2.2. BASES TEÓRICAS-CIENTÍFICAS
2.2.2 Conocimiento lingüístico y conciencia metalingüística
These are p-n junctions which emit visible light when forward biased (Fig.5.34).
Figure 5.34 Biasing circuit for a light emitting diode.
When a p-n junction is forward biased, the majority carriers (electrons) diffuse from the conduction band of n-type region into the conduction band of the p-type region. This is referred to as minority carrier injection since electrons are minority carriers in the p-type region. This disturbs the equilibrium between electrons and holes in the p- type region and in order to regain equilibrium concentration, the electrons recombine with holes thereby releasing energy. In all p-n junctions, some of this energy is given off as heat and some energy as light depending on the energy released. In indirect band gap semiconductors like germanium and silicon, greater percentage of the energy is given off as heat. In gallium arsenide phosphide (GaAsP), a large fraction of the energy is released as light. The process of emission of light from a forward biased
p-n junction is called ‘injection electroluminescence’. A diode made of GaAs with a band gap of 1.43 eV emits infrared radiation of about 9000A◦. When GaP (band gap = 2.26 eV) is used, the emitted radiation is in the green region. A mixed compound of the type GaAs1−xPx with varying concentration of As and P may be used to select the emitted radiation in the visible region.
A commercial light emitting diode is made out of a suitable material and the diode is embedded in a clear plastic cover with a parabolic reflector provided at the back to direct the emitted radiation. The emitted intensity is found to be linearly related to the forward current flowing across the junction. The application of a forward bias is a necessary condition for the operation of a light emitting diode because minority carrier injection takes place in a forward biased junction. The increase in minority carriers will disturb the equilibrium and result in recombination associated with emission of radiation. The recombination probability is high in the transition region, i.e., the region just outside the depletion region where the carrier concentration is modified due to carrier injection. The emission can be concentrated by keeping the depletion region to a minimum using a forward bias.
Light emitting diodes are used in visual display units, calculators, clocks, panel meters, pilot lamps, etc. They are used in optical communication applications using fiber optics like data transmission and telephones. They are also used in card and paper tape readers, burglar alarms, etc.
Numerical Examples
5.1 Calculate the conductivity of silicon doped with 1021 atoms m−3of boron if the
mobility of holes is 0.048m2v1s1.
Solution: Conductivity of the extrinsic semiconductor with an acceptor con- centration of Nais given by
σ = Naeµh
= 1021× 1.6 × 10−19× 0.048 = 7.68 ohm−1m−1(Ans.).
5.2 Calculate the resistivity of intrinsic germanium if the intrinsic carrier density is 2.5 × 1019m−3 assuming electron and hole mobilities of 0.38 and 0.18 m2v−1s−1
respectively.
Solution: Resistivity is given by ρ = 1
σ =
1 ne(µe + µh)
= 1
2.5 × 1019× 1.6 × 10−19 × (0.38 + 0.18)
= 0.45 ohm m (Ans.).
5.3 Calculate the electron and hole densities in n-Si doped with 1019 donors m−3if the intrinsic carrier concentration in silicon is 1.4 × 1016m−3.
Solution: Density of electrons,
n = 1019+ 1.4 × 1016 ≈ 1019m−3 Density of holes,
p = n2i/n where niis the intrinsic concentration.
∴ p = (1.4 × 10 16)2 1019
= 1.96 × 1013m−3(Ans.).
5.4 Calculate the conductivity of germanium with the given data. What is the effect of doping germanium with donor - type impurity to the extent of one atom per 108germanium atoms?
Given data: Intrinsic carrier concentration in germanium at 300 K, ni = 2.4 × 1019 m−3
Electron mobility,µe = 0.39m2v−1s−1 Hole mobility,µh = 0.19m2v−1s−1 Number of atoms/m3 = 4.4 × 1028
Electron Charge,e = 1.6 × 10−19C Solution: Conductivity of intrinsic germanium is
σi = nie(µe+µh)
= 2.4 × 1019× 1.6 × 10−19× (0.39 + 0.19) = 2.23 ohm−1m−1
Conductivity of n-type germanium is σn = neµe + peµh
n ≈ Nd ≈ 4.4 × 1020m−3; p = n2i/Nd = 1.3 × 1018m−3
Since p is very small compared to n, the second term in the conductivity equa- tion may be neglected.
∴σn= Ndeµe = 27.46 ohm−1m−1(Ans.).
5.5 The electrical conductivity of an intrinsic semiconductor increases from 19.96 ohm−1m−1 to 79.44 ohm−1m−1 when the temperature is increased from 60 to
100◦C. Find the band gap energy of the semiconductor.
Solution: The electrical conductivity of an intrinsic semiconductor is given by
σ = k.exp(Eg/2kT )
Neglecting its weak temperature dependence, k can be considered to be a con- stant. A graph of ln σ versus (1/T ) will have a slope given by
△(lnσ) △(1/T ) = −Eg 2k Eg = − (lnσ2− lnσ1) (1/T2− 1/T1) × 2k = (−4.3750 + 2.9937) (3.003 − 2.681) × 10−3 × 2 × 1.38 × 10 −23J = 1.184 × 10−19J. = 1.184 × 10 −19 1.6 × 10−19 eV = 0.74 eV (Ans.).
5.6 A semiconductor sample of thickness 1.2 × 10−4m is placed in a magnetic field of
0.2T acting perpendicular to its thickness. Find the Hall voltage generated when a current of 100 mA passes through it. Assume the carrier concentration to be 1023m−3.
Solution: Hall voltage generated is given by the formula VH =
3π 8 ·
BI net
= 3 × 3.14 × 0.2 × 100 × 10
−3
8 × 1023× 1.6 × 10−19× 1.2 × 10−4
VH = 0.0123 Volt or 12.3 mV (Ans.).
5.7 Calculate the barrier potential existing at the junction of p-type silicon with an acceptor concentration of 1023m−3and n-type silicon with a donor concentration
of 1020m−3if the intrinsic carrier concentration at 300 K is 1.4 × 1016m−3. Solution: The barrier potential is given by
VB = kT e ln NaNd n2 1 = 1.38 × 10 −23 × 300 1.6 × 10−19 ln 1023× 1020 (1.4 × 1016)2 VB = 0.64 V (Ans.).
5.8 Intrinsic silicon has a carrier concentration of 1.1 × 1016m−3. If the mobilities of
electrons and holes are 0.17 and 0.035m2v−1s−1respectively at room temperature, compute the resistivity of silicon.
Solution: Conductivity is given by σi = nie(µe+µh)
= 1.1 × 1016× 1.6 × 10−19(0.17 + 0.035) = 3.608 × 10−4ohm−1m−1
Resistivity = 1/σ = 2.77 × 103ohm m (Ans.).
5.9 The compound gallium arsenide has an intrinsic conductivity of 10−6 ohm−1m−1 at 20oC. How many electrons have jumped the forbidden energy gap? [given: µe = 0.88 m2V−1s−1and µh = 0.04 m2V−1s−1]
Solution: Number of electrons that have crossed the forbidden energy gap is the number of electrons contributing to conductivity, which is the carrier concentration, ni.
ni =σ/e(µe +µh)
= 1 × 10−6/1.6 × 10−19(0.88 + 0.04) 6.79 × 1012 m−3( Ans.).
5.10 An electric field of 100 Vm−1 is applied to a sample of n-type semiconductor
having a Hall coefficient - 0.0125 m3C−1. Determine the current density in the sample if the electron mobility is 0.36 m2V−1s−1.
Solution:
Current density J = nev = neµE = −µE/RH = 0.36 × 100/0.0125 = 2880 Am−2(Ans.).
5.11 The resistivity of germanium at 20◦C is 0.5 ohmm. What will be its resistivity at 40◦C if the band gap of germanium is 0.7 eV?
Solution: We have the relation, △(lnρ) △(l/T ) = Eg 2k △(lnρ) = Eg.△(1/T )/2k = 0.7 × 1.6 × 10 −19× (1/313 − 1/293) 2 × 1.38 × 10−23 lnρ40− lnρ20= −0.885 lnρ40= lnρ20− 0.885 = −0.693 − 0.885 = −1.578 ρ40= 0.206 ohmm (Ans.).
5.12 sample of silicon is doped with 1022 phosphorus atoms m−3. What would you
expect to measure for its resistivity? What Hall voltage would you expect in a sample 100 µm thick when a current of 1 mA is passed perpendicular to a magnetic field of 0.1 T. [Given: µe = 0.07m2V−1s−1]
Solution:
Conductivity σ = Ndeµe (5.1)
= 1022× 1.6 × 10−19× 0.07 (5.2)
= 112 ohm−1m−1. (5.3)
Resistivity ρ = 1/ρ = 8.93 × 10−3 ohmm (Ans.). (5.4) Hall voltage VH =
3π 8
BI
= 3 × 3.14 × 0.1 × 1 × 10
−3
8 × 1022× 1.6 × 10−19× 100 × 10−6 (5.6)
= 0.735 mV (Ans.). (5.7)
5.13 An intrinsic semiconductor has an energy gap of 0.7 eV. Calculate the probability of occupation of the lowest level in conduction band at 0◦, 50◦and 100◦C.
Solution: The lowest level in conduction band corresponds to (E − EF) = 0.35 eV.
The probability of occupation of this level is given by F(E) = 1
1 + exp (E − EF)/kT Substituting the values, we get
At T = 00C = 273K, F(E) = 3.5 × 10−7 T = 500C = 323K, F(E) = 3.5 × 10−6
T = 1000C = 373K, F(E) = 1.88 × 10−5(Ans.).
Exercise
5.1 Explain the formation of a potential barrier in a p-n junction. (March 1999). 5.2 What is the difference between p-type and n-type semiconductors?
(March 1999). 5.3 What are zener diodes? Explain breakdown voltage in the case of such diodes.
(August 1999). 5.4 Distinguish between conductors, insulators and semiconductors on the basis of band theory. (August 1999, August 2000, March 2001). 5.5 Describe a p-n junction and explain the effect of forward and reverse biasing on the barrier potential of the junction. (August 1999). 5.6 The band gap in germanium is 0.68 eV. Assuming that the number of electron- hole pair is proportional to exp(−Eg/2kT ), find the percentage increase in the number of charge carriers, when the temperature increases from 300◦K to 320◦K.
5.7 Distinguish between avalanche diode and zener diode. (March 2000). 5.8 What is Hall effect? How is it useful in the study of conduction in materials?
(March 2000). 5.9 What is Hall effect? Derive an expression for the Hall coefficient. (August 2000). 5.10 Derive an expression for the conductivity of an intrinsic semiconductor. The electron and hole mobility of silicon are 0.14 and 0.05 m2v−1s−1 respectively at a given temperature. If the electron density is 1.5 × 1016 m−3, calculate the
resistivity of silicon. (August 2000).
5.11 Discuss the I-V characteristics of a zener diode. Discuss the different types of
breakdown at the p-n junction. (March 2001).
5.12 Describe a p-n junction and explain its voltage-current characteristics. What is the effect of temperature on the conductivity of semiconductors.
(August 2001). 5.13 Explain how a potential barrier is formed in a p-n junction. (March 2002). 5.14 Distinguish between direct band gap and indirect band gap semiconductors.
(March 2002). 5.15 Distinguish between zener breakdown and avalanche breakdown.
Dielectric Properties Of Materials
6.1
Introduction
In this chapter we study the properties of insulator materials in static electric fields. In- sulators possess very few free charge carriers and hence are bad conductors of electric current. However, application of an electric field results in some interesting processes involving rearrangement of charges in the material. These characteristic features of insulators are called dielectric properties. These properties and the behaviour of di- electric materials in a static electric field may be explained reasonably well with the help of a simple atomic theory.
Gauss theorem is one of the most fundamental theorems related to static electric fields. The theorem may be explained by considering a closed surface containing a large number of charges. If Q1, Q2, Q3, . . . Qn represent the charges enclosed by a surface, then the total electric flux emerging from the surface is given by
Φ = n X i=1 Qi = Z D · ds (6.1)
We note that the total flux is expressed in units of charge, namely coulombs which may also be written as equal to the surface integral where D represents the flux density expressed in coulombs m−2 and ds is the surface element where the flux density is
evaluated. The flux density in vacuum is proportional to the electric field strength E according to the relation
D = ǫoE (6.2)
Since the electric field is measured in units of volts m−1, ǫ
owill have the units farad m−1. This constant of proportionality is called the permittivity of vacuum. If the electric field and the flux density are considered in any other medium, the relation between them will be modified as
D = ǫE (6.3)
where ǫ is the permittivity of the medium. This relation may also be expressed as D = ǫoǫrE (6.4)
where ǫr is a dimensionless quantity called “relative permittivity” or “dielectric con- stant” of the medium. For a dielectric medium, this is an important parameter and can be easily evaluated by simple measurements. Consider a parallel plate condenser of area A and separation between the plates, d (Fig.6.1).
− − − − − − − − − − − − − − + + + + + + + + + + + + + +
E
Figure 6.1 A parallel capacitor with an applied electric field E. The flux lines will be directed from positive plate to the negative plate.
Let q be the charge density on the plates which will be numerically equal to the flux density D. Hence, we have the electric field existing between the plates given by
E = D/ǫ◦ǫr = q/ǫ◦ǫr (6.5) The capacitance of the parallel plate condenser is given by
C = Total charge on the plate Potential difference =
qA
Ed (6.6)
Substituting for E from equation (6.5) we get
C = ǫ◦ǫrA/d (6.7) If the dielectric medium between the two parallel plates is removed and the space evacuated, the capacitance will be
Cvac = ǫ◦A/d (6.8)
From equation (6.7) and (6.8), the dielectric constant of the medium is given by
ǫr = C/Cvac (6.9)
Thus, the value of ǫr can be determined experimentally by measuring the capacitance of a parallel plate condenser with and without the dielectric.