Anna University Solved Problems

ANNA UNIVERSITY SOLVED PROBLEMS

Carrier concentration in an intrinsic semiconductor

Problem 3.1

Find the resistance of an intrinsic germanium rod 1 cm long, 1 mm wide and 1 mm thick at 300 K.

For germanium ni = 2.5 × 1019m-3

μe = 0.39 m2 V-1s-1

μh = 0.19 m2 v-1 s-1 at 300 K

Given data

Intrinsic carrier concentration ni = 2.5 × 1019m-3

Electron mobility μe = 0.39 m2 V-1s-1

Hole mobility μh = 0.19 m2 v-1 s-1 

l = length of the rod = 1 cm = 1 × 10-2m

A = Area of cross-section (width × thickness)

A = (1 × 10-3) (1 × 10-3)

Solution

Electrical conductivity of an intrinsic semiconductor

σ = nie (μe + μh)

Substituting given values, we have

σ = 2.5 × 1019 × 1.6 × 10-19 × (0.39 + 0.19)

σ = 2.32 Ω-1 m-1

Extrinsic semiconductor

Problem 3.2

Find the concentration of holes and electrons in n-type silicon at 300 K, if the conductivity is 3 × 104 ohm-1 m-1. Also find these values for p-type silicon.

Given data

For silicon at 300 K, ni = 1.5 × 1016 m-3

μe = 1300 × 10-4 m2 V-1 s-1

μh = 500 × 10-4 m2 V-1 s-1

Solution

(a) Concentration in n-type silicon

σ = neμe

We know that np = ni2

(From the law of mass action)

(b) Concentration in p-type silicon

σ = реμh

Problem 3.3

A silicon material is uniformly doped with phosphorus atoms at a concentration of 2 × 1019 m-3. The mobilities of holes and electrons are 0.05 and 0.12 m2 V-1s-1 respectively, ni = 1.5 × 1016m-3. Find the electron and hole concentrations and electrical conductivity.

Solution

Problem 3.4

Find the hole and electron concentrations in a p-type semiconductor, if the acceptor density is 1020 atoms/m3 and the intrinsic concentration is 2.5 × 1019 per m3 at 300 K.

Solution

In a p-type semiconductor, the hole concentration is equal to the acceptor density.

Problem 3.5

The Hall coefficient of a specimen of a doped silicon is found to be 3.66 × 10-4 m3/C. The resistivity of the specimen is 8.93 × 10-3 Ω m. Find the mobility and density of the charge carriers.

Given data

Hall coefficient of the specimen RH = 3.66 × 10-4m3/C

Resistivity of the specimen ρ = 8.93 × 10-3 Ω m

Mobility of the carrier μh = ?

Density of charge carriers nh = ?

Solution

We know that density of charge carriers

Substituting the given values, we have

nh = 1.708 × 1022 m-3

Problem 3.6

Find the Hall coefficient and electron mobility of germanium for a given sample (length 1 cm, breadth 5 mm, thickness 1 mm). A current of 5 milliampere flows from a 1.35 volt supply and develops a Hall voltage of 20 millivolt across the specimen in a magnetic field of 0.45 Wb/m2. 

Given data

Current through the specimen I = 5 mA or 5 × 10-3 A

Voltage across the specimen V = 1.35 V

Length of the sample L = 1 cm or 1× 10-2 m

Breadth of the sample b = 5 mm or 5 × 10-3 m

Thickness of the sample t = 1 mm or 1 × 10-3 m

Hall voltage Vy = 20 × 10-3 V

Magnetic field H = 0.45 Wb/m2

Solution:

Hall coefficient

Electron mobility

Problem 3.7

A copper strip 2.0 cm wide and 1.0 mm thick is placed in a magnetic field with B = 1.5 weber/m2 perpendicular to the strip. Suppose a current of 200 A is set up in the strip. What Hall potential difference would appear across the strip?

Given N = 8.4 × 1028 electrons /m3.

Given data

Current flowing Ix = 200 A

Applied magnetic field Hz = 1.5 Wb m-2

Number of electrons, per unit volume n = 8.4 × 1028 electrons m-3

Thickness of the strip t = 1.0 × 10-3m

Solution

Note: This problem is important in the sense that it shows that Hall voltage can be also observed in metals besides semiconductor.

In semiconductors, Hall voltage is comparatively much larger; it is of the order of milli-volts as compared to the order of micro-volts in metals.

Moreover, to observe Hall voltage in metals, current of the order of amperes is needed when compared to the order of milliamperes as in the case of semiconductors.