Semiconductor: Important Example Solved Problems

Electron Devices

Chapter 1: Semiconductor

Important Example Solved Problems

Mobility of Charged Particle

Ex.1 A bar of intrinsic germanium 6 cm long is subjected to a potential difference of 12 V. If the velocity of electrons in bar is 73 m/s, determine the mobility of electrons.

Solution: The given values are,

 1 = 6 cm = 6×10^‒2 m,

 V = 12 V,

v = 73 m/sec

E=Electric field intensity

= V / l =  12 / [6×10^‒2] = 200 V/m

Now

 v = μ E

73 = μ. × 200

Μ = Mobility = 73/200

= 0.3650 m²/V-sec

= 3650 cm²/V-sec

General Expression for Conductivity (σ)

Ex. 2: A bar of n type silicon has length of 4 cm and circular cross‒section of 10 mm². When it is subjected to a voltage of 1 V applied across its length, the current flowing through it is 5 mA. Calculate i) Concentration of free electrons. ii) Drift velocity of electrons.

Assume ‒ Charge on one electron as 1.6 × 10^‒19 C.

Mobility of free electron as 1300 cm²/V‒s.

Solution:

The bar is shown in Fig. 1.5.3.

 Current density J = I/A = 5×10^‒3 / 10×10^‒6 = 500 A/m²

The current density is also given by,

J = nqμE         ………..(1)

where

E = Field intensity = V/L

= 1 / 4×10^‒2

 = 25 V/m

q = Charge on electron = 1.6×10^‒19 C

 μ = Mobility = 1300 cm²/V‒s = 0.13 m²/V‒s

Substituting all the values in equation (1),

500 = n×1.6×10^‒19 × 0.13×25

i.e.

 n = 0.9615×10²¹ per m³

Drift velocity v = μ E = 0.13×25 = 3.25 m/s        ... Drift velocity

Intrinsic Concentration and Conductivity of Intrinsic Semiconductor

Ex. 3: Find the resistivity of an intrinsic silicon at 300 °K if intrinsic concentration of silicon at 300 °K is 1.5×10¹⁰ cm³ per while μ_n = 1300 cm²/V‒sec and μ_p= 500 cm²/V‒sec. Assume q=1.6×10^‒19 C.

Solution:

The given values are, n_i = 1.5×10¹⁰/cm³

n_i = 1.5× 10¹⁰ / 10^‒6 /m³

 = 1.5×10¹⁶/m³

And

μ_n = 1300×10^‒4 m²/V‒sec,

μ_р = 500×10^‒4 m²/V‒sec

Now

 σ_i = n_i (μ_n + μ_p) q

= 1.5 × 10¹⁶ [1300 + 500] × 10^-4 × 1.6×10^‒19

 = 0.000432 (Ω-m)^-1              ………. Conductivity

 ρ = 1/σ_i = 1/0.000432 = 2314.8148 Ω‒m          …….Resistivity

The resistivity is reciprocal of conductivity.

Ex. 4: Calculate the intrinsic concentration of Germanium in carriers/m³ at a temperature of 320 °K given that ionization energy is 0.75 eV and Boltzmann's constant k = 1.374× 10^‒23 J/°K. Also calculate the intrinsic conductivity given that the motilities of electrons and holes in pure germanium are 0.36 and 0.17 m²/volt ‒ sec respectively.

Solution:

The temperature dependence of n_i is,

n_i  = BT^3/2 e^‒EG0/2kT

B = Constant, E_G0 = Energy gap at absolute zero,

k = Boltzmann's constant

At 300°K,

 n_i = 2.5 × 10¹³/cm³ = 2.5 × 10¹⁹/m³

 k = 1.374 × 10^‒23 J/°K

and

 1 eV = 1.374×10^‒23 / 1.6×10^‒19  eV/°K = 8.58 × 10^‒5 eV/°K

 2.5 × 10¹⁹ = B (300)^3/2 e^‒0.75/2 × 8.58 × 10^‒5 × 300

B = 1.0218 × 10²²

At T = 320°K,

 n₁ = 1.0218 × 10²² × (320)^3/2 e^‒0.75/2 × 8.58 × 10^‒5 × 320 = 6.84 × 10¹⁹/m³

 σ_i = n_i (μ_n +μ_p) q = 6.84 × 10¹⁹ (0.36 + 0.17) × 1.6 × 10^‒19

= 5.8 (Ω-m)^‒1

……. Conductivity at 320°K

Conductivity of Extrinsic Semiconductor

Ex. 5: If a donor impurity is added to the extent of one atom per 10⁸ germanium atoms, calculate its resistivity at 300 °K. If its resistivity without addition of impurity at 300 °K is 44.64 Ω‒cm, comparing two values, comment on the result.

Assume: μ_n = 3800 cm²/V‒sec.

Solution:

Referring to Table 1.13.1 of properties of germanium, germanium has

 4.4× 10²² atoms/cm³.

For 10⁸ germanium atom there is 1 atom impurity added, as given.

Thus, for 4.4× 10²² germanium atoms, we have,

= 4.4×10²² / 10⁸⁰

= 4.4× 10¹⁴ atoms of impurity/cm³

This is nothing but concentration of donor atoms i.e. N^D

 N_D = 4.4 × 10¹⁴ per cm³

 = 4.4 × 10¹⁴ / 10^‒6

= 4.4× 10²⁰ per m³

Now as donor impurity is added, n‒type material will form,

 σ_n = n_n μ_n q = N_D μ_n q

where

n_n = N_D

and

μ_n = 3800 cm²/V‒sec

= 3800 × 10^‒4 m²/V‒sec

σ_n = 4.4× 10²⁰ × 3800 × 10^‒4 × 1.6 × 10^‒19

= 26.752 (Ω‒m)^‒1

Resistivity = ρ_n = 1/σ_n

= 1 / 26.752

= 0.0373 Ω‒m = 3.73 Ω ‒ cm

Comment: Comparing this with resistivity of intrinsic germanium it can be observed that resistivity reduces considerably due to addition of impurity. Hence conductivity of n‒type material is much higher and hence it can carry more current as compared to the intrinsic semiconductor. By controlling amount of doping we can control the conductivity.

Law of Mass Action

Ex. 6: Calculate the majority and minority carrier concentrations in silicon at room temperature of 27° C if

a) N_A =10¹⁷/cm³ and b) N_D = 5×10¹⁵/cm³

Solution:

a) As impurity is acceptor, the material is p‒type.

Pp = Majority carrier concentration = N_A =10¹⁷/cm³

For p‒type,

n_p×P_p = n_i²

                .... Law of mass action

n_i = 1.5 × 10¹⁰ /cm³

                …….From Table 1.13.1

n_p = n_i² / P_p = 2.25 × 10³ /cm³

                Minority carrier concentration

b) As impurity is donor, the material is n‒type

 n_n = Majority carrier concentration = N_D = 5 × 10¹⁵ /cm³

For n‒type,

 n_n × P_n = n_i²

               Law of mass action

 n_i = 1.5 × 10¹⁰ /cm³

               Basic material same

 P_n = n_i² / n_n = 45 × 10³ /cm³

               Minority carrier concentration

Ex. 7: A bar of silicon 0.1 cm long has a cross‒sectional area of 8 × 10^‒8 m², heavily doped with phosphorous. What will be the majority carrier density resulting from doping if bar is to have a resistance of 1.5 kΩ?

Given: For silicon at room temperature, μ_n = 0.14 m²/ V‒sec, μ_p = 0.05 m²/V‒sec, n_i = 1.5×10¹⁰ per cm³.

Solution: :

 R = ρl / A

where

 l = 0.1 cm = 0.1× 10⁻² m,

 A = 8×10^‒8 m², R = 1.5 kΩ

 ρ = RA/l = [ 1.5×10³ × 8×10^‒8  ]  / 0.1× 10^‒2

 = 0.12 Ω ‒ m

Conductivity.

σ = 1/ ρ  = 1/ 0.12 = 8.333 (Ω‒m)^‒1

But

σ = σ_n = (n_nμ_n+P_pμ_P)q

This is because phosphorous is donor impurity and will form n‒type material.

According to law of mass action for n‒type material,

n_n P_n = n_i²

p_n = n_i² / n_n

σ_n = (n_nμ_n+P_pμ_P)q 

σ_n = (n_nμ_n+( n_i² / n_n)μ_P)q 

σ_n = (n_n²μ_n+ n_i²μ_P)q 

 n_n × 8.333 = [ n_n² × 0.14 + (1.5 × 10¹⁰ / 10^‒6)² × 0.05 ] 1.602 × 10^‒19

  0.14n_n² ‒5.201× 10¹⁹ n_n + 1.125× 10³¹ = 0

Solving,

 n_n = 3.715 × 10²⁰ per m³ (neglecting other value as comparable to n_i)

n_n = N_D        

This is majority carrier density.

Ex. 8: Find the concentration of holes and electrons in a p type silicon at 300 °K assuming its resistivity in a p type silicon as 300 °K assuming its resistivity as 0.02 Ω‒cm, μ_p = 475 cm² / V ‒ sec, n_i = 1.45 ×10¹⁰ per cm³.

Solution:

 ρ = Resistivity = 0.02 Ω‒cm = 0.02 × 10^‒2 Ω‒m

The material is p type and its conductivity is,

 σ_P = 1/ρ = 1 / 0.02×10^‒2 = 5×10³ (Ω‒m) ^‒1

But

σ_P = N_Aμ_p q

where

 q = 1.6 x 10^‒19 C,

μ_p = 475 cm²/V‒sec = 475 × 10^‒4 m² / V‒sec

5×10³ = N_A × 475 × 1.6 × 10⁻¹⁹ ×10^‒4

 N_A = 6.5789 × 10²³ per m³

But

 P_p = N_A = 6.5789 × 10²³ per m³

Concentration of holes

Using law of mass action, Pp × n_p = n_i²

And

n_i =  1.45 × 10¹⁰ / 10^‒6 per m³

n_p = n_i² / Pp = (1.45 × 10¹⁶)²  / 6.5789×10²³

= 3.1958 × 10⁸ per m³

Concentration of electrons

Ex. 9: Find the concentration of holes and electrons in a p‒type Germanium at 300 °K, if the conductivity is 100 per ohm‒cm. Also find these values for n‒type silicon, if the conductivity is 0.1 per ohm‒cm. Given that

For Germanium n_i = 2.5×10¹³/cm, μ_n = 3800 cm²/v‒s, μ_p = 1800 cm²/v ‒ s

For silicon, n_i = 1.5×10¹⁰ per cm³, μ_n = 1300 cm²/v ‒ s and μ_p = 500 cm²/v ‒ s

Solution:

Case 1: p‒type Germanium

n_i = 2.5×10¹³/cm³ = 2.5×10¹³ / 10^-6 /m³

= 2.5×10¹⁹/m³

μ_n = 3800 cm²/V‒s = 3800×10^-4 m²/V‒s

μ_р= 1800 cm²/V‒s = 1800×10^‒4 m²/V‒s

σ_р = N_Aμ_pe  where σ = 100 (Ω‒cm)^‒1

100 / 10^‒2 = N_A×1800×10^‒4×1.6×10^‒19     i.e. N_A = 3.47×10²³/m³

P_p = N_A = 3.47 × 10²³/m³

       ... Concentration of holes

n_p = n_i² / P_P = n_i² / N_A

=  (2.5×10¹⁹)² /  3.47×10²³

= 1.8×10¹⁵/m³

... Concentration of electrons

case 2: n‒type Silicon

n_i = 1.5×10¹⁰/cm³ = 1.5×10¹⁰ / 10^-6 /m³

= 1.5×10¹⁶/m³

μ_n = 1300 cm²/V‒s = 1300×10^-4 m²/V‒s

μ_р= 500 cm²/V‒s = 500×10^‒4 m²/V‒s

σ_n = N_Dμ_nq  where σ_n = 0.1 (Ω‒cm)^‒1

0.1 / 10^‒2 = N_D×1300×10^‒4×1.6×10^‒19     i.e. N_D = 4.807×10²⁰/m³

n_n = N_D = 4.807 × 10²⁰/m³

       ... Concentration of electrons

p_n = n_i² / n_n = n_i² / N_D

=  (1.5×10¹⁶)² /  4.807×10²⁰

= 4.68×10¹¹/m³

... Concentration of holes

Equation of Charge Neutrality

Ex. 10: In a sample of germanium at 300 °K, it is found that donor concentration is 2×10¹⁴ atoms/cm³ and acceptor concentration is 3× 10¹⁴ atoms/cm³.

Determine the actual concentrations of free electrons and holes in the sample. Will it behave as p‒type or n‒type? Assume n_i at 300 °K = 2.5 × 10¹³ per cm³.

Solution:

In this sample,

 N_D = 2×10¹⁴ atoms/cm³ = 2×10¹⁴ / 10^‒6 = 2×10²⁰ m³

 N_A = 3×10¹⁴ atoms/cm³ = 3×10¹⁴ / 10^‒6 = 3×10²⁰ m³

Using equation of charge neutrality, p + N_D = n + N_A.

N_A ‒ N_D = p‒n

3×10²⁰  ‒ 2×10²⁰ = p‒n

i.e. p = n + 1×10²⁰

where p and n are actual concentrations of holes and electrons.

 n_i = 2.5×10¹³ per cm³

 = 2.5×10¹³ / 10^‒6

= 2.5×10¹⁹ per cm³

According to law of mass action, np = n_i²

 n (n + 1×10²⁰) = (2.5 × 10¹⁹)²      i.e. n² + 1×10²⁰n − 6.25×10³⁸ = 0

Solving

n = 5.901×10¹⁸ electrons/m³

p = 1.059×10²⁰ holes/m³

 As holes are much more than electrons, sample will behave as p‒type.

Ex. 11: A sample of Ge is doped to the extent of 10¹⁴ donor atoms/cm³ and 7 × 10¹³ acceptor atoms/cm³. At the temperature of the sample, the resistivity of pure Ge is 60 ohm‒cm. If the applied electric field is 2 V/cm, find the total conduction current density.

Solution:

 N_D = 10¹⁴/cm³ and N_A =7×10¹³/cm³

 ρ_i = 60 Ω‒cm

 J = (nμ_n + Pμ_p) qE

   …….. μ_n = 3800, μ_p = 1800

According to law of electrical neutrality,

 N_D + P = N_A + n

i.e. n‒p = 3 × 10¹³

According to law of mass action, np =n_i²

σ_i = n_i(μ_n+μ_p) q

i.e. 1/ρ_i = n_i (μ_n+ μ_p) q

1/60 = n_i (3800 + 1800) × 1.6 × 10^‒19

i.e. n=1.86 × 10¹³/cm³

Solve equations (1) and (2) to obtain n and p as,

n = 3.88×10¹³/cm³ and

p=0.88×10¹³/cm³

 J = (3.88 × 10¹³ × 3800 + 0.88 × 10¹³ × 1800) × 1.6 × 10^‒19 × 2 = 0.0522 A/cm

Diffusion Current and Diffusion Current Density

Ex. 12: Find out the diffusion constant of holes if their mobility is given as 0.039 m²/v‒sec.

Solution:

 μ_p = 0.039 m²/V‒s

According to Einstein's relation, D_p / μ_р = KT

 At 300 °K,

D_p / μ_р = D_p / 0.039 = 26 × 10^‒3

i.e. Dp = 1.01 × 10^‒3

Ex. 13: The phosphrous (donor) concentration in a region of a silicon crystal varies linearly from a concentration of n₀ = 10¹⁴ cm^‒3 at x = 0 mm to a concentration of n₁ = 10¹⁷ cm^-3 at x = 1 mm. The diffusion constant for electrons is  D_n = 22.5 cm²/s, the diffusion constant for holes is D_p = 5.2 cm²/s and the temperature is 300 °K. What is the diffusion  current density in the positive x‒direction?

Solution:

 n₀ = 10¹⁴ cm^-3

 n₁ = 10¹⁷ cm^-3,

x₁ = 1 mm,

x₀ = 0 mm,

 D_n = 22.5 × 10^-4 m²/s,

D_p = 5.2 × 10^-4 m²/s.

For silicon, n_i = 1.5 × 10¹⁰ cm^‒3

 n_i² = np

        ... Law of mass action

 P₀ = n_i² / n₀ = (1.5×10¹⁰)² / 10¹⁴ = 2.15 × 10⁶ cm^-3

P₁ = n_i² / n₁ = (1.5×10¹⁰)² / 10¹⁷ = 2.15 × 10³ cm^-3

dp/dx = [P₁‒P₀] / [x₁‒x₀] = [ 2.25×10³ ‒ 2.25×10⁶] / 10^‒1 = ‒2.248 × 10⁷ cm^‒4

 dn/dx = [n₁‒n₀] / [x₁‒x₀] = [10¹⁷ ‒ 10¹⁴] / 10^‒1 = 9.99 × 10¹⁷ cm^‒4

 J = J_n + J_p = [ qD_n dn/dx ] ‒ [ qD_p dp/dx ]

= 1.602×10^‒19 [ 22.5×9.99×10¹⁷ ‒ 5.2×(‒2.248×10⁷)]

= 3.6 A/cm²

= 3.6×10⁴ A/m²