ee145hmwk5sol

EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri

Second Edition ( 2001 McGraw-Hill)

Chapter 5 5.1 Bandgap and photodetection

a. Determine the maximum value of the energy gap that a semiconductor, used as a photoconductor, can have if it is to be sensitive to yellow light (600 nm).

b. A photodetector whose area is 5 10-2 cm2 is irradiated with yellow light whose intensity is 2 mW cm-2. Assuming that each photon generates one electron-hole pair, calculate the number of pairs generated per second.

c. From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the primary wavelength of photons emitted from this crystal as a result of electron-hole recombination.

d. Is the above wavelength visible?

e. Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?

Solution a We are given the wavelength = 600 nm, therefore we need Eph = h = Eg so that, Eg = hc/ = (6.626 10-34 J s)(3.0 108 m s-1) / (600 10-9 m) Eg = 3.31 10-19 J or 2.07 eV b Area A = 5 10-2 cm2 and light intensity Ilight = 2 10-3 W/cm2. The received power is: P = AIlight = (5 10-2 cm2)(2 10-3 W/cm2) = 1.0 10-4 W Nph = number of photons arriving per second = P/Eph

Nph = (1.0 10-4 W) / (3.31 10-19 J) = 3.02 1014 Photons s-1 Since the each photon contributes one electron-hole pair (EHP), the number of EHPs is then:

NEHP = 3.02 1014 EHP s-1 c For GaAs, Eg = 1.42 eV and the corresponding wavelength is

= hc/Eg = (6.626 10-34 J s)(3.0 108 m s-1) / (1.42 eV 1.602 10-19 J/eV) = 8.74 10-7 m or 874 nm The wavelength of emitted radiation due to electron-hole pair (EHP) recombination is therefore 874 nm.

d It is not in the visible region (it is in the infrared).

e From Table 5.1 (in the textbook), for Si, Eg = 1.10 eV and the corresponding cut-off wavelength is,

g = hc/Eg = (6.626 10-34 J s)(3.0 108 m s-1) / (1.1 eV 1.602 10-19 J/eV) g = 1.13 10-6 m or 1130 nm

5.1

EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri Since the 874 nm wavelength of the GaAs laser is shorter than the cut-off wavelength of 1130 nm, the Si photodetector can detect the 874 nm radiation (Put differently, the photon energy corresponding to 874 nm, 1.42 eV, is larger than the Eg, 1.10 eV, of Si which means that the Si photodetector can indeed detect the 874 nm radiation).

5.2 Minimum conductivity a. Consider the conductivity of a semiconductor, = ene + eph. Will doping always increase the

conductivity?

b. Show that the minimum conductivity for Si is obtained when it is p-type doped such that the hole concentration is

pm = ni eh

and the corresponding minimum conductivity (maximum resistivity) is

min = 2eni eh c. Calculate pm and min for Si and compare with intrinsic values.

Solution a Doping does not always increase the conductivity. Suppose that we have an intrinsic sample with n = p but the hole drift mobility is smaller. If we dope the material very slightly with p-type then p > n. However, this would decrease the conductivity because it would create more holes with lower mobility at the expense of electrons with higher mobility. Obviously with further doping p increases sufficiently to result in the conductivity increasing with the extent of doping.

b To find the minimum conductivity, first consider the mass action law: np = ni2

isolate n: n = ni2/p

Now substitute for n in the equation for conductivity:

= ene + eph = eni

2 ep

+ hep To find the value of p that gives minimum conductivity (pm), differentiate the above equation with respect to p and set it equal to zero:

d = eni

2 e2 + he dp p

eni2 e

2pm+ he = 0

Isolate pm and simplify,

5.2


pm = ni eh

Substituting this expression back into the equation for conductivity will give the minimum conductivity:

min = eni2 e

pm+ hepm = eni

2eni e h + heni

eh

min = enie he + eni e h = eni eh + eni eh

min = 2eni eh c From Table 5.1, for Si: e = 1350 cm2 V-1 s-1, h = 450 cm2 V-1 s-1 and ni = 1.45 1010 cm-3. Substituting into the equations for pm and min: pm = ni eh = 1.45 10

10 cm3( ) 1350 cm 2 V1 s1450 cm2 V1 s1

= 2.51 1010 cm-3

min = 2eni eh min = 2 1.602 1019 C( )1.45 1010 cm3( ) 1350 cm2 V1 s1( ) 450 cm2 V1 s1( ) min = 3.62 10-6 -1 cm-1 The corresponding maximum resistivity is:

max = 1 / min = 2.76 105 cm The intrinsic value corresponding to pm is simply ni (= 1.45 1010 cm-3). Comparing it to pm:

pmni

= 2.51 1010 cm 3

1.45 1010 cm3 =1.73

The intrinsic conductivity is:

int = eni(e + h) int = (1.602 10-19 C)(1.45 1010 cm-3)(1350 cm2 V-1 s-1 + 450 cm2 V-1 s-1) int = 4.18 10-6 -1 cm-1 Comparing this value to the minimum conductivity:

intmin =

3.62 106 W1 cm14.18 106 W-1 cm-1 = 0.866

Sufficient p-type doping that increases the hole concentration by 73% decreases the conductivity by 15% to its minimum value.

5.3 Compensation doping in Si a. A Si wafer has been doped n-type with 1017 As atoms cm-3.

1. Calculate the conductivity of the sample at 27 C. 5.3


2. Where is the Fermi level in this sample at 27 C with respect to the Fermi level (EFi) in intrinsic Si?

3. Calculate the conductivity of the sample at 127 C. b. The above n-type Si sample is further doped with 9 1016 boron atoms (p-type dopant) per

centimeter cubed.

1. Calculate the conductivity of the sample at 27 C. 2. Where is the Fermi level in this sample with respect to the Fermi level in the sample in (a) at 27

C? Is this an n-type or p-type Si?

Solution a Given temperature T = 27 C = 300 K, concentration of donors Nd = 1017 cm-3, and drift mobility e 800 cm2 V-1 s-1 (from Figure 5Q3-1). At room temperature the electron concentration n = Nd >> p (hole concentration).

50

100

1000

2000

1015 1016 1017 1018 1019 1020-3

ElectronsHoles

Dopant Concentration, cm Figure 5Q3-1 The variation of the drift mobility with dopant concentration in Si

for electrons and holes at 300 K.

(1) The conductivity of the sample is:

= eNde (1.602 10-19 C)(1017 cm-3)(800 cm2 V-1 s-1) = 12.8 -1 cm-1 (2) In intrinsic Si, EF = EFi,

ni = Ncexp[(Ec EFi)/kT] (1) In doped Si, n = Nd, EF = EFn,

n = Nd = Ncexp[(Ec EFn)/kT] (2) Eqn. (2) divided by Eqn. (1) gives,

Nd = exp EFn EFi (3) ni kT

ln Ndni

= EFn EFi kT 5.4

EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri EF = EFn EFi = kT ln(Nd/ni) (4) Substituting we find (ni = 1.45 1010 cm-3 from Table 5.1 in the textbook), EF = (8.617 10-5 eV/K)(300 K)ln[(1017 cm-3)/ (1.45 1010 cm-3)] EF = 0.407 eV above Efi

10

100

1000

10000

50000

70 100 800

Ge

Nd =1013

Si

LT1.5Nd =10

14

Nd =1016

Nd =1017

Nd =1018

Nd =1019

T1.5

Temperature (K) Figure 5Q3-2 Log-log plot for drift mobility versus temperature for n-type Ge and n-type Si samples. Various donor concentrations for Si are shown, Nd are in cm-3. The upper right insert is the simple theory for lattice limited mobility whereas the lower left inset is the simple theory for impurity scattering limited mobility.

(3) At Ti = 127 C = 400 K, e 450 cm2 V-1 s-1 (from Figure 5Q3-2). The semiconductor is still n-type (check that Nd >> ni at 400 K), then

= eNde (1.602 10-19 C)(1017 cm-3)(450 cm2 V-1 s-1) = 7.21 -1 cm-1 b The sample is further doped with Na = 9 1016 cm-3 = 0.9 1017 cm-3 acceptors. Due to compensation, the net effect is still an n-type semiconductor but with an electron concentration given by,

n = Nd Na = 1017 cm-3 0.9 1017 cm-3 = 1 1016 cm-3 (>> ni) We note that the electron scattering now occurs from Na + Nd (1.9 1017 cm-3) number of ionized centers so that e 700 cm2 V-1 s-1 (Figure 5Q3-1). (1) = eNde (1.602 10-19 C)(1016 cm-3)(700 cm2 V-1 s-1) = 1.12 -1 cm-1 (2) Using Eqn. (3) with n = Nd Na we have

Nd Nani

= exp E Fn EFikT

so that E = EFn EFi = (0.02586 eV)ln[(1016 cm-3) / (1.45 1010 cm-3)] E = 0.348 eV above EFi The Fermi level from (a) and (b) has shifted down by an amount 0.059 eV. Since the energy is still above the Fermi level, this an n-type Si.

5.5


5.4 Temperature dependence of conductivity An n-type Si sample has been doped with 1015 phosphorus atoms cm-3. The donor energy level for P in Si is 0.045 eV below the conduction band edge energy.

a. Calculate the room temperature conductivity of the sample.

b. Estimate the temperature above which the sample behaves as if intrinsic.

c. Estimate to within 20% the lowest temperature above which all the donors are ionized.

d. Sketch schematically the dependence of the electron concentration in the conduction band on the temperature as log(n) versus 1/T, and mark the various important regions and critical temperatures. For each region draw an energy band diagram that clearly shows from where the electrons are excited into the conduction band.

e. Sketch schematically the dependence of the conductivity on the temperature as log() versus 1/T and mark the various critical temperatures and other relevant information.

Solution

Si

Ge

GaAs

0C200C400C600C 27C

1018

1 1.5 2 2.5 3 3.5 4

1015

1012

103

106

1091.451010 cm-3

2.41013 cm-3

2.1106 cm-3

1000/T (1/K)Figure 5Q4-1 The temperature dependence of the intrinsic concentration.

a The conductivity at room temperature T = 300 K is (e = 1350 10-4 m2 V-1 s-1 can be found in Table 5.1 in the textbook):

= eNde = (1.602 10-19 C)(1 1021 m-3)(1350 10-4 m2 V-1 s-1) = 21.6 -1 m-1 b At T = Ti, the intrinsic concentration ni = Nd = 1 1015 cm-3. From Figure 5Q4-1, the graph of ni(T) vs. 1/T, we have:

5.6

EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri 1000 / Ti = 1.9 K-1

Ti = 1000 / (1.9 K-1) = 526 K or 253 C c The ionization region ends at T = Ts when all donors have been ionized, i.e. when n = Nd. From Example 5.7, at T = Ts:

n = Nd = 12 Nc Nd

12exp

E2kTs

Ts = E2k ln

Nd12 NcNd

= E2k ln 2Nd

Nc

Ts = Ek ln Nc

2Nd

Take Nc = 2.8 1019 cm-3 at 300 K from Table 5.1 (in the textbook), and the difference between the donor energy level and the conduction band energy is E = 0.045 eV. Therefore our first approximation to Ts is:

Ts = Ek ln Nc

2Nd

= 0.045 eV( ) 1.602 1019 J/eV( )

1.3811023 J/K( )ln 2.8 1019 cm 3( )2 1015 cm 3( )

= 54.68 K

Find the new Nc at this temperature, Nc:

N c = Nc Ts300

32 = 2.8 1019 cm3( ) 54.68 K

300 K

32= 2.179 1018 cm-3

Find a better approximation for Ts by using this new Nc:

T s = Ek ln N c

2Nd

= 0.045 eV( ) 1.602 1019 J/eV( )

1.3811023 J/K( )ln 2.179 1018 cm 3( )2 1015 cm 3( )

= 74.64 K

N c = Nc T s300

32 = 2.8 1019 cm3( ) 74.64 K

300 K

32= 3.475 1018 cm-3

A better approximation to Ts is:

T s = Ek ln N c

2N

= 0.045 eV( ) 1.602 1019 J/eV( )

1.3811023 J/K( )ln 3.475 1018 cm3( )2 1015 cm3( )

= 69.97 K

d

N c = Nc T s300

32

= 2.8 1019 cm3( ) 69.97 K

300 K

32 = 3.154 1018 cm-3

5.7


T s = Ek ln N c

2Nd

= 0.045 eV( ) 1.602 1019 J/eV( )

1.3811023 J/K( )ln 3.154 1018 cm 3( )2 1015 cm 3( )

= 70.89 K

We can see that the change in Ts is very small, and for all practical purposes we can consider the calculation as converged. Therefore Ts = 70.9 K = 202.1 C. d and e See Figures 5Q4-2 and 5Q4-3.

ln(n)

1/T

Ti

Ts

Intrinsic

Extrinsic Ionization

ni(T)

ln(Nd) slope = E/2k

slope = Eg/2k

Figure 5Q4-2 The temperature dependence of the electron concentration in an n-type semiconductor.

log(n)

INTRINSIC

EXTRINSIC

IONIZATION

log()

log() T 3/2 T 3/2

Latticescattering

Impurityscattering

1/TLow TemperatureHigh Temperature

T

Metal

Semiconductor

Figure 5Q4-3 Schematic illustration of the temperature dependence of electrical

conductivity for a doped (n-type) semiconductor.

5.5 GaAs Ga has a valency of III and As has V. When Ga and As atoms are brought together to form the GaAs crystal, as depicted in Figure 5Q5-1, the 3 valence electrons in each Ga and the 5 valence electrons in each As are all shared to form four covalent bonds per atom. In the GaAs crystal with some 1023 or so equal numbers of Ga and As atoms, we have an average of four valence electrons per atom, whether Ga or As, so we would expect the bonding to be similar to that in the Si crystal: four bonds per atom. The crystal structure, however, is not that of diamond but rather that of zinc blende (Chapter 1 of the textbook).

5.8


a. What is the average number of valence electrons per atom for a pair of Ga and As atoms and in the GaAs crystal?

b. What will happen if Se or Te, from Group VI, are substituted for an As atom in the GaAs crystal?

c. What will happen if Zn or Cd, from Group II, are substituted for a Ga atom in the GaAs crystal?

d. What will happen if Si, from Group IV, is substituted for an As atom in the GaAs crystal?

e. What will happen if Si, from Group IV, is substituted for a Ga atom in the GaAs crystal? What do you think amphoteric dopant means?

f. Based on the above discussion ,what do you think the crystal structures of the III-V compound semiconductors AlAs, GaP, InAs, InP, and InSb will be?

As GaGa As

As GaAsGa

As GaGa As

As GaAsGa

AsGa

Ga atom (Valency III) As atom (Valency V)

Figure 5Q5-1 The GaAs crystal structure in two dimensions. Average number

of valence electrons per atom is four. Each Ga atom covalently bonds with four neighboring As atoms and vice versa.

Solution

As GaGa As

As GaAsGa

As GaGa As

As GaAsGa

As Ga

Ga atom (Valency III)As atom (Valency V)

hyb orbitals

As ion core (+5e)

Valenceelectron

hyb orbitals

Ga ion core (+3e)

Valenceelectron

Explanation of bonding in GaAs: The one s and three p orbitals hybridize to form 4 hyb orbitals. In As there are 5 valence electrons. One hyb has two paired electrons and 3 hyb have 1 electron each as shown. In Ga there are 3 electrons so one hyb is empty. This empty hyb of Ga can overlap the full hyb of As. The overlapped orbital, the bonding orbital, then has two paired electrons. This is a bond between Ga and As even though the electrons come from As (this type of bonding is called dative bonding). It is a bond because the electrons in the overlapped orbital are shared by both As and Ga. The other 3 hyb of As can overlap 3 hyb of neighboring Ga to form "normal bonds".

5.9

EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri Repeating this in three dimensions generates the GaAs crystal where each atom bonds to four neighboring atoms as shown. Because all the bonding orbitals are full, the valence band formed from these orbitals is also full. The crystal structure is reminiscent of that of Si. GaAs is a semiconductor.

a The average number of valence electrons is 4 electrons per atom.

b Se or Te replacing As will have one additional electron that cannot be involved in any of the four bonds. Hence Se and Te will act as a donor.

c Zn or Cd replacing Ga will have one less electron than the substituted Ga atom. This creates a hole in a bond. Zn and Cd will act as acceptors.

d The Si atom has 1 less electron than the As atom and when it substitutes for an As atom in GaAs there is a "hole" in one of the four bonds. This creates a hole, or the Si atom acts as an acceptor.

e The Si atom has 1 more electron than the Ga atom and when it substitutes for a Ga atom in GaAs there is an additional electron that cannot enter any of the four bonds and is therefore donated into the CB (given sufficiently large temperature). Si substituting for Ga therefore acts as a donor.

f All these compounds (AlAs, GaP, InAs, InP, InSb) are compounds of III elements and V elements so they will follow the example of GaAs.

5.10

Chapter 5

ee145hmwk5sol

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