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Class notes for the course EE3310 on Solid State Electronic Devices at UTD. The notes cover the introduction to the course, homework sets, and the devices that will be studied. The document also provides a brief overview of semiconductors and their conductivity. The notes are based on the handwritten notes of Larry Overzet and were typed up by Matthew Goeckner. expected to be modified as time goes on.
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Version: Fall 2002 These class notes were originally based on the handwritten notes of Larry Overzet. It is expected that they will be modified (improved?) as time goes on. This version was typed up by Matthew Goeckner.
Homework Set 1 Streetman Chap 1 # 1,3,4,12, Chap. 2 # 2,5 Assigned 8/22/02 Due 8/29/
Q: Why study electronic devices? A: They are the backbone of modern technology
Q: Why study the physical operation? A: This is an engineering class. You are studying so that you know how to make better devices and tools. If you do not understand how a tool works, you cannot make a better tool. (Technicians and electricians can make a tool work but they cannot significantly improve it. They, however, are not trained to understand the basic operation of the tool.)
Q: What devices will we study? A:
By course end, the students should know:
Let us start the class by describing just what is a ‘semiconductor’.
The conductivity of semiconductors occupy the area between conductors and insulators. This implies that the conductivity can range over many orders of magnitude. Further, the conductivity of semiconductors can be adjusted through a number of means, each related to the physical properties of semiconductors. Typical methods for adjusting the conductivity of a semiconductor are: a. Temperature b. Purity (Doping) c. Optical excitation d. Electrical excitation.
Materials come from Ia IIa IIIb IVb Vb VIb VIIb VIII VIII VIII Ib IIb IIIa IVa Va VIa VIIa VIIIa Hydroge1_ H_1.
Helium2_ He_4. Lithium3_ Li_6.939_
Berylliu4_ Be_9.0122_
Boron_5_ B_
Carbon6_ C_
Nitroge7_ N_
Oxyge8_ O_15.
Fluorin9_ F_18.
Neon_10_ Ne_20. Sodium11_ Na_22.
Magnesiu12_ Mg_24.312_
Aluminu13_ Al_
Silicon14_ Si_ 28.086_
Phospho15_ P_ 30.9738_
Sulfur16_ S_32.
Chlorin17_ Cl_35.453_
Argon18_ Ar_39. Potassiu19_ K_39.
Calcium20_ Ca_40.078_
Scandiu21_ Sc_44.
Titanium22_ Ti_47.867_
Vanadium23_ V_ 50.9415_
Chromiu24_ Cr_ 51.9961_
Mangane25_ Mn_
Iron_26_ Fe_ 55.847 _
Cobalt_27_ Co_ 58.933 _
Nickel_28_ Ni_ 58.71_
Copper29_ Cu_ 63.54_
Zinc_30_ Zn_ 65.37_
Gallium31_ Ga_ 69.72_
Germani32_ Ge_ 72.59_
Arsenic33_ As_ 74.922_
Seleniu34_ Se_78.96_
Bromin35_ Br_79.909_
Krypon36_ Kr_83.80_ Rubidiu37_ Rb_85.
Strontiu38_ Sr_87.62_
Yitrium39_ Y_88.
Zirconiu40_ Zr_91.22_
Niobium_41_ Nb_ 92.906 _
Molybden42_ Mo_ 95.94_
Techneti43_ Tc_ [98]_
Rutheniu44_ Ru_ 101.07 _
Rhodiu45_ Rh_
Palladiu46_ Pd_ 106.4 _
Silver_47_ Ag_
Cadmiu48_ Cd_ 112.40 _
Indium49_ In_
Tin_50_ Sn_ 118.69_
Antimo51_ Sb_ 121.75_
Telluriu52_ Te_127.60_
Iodine_53_ I_126.
Xenon54_ Xe_131. Caesium55_ Cs_132.
Barium56_ Ba_137.
Lanthiu57_ La_ 138.91 _
Hafnium72_ Hf_ 178.49 _
Tantalum73_ Ta_ 180.948_
Tungsten_74_ W_ 183.85_
Rhenium75_ Re_ 186.2 _
Osmium76_ Os_ 190.2 _
Iridium77_ Ir_ 192.2 _
Platinum78_ Pt_ 195.09_
Gold_79_ Au_
Mercur80_ Hg_ 200.59_
Thalliu81_ Tl_ 204.37 _
Lead_82_ Pb_ 207.19_
Bismut83_ Bi_
Poloniu84_ Po_[210]_
Astatin85_ At_[210]_
Radon86_ Rn_[222]_ Franciu87_ Fr_[223.
Radium88_ Ra_[226.]_
Actiniu89_ Ac_ [227]_
Rutherford104_ Rf_ [] _
Dubniium105_ Db_ []_
Seaborgi106_ Sg_ [] _
Bohrium107_ Bh_ [] _
Hassium108_ Hs_ [] _
Meithner109_ Mt_ []_
Ununnilli110_ Uun_ [] _
Unununi111_ Uuu [] _
Ununbiu112_ Uub []_
Ununquad114_ Uuq_ []_
Ununhex115_ Uuh_[]_ Lanthanides Cerium58_ Ce_140.
Preseedymiu59_ Pr_ 140.907_
Neodymi60_ Nd_ 144.24_
Promethi61_ Pm_ [147]_
Samariu62_ Sm_ 150.35_
Europiu63_ Eu_ 151.96 _
Gadolini64_ Gd_ 157.25 _
Terbium65_ Tb_
Dysprosi66_ Dy_ 162.50 _
Holmiu67_ Ho_
Erbium68_ Er_ 167.26_
Thulium69_ Tm168.
Ytterbiu70_ Yb_173.04_
Lutetiu71_ Lu_174.97_ Actinides Thorium90_ Th_
Profactinium91_ Pa_ [231]_
Uranium92_ U_ 238.03 _
Neptuniu93_ Np_ [237]_
Plutoniu94_ Pu_ [242]_
Americiu95_ Am_ [243]_
Curium96_ Cm [247]_
Berkeliu97_ Bk_ [247]_
Californi98_ Cf_ [249]_
Einsteini99_ Es_ [254]_
Fermiu100_ Fm_ [253]_
Mendelev101_ Md_[256]_
Nobeliu102_ No_[254]_
Lawrenci103_ Lr_[257]_
To understand the distinction between these solid material types, we must first understand the concept of order. Order can be described as the repetition of identical structures or identical placement of atoms. An example of this would be an atom that has six nearby atoms, each 5 Å away, arranged in a pattern as such.
STRUCTURES. These structures are:
Here a , b , and c are the BASIS VECTORS along the edges of the standard SC cell.
Here the ‘new’ atom is at a /2 + b /2 + c /
Here the ‘new’ atoms are at ( a /2 + b /2), ( b /2 + c /2), ( a /2 + c /2), ( a + b /2 + c /2), ( a /2 + b + c /2), ( a /2 + b /2 + c ).
A Diamond lattice starts with a FCC and then adds four additional INTERAL atoms at locations a /4 + b /4 + c /4, away from each of the atoms.
Now that we have described a few of the simple crystal types, we need to figure out how to describe a location in the crystal. We could use our basis vectors, a, b and c, but it has been found that this is not the most advantageous description. For that we turn to MILLER INDICES. Miller Indices define both planes in the crystal and the direction normal to said plane. As we know, all planes are defined by three points. Thus, one can pick three Lattice points in the crystal and hence define a plane. From these three
neighbor. This is known as the “HARD PACK” approximation. Now each of the sides of the SC have a length of a. (‘a’ is not to be confused with the vector ‘ a ’.) Thus the volume of the cube is a^3. Now we need to determine how much of each atom is inside the cubic volume. For this we need to look at our picture of the SC lattice.
Let us look in more detail at the atom at the origin.
We see that 1/8 of each atom is inside the cube. Thus the total volume of atoms in the cube is
81/8volume of an atom = (^433 )
3 1 6
3 2
π = π
r (^) = π a a. This means that the fraction of the volume filled by
the atoms is 16 π ≈ 0 52. = 52 %.
Read Sections 9.1 and 9.
The late 1800s and the early 1900s set the stage for modern electronic devices. A number of experiments showed that classical mechanics was not a good model for processes on the very small scale. Among these experiments were the following:
A rapid series of new models were developed which began to explain these observed phenomena.
hν
E=eVapplied
Φ - work function
Einstein explained this by hypothesizing that light is made up of localized bundles of electromagnetic energy called photons. Each of these photons had the same amount of energy, namely hν, where ν is the
frequency of the light and h is a constant, the slope of the line, known as Planck’s constant. Sommerfield later proposed a model of a conductor that looks like
Φ - work function Free electrons (Fermi Sea)
hν E=eV (^) applied
Thus, one finds that the electrons in the metal are ‘stuck’ in a potential energy well. The photons then supply all of their energy to a single electron. The electron uses the first part of the energy to overcome the potential energy well, and the rest remains as kinetic energy.
Bohr model of the Hydrogen atom Bohr’s model of the Hydrogen atom was perhaps the first ‘true’ quantum model. It does a wonderful job of predicting the then measured frequency of light emitted from an atom. (It misses some ‘splitting’ of the lines that later improvements to the experiments found and later improved versions of the model deal with correctly.)
The basis of the model is that the path integrated angular momentum of the electron, while in orbit around an atom, is in discrete states that vary as integer multiples of h. Namely, p mvr nh n
v n mr
θ = = π = ⇓
=
h
h
We now have two other equations to work with The energy of the electron E K E P E
mv
e Kr
1 2
2 2
The centripetal force on the electron
F mv r
e Kr
r e Kmv
r
K n me
centripetal
n
2 2 2
2 2
2 2 2
h
From this we note that r is a function of n. For n = 1, ‘ground’ state, we find
r a
me
2 = = 2 =0 529 h .
where a 0 is the Bohr radius and is the smallest radius at which the electron orbits the proton in the Hydrogen atom. Finally plugging both velocity and radius into our energy equation we find the energy of the electron,
0
100
200
300
400
500
600
700
800
900
0 5 10 15 20 25 30 35 40 45 n
radius (Å)
0
Energy (eV)
Radius Energy
If we look at the potential well the electron is trapped in, we see that the higher the energy, the higher the expected radius.
In a true Hydrogen atom, the electron is trapped between the repulsive ‘strong force’ and the attractive electromagnetic force. The potential well that is created between these forces looks like
Proton (not to scale!)
‘Strong’
‘electromagnetic’
total
The one major item that Bohr’s model missed is a splitting of the levels, or ‘shells’. This splitting is due to a splitting in the allowed angular momentum and particle spin (internal angular momentum) in each shell. Thus we find each shell given by a label n has an allowed set of angular momentums, given by a labels l, and labels m, as well as spin given by label s. The overall requirements are n≥ 1 L≤n- -L≤m≤L s=±1/ The label ‘l’ is often replaced with l=0 => ‘s’, l=2 => ‘p’, l=3 => ‘d’, l=4 => ‘f’, (and then follow the alphabet). Thus an electron in shell n=3, l=3 can be labeled 3d. The higher the quantum numbers n and L, the higher the energy. This means that our picture of the potential well now looks like
Proton (not to scale!)
1s
2s
2p
3d 3p 3s
We can have up to 2(2L+1) electrons in that state because of the possible m’s and ‘s’s. We often add a superscript to our label to tell us how many electrons are in a given state thus 3d => 3d^5 or 3d 2 etc.
Usually the lowest energy states are the first to fill -
This is in fact why the periodic table is the shape that it is. The Noble gases are on the right hand side and have completely filled – or closed – outer shells. The element on the farthest left will have a [noble]ns^1 configuration, i.e. [He]2s 1 is Lithium while [Ne]3s^1 is Sodium (Na).
At the close of the 1920, two versions of full fledge Quantum Mechanics were proposed, a wave version of QM by Schrödinger and a particle version, employing matrices, by Heisenberg. These are equivalent yet different and can be used to independently solve problems. For what little QM we do do, we will predominately use Schrödinger’s version is this class.
K E P E E
m
V t j t t
h 2 2 h 2
Ψ r Ψ r
h^22 2
m x V x^ E^ ψ^ r
Outside the well, the wave function must be zero, as the potential is infinite. (Or else the second derivative is infinite which is unphysical.) Thus we find h^22 2
m
E x x L
x elsewhere
∂ + x
ψ
ψ
We start by looking at the 0 to L part and integrate twice to find h
h
2 2
0
2
2 m
x E x
x e
x
i mE x
ψ ψ
ψ ψ /
or
Now our wave function must be continuous in both zeroth and first order derivatives, so that at x = 0 we
ψ
ψ
L B mE L
mE L n n
n mL
x B n x L
n
= π =
⇓
=
π
π
0
2 2 2 2
0
sin /
sin
h
h
h
Finally, we typically normalize the wavefunction to 1, so that our total probability is ‘1’. This is done by integrating
π
π
π
π
−∞
∞
−∞
∞
∫
∫
∫
∫
ψ * ψ
sin
cos
x x dx
n x L
n x L
dx
n x L
dx
n x L
dx
L
L
0 0
0
2 2 0
0
(^2 ) 2 0
0
2
Thus,
ψ x L
n x L
π
sin.
How is this related to our Hydrogen Atom? First the higher the value of n the higher we move up the sides of the potential well. Now, if we look at both positive and negative direction of our potential well around the core of the Hydrogen atom, we see a shape that looks like
Energy states
core
Where we have ignored the core area where the electron is not allowed. This, we can model Hydrogen in a way that is very similar to the above. Further, we would expect to see that higher energy states correspond to being higher up the potential well. Because of the shape of the well, we expect the more energetic electrons to orbit at a distance further from the core. This is indeed what we see.
Day 3 Homework set 2 Chapter 3 # 4,7,8,9 Due Sept 10 th^ , 2002
Recap
We bring this idea up because we are dealing with solid-state devices. Thus the interaction of multiple atoms and atomic species is important to our understanding of this topic. How these atoms bond together is critical to the characteristics of the devices.
We will now examine bonds between atoms. They fall into four main categories.
Ionic: The first of these types of material is related to the complete transfer of an electron from one atom to another. Cl for example would like to have a closed top shell and thus it takes an electron from the Na to produce a [Ar] electron cloud. Sodium on the other hand would like to give up an electron, so to also have a closed shell, in this case [Ne]. [THESE OUTER SHELL ELECTRONS ARE KNOW AS VALANCE ELECTRONS.] Both of these acceptor/donor processes provide lower energy states. This means that the two particles Na+^ and Cl-^ are electrostatically pulled together or bonded. The electrons in question, are not shared by the atoms. Picture wise, this looks like
e- Na+ e-
e-
e-
e- e-
Na = [Ne] 3s^1 => Na +^ = [Ne] Cl = [Ne] 3s^2 3p^5 => Cl-^ = [Ne] 3s 2 3p^6 = [Ar]
Na+
Na+
Na+
Na+
Na+
Na+
Na+^ Na+ Na+
Metallic: The second of these comes in two forms. The first form has only a few valance electrons in the outer orbital. These outer valance electrons thus tend to be weakly bound to the atoms and are ‘free’ to move around. An example of this type would be Sodium, Na = [Ne]3s^1.
Background electron cloud
Na+^ Na+ Na+^ Na+ Na+^ Na+
Na+^ Na+ Na+^ Na+ Na+^ Na+
Na+^ Na+ Na+^ Na+ Na+^ Na+
Na+^ Na+ Na+^ Na+ Na+^ Na+
Na+^ Na+ Na+^ Na+ Na+^ Na+
Na+^ Na+ Na+^ Na+ Na+^ Na+
Covalent: In the covalent bond, two atoms share one or more valance electrons. In this way, each atom thinks that it has a closed outer shell. Because the outer shell is closed, these materials are typically insulators – although some might also be semiconductors. (This in part depends on the size of the atoms. The smaller it is, the more likely it is to be an insulator.) An example of this is Carbon, C=[He]2s^2 2p^2.
C
e-
e-
e- e-