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This mock test of Electronic Devices 1 for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam.
This contains 10 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Electronic Devices 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

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: 2

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

(Take kT = 25 mV)

Solution:

Using the relation

D = 0.025 × 1900 cm^{2}/sec

= 47.5 cm^{2}/sec

QUESTION: 3

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.

*Answer can only contain numeric values

QUESTION: 4

A hypothetical semiconductor has bandgap of 0.56 eV at 300 K. If the effective mass of electron in such semiconductor is 4 times that of hole. Then the probability of occupancy of top of valence band (upto 2 decimal) at 300 K is

(Take kT = 26 mV)

Solution:

Position of Fermi level

E_{F} = 0.28 – 0.027

E_{F} = 0.252 eV

Probability of occupancy

≈ 0.99

**Note:**

This signifies that at room temperature the top of valence band is almost filled

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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