EEE 455 1

Hetero-junction and Compound Semiconductor

Assignmentsubmitted to:

Md. Hasibul AlamAssistant professor

submitted by:MD.JUBAYER AL-MAHMUD

ID:1006164L/T:4/1

Dept. of Electrical and Electronic Engineering

.

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I. BRILLOUIN ZONE

A. Brillouin Zone For BCC crystal

For the Brillouin Zone demonstration”BrillouinzoneV iewer”IX has been used.

Fig. 1. Real space structure and unit cell

Fig. 2. Brillouin Zone For BCC crystal

B. Brillouin Zone For FCC crystal

Fig. 3. Real space structure and unit cell

For FCC crystal the symmetry points are shown on fig-ure 5,10,7,9,8,6. For demonstration BandStructureLabIX isused.

II. ISO-ENERGY SURFACE CALCULATION

A. GaAs

degeneracy number N=1ml = 0.06653me

Fig. 4. Brillouin Zone For FCC crystal

Fig. 5. Gamma

mt = 0.06653me where, me = 9.11 × 10−31 As ml and mtare same ,the constant energy surface will be spherical.

MATLAB CODE:

clc;close all;clear all;h=6.626e-34;e=1.6e-19hReduced=h/(2*pi);me=9.11e-31;mlong=0.06653;mTrans=0.06653;N=1;mC=Nˆ(2/3)*(mlong*mTransˆ2)ˆ(1/3);E0=1.42;a=5.654e-10;del=(-1:.07:1);[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me)).*((kx.ˆ2+ky.ˆ2+kz.ˆ2)/mC);E=E/e;isosurface(kx,ky,kz,E,E0);grid on;xlabel('X');

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Fig. 6. X

Fig. 7. L

ylabel('Y');zlabel('Z');title('GaAs Iso-energy surface');

B. Silicon

Degeneracy number N=6ml = 0.9041me mt = 0.20061me where, me = 9.11×10−31

As ml and mt are different constant energy surface will beellipsoidal.

MATLAB CODE:

clc;close all;clear all;h=6.626e-34;hReduced=h/(2*pi);me=9.11e-31;

Fig. 8. W

Fig. 9. U

E0=0.25;a=5.4306e-10;del=(-1:0.05:1);% X Axis[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me*1.6e-19)).*(((kx-pi*(aˆ-1)).ˆ2).../.90+(ky.ˆ2+kz.ˆ2)/.20);% Energy Formulaisosurface(kx,ky,kz,E,E0);

grid on;xlabel('X');ylabel('Y');zlabel('Z');title('Silicon Iso-energy Surface');hold on;

[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me*1.6e-19)).*(((kx+pi*(aˆ-1)).ˆ2).../.90+(ky.ˆ2+kz.ˆ2)/.20);

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Fig. 10. K

Fig. 11. Iso-energy Surface of GaAs

isosurface(kx,ky,kz,E,E0);% y Axis[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me*1.6e-19)).*(((ky-pi*(aˆ-1)).ˆ2).../.90+(kx.ˆ2+kz.ˆ2)/.20);

isosurface(kx,ky,kz,E,E0);[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me*1.6e-19)).*(((ky+pi*(aˆ-1)).ˆ2).../.90+(kx.ˆ2+kz.ˆ2)/.20);isosurface(kx,ky,kz,E,E0);%z Axis[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me*1.6e-19)).*(((kz-pi*(aˆ-1)).ˆ2).../.90+(ky.ˆ2+kx.ˆ2)/.20);isosurface(kx,ky,kz,E,E0);[kx,ky,kz]=meshgrid(2*pi*(aˆ-1)*del,...2*pi*(aˆ-1)*del,2*pi*(aˆ-1)*del);E=((hReducedˆ2)/(2*me*1.6e-19)).*(((kz+pi*(aˆ-1)).ˆ2).../.90+(ky.ˆ2+kx.ˆ2)/.20);isosurface(kx,ky,kz,E,E0);

III. BAND-STRUCTURE CALCULATION

The BandstructureLabIX is used for the calculations.1) No strain effect considered2) 150 points used for the calculation

Fig. 12. Silicon iso-energy Surface

3) Spin effect considered.

Band structure for Si,GaAs and Ge are shown on figure13,14and 15 respectively.

W L Gamma X K UW L Gamma X K U

E1E2E3E4E5E6

E7E8E9E10E11E12

E13E14E15E16E17E18

E19E20E21E22E23E24

E25E26E27E28E29E30

E31E32E33E34E35E36

E37E38E39E40

k ((nm)−1)

0 10 20 30 40

E (

eV)

-10

0

10

20

30

Fig. 13. Silicon band-structure

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W L Gamma X K UW L Gamma X K U

E1E2E3E4E5

E6E7E8E9E10

E11E12E13E14E15

E16E17E18E19E20

E21E22E23E24E25

E26E27E28E29E30

E31E32E33E34E35

E36E37E38E39E40

wave Vector

0 10 20 30 40

E (

eV)

0

20

Fig. 14. GaAs band-structure

IV. EFFECTIVE MASS CALCULATION

Figure 16,17 and 18 show the effective mass of electron andhole for Si,GaAs adn Ge.

1) Green rectangles identify the hole effective mass.2) Blue rectangles identify energy band gap3) red rectangles identify the transverse and longitudinal

effective mass of electron.

V. DENSITY OF STATE EFFECTIVE MASS ANDCONDUCTIVITY EFFECTIVE MASS

A. Electron

1) Density of State effective Mass: EquationIX for desity

of effective mass of electron is mc = V23

d (ml ×mt)

1

3

Where Vd = Degeneracy number ml= Longitudinal mass ofelectron mt= Transverse mass of electronred rectangles of figure 16,17,18 produces the value of lon-gitudinal and transverse effective mass of electron. Using theequation above the density of sate effective mass for

1) Si:1.094 me2) GaAs:0.06653 m3) Ge: 0.8766 me

W L Gamma X K UW L Gamma X K U

E1E2E3E4E5E6E7E8

E9E10E11E12E13E14E15E16

E17E18E19E20E21E22E23E24

E25E26E27E28E29E30E31E32

E33E34E35E36E37E38E39E40

wave Vector

0 10 20 30 40

E (

eV)

-10

0

10

20

30

Fig. 15. Ge band-structure

Fig. 16. Effective mass: Silicon

2) Conductivity effective mass: Formula for conductivityeffective mass calculation can be found on IX.me,cond =

31

ml+

2

mtForm the figure 16,17 and 18 and using the above equation

conductivity effective mass for:1) Si:0.27 me2) GaAs:0.06653 m3) Ge: 0.119 me

B. Hole

1) Density of state effective mass: From IX we find theequation for DOS effective mass for hole mh = (m

32

hh+m32

lh)23

Where mlh=Light hole effective mass mhh=Heavy holeeffective

Average value of heavy hole and light hole effective mass:

siliconmhh = 0.5302meandmlh = 0.17011me Germaniummhh = 0.35719meandmlh = 0.04407me GaAsmhh = 0.62421meandmlh = 0.07797me

using aforementioned equation:1) Si = 0.5926 me2) Ge= 0.36744 me3) GaAs= 0.6424 me

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Fig. 17. Effective mass: GaAs

Fig. 18. Effective mass: GeE

2) Conductivity effective mass: Form the figure 16,17,18.For Gemhh= 0.1732 me and mlh=0.0482 meFor Simhh=0.2758 me and mlh=0.21409 meFor GaAsmhh=0.37686 me and mlh=0.08339 me

calculation provides,Si Hole effective mass = 0.19 meGe Hole effective mass= 0.39 meGaAs Hole effective mass= 0.4025 meAssuming ml =mt=Hole effective mass Putting these values

in me,cond =3

1

ml+

2

mt

we have Si Conductivity effective

mass = 0.19 meGe Conductivity effective mass= 0.39 meGaAs Conductivity effective mass= 0.4025 me

VI. DETERMINATION OF BAND GAP

3) For Si: we see its conduction band minima is at X pointand the valance band maxima is at γ point .The differenceis 1.13 eV which is less than Egap = 3.3985(figure: 16).therefore, The band gap is INDIRECT. Silicon’s Band Gap= 1.13218 eV.

4) For GaAs: For GaAs we see its conduction band minimais at point while the valance band maxima is also at Γ .Thedifference is 1.42 eV equal to the Egap=1.42eV. GalliumArsenide’s Band Gap= 1.42416 eV and band-gap is DIRECT.

5) For Ge: In case of Ge we see its conduction band min-ima is at L point while the valance band maxima is at Γ .Thedifference is 0.69 eV less Egap=0.81 eV value. Germanium’sBand Gap= 0.68091 eV and bandgap is INDIRECT

VII. DEPENDENCE OF BANDGAP AND EFFECTIVE MASSON TEMPERATURE

Relation E(g) = Eg(0) − AT 2

B+T from reference :IX is usedand also Table 2.3IX is used for Constants.Only γ points forSi,GaAS and Ge are plotted on figure: 19.

MATLAB CODE

clc;

0 500 1000 1500 2000 2500 3000−1

0

1

2

Temperature in K

Ban

dgap

Ene

rgy

in e

V

Temperature Dependence of Bandgap

SiliconGallium ArsenideGermanium

Fig. 19. Dependence of Bandgap on temperature

close allclear all;addpath('../lib');

% Sia=0.391*10ˆ-3;b=125;Eg0=1.1557;dT=10;T_initial=0;T_final=3000;T=T_initial:dT:T_finall=length(T)Eg=zeros(1,l);EgG=zeros(1,l);Egx=zeros(1,l);for i=1:lEg(i)=Eg0-a*T(i)ˆ2/(b+T(i));endfigure(1)plot(T,Eg)xlabel('Temperature in K')ylabel('Bandgap Energy in eV')title('Temperature Dependence of Bandgap')hold ongrid on%GaAsaG=0.5405*10ˆ-3;bG=204;EgG0=1.519;for i=1:lEgG(i)=EgG0-aG*T(i)ˆ2/(bG+T(i));endplot(T,EgG)%%Geax=0.6842*10ˆ-3;bx=398;Egx0=0.8893;for i=1:lEgx(i)=Egx0-ax*T(i)ˆ2/(bx+T(i));endplot(T,Egx)legend('Silicon','Gallium Arsenide','Germanium')opt.Colors = [

1, 0, 0;0, 1, 0;0, 0, 1;];

opt.LineWidth = [2, 1, 1];opt.LineStyle = {'-', '-', '-'};opt.FileName = 'temp.eps';setPlotProp(opt);

From the reference IX. me(T ) = me(0)(1 + 5αKBT )

KB= Boltzman constant and α is 0.6 /eV for Si.

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0 500 1000 1500 2000 2500 30000.25

0.3

0.35

0.4

0.45

0.5

Temperature in K

Con

duct

ivity

Effe

ctiv

e m

ass

M/M

e

Effective mass(electron) for Silicon vs Temprature

Fig. 20. Dependence of Effective mass on temperature

MATLAB CODE:

clc;close all;clear all;

%for Si

KB= 1.38e-23alpha=0.6/1.6e-19;me0=0.26;T=0:10:3000l=length(T)me=zeros(1,l)for i=1:lme(i)=me0*(1+5*alpha*KB*T(i));endplot(T,me)xlabel('Temperature in K')ylabel('Conductivity Effective mass M/Me')title('Effective mass(electron) for Silicon vs Temprature')grid on

VIII. APPENDIX

IX. REFERENCE:

1.www.nanohub.org2.JoachIM Piprek,Semiconductor Optoelectronic

Devices:Introduction to Physics and Simulation3.ecee.colorado.edu

4.D. Mark Riffe,Temperature dependence of silicon carriereffective masses with application to femtosecond reflectivitymeasurements.