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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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