Sadia Lecture 3

8/18/2019 Sadia Lecture 3

1/35

•Miller index is used to describe directions and planes in acrystal.

•Directions - written as [u v w] where u, v, w.• Integers u, v, w represent coordinates of the vector in realspace.

•A family of directions which are equivalent due tosymmetry operations is written as

•Planes: Written as (h k l ).

•Integers h , k , and l represent the intercept of the plane withx -, y -, and z - axes, respectively.

• Equivalent planes represented by {h k l}.

Miller Index For Cubic Structures


2/35

The intercepts of a crystal plane with the axis defined by a set of unit

vectors are at 2a, -3b and 4c. Find the corresponding Miller indices of this

and all other crystal planes parallel to this plane.

The Miller indices are obtained in the following three steps:

1. Identify the intersections with the axis, namely 2, -3 and 4.

2. Calculate the inverse of each of those intercepts, resulting in 1/2, -

1/3 and 1/4.

3. Find the smallest integers proportional to the inverse of theintercepts. Multiplying each fraction with the product of each of the

intercepts (24 = 2 x 3 x 4) does result in integers, but not always

the smallest integers.

4. These are obtained in this case by multiplying each fraction by 12.

5. Resulting Miller indices is6. Negative index indicated by a bar on top.

346


3/35

z

y

x

z=

y=

x=a

x y z

[1] Determine intercept of plane with each axis a ∞ ∞

[2] Invert the intercept values 1/a 1/∞ 1/∞

[3] Convert to the smallest integers 1 0 0

[4] Enclose the number in round brackets (1 0 0)

Miller Indices of Planes


4/35

4


5/35

5


6/35

• Planes (100), (010), (001), (100), (010), (001) areequivalent planes. Denoted by {1 0 0}.

• Atomic density and arrangement as well as electrical,optical, physical properties are also equivalent.

z

yx

(100)

plane

(010)plan

e

(001) plane

Equivalent Planes


7/35

The (110) surface

Assignment

Intercepts : a , a ,

Fractional intercepts : 1 , 1 , Miller Indices : (110)


8/35

z

y

x

x y z

[1] Determine intercept of plane with each axis 2a 2a 2a

[2] Invert the intercept values 1/2a 1/2a 1/2a

[3] Convert to the smallest integers 1 1 1

[4] Enclose the number in round brackets (1 1 1)

Miller Indices of Planes


9/35

The (111) surface

Assignment Intercepts : a , a , a

Fractional intercepts : 1 , 1 , 1

Miller Indices : (111)

The (210) surface

Assignment

Intercepts : ½ a , a , Fractional intercepts : ½ , 1 ,

Miller Indices : (210)


10/35

z

yx

Planes with Negative Indices

x y z

[1] Determine intercept of plane with each axis a -a a

[2] Invert the intercept values 1/a -1/a 1/a

[3] Convert to the smallest integers 1 -1 -1

[4] Enclose the number in round brackets 111


11/35


12/35

12


13/35


14/35

x y z

[1] Draw a vector and take components 0 2a 2a

[2] Reduce to simplest integers 0 1 1

[3] Enclose the number in square brackets [0 1 1]

z

y

x

Miller Indices: Directions


15/35

z

y

x

x y z[1] Draw a vector and take components 0 -a 2a

[2] Reduce to simplest integers 0 -1 2

[3] Enclose the number in square brackets 210

Negative Directions


16/35

Miller Indices: Equivalent Directions

z

yx

12

3

1: [100]

2: [010]3: [001]

Equivalent directions due to crystal symmetry:

Notation used to denote all directions equivalent to [100]


17/35

Directions

18


18/35

18


19/35

In the cubic system the (hk l ) plane and thevector [hk l ] are normal to one another.

This characteristic is unique to the cubic

crystal system and does not apply to crystal

systems of lower symmetry.

20


20/35

20

Importance of Miller Indices

• They are important because properties aredifferent along different directions

• The planes influence• Optical Properties• Reactivity• Surface Tension

• Dislocation

21


21/35

21

Miller Indices of Wafers

22


22/35

22

Angle between (100) and

(011) planes

cos(θ)=(1x0+0x1+0x1)/(√1x√2 ) = 0so θ=90 degrees(011) surface is normal to (100) surface

1 2 1 2 1 2

2 2 2 2 2 2

1 1 1 2 2 2

½ ½

u u v v w wcos

(u v w ) (u v w )

23


23/35

23

24


24/35

24

25


25/35

25

26


26/35

26

Sketch and Find Planar Density or Atomic

Density (atoms/unit area) For Simple Cubic (SC)

unit cell. If lattice constant is a and radius is R.

• (100) plane• (110) plane• (111) plane

27


27/35

27

Simple Cubic(SC) Planer/Areal Density of

(100) plane

28


28/35

28

Simple Cubic(SC) –Planer/Areal density (110)

plane


29/35

STM image of Si (111)

STM image of Gold (111)

Miller Index For Cubic Structures

30


30/35

30

Sketch and Find Planar Density or Atomic

Density (atoms/unit area) For Face Centered

Cubic (FCC) unit cell. If lattice constant is a and

radius is R.


31


31/35

31

32

http://www.google.com.pk/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&docid=jCnxNKfwBLjfVM&tbnid=sVB7IeBcwL_rkM:&ved=0CAUQjRw&url=http%3A%2F%2Fwww.chem.wisc.edu%2Fcourses%2F801%2FSpring00%2Fchemlecnotes%2Fchemln2.html&ei=SkDwU_vqOumd0QXM-oCYCg&bvm=bv.73231344,d.d2k&psig=AFQjCNGk0dVfMPxjmmz2SS4w4ArTLjSJOw&ust=1408340299658215


32/35

32

32

1)(

24

1)(

4

12)(

2111

2110

22100

R FCC PD

R FCC PD

Ra

FCC PD

Planar Density or Atomic Density (atoms/unit

area) For Face Centered Cubic (FCC) unit cell

33


33/35

33

Home Work:

Sketch the plane and Find Planar Density or Atomic Density

(atoms/unit area) for Body centered cubic Unit cell(BCC) and

Diamond Lattice for


34


34/35

34

Why MOS Devices are made on (100)

wafers and Bipolar Devices are made on

(111) wafers?Why we do not have (110) Wafers?

35


35/35

1. (111) ingot is easy to pull therefore initially(111) ingots used to be made in olden

timings.

2. As BJT are old devices therefore they were

made on (111) wafers.

3. But now a days small scale ICs of BJT aremade on (100) wafers.

4. For (100) surfaces the interface charge

densities at Si-SiO2 interfaces is 1010

charges/cm2

for (111) wafer is is ten timeshigher therefore MOS devices and other

surface devices are made on (100) wafers.

Documents

Sadia Lecture 3