Problem Set Solution # 2

ECEN 3320 Fall 2013

Semiconductor Devices

September 4, 2013 Due September 9, 2013

For all problems in this homework set and future ones, use the values of physical constants andmaterial parameters listed in Appendices 2 and 3 of VZ. Also, assume the temperature is T=300Kunless otherwise specified.

1. (a) Derive an expression for the total number of states in a semiconductor material (per unitvolume) between Ec and Ec + kT , where Ec is the conduction band edge (bottom of theconduction band), k is Boltzmanns constat and T is the temperature. You can solvethis problem by simply integrating the density of states over the energy range given inthis problem.

Solution: The density of states represents the number of states per unit volume. Forthis problem, we simply need to integrate the density of states, g(E)

g(E) =1

2pi2

(2meh2

)3/2E Ec

from Ec to Ec + kT : Ec+kTEc

gc(E)dE =1

2pi2

(2meh2

)3/2 EC+kTEc

(E Ec)1/2 dE

=1

2pi2

(2meh2

)3/2 kT0

x1/2dx

=1

2pi2

(2meh2

)3/2 23

(kT )3/2

=1

3pi2

(2mekTh2

)3/2=

(mem0

)3/2 1

3pi2

(2m0kT

h2

)3/2.

For values of

m0 = 9.11 1031 kgkT = 0.0258 1.6 1019 Jh = 1.05 1034 J s

1

we find Ec+kTEc

gc(E)dE = 1.97 1025 m3

= 1.97 1019 cm3

(b) Evaluate the expression you derived in (a) for GaAs and Si.

Solution: For GaAs, using me=0.067 m0, we find 3.411017 cm3.For Si, using me=1.08 m0, we find 2.211019 cm3.

2. Consider a silicon crystal whose band gap energy is Eg =1.12 eV and whose temperature iskept at T=300K.

(a) If the Fermi level, Ef , is located in the middle of the band gap, what is the probabilityof finding an electron (or equivalently, the probablility of a state being occupied) atE = Ec + kT .

Solution: The probability is given by the Fermi-Dirac function. Since Ef = EcEg/2,

f(E) =1

exp [(E Ec + Eg/2) /kT ] + 1 .

As E = Ec + kT , we find

f(Ec + kT ) =1

exp [(kT + Eg/2) /kT ] + 1

=1

exp [(.0258 + 0.56)/.0258] + 1

exp[22.6] 1.53 1010.

(b) If the Fermi level, Ef is located at the conduction band edge, EF = Ec, what theprobablility of finding an electron at E = Ec + kT .

Solution: The probability is given by evaluating the Fermi-Dirac probability density,with Ef = Ec

F (E) =1

exp [(E Ec)/kT ] + 1 .

With E = Ec + kT we find

F (Ec + kT ) =1

exp [1] + 1

0.27

3. The equilibrium electron concentration is given by the product of the density of states andthe probablility function, n(E) = gc(E)F (E). If E EF kT , the Fermi-Dirac probabilityfunction can be approximated with the Maxwell-Boltzmann function

F (E) =1

exp [E Ef )/kT ] + 1 exp [(E Ef )/kT ]

2

(a) Using this approximation, find the energy relative to the conduction band edge, EEc,at which the electron concentration becomes maximum.

Solution: The electron density, n(E) = g(E)f(E), can be written in the form

n(x) = Constantx exp(x)where x = (E Ec)/kT , for purposes of finding the maxima. Taking the derivativeand setting equal to zero

dn(x)

dx= Constant

[1

2xx

]exp(x) = 0

we find that the distribution peaks at E Ec = kT/2.(b) Using this approximation, calculate the electron concentration per unit energy interval

(in units of cm3 eV1) in silicon at energy E = Ec kT . Assume the Fermi level islocated at the center of the band gap, EF = Ec Eg/2.Solution: We want to evaluate

n(E) = g(E)f(E) =1

2pi2

(2meh2

)3/2E Ec exp [(E Ef )/kT ]

at E = Ec + kT for Ef = g/2. The result is

n(Ec + kT ) = g(E)f(E)

=1

2pi2

(2meh2

)3/2kT exp [(kT + Eg/2)/kT ]

=1

2pi2

(2meh2

)3/20.0258 exp [(0.0258 + 0.56)/0.0258]

= 1.84 1011cm3eV1(c) Repeat the calculation in (b) without using the approximation.

Solution: The answer is really close, to within a part in roughly 1010.

4. (a) Read the example given in Section 2.5.2 of Van Zeghbroeck carefully. Find all 24 possibleconfugurations and construct Fig. 2.5.3 yourself.

Solution: The result is in figure 1.

(b) Discuss how the distribution function would change if you change the total energy or ifyou change the total number of electrons in the system.

Solution: When the total energy is increased (or decreased), overall electron distribu-tion shifts toward higher (or lower) energy. But more importantly, if the energy isincreased there are more possibilities of electron configurations with high electronenergy. This tends to flatten the distribution function. This is the same as the effectof temperature on the Fermi-Dirac distribution function. That is, as the tempera-ture is increased, the FD function tends to get flattened, and vice versa.

When the total number of electrons is increased (or decreased), the overall dis-tribution function shifts accordingly. However, it doesnt necessarily accompanyflattening of distribution function, which is determined by the total energy. There-fore the effect of the number of electrons in the system is to shift the position ofFermi level.

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Figure 1: Figure 2.5.3 reconstructed

4