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Test: Carrier Transport - 1 - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - Test: Carrier Transport - 1

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Test: Carrier Transport - 1 - Question 1

Gradual flow of charge from a region of high density to a region of low density is called _________.

Detailed Solution for Test: Carrier Transport - 1 - Question 1

Drift:

  • Directed motion of charge carriers in semiconductors occurs through two mechanisms:
  • Charge drift under the influence of applied electric field and
  • Diffusion of charge from a region of high charge density to one of low charge density.

Phenomenon:

  • When no electric field is applied to the semiconductor which is above 0º K, the conduction electrons move within the crystal with random motion and repeatedly collide with each other and the fixed ions.
  • Due to the randomness of their motion, the net average velocity of these charge carriers in any given direction is zero. Hence, no current exists in the crystal under this condition of no field.
  • Now, consider the case when an electric field is applied to the crystal. Under the influence of this field, the charge carriers attain a directed motion which is superimposed on their random thermal motion.
  • This results in a net average velocity called drift velocity in the direction of the applied electric field.
  • Electrons and holes move in opposite directions but because of their opposite charges, both produce current in the same direction.
  • In extrinsic semiconductors, this current is essentially a majority carrier flow.
  • The drift velocity is proportional to electric field strength E, the constant of proportionality being called mobility µ.

Diffusion:

  • It is the gradual flow of charge from a region of high density to a region of low density.
  • It is a force-free process based on the non-uniform distribution of charge carriers in a semiconductor crystal.
  • It leads to an electric current without the benefit of an applied field.
  • This flow or diffusion of carriers is proportional to the carrier density gradient, the constant of proportionality being called diffusion constant or diffusion coefficient (D) which has a unit of m2/s.

Recombination:

  • It is the collision of an electron with a hole.
  • The process is essentially the return of a free conduction electron to the valence band and is accompanied by the emission of energy.
  • The recombination rate is directly proportional to the carrier concentration for the simple reason that the larger the number of carriers, the more likely is the occurrence of electron-hole recombination.
  • This phenomenon is important in describing minority carrier flow.
  • In a semiconductor, the thermal generation of electron-hole pairs also takes place continuously.
  • Hence, there is a net recombination rate given by the difference between the recombination and generation rates.
Test: Carrier Transport - 1 - Question 2

An n–type silicon sample of 10–3 m length and 10–10 m2 cross sectional area has an impurity concentration of 5 × 1020 atom/m3. If mobility of majority carries is 0.125 m2/v-sec, then the resistance of the sample will be ________.

Detailed Solution for Test: Carrier Transport - 1 - Question 2

Concept:

The conductivity of the silicon sample is given by σ 

⇒ σ = n q μn,

Here,

  • n is the impurity concentration
  • μ is the mobility of electrons
  • q is the electronic charge in coulomb

The resistance of the silicon sample is given by

⇒ R = ρL/A   

Here, 

  • L is the length of the sample
  • A is the cross-sectional area of the sample
  • ρ is the resistivity of the sample

Calculation

Given:

n = 5 × 1020 atom/m3, μn = 0.125 m2/v-sec, q = 1.6 × 10-19 C

⇒ σ = ( 5 × 1020) × (1.6 × 10-19) × (0.125)

⇒ σ = 10 

Therefore Resistivity (ρ) = 1/σ = 0.1

Also given: L = 10–3 m, A = 10–10 m2

⇒ R = 10Ω = 1MΩ 

Hence the resistance of the silicon sample is 1MΩ 

Therefore the correct answer is option b. 

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Test: Carrier Transport - 1 - Question 3

Calculate the hall voltage when the magnetic field is 8 A/m, current is 4 A, width is 5 m and the concentration of carrier is 100000.

Detailed Solution for Test: Carrier Transport - 1 - Question 3

Concept: 

The hall voltage VA is given by:

ρc: Charge density = ne

RH: Hall coefficient

W is the side across which the magnetic field is applied.

n is the carrier concentration

e is the electron charge

Calculation:

W = Thickness = 5 m, I = 4 A, B =  8 A/m, n = 100000

ρc = ne

ρc = 100000 × 1.6 × 10-19

ρc = 1.6 × 10-14

Test: Carrier Transport - 1 - Question 4

In a semiconductor, Drift current is due to:

Detailed Solution for Test: Carrier Transport - 1 - Question 4

Drift Current:

  • Drift current in the diode is the combined effect of movement of the minority charge carrier and majority charge carriers and also on the electric field applied to the diode.
  • Drift current is due to the motion of charge carriers when a force exerted on them by an electric field.
  • In the p-n junction diode, electrons are the majority charge carriers in the n-region and holes are the majority charge carriers in the p-region.
  • When an electric field is applied to the diode there is more number of covalent bonds break and the concentration of charge carriers also increases in both region (p-type, n-type), and hence they affect the drift current in the diode.

Drift current Idrift = Jdrift × A

Where, Jdrift = Drift current density

A = Area of semiconductor

Now, Jdrift = σ ⋅ E

= (nqμn + p.qμp)

σ = Conductivity

E = Electric field

⇒ Idr = (nq.μn + p.qμp) ⋅ E A,

n = Number of electrons in n region

p = Number of holes in p-region

q = Charge on electrons and holes

μp = Mobility of holes

μn = Mobility of electrons

Additional Information

Diffusion

1) Diffusion is a natural phenomenon.

2) The migration of charge carriers from higher concentration to lower concentration or from higher density to lower density is called diffusion.

3) Diffusion is mainly due to the concentration gradient and is always negative.

It is given by:

dp/dx for holes and

dn/dx for electrons.

Diffusion current is calculated by:

where,

Dis hole diffusion constant in cm2/sec

and q is charge in Coulomb.

Test: Carrier Transport - 1 - Question 5

According to Einstein’s relationship for a semiconductor, the ratio of the diffusion constant to the mobility of the charge carriers is

Detailed Solution for Test: Carrier Transport - 1 - Question 5

According to Einstein's relation, the ratio of Diffusion constant to mobility remains constant, at fixed

Temperature, i.e.

VT – Temperature equivalent

K – Boltzman’s constant

T – Temperature in Kelvin

Test: Carrier Transport - 1 - Question 6

Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016 atoms/cm3 at 300 K temperature. Determine the approximate resistivity of the slab.

(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010 /cm3; Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).

Detailed Solution for Test: Carrier Transport - 1 - Question 6

Concept:

For a semiconductor slab doped with excess carriers and placed in an electric field, the current equation is given by:-

J = σ E

Where J = Current Density

σ = Conductivity of the semiconductor defined as:

σ = qnμn + qpμp

n = excess electrons contributing to the current conduction.

p = excess holes contributing to the current conduction.

Also, the Resistivity is defined as the reciprocal of conductivity.

Calculation:

The given Si semiconductor is doped with Boron impurity.

So, Na = 1016 /cm(p-type)

q = 1.6 × 10-19 C

μn = 1300 cm2/V-s and μp = 500 cm2/V-s

The excess electrons is given by; 

Since, Na ≫ ni, we can assume that the conduction current exists only because of acceptor impunities, i.e. because of Na.

So, σ = q Na μp

ρ = Resistivity of the given Slab
Putting on the respective values, we get:

Test: Carrier Transport - 1 - Question 7

The unit of mobility is

Detailed Solution for Test: Carrier Transport - 1 - Question 7

Analysis:

Drift velocity is directly proportional to the electric field, i.e.

Vd ∝ E

Vd = μE

The proportionality constant 'μ' is called as the mobility and is given by:

Substituting the respective units, we get:

Test: Carrier Transport - 1 - Question 8

If temperature will increase, the conductivity of semiconductor will:

Detailed Solution for Test: Carrier Transport - 1 - Question 8

Explanation:

Current in a material can be expressed as:

I = NqAvd

Where

N = Carrier concentration

q = electrons charge (magnitude)

A = cross-section area

The current density (J) will be:

Also, J = σ E     ----(2)

σ = conductivity

E = Electric field

J = Current density

vd = μ E    ----(3)

μ = mobility of carriers

E = Electric field

Analysis:

On using equation (1), (2) & (3) we get,

J = NqVd = NqμE = σE

⇒ σ = Nqμ  

So as conductivity is directly proportional to the mobility of ions and the number of EH pair ions.

The conductivity of an intrinsic semiconductor depends upon the number of hole electron pairs and mobility. The number of hole electron pairs increases with an increase in temperature, while its mobility decreases. However, the increase in hole electron pairs is greater than the decrease in their mobility.

So with an increase in temperature, conductivity in semiconductors also increases.

Hence option (d) is the correct answer.

Test: Carrier Transport - 1 - Question 9

Hall effect can be used to measure

Detailed Solution for Test: Carrier Transport - 1 - Question 9

Hall Voltage states that if a specimen (metal or semiconductor) carrying a current I is placed in transverse magnetic field B, an electric field is induced in the direction perpendicular to both I and B.

Hall Voltage is given by:

ρ = Charge density

Hall coefficient can be written as:

Where,

ρ = charge density = σ / μ 

n = charge concentration

σ = conductivity

μ = mobility constant

Hence, the Hall coefficient becomes

The Hall effect provides information on the sign, concentration, and mobility of charge carriers in the normal state.

A positive sign for the Hall coefficient indicates that the majority carriers are holes and the semiconductor is P-type.

A negative sign for the Hall coefficient indicates that the majority carriers are electrons and the semiconductor is N-type.

Applications of Hall-effect:

Hall effect can be used to find:

1. Carrier concentration

2. Type of semiconductor

3. Conductivity

4. Mobility

It cannot be used to find a magnetic field.

Common Confusion Point:

Looking at the formula one can think that the magnetic field can be calculated but in the HALL Experiment, perpendicular MAGNETIC field and electric field are applied on the material and other parameters are measured.

Test: Carrier Transport - 1 - Question 10

On applying an electric field of intensity 10 V/cm across a semiconductor at a certain temperature the average drift velocity of free electrons is measured to be 70 m/s. Then the electron mobility is

Detailed Solution for Test: Carrier Transport - 1 - Question 10

Concept:

Mobility: It denotes how fast is the charge carrier is moving from one place to another.

It is demoted by μ.

Mobility is also defined as:

Where,

Vd = Drift velocity

E =  field intensity

Calculation:

E = 10 V/cm

Vd = 70 m/s = 7000 cm/s

From equation (1):

μ = 7000/10

μ = 700 cm2/Vs

Note: 

Electron mobility is always greater than hole mobility, 

Hence electron can travel faster and contributes more current than a hole.

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