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Solutions to EDC Tutorial - 1 on Semiconductors

Electronic Devices & Circuits
Solutions to Tutorial – 1 
Topic : Semiconductors 

Short Questions:
1. Explain why every semiconductor is electrically neutral in nature?
Ans: Every atom is electrically neutral in nature as they consists of equal amount of protons (positive charge) and electrons (negative charge). We know the charge of an electron is -1.6 x 10-19coulombs and the charge of a proton is +1.6 x 10-19coulombs.
Intrinsic semiconductor is a material which consists of all identical atoms arranged in a particular dimension. Since atom is electrically neutral, so the intrinsic semiconductor.
Extrinsic semiconductor is formed when impurity is added to intrinsic semiconductor. The impurity may be either pentavalent impurity atom (to get N-type ) or trivalent impurity atom(to get P-type). Since every atom is neutral, impurity atom is also electrically neutral so extrinsic semiconductor is also electrically neutral.
Hence every semiconductor is electrically neutral.
Note: Ions are atoms with extra electrons (or) missing electrons.
Explanation Video:  https://www.youtube.com/watch?v=-fs1FkT51bM

2. Define free electron, bound (valence) electron, hole, electron current and hole current in a semiconductor?
Ans: electron in outermost orbit or valence band is called valence electron or bound electron, as it is in the control of parent atom. When external energy (like temprature or light) is applied, some covalent bonds break, so that many electrons will move from valence band to conduction band. The electrons in conduction band are called free electrons.
In semiconductor, current flows due to movement of both charge carriers, electrons and holes. Electron current is defined as movement of electrons in conduction band, where as hole current is defined as the opposite direction of current results from movement of electrons in valence band.
Note: hole is an empty energy level in valence band, which can't move itself. But as electron in valence band tries to occupy that vacancy, it looks like hole is moving in the direction opposite to the direction of electron in valence band.
Explanation Video:  https://www.youtube.com/watch?v=I_meqVX3log

3. why intrinsic concentration of germanium is more than silicon at any temperature?
Ans: Intrinsic concentration (ni) is referred as the number of covalent bonds break at a given temperature per unit volume. It is a function of temperature, so as temperature increases, ni increases.
We know, energy gap is defined as the difference between lowest energy level of conduction band and highest energy level of valence band, and is the minimum energy required to break a covalent bond so that an electron from valence band can reach conduction band. When a covalent bond breaks, an electron in conduction band (free electron) and a vacancy in valence band (hole) is created.
Germanium atom has less energy gap than Silicon atom, so at a given temperature ie. For the given thermal energy, more covalent bonds will break in germanium than in silicon, so ni is more in Ge than Si.
Explanation Video: https://www.youtube.com/watch?v=A6ZurAx90bE

4. Why mobility of electron is more than mobility of hole in a semiconductor?
Ans: we know, conduction band consists of only free electrons, remaining are all empty energy levels where as valence band consists of valence electrons and holes (empty energy levels in valence band). So more empty energy levels are available in conduction band than valence band so electron in conduction band can move easily than electron in valence band, that is the reason why mobility of electron is more than hole.
Explanation Video: https://www.youtube.com/watch?v=MzFc4WIlvIA

5. Justify : “Majority carriers in extrinsic semiconductor depends on doping concentration, while minority carriers depends on externally applied energy like temperature, light etc.. (but not on voltage)”.
Ans: Take N-type semiconductor to justify.
N-type semiconductor is formed when pentavalent impurity is added to pure semiconductor. Assume silicon crystal as intrinsic semiconductor and phosphorous as pentavalent impurity atom. Phosphorous has five valence electrons, four out of them covalently bonded with neighboring electrons of silicon and one electron is available as extra electron, which occupies the energy level just below lowest energy level of conduction band of silicon.
At T = 0ok, no thermal energy is applied, so no covalent bonds will break. Hence conduction band is completely empty and valence band is completely full.
no. of free electrons in conduction band = electron concentration (nn) = zero
no. of holes in valence band = hole concentration (pn) = zero
no. of extra electrons = ND
At T = 50ok, all extra electrons get excited as small amount of thermal energy is applied and they all moved into conduction band, so become free electrons. But no covalent bond will break at this temperature i.e. no EHPs are generated.
no. of free electrons in conduction band = electron concentration (nn) = ND
no. of holes in valence band = hole concentration (pn) = zero
no. of extra electrons = zero
At T = 300ok, some covalent bonds will break and that many EHPs will be generated. Say 'x' no. of covalent bonds are broken at 300ok, so x no. of free electrons in conduction band and x no. of holes in valence band gets created. Now
no. of free electrons in conduction band = electron concentration (nn) = ND+ x ≈ ND
no. of holes in valence band = hole concentration (pn) = zero + x = x
so nn depends directly on ND, impurity doping concentration and pn depends on EHPs generated which depends directly on applied external energy in the form of light or temperature.


Long Questions:
1. Find the electrid field required to move an electron from valence band to conduction bond? Assume Silicon atom.
Ans: ε = 45.6 kV/cm.
Solution: The energy of an electron having mass 'm' moving with an average drift velocity 'v' is
w = ½ mv2 = ½ m(μ.ε)2 joules
we know 1eV = 1.6 x 10-19 Joulesand Eg of Si at room temperature ≈ 1 eV = 1.6 x 10-19 Joules
equate the energy of electron to the energy gap of silicon, and find required electric field, ε.
after solving for ε, ε = 45.6 kV/cm.

2. Find the conductivity of intrinsic silicon at room temperature? If it is doped with phosphorous so that majority carrier concentration becomes 10 16 donors/cm3. Find the conductivity of doped semiconductor? Give your comments from the results obtained.
Ans:
Conductivity of intrinsic silicon = 4.32 x 10-6 (Ω-cm)-1
Conductivity of doped semiconductor = 2.08 (Ω-cm)-1
Hence after doping, conductivity is increased by around 5,00,000 times.
Note: Intrinsic semiconductor is not used practically because of its low conductivity, which is not sufficient for practical applications. So impurity is added to intrinsic semiconductor to improve conductivity i.e. extrinsic semiconductor has higher conductivity than intrinsic semiconductor, which is used in real world applications.

3. An N type silicon bar 0.1 cm long and µm2 in cross sectional area has a majority carrier concentration of 5 x 1020 m-3 and the carrier mobility is 0.13 m2/V-sec at 300oK. If the charge of an electron is 1.6 x 10-19 coulomb. Find the resistance of the bar..
Ans: Resistance of the N-type semiconductor bar = 96 MΩ

4. Define Mass Action Law and also find the minority carrier concentration of an n-type silicon, if it is doped with 1 phosphorous in every 1000 silicon atoms per cubic centimeter.

Mass Action Law: under thermal equilibrium, the product of total positive charge and total negative charge is always constant and is equal to square of intrinsic concentration.
n.p = ni2
where n = electron concentration in conduction band and p = hole concentration in valence band.
Ans: Minority carrier concentration of N-type semiconductor given (pn) = 4.5 holes/cm3
Note: Using Mass Action Law, we can find minority carrier concentration in a semiconductor if majority carrier concentration is known and vice versa.

5. Find Minority carrier concentration of p-type silicon sample, whose resistivity is 1 kΩ-cm.
Ans: Minority carrier concentration of P-type semiconductor given (np) = 5 x 1012 electrons/cm3