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This mock test of Electronic Devices 2 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 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Which of the following quantities cannot be measured/determined using Hall Effect?

Solution:

If a semiconductor carrying current is placed in a transverse magnetic field B, an electric field E is induced in the direction perpendicular to both I and B. This phenomenon is known as the Hall Effect, it can be used to determine whether semi-conductor is n-or p-type and to find the carrier concentration and mobility.

Diffusion constant is found from electron mobility using Einstein relationship.

*Answer can only contain numeric values

QUESTION: 2

The ratio of hole diffusion current to the electron diffusion current in an infinite p-n junction is 4. If conductivity on p-side is 1.4 S/cm and on n-side is 2.8 S/cm, then the ratio of diffusion length L_{n} and L_{p} (ie L_{n}/L_{p}) is _______.

Solution:

For diffusion current

QUESTION: 3

Under high electric fields, in a semiconductor with increasing electric field,

Solution:

we known that

V = μE

μ = constant when E < 10^{3} V/cm

when 10^{3} < E < 10^{4} V/cm

when E > 10^{4} V/cm

∴ When E Increase mobility decrease.

*Answer can only contain numeric values

QUESTION: 4

The energy band structure of n-type semiconductor is as shown below

The probability that the donor energy state is empty is (up to 3 significant digits)

Solution:

According to Fermi function,

Probability it is empty is 1 – F(E_{D}) = 1 – 0.014 = 0.986

QUESTION: 5

In a hypothetical semiconductor hole concentration is 2 × 10^{15}/cm^{3} and electron concentration is 1 × 10^{15}/cm^{3}. The mobility of holes is and mobility of electrons is

If the specimen is subjected to hall measurement. Then the measured hall coefficient (R_{H}) is

Solution:

For a semiconductor with significant concentration of both type of carriers

*Answer can only contain numeric values

QUESTION: 6

In a germanium semiconductor at 127°C has a hole concentration gradient between two points 1.5 × 10^{22}/m^{3}. If the hole mobility in germanium is 800 cm^{2}/v-s, then the diffusion current density is ______ A/m^{2}.

Solution:

Given,

T = 127°C = 400°K

μ_{P} = 800 cm^{2}/v-s

D_{P} = μ_{P }V_{T}

V_{T} = kT/2 =

⇒ D_{P} = 800 × 0.0345 cm^{2}/s = 27.6 cm^{2}/s

= -1.5 × 10^{22} × 10^{-8} × 1.6 × 10^{-19} × 27.6 A/cm^{2}

= -6.624 A/m^{2}

QUESTION: 7

2 Consider two energy levels: E_{1}, E eV above the Fermi level and E_{2}, EeV below the Fermi level. P_{1} and P_{2} are respectively the probabilities of E_{1} being occupied by an electron and E_{2} being empty. Then

Solution:

**Case 1:**

E_{1} ⇒ EeV above the fermi level probability of E_{1} being occupied by on electron = P(E_{1})

**Case II:**

E_{2} → EeV below the fermi level probability of E_{2} being occupied

Probability of E_{2} being empty:

*Answer can only contain numeric values

QUESTION: 8

A silicon sample is dope with unknown concentration of Arsenic, such that the fermi level (E_{F}) is 0.407 eV relative to intrinsic level (E_{i}). The conductivity of the silicon sample after doping is______(Ω – cm)^{-1}.

The mobility of electrons μ_{n} = 3800 cm^{2}/v-s and holes μ_{p} = 1800 cm^{2}/v-s. Take kT = 0.0259 eV at 300°K and intrinsic concentration n_{i} = 1.5 × 10^{10} cm^{-3}

Solution:

Given, Si sample is doped with As (donor impurity) and the band structure is

Let n_{d} be doping concentration, hence the electron concentration also is n_{d}.

The extrinsic electron concentration is related to intrinsic concentration by

∴ Conductivity σ = μn_{d}q

= 3800 × 1 × 10^{17} × 1.6 × 10^{-19}

= 60.8 (Ω - cm)^{-1}

QUESTION: 9

The effective density of states in conduction and valence band N_{C} and N_{v} at 300K is 4.7x10^{17 }cm^{−3 }and 7x10^{18}cm^{−3} respectively. The bandgap of the mentioned semiconductor is 1.42eV at 300K. The intrinsic carrier concentration is:

Use Boltzmann constant k = 8.62 x 10^{−5 }eV/K

Solution:

Given

N_{c} = 4.7 x 10^{17} cm^{−3}

N_{v} = 7 x 10^{18} cm^{−3}

Eg = 1.42 eV

Now,

*Answer can only contain numeric values

QUESTION: 10

Holes are being steadily injected into a region of n-type silicon. In the steady state, the excess hole concentration profile is shown. [The term excess means over and above concentrations p_{no}]

If D_{p} = 12 cm^{2}/sec N_{D} = 10^{16}/cm^{3}, n_{i} = 1.5 × 10^{10}/cm^{3} and w = 5 μm.

The current density that will flow in x-direction is _____ nA/cm^{2}.

Solution:

The diffusion current density is given by

= -4.49 × 10^{10}/cm^{4}

= -1.6 × 10^{-19} × 12 × (-4.49 × 10^{10})

= 8.64 × 10^{-8} A/cm^{2}

= 86.4 × 10^{-9} nA/cm^{2}

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