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QUESTION: 1

In a semiconductor which of the following carrier can contribute to the current?

Solution:

In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons take part in the flow of the current.

QUESTION: 2

The Fermi level in a semiconductor bar should:

Solution:

__Under equilibrium____:__

- The fermi level across the entire material will be sum and does not vary with distance.
- If there is any disturbance in the material, like junction contact, injection of impurities at any point, the charge carriers redistribute themselves such that the fermi-potential is same in entire material.

__Under non-equilibrium:__

- The fermi level is uneven with gradient of charge distribution across distance, in material.
- It can be studied using quasi-fermi states, related to charge distribution.

*Answer can only contain numeric values

QUESTION: 3

Hole mobility in Ge at room temperature is 1900 cm^{2}/V-sec. The diffusion coefficient is ________cm^{2}/sec.

(Write answer to one decimal point.)

(Take kT = 25 mV)

Solution:

Using the relation, D = 0.025 × 1900 cm^{2}/sec

= **47.5 cm ^{2}/sec**

QUESTION: 4

In the band diagram of a semiconductor, the fermi level is 0.3 eV above the intrinsic level. The energy level E in the diagram represents:

Solution:

- Given that fermi level is 0.3 eV above the intrinsic level, hence the type of semiconductor is n-type.
- In n-type the donor level is just below the conduction band and above the fermi level.

QUESTION: 5

The Difference between the donor energy level and fermi level in a n-type semiconductor in where 25% of the atoms are ionised at 300 k is:

Solution:

As 25% of donor atoms are ionised, the occupation probability of donor level is 0.75.

E_{D} - E_{F} = -0.028 eV

*Answer can only contain numeric values

QUESTION: 6

A semiconductor with intrinsic carrier concentration 1 × 10^{10} cm^{-3} at 300°K has both valence and conduction band effective densities of states N_{C} and N_{V} equal to 10^{19} cm^{-3}. The band gap E_{g} is _____ eV.

Solution:

The intrinsic concentration is related to bandgap by:

E_{g} = 0.0518 ln (10^{8}) eV

=** 0.954 eV**

*Answer can only contain numeric values

QUESTION: 7

In a very long p-type Si bar with doping concentration N_{a} = 10^{17} cm^{-3}, excess holes are injected such that excessive concentration of holes at x = 0 is 5 × 10^{16} cm^{-3}. The hole concentration at x = 1 μm is _____ × 10^{17} cm^{-3}. Take μ_{p} = 500 cm^{2}/v-s and recombination time constant τ_{p} = 10^{-8} s, kT = 0.0259 eV

Solution:

Diffusion constant:

Diffusion length:

- Hole concentration at any distance x is sum of equillibrium and excess hole concentration excess hole concentration varies with the distance x as Δp e
^{-x/Lp} **p = p**_{o}+ Δp e^{-x/Lp}

Where p_{o}= Doping concentration

Δp = extra hole concentration

= (1+.379) x 10^{17}

**1.379 × 10**^{17}cm^{-3}

QUESTION: 8

In an intrinsic semiconductor, the Fermi energy level EF doesn’t lie in the middle of the band gap cause:

Solution:

- In an intrinsic semi-conductor the Fermi energy:Where N
_{C}and N_{V}are density of state in valence and conduction bond respectively. - Since N
_{C}∝ m_{n}, effective mass of electron and N_{v}∝ m_{p}, effective mass of hole. Since m_{n}and m_{p}are not equal:

QUESTION: 9

A silicon bar is doped with donor impurities ND = 2.25 × 10^{15} cm^{-3}. If the electron mobility μn = 1000 cm2/v-s then the approximate value of resistivity of silicon bar assuming partial ionization of 55% is __________ (Ω - cm)

Solution:

Given ND = 2.25 × 1015 cm^{-3}

__Partial ionization of 55%, number of electrons is:__

⇒ n = 0.55 × 2.25 × 1015 cm-3

= 1.2375 × 1015 cm-3

Electron mobility μn = 1000 cm2/v-s

Resistivity of n-type silicon =

= **5.05 Ω – cm**

QUESTION: 10

Holes are injected into n-type Ge so that the at the surface of the semiconductor hole concentration is 10^{14}/cm^{3}. If diffusion constant of a hole in Ge is 49cm^{2}/sec and minority carrier lifetime is τ_{p} = 10^{-3} sec. Then the hole concentration Δp at a distance of 4mm from the surface is ______10^{14}/cm^{3}.

Solution:

= 1.6 × 10^{13}/cm^{3}

= 0.16 × 10^{14}/cm^{3}

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