current density J_x = pqv_x, where p is hole density. So v_x = J_x / pq.
8. Substitute v_x: V_H = (J_x / pq) B_zw.
9. Since J_x = I_x / (w·t) (where 't' is thickness), V_H = (I_x / (w·t·pq)) B_zw.
10. This simplifies to: V_H = (B_zI_x) / (pqt).
11. We define the Hall Coefficient R_H = 1 / (pq).
12. Therefore, V_H = (R_HB_zI_x) / t. (For n-type, R_H = -1/nq).

Experimental Method to Measure Hall Coefficient:

1. Prepare a thin, rectangular sample of the specimen with known thickness 't'.
2. Pass a known, constant current (I_x) through the length of the sample.
3. Place the sample in a uniform, perpendicular magnetic field (B_z) of known strength.
4. Using a high-impedance voltmeter, measure the Hall Voltage (V_H) that develops across the width of the sample.
5. The Hall Coefficient can then be calculated using the rearranged formula: R_H = (V_H · t) / (B_zI_x).

Uses of Hall Effect:

• Determine Semiconductor Type: The sign of V_H (and R_H) tells if it's n-type (negative) or p-type (positive).
• Calculate Carrier Concentration: n = 1 / |R_H q| or p = 1 / |R_H q|.
• Calculate Carrier Mobility: Mobility (μ) can be found using μ = σR_H, where σ is the conductivity.
• Magnetic Field Sensors: Hall effect probes are used to measure magnetic field strengths.
• Contactless Current Sensors: Can measure a current in a wire by measuring the magnetic field it produces.

5. (i) Draw the band diagram for an ohmic contact and explain its principle, theory, V-I characteristics... (10) (ii) The Hall coefficient of certain silicon specimen... (6)