Semiconductor: Two Marks Important Questions and Answers

Electron Devices

Chapter 1: Semiconductor

Two Marks Important Questions and Answers

1. Why silicon is widely used than germanium ?

Looking at the structure of silicon and germanium atom, it can be seen that valence shell of silicon is 3rd shell while valence shell of germanium is 4th shell. Hence valence electrons of germanium are at larger distance from nucleus than valence electrons of silicon. Hence valence electrons of germanium are more loosely bound to the nucleus than those of silicon. Thus valence electrons of germanium can easily escape from the atom, due to very small additional energy imparted to them. So at high temperature, germanium becomes unstable than silicon and hence silicon is widely used semiconductor material.

2. Find forbidden energy gap for germanium and silicon at 40°C.

T = 40 °C = 313 °K

For S_i,

E_G = 1.21 ‒ 3.6 × 10^‒4 × T

 = 1.21 ‒ 3.6×10^‒4 × 313 = 1.097eV

For Ge, 

E_G = 0.785 ‒ 2.23 × 10^‒4 × T

= 0.785 ‒2.23 × 10^‒4 × 313

= 0.7152 eV

3. Define one electron‒volt.

The energy required by an electron to fall through a potential of one volt is called one electron‒volt (eV).

 1 eV = 1.6 × 10^‒19 J

4. What is intrinsic semiconductor?

A sample of semiconductor in its purest form is called an intrinsic semiconductor. The impurity content in an intrinsic semiconductor is very very small, of the order of one part in 100 million parts of semiconductor.

5. Why the intrinsic semiconductors are not used in practice for manufacturing of electronic devices ?

In intrinsic semiconductor, very few electron‒hole pairs pairs get generated at room temperature. Hence very small current can be constituted, due to the application of voltage to an intrinsic semiconductor. Thus the conductivity of an intrinsic semiconductor at room temperature is very low. Such a low conductivity has very little practical significance.

Hence the intrinsic semiconductors are not used for the manufacturing of electronic devices.

6. Draw the energy band diagram for intrinsic and extrinsic semiconductors.

7. What is diffusion current in PN junction diode ?

When a semiconductor is nonuniformly doped, then there exists concentration gradient. On one side there is high carrier concentration while on the other there is low carrier concentration. Due to this, charges start moving from higher to lower concentration area. This process is called diffusion. When the charges move charges move due to diffusion, the current gets established in a semiconductor which is called a diffusion current.

8. Define drift and diffusion current.

When a voltage is applied to a material, the free electrons move towards the positive of the battery. While moving they collide with the adjacent atoms and keep changing their directions randomly. Still they keep drifting towards the positive of the battery. This is called drifting of charge carriers and the current due to such drifting of charge carriers is called drift current.

In case of diffusion current, the external voltage is not required. Due to nonuniform doping of the material, a concentration gradient is created across the material due to which the charge carriers move from higher to lower concentration area. This is called diffusion and the corresponding current is called diffusion current.

9. Name some donor and acceptor impurities.

The donor impurities are arsenic, bismuth, phosphorous while acceptor impurities are gallium, indium and boron.

10. What is the value of V_T (Volt equivalent of temperature) at a temperature of 300 °K?

V_T = KT = 8.62 × 10^‒5 × (300) = 0.02586 V

11. Define mass action law.

If n is the concentration of free electrons and P is the concentration of holes then the law of mass action states that the product of concentrations of electrons and holes is always constant, at a fixed temperature.

Mathematically it is expressed as,

 Np = n_i² where n_i is intrinsic concentration

12. Consider a silicon pn junction at T = 300 ° K so that n₁ = 1.5×10¹⁰ cm^‒3. The n type doping is 1× 10¹⁶ cm^‒3 and a forward bias of 0.60 V is applied to the pn junction. Calculate the minority hole concentration at the edge of the space charge region.

The hole concentration at the edge of the space charge region is,

 P_n = P_n0 e^V/VT

      …….Refer equation (1.18.8)

P_n0 = n_i² / N_D = (1.5×10¹⁶)² / 1×10²²

P_n = 2.25 × 10¹⁰ e^0.6/0.0259

 = 2.588 × 10²⁰ /m³

Note that n_i = 1.5 × 10¹⁰ / cm³ = 1.5 × 10¹⁶ / m³

13. Consider a gallium arsenide sample at T = 300 K with doping concentration of N_a=0, N_d=10¹⁶ cm^‒3 and μ_n = 8500. Calculate the drift current density if the applied electric field is E = 10 V/cm.

N_d=10¹⁶ cm^‒3

μ_n = 8500,

E = 10 V/cm.

 J = N_d μ_n qE

= 10¹⁶ × 8500 × 1.6 × 10^‒19 × 10

 J = 136 A/cm²