Semiconductor Physics: Assignment | Solid State Physics, Devices & Electronics PDF Download

Q.1. Pure silicon at 300 K has equal electron and hole concentration of 2 x 10¹⁶ m^-3.  It is doped by nitrogen to the extent one part in 10⁶ silicon atom. If the density of silicon is 4 x 10²⁹ m^-3, then find the electron and hole concentration in the doped silicon.

Donor ion concentration
According to Law of Mass Action, n.p = n_i²

Q.2. In a p-type semiconductor, the Fermi level is 0.27 eV above the valance band at room temperature of 300 K . If the temperature is increased to 400 K , then find the new position of the Fermi-level. (Assume effective density of states to be independent of temperature).

E_F-E_V = kT ln NV/NA ⇒ 0.27 = 300k In NV/NA and E'_F-E_V = 400K In N_V/N_A

Q.3. The acceptor concentration in a sample of p -type silicon is increased by a factor of 50. Find the shift in the position of the Fermi level at 350K. 

E_F-E_V = kT ln (N_V/N_A)  and E'_F - E_V = kT In (N_V/50N_A) = kT In (N_V/N_A) - kT In (50)
Thus shift is 

⇒ ΔE = 486 x 10^-4 eV = 48.6 meV

Q.4. A Si sample is doped with 2.25x10¹⁷ Al atoms/cm³ at 300 K. The intrinsic carrier concentration at 300 K is 1.5 x 10¹⁰ cm^-3.
(a)What is the equilibrium electron and hole concentration at 300 K?
(b) Where is E_F relative to E_i?

Since N_A>>n_i we can approximate n_i and p = 2.25 x 10 ¹⁷Al

Q.5. For a given semiconductor the effective mass of the electron is 1.25m_e and the Fermi level is 0.3 eV below the conduction at 300 K . Determine the concentration of electrons in conduction band.

At 300 K,
⇒ n = 3.5 x 10²⁵ exp[-0.3x1.6x10^-19 / 1.38x10^-23x 300] = 32 x 10¹⁹m^-3

^{Q.6. Consider an intrinsic semiconductor with energy band gap of 0.72 eV at 300 K . What will be shift in position of Fermi-level from the middle of the band gap if effective mass of hole is five times the effective mass of electrons?}

Q.7. Consider an intrinsic semiconductor with energy band gap of 1.43 eV , effective density of states in conduction band is 1.54 x 10²⁴ m^-3 and effective density of states in valance band is 1.3 x 10²⁵ m^-3 at 300 K.
(a) Determine the intrinsic carrier concentration of the semiconductor.
(b) Determine the effective masses of electrons and holes.

(b) At 300 K,
 

Q.8. A sample of Si has electron and hole mobilities of 0.13 and 0.05 m²V^-1s^-1 respectively at 300K.
(a) It is doped with P and Al with doping densities of 3.5 x 10²¹/m³ and 1.5 x 10²¹/m³ respectively. Find the resistivity of doped Si sample at 300K.
(b) It is doped with Ga and N with doping densities of 3.5 x 10²¹/m³ and 1.5 x 10²¹/m³ respectively. Find the conductivity of doped Si sample at 300 K.

(a) Resulting doped crystal is n-type and n_n = (3.5 - 1.5) x 10²¹/m³ = 2 x 10²¹/m³

σ = e(n_nμ_n + pnμ_p) ≈ en_nμ_n = 1.6 x 10^-19 x 2 x 10²¹ x 0.13 = 41.6Ω^-1m^-1

(b) Resulting doped crystal is p-type and p_p = (3.5-1.5) x10²¹/m³ = 2 x 10²¹/m³
 1.6 x 10^-19 x 2 x 10²¹ x 0.05 = 16Ω^-1m^-1.

Q.9. Consider an extrinsic semiconductor with intrinsic concentration of n_i. If μ_p and μ_n are mobility of holes and electron where μ_p = 3μ_n then
(a) Find the electron concentration at which semiconductor has minimum conductivity
(b) Find the hole concentration at which semiconductor have minimum conductivity
(c)  Find minimum conductivity σ_min.

(a) Conductivity
For minimum conductivity,
(b) Conductivity
For minimum conductivity,
(c) Thus, σ_min =

Q.10. In an intrinsic semiconductor, the free carrier concentration n (in cm^-3) varies with temperature T (in Kelvin) as shown in the figure below.
If n1 = 3 x 10¹⁴m^-3 and n₂ = 1 x 10⁸m^-3 and 1/T₁ = 2 x 10^-3K^-1 and 1/T₂= 4 x 10^-3K^-1.
Find the band gap energy of the semiconductor material. (Assume effective density of states and band gap energy to be independent of temperature).

Since

Q.11. Find the ratio of mobility (μ) and diffusion coefficient (D) for electrons in a semiconductor at temperature T = 27⁰C and T = 127⁰C.

 where V_T is the ‘Volt-equivalent of temperature’.

At temperature
At temperature

Q.12. The electron concentration in a sample of uniformly doped n-type silicon 300 K varies linearly from 10¹⁷/cm³ at x = 0 to 8x10¹⁶/cm³ at x = 2μm. Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronics charge is 1.6 x 10^-19 C and the diffusion constant D_n = 35cm²/s, find the current density in the silicon, if no electric field is present.

Q.13. Determine the wavelength of light emitted from LED which is made up of GaAsP semiconductor whose forbidden energy gap is 1.875 eV. Mention the colour of the light emitted (Take h = 6.6 × 10^-34 Js).

E_g = h_c / λ
Therefore
λ = h_c / E_g  = 6.6 × 10⁻³⁴ × 3 × 10⁸  / 1.875 × 1.6 × 10⁻¹⁹ 
= 660 nm
The wavelength 660 nm corresponds to red colour light.

Q.14. In a transistor connected in the common base configuration, α = 0.95 , I_E = 1 mA . Calculate the values of I_C and I_B.

α = I_C/I_E
I_C = α I_E = 0.95 × 1 = 0.95 mA
I_E = I_B + I_C 
∴ I_B = I_C − I_E = 1 − 0.95 = 0.05 mA

Q.15. Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 μH. The frequency band is from 500 kHz to 1500 kHz.

Q.16. In the given figure of a voltage regulator, a Zener diode of breakdown voltage 15V is employed. Determine the current through the load resistance, the total current and the current through the diode. Use diode approximation.

Voltage across R_L(V_O) = V_z = 15V
Voltage across R_S(V_RS) = 25 - 15 = 10V
current through R_L is
I_L = V₀/R_L = 15 / 3 × 10³ = 5 × 10^-3 A
IL = 5 mA
Current through RS is
I = VRS/RS = 10/500 = 20 × 10^-3A
I = 20mA
Current through Zener diode is
IZ = I-IL = (20 - 5) × 10^-3 Iz = 15mA