Transcript CHAPTER 1 Atoms and bonding

‫מבוא למצב מוצק‬
‫ולמוליכים למחצה‬
‫מעבדה ‪ 4‬מח'‬
‫מטרה‪:‬‬
‫‪ ‬להסביר בקצרה את עיקרי החומר התאורטי‬
‫בתחום הפיסיקה של המצב המוצק וכן הפיסיקה‬
‫של מוליכים למחצה‪ ,‬הנדרשים להבנת הניסויים‪:‬‬
‫•‬
‫•‬
‫•‬
‫•‬
‫הולכה חשמלית במוצקים‪.‬‬
‫אפקט הול‪.‬‬
‫תא פוטוולטאי‪.‬‬
‫פוטולומינסנציה‪.‬‬
References:
• Paul A. Tipler, Modern Physics, Chapter 9
• Charles Kittel, Introduction to Solid State
Physics, Chapters 7,8 (Sixth edition)
• ‫ מוליכים למחצה‬- ‫ גדי גולן‬,‫לב‬-‫אדיר בר‬
• Prof. Dr. Beşire GÖNÜL presentation.
• E.E. Technion s.c. course.
Outlines:
• Atoms and bonding
• Energy bands and effective mass
• Conduction in solids
• Temperature dependant of solid conductance
• p – n junction
Atoms and bonding
• The periodic table
• Ionic bonding
• Covalent bonding
• Metallic bonding
• van der Waals bonding
Atoms and bonding
• In order to understand the physics of semiconductor (s/c) devices,
we should first learn how atoms bond together to form the solids.
• Atom is composed of a nucleus which contains protons and
neutrons; surrounding the nucleus are the electrons.
• Atoms can combine with themselves or other atoms. The valence
electrons, i.e. the outermost shell electrons govern the chemistry of
atoms.
• Atoms come together and form gases, liquids or solids depending
on the strength of the attractive forces between them.
• The atomic bonding can be classified as ionic, covalent, metallic, van
der Waals,etc.
• In all types of bonding the electrostatic force acts between charged
particles.
The periodic table
3
4
5
6
7
Valance electrons
4 out of 8 Electrons
In outer shell
First
Shell
He
H
Second
Shell
Li
Be
Na
Mg
B
C
Third
Shell
Al
Si
N
O
F
Ne
The periodic table
•
Ionic solids
Alkali metals contains lithium (Li), sodium (Na), potassium (K),... and
these combine easily with Halogens like fluorine (F), chlorine (Cl),
bromine (Br),.. . and produce ionic solids of NaCl, KCl, KBr, etc.
•
Rare (noble) gases
Elements of noble gases of helium(He), neon (Ne), argon (Ar),… have
a full complement of valence electrons and so do not combine
easily with other elements.
•
Elemental semiconductors
Silicon(Si) and germanium (Ge) have 4 valance electrons.
•
Compound semiconductors
1) III-V compound s/c’s; GaP, InAs, AlGaAs etc.
2) II-VI compound s/c’s; ZnS, CdS etc.
Ionic bonding
The metallic elements have only up to the valence electrons in
their outer shell will lose their electrons and become positive
ions, whereas electronegative elements tend to acquire
additional electrons to complete their octed and become
negative ions, or anions.
Na
Cl
Ionic bonding
•
•
•
•
•
Ionic bonding is due to the electrostatic force of attraction between
positively and negatively charged ions.
This process leads to electron transfer and formation of charged ions;
a positively charged ion for the atom that has lost the electron and a
negatively charged ion for the atom that has gained an electron.
All ionic compounds are crystalline solids at room temperature.
Ionic crystals are hard, high melting point, brittle and can be dissolved
in ordinary liquids.
NaCl and CsCl are typical examples of ionic bonding.
Covalent bonding
• The bonding is due to the sharing of electrons.
• Covalently bonded solids are hard, high melting
points, and insoluble in all
ordinary solids.
• Elemental semiconductors of Si, Ge and diamond
are bonded by this mechanism and these are purely
covalent.
• Compound s/c’s exhibit a mixture of both ionic and
covalent bonding.
Comparison of Ionic and Covalent Bonding
Metallic bonding
• Valance electrons are relatively bound to the nucleus and
therefore they move freely through the metal and they are
spread out among the atoms in the form of a low-density
electron cloud.
• A metallic bond result from the
sharing of a variable number of
electrons by a variable number of

atoms. A metal may be described
as a cloud of free electrons.
• Therefore, metals have high
electrical and thermal conductivity.
+
+
+
+
+
+
+
+
+
Metallic bonding
• All valence electrons in a metal combine to form a “sea” of
electrons that move freely between the atom cores. The more
electrons, the stronger the attraction. This means the melting
and boiling points are higher, and the metal is stronger and
harder.
• The positively charged cores are held together by these
negatively charged electrons.
• The free electrons act as the bond (or as a “glue”) between the
positively charged ions.
• This type of bonding is nondirectional and is rather insensitive to
structure.
• As a result we have a high ductility of metals - the “bonds” do
not “break” when atoms are rearranged – metals can experience
a significant degree of plastic deformation.
Van der Waals bonding
• It is the weakest bonding mechanism.
• It occurs between neutral atoms and molecules.
• The explanation of these weak forces of attraction is that
there are natural fluctuation in the electron density of all
molecules and these cause small temporary dipoles within
the molecules. It is these temporary dipoles that attract one
molecule to another. They are as called van der Waals' forces.
• Such a weak bonding results low melting and boiling points
and little mechanical strength.
Van der Waals bonding
The dipoles can be formed
as a result of unbalanced
distribution of electrons in asymettrical molecules. This is
caused by the instantaneous location of a few more electrons
on one side of the nucleus than on the other.
symmetric
asymmetric
Therefore atoms or molecules containing dipoles are attracted
to each other by electrostatic forces.
Classification of solids
SOLID MATERIALS
CRYSTALLINE
Single Crystal
POLYCRYSTALLINE
AMORPHOUS
(Non-crystalline)
Crystalline Solid
• Crystalline Solid is the solid form of a
substance in which the atoms or
molecules are arranged in a definite,
repeating pattern in three dimension.
Crystalline Solid
•
Single crystal has an atomic structure that repeats
periodically across its whole volume. Even at infinite length
scales, each atom is related to every other equivalent
atom in the structure by translational symmetry
Single Pyrite
Crystal
Amorphous
Solid
Single Crystal
Polycrystalline Solid
•
•
Polycrystal is a material made up of an aggregate of many small single
crystals (also called crystallites or grains).
The grains are usually 100 nm - 100 microns in diameter. Polycrystals with
grains that are <10 nm in diameter are called nanocrystalline.
Polycrystalline
Pyrite form
(Grain)
Polycrystal
Amorphous Solid
•
Amorphous (non-crystalline) Solid
is composed of
randomly orientated atoms, ions, or molecules that do
not form defined patterns or lattice structures.
‫מבני גבישים‬
Simple Cubic
Cubic Close Packed
Body Centered Cubic
Hexagonal Close Packed
http://cst-www.nrl.navy.mil/lattice
‫מודל הפסים במוצקים‬
‫• מודל הפסים‪.‬‬
‫• פס ערכיות‪ ,‬פס הולכה ואנרגיית הפער‪.‬‬
‫• איכלוס הפסים במבודד‪ ,‬מוליך ומוליך למחצה‪.‬‬
‫• זיהומים נוטלים וזיהומים תורמים‪.‬‬
‫• רמת פרמי ‪.EF‬‬
‫• מסה אפקטיבית‪.‬‬
‫מבנה רמות אנרגיה במוצקים מודל הפסים‬
‫• מודל הפסים מתאר את מבנה רמות האנרגיה‬
‫המותרות לאלקטרונים‪ ,‬הנוצרות במוצק‪.‬‬
‫• מהו התנאי לקיום פסים אחידים?‬
‫• האטומים בגביש חייבים להיות מסודרים במבנה‬
‫מרחבי מחזורי (האטומים יושבים על סריג מרחבי)‪.‬‬
‫• בשקפים הבאים נראה מדוע נוצרים פסי אנרגיה‬
‫בגביש מסודר‪.‬‬
‫נתחיל עם רמות האנרגיה של אטום בודד‪:‬‬
‫כפי שלמדנו לאטום רמות אנרגיה בדידות (דיסקרטיות)‪ ,‬וחלקן מאוכלסות‬
‫באלקטרונים (האטום מיוצג ע"י בור פוטנציאל)‪.‬‬
‫אין שינוי ברמות האנרגיה כאשר יש שני אטומים רחוקים אחד מהשני‪.‬‬
‫‪E‬‬
‫‪E‬‬
‫‪0‬‬
‫‪0‬‬
‫‪Eex2‬‬
‫‪Eex1‬‬
‫‪EV‬‬
‫‪Eex2‬‬
‫‪Eex1‬‬
‫‪EV‬‬
‫‪E2‬‬
‫‪E2‬‬
‫‪E1‬‬
‫‪E1‬‬
‫כעת האטומים מתקרבים‪:‬‬
‫רמות האנרגיה משתנות במקצת בגלל האינטראקציה בין האטומים‪.‬‬
‫סביב כל רמת אנרגיה של האטום הבודד יש כעת שתי רמות אנרגיה‪.‬‬
‫מספר המקומות הניתנים לאכלוס ע"י אלקטרונים נותר ללא שינוי‪.‬‬
‫‪E‬‬
‫‪E‬‬
‫‪0‬‬
‫‪0‬‬
‫‪Eex2‬‬
‫‪Eex1‬‬
‫‪EV‬‬
‫‪Eex2‬‬
‫‪Eex1‬‬
‫‪EV‬‬
‫‪E2‬‬
‫‪E2‬‬
‫‪E1‬‬
‫‪E1‬‬
‫מה קורה לרמות האנרגיה בגביש מוצק ?‬
‫בכל סמ"ק של חומר מוצק יש כ ‪ 1022‬אטומים‪ .‬לכן בגביש מסודר כל רמה‬
‫מתפצלת ל ‪ 1022‬רמות צפופות מאוד‪ .‬כך שכל רמת אנרגיה בדידה באטום‬
‫המקורי הופכת להיות‪ ,‬באופן אפקטיבי‪ ,‬פס אנרגיה‪ .‬בין כל שני פסי אנרגיה‬
‫יש תחום אנרגיות שבו אסור לאלקטרונים להימצא – אנרגית הפער‪.‬‬
‫פס הערכיות‪ ,‬פס ההולכה‬
‫ופער האנרגיה בינם (‪)Eg‬‬
‫חשובים להולכה החשמלית‪.‬‬
‫גביש‬
‫‪0‬‬
‫פס הולכה‬
‫‪Conductance band‬‬
‫אנרגיית הפער‬
‫)‪Eg (gap‬‬
‫‪E‬‬
‫אטום בודד‬
‫פס הולכה‬
‫פס ערכיות‬
‫‪Eex2‬‬
‫‪Eex1‬‬
‫‪EV‬‬
‫‪E‬‬
‫‪0‬‬
‫‪X 1022‬‬
‫‪Eex2‬‬
‫‪Eex1‬‬
‫‪EV‬‬
‫פס ערכיות‬
‫‪Valance band‬‬
‫‪E2‬‬
‫‪E2‬‬
‫‪E1‬‬
‫‪E1‬‬
:‫חישוב לדוגמא‬
• Consider 1 cm3 of Silicon. How many atoms does this contain ?
•
Solution:
The atomic mass of silicon is 28.1 g which contains Avagadro’s number of
atoms.
Avagadro’s number N is 6.02 x 1023 atoms/mol .
The density of silicon: 2.3 x 103 kg/m3
so 1 cm3 of silicon weighs 2.3 gram and so contains
6.02  1023
 2.3  4.93  1022 atoms
28.1
This means that in a piece of silicon just one cubic centimeter in
volume , each electron energy-level has split up into 4.93 x 1022
smaller levels !
‫מבודדים‬
• The magnitude of the band gap
determines
the
differences
between insulators, s/c‘s and
metals.
• The excitation mechanism of
thermal is not a useful way to
promote an electron to CB even
the melting temperature is
reached in an insulator.
• Even very high electric fields is
also unable to promote electrons
across the band gap in an
insulator.
E
CB (completely empty)
Eg~several electron volts
VB (completely full)
Wide band gaps between VB and CB
)‫מוליכים (מתכות‬
E
‫• פס ההולכה צמוד או‬
.‫חופף לפס הערכיות‬
E
CB
VB
Overlapping VB and CB
CB
‫• הפסים לא מלאים ולכן‬
‫מתאפשרת תנועת‬
.‫המטענים‬
VB
Touching VB and CB
• No gap between valance band and conduction band
‫מבודד לעומת מוליך‬
‫פס‬
‫הולכה‬
‫פס הולכה‬
‫פס‬
‫הולכה‬
‫פס‬
‫ערכיות‬
‫פס ערכיות‬
‫פס‬
‫ערכיות‬
‫מבודד‬
‫מוליך‬
‫(לדוגמא ‪)αSn cubic‬‬
‫מוליך‬
‫(לדוגמא ‪)HgTe‬‬
‫מוליכים למחצה‬
•
•
•
At 0K valance band full and
conduction
band
emptylike
Insulator.
When enough energy is supplied to
the e- sitting at the top of the valance
band, e- can make a transition to the
bottom of the conduction band.
When electron makes such a
transition it leaves behind a missing
Empty
conduction
band
electron state.
•
•
•
This missing electron state is called
as a hole.
Hole behaves as a positive charge
carrier.
Magnitude of its charge is the same
with that of the electron but with an
opposite sign.
Forbidden
energy gap [Eg]
energy
e+- e+- e+- e+Full
valance
band
? ‫ כיצד ניתן לעורר את האלקטרונים‬- ‫מוליכים למחצה‬
Answer :
• Thermal energy
• Electrical field
• Electromagnetic radiation
Partly filled
CB
Eg
Partly filled
VB
Energy band diagram of a
s/c at a finite temperature.
To have a partly field band configuration in a s/c ,
one must use one of these excitation mechanisms.
1-Thermal Energy
Thermal energy = k x T = 1.38 x 10-23 J/K x 300 K =25 meV
Excitation rate = constant x exp(-Eg / kT)
Although the thermal energy at room temperature, RT, is very small, i.e. 25
meV, a few electrons can be promoted to the CB.
Electrons can be promoted to the CB by means of thermal energy.
This is due to the exponential increase of excitation rate with increasing
temperature.
Excitation rate is a strong function of temperature.
2- Electric field
• For low fields, this mechanism doesn’t promote electrons to
the CB in common s/c’s such as Si and GaAs.
• An electric field of 1018 V/m can provide an energy of the
order of 1 eV. This field is enormous.
So , the use of the electric field as an excitation
mechanism is not useful way to promote electrons in
s/c’s.
3- Electromagnetic Radiation
c
1.24
34
8
E  h  h  (6.62 x10 J  s) x(3x10 m / s) /  (m)  E (eV ) 

 (in  m)
h = 6.62 x 10-34 J-s
c = 3 x 108 m/s
1 eV=1.6x10-19 J
for Silicon
Eg  1.1eV
Near
infrared
1.24
 (  m) 
 1.1 m
1.1
To promote electrons from VB to CB Silicon , the wavelength
of the photons must 1.1 μm or less
3- Electromagnetic Radiation
Conduction Band
• The converse transition can also
happen.
e-
• An electron in CB recombines with a
hole in VB and generate a photon.
photon
+
Valance Band
• The energy of the photon will be in
the order of Eg.
• If this happens in a direct band-gap
s/c, it forms the basis of LED’s and
LASERS.
Intrinsic semiconductor
• The conductivity of a pure (intrinsic) s/c is low due to the
low number of free carriers.
• The number of carriers are generated by thermally or
electromagnetic radiation for a pure s/c.
• For an intrinsic semiconductor
n = p = ni
n = concentration of electrons per unit volume
p = concentration of holes per unit volume
ni = the intrinsic carrier concentration of the semiconductor under
consideration.
Intrinsic semiconductor
The intrinsic carrier concentration ni depends on;
• the semiconductor material, and
• the temperature.


For silicon at 300 K, ni has a value of 1.4 x 1010 cm-3.
Clearly , equation (n = p = ni) can be written as
n.p = ni2
Donors and Acceptors
What is doping and dopants impurities ?
• To increase the conductivity, one can
dope pure s/c with atoms from column lll
or V of periodic table. This process is
called as doping and the added atoms
are called as dopants impurities.
 n-type
 p-type
Addition of different atoms modify the conductivity of
the intrinsic semiconductor.
p-type doped semiconductor
Si + Column lll impurity atoms
Electron
Have
four
valance
e-’s


Boron (B)
has three valance
e-’ s
Hole
Boron bonding in Silicon
Boron sits on a lattice side
Si
Si
Bond
with
missing
electron
B
Si
Si

p >> n
Normal
bond with
two
electrons
p-type doped semiconductor
• Boron(column III) atoms have three valance
electrons, there is a deficiency of electron or
missing electron to complete the outer shell.
• This means that each added or doped boron
atom introduces a single hole in the crystal.
There are two ways of producing hole
1) Promote e-’s from VB to CB,
2) Add column lll impurities to the s/c.

p-type Energy Diagram
CB
Ec = CB edge energy level
acceptor
(Column lll) atoms
Eg
EA= Acceptor energ level
Ev = VB edge energy level
VB
Electron
Hole
p-type doped semiconductor

1.
2.



The impurity atoms from column lll occupy at an
energy level within Eg . These levels can be
Shallow levels which is close to the band edge,
Deep levels which lies almost at the mid of the band
gap.
If the EA level is shallow i.e. close to the VB edge,
each added boron atom accepts an e- from VB
and have a full configuration of e-’s at the outer
shell.
These atoms are called as acceptor atoms since
they accept an e- from VB to complete its bonding.
So each acceptor atom gives rise a hole in VB.
The current is mostly due to holes since the number
of holes are made greater than e-’s.
n-type doped semiconductor
Si + Column V impurity atoms
Have
four
valance
e-’s
Arsenic (As)
has five valance e-’ s
Electron
Si
Weakly
bound
electron
Conduction band
Ec
Si
Ed
Eg
As
Si
Band gap is 1.1 eV for silicon
Valance band
Ev
Electron
Si

n >> p
Normal
bond with
two
electrons
n-type, p- type :‫מוליך למחצה‬
Ge
Ge
n-type
Ge
E
‫פס הולכה‬
+
Ge
Ge
−
As
Ge
Excess
+charge
Eg
‫פער האנרגיה‬
Ge
Excess electron from arsenic atom
n-type germanium.
Ge
Ed≈0.012eV
Ge
Ge
p-type
0
‫פס ערכיות‬
E
−
Ge
Ge
Ge
+
B
Ge
Excess
−charge
‫פס הולכה‬
Eg
Ge
Positive hole, as one electron was
removed from a bond to complete the
tetrahedral bonds of the boron atom.
0
Ea≈0.01eV
‫פער האנרגיה‬
‫פס ערכיות‬
‫מוליך למחצה‪ :‬אינטרינסי ואקסטרינסי‬
‫פס‬
‫הולכה‬
‫פס‬
‫הולכה‬
‫פס‬
‫הולכה‬
‫‪Ed‬‬
‫‪Ea‬‬
‫פס‬
‫ערכיות‬
‫פס‬
‫ערכיות‬
‫פס‬
‫ערכיות‬
‫מוליך למחצה‬
‫אינטרינסי‬
‫מוליך למחצה‬
‫אקסטרינסי‬
‫מוליך למחצה‬
‫אקסטרינסי‬
‫‪n- type‬‬
‫‪p- type‬‬
Fermi level , EF
• This is a reference energy level at which the probability of
occupation by an electron is ½.
• Since Ef is a reference level therefore it can appear anywhere in
the energy level diagram of a S/C .
• Fermi energy level is not fixed.
• Occupation probability of an electron and hole can be
determined by Fermi-Dirac distribution function, FFD ;
FFD
EF = Fermi energy level
kB = Boltzman constant
T = Temperature
1

E  EF
1  exp(
)
k BT
Fermi level , EF
FFD
1

E  EF
1  exp(
)
k BT
• E is the energy level under investigation.
• FFD determines the probability of the energy level E being
occupied by electron.
if E  EF 
 f FD 
•
1
1

1  exp0 2
1  f FD determines the probability of not finding an
electron at an energy level E; the probability of finding a
hole .
Carrier concentration equations
The number density, i.e., the number of electrons
available for conduction in CB is
3/ 2
 2 m kT 
EC  EF
n  2
exp

(
)

h
kT


E  EF
E  Ei
n  N C exp  ( C
)
n  ni exp( F
)
kT
kT
*
n
2
The number density, i.e., the number of holes available
for conduction in VB is
3/ 2
 2 m*p kT 
EF  EV
p  2
exp

(
)

 h2 
kT


E  EV
E  EF
p  NV exp ( F
)
p  ni exp( i
)
kT
kT
‫(‪)11‬‬
‫ריכוז נושאי מטען – מל"מ אינטרינסי‬
‫) ‪3/4 (  Eg 2kT‬‬
‫‪ni  pi = 2( 2kT h ) ( me mh ) e‬‬
‫*‬
‫*‬
‫‪2 32‬‬
‫‪ -M*e‬מסה אפקטיבית אלקטרונים‪.‬‬
‫‪ -M*h‬מסה אפקטיבית חורים‪.‬‬
‫‪ - Eg‬היא אנרגית הפער בין פס הערכיות לפס ההולכה‪.‬‬
‫‪2(2πmkT/h2)3/2 ≈ 1019 per cm3‬‬
‫‪T = 300K‬‬
‫• לדוגמא ריכוזי האלקטרונים בפס הערכיות של מל"מ אינטרינסיים ב ‪: T=300K‬‬
‫‪ni )Si( ≈ 1.2 x 1010 cm-3‬‬
‫‪ni (Ge( ≈ 2.4 x 1013 cm-3‬‬
‫‪ni (GaAs( ≈ 2.2 x 106 cm-3‬‬
‫• ריכוזים אלו קטנים באופן משמעותי מצפיפות האטומים בגביש שהיא כ ‪.5 x 1022 cm-3‬‬
‫ריכוז נושאי המטען כתלות בטמפרטורה‬
‫במל"מ אקסטרינסי מקובל לחלק לארבעה תחומים‪:‬‬
‫• תחום הקיפאון (‪)Freeze-out‬‬
‫• תחום היינון החלקי של אטומי הזיהום (‪)Partial ionization‬‬
‫• התחום האקסטרינסי (‪)Saturation‬‬
‫• התחום האינטרינסי‬
‫במל"מ אינטרינסי ‪ -‬נמצאים תמיד בתחום האינטרינסי‬
‫ריכוז נושאי המטען כתלות בטמפרטורה‬
‫תחום היינון החלקי של אטומי הזיהום (‪)Partial ionization‬‬
‫‪kT << Ed, Ea << Eg‬‬
‫• תחום זה מתחיל קצת מעל האפס המוחלט‪.‬‬
‫• האנרגיה התרמית מספיקה כדי להעביר חלק מנושאי המטען מרמות‬
‫האנרגיה של הזיהומים‪.‬‬
‫)‪kT‬‬
‫) ‪kT‬‬
‫‪n = (2Nd )1 2 (2me kT h 2 )3 2 e( Ed‬‬
‫‪n-type‬‬
‫‪p = (2Na )1 2 (2mh kT h 2 )3 2 e(Ea‬‬
‫‪p-type‬‬
‫‪ Na, Nd‬ריכוז האטומים הזיהומים הנותנים והנוטלים הכללי‪.‬‬
‫‪ Ed‬נמדד מתחתית פס ההולכה‪.‬‬
‫‪ Ea‬נמדד מתקרת פס הערכיות‪.‬‬
‫ריכוז נושאי המטען כתלות בטמפרטורה‬
‫התחום האקסטרינסי (‪)Saturation‬‬
‫‪Ed, Ea < kT << Eg‬‬
‫• כמעט כל נושאי המטען שמקורם בזיהומים מיוננים‪.‬‬
‫• ריכוז נושאי מטען האינטרינסיים עודנו זניח‪.‬‬
‫‪p=0‬‬
‫‪n = Nd‬‬
‫‪n-type‬‬
‫‪n=0‬‬
‫‪p = Na‬‬
‫‪p-type‬‬
‫‪ Na, Nd‬ריכוז האטומים הזיהומים הנותנים והנוטלים הכללי‪.‬‬
‫ריכוז נושאי המטען כמעט קבוע בטמפרטורה‪.‬‬
‫ריכוז נושאי המטען כתלות בטמפרטורה‬
‫התחום האינטרינסי‬
‫‪Ea, Ed < kT < Eg‬‬
‫• האנרגיה התרמית מספיקה להעלות אלקטרונים מפס הערכיות לפס‬
‫ההולכה‪.‬‬
‫• מספר נושאי המטען האינטרינסיים גדול ממספר נושאי המטען שמקורם‬
‫בסימום‪.‬‬
‫)‪n= Nd + N(T‬‬
‫)‪p = N (T‬‬
‫‪n-type‬‬
‫)‪p= Na + N(T‬‬
‫)‪n = N (T‬‬
‫‪p-type‬‬
‫‪ Na, Nd‬ריכוז האטומים הזיהומים הנותנים והנוטלים הכללי‪.‬‬
‫) ‪(  E g 2kT‬‬
‫‪N (T )  ni  pi = 2( 2kT h ) ( me mh ) e‬‬
‫‪3/4‬‬
‫*‬
‫*‬
‫‪2 32‬‬
‫ריכוז נושאי המטען כתלות בטמפרטורה (‪.)Si‬‬
The Concept of Effective Mass
• If the same magnitude of electric field is
applied to both electrons in vacuum and
inside the crystal, the electrons will
accelerate at a different rate from each other
due to the existence of different potentials
inside the crystal.
Comparing
Free e- in vacuum
In an electric field
mo =9.1 x 10-31 kg
Free electron mass
• The electron inside the crystal has to try to
make its own way.
• So the electrons inside the crystal will have a
different mass than that of the electron in
vacuum.
• This altered mass is called as an effectivemass.
An e- in a crystal
In an electric field
In a crystal
m = ?
m*
effective mass
What is the expression for m*
• Particles of electrons and holes behave as a wave
under certain conditions. So one has to consider the
de Broglie wavelength to link partical behaviour with
wave behaviour.
• Partical such as electrons and waves can be diffracted
from the crystal just as X-rays . (Bragg diffraction)
• Certain electron momentum is not allowed by the
crystal lattice. This is the origin of the energy band
gaps.
n  2d sin
n = the order of the diffraction
λ = the wavelength of the X-ray
d = the distance between planes
θ = the incident angle of the X-ray beam
n = 2d
(1)
The waves are standing waves
2
=
k
is the propogation constant
The momentum is
By means of equations (1) and (2)
certain e- momenta are not allowed
by the crystal. The velocity of the
electron at these momentum values
is zero.
Energy
P = k
(2)
The energy of the free electron
can be related to its momentum
E=
P
2
P=
2m0
h

free e- mass , m0
2 1
2 k2
h
h
E

2m  2
2m (2 ) 2
h
=
2
E
2k 2
2m
The energy of
the free e- is
related to the k
k
momentum
E versus k diagram is a parabola.
Energy is continuous with k, i,e, all energy
(momentum) values are allowed.
E versus k diagram
or
Energy versus momentum diagrams
Find effective mass , m*
We will take the derivative of energy with respect to k ;
2
dE
k

dk
m
- m* is determined by the curvature of the E-k curve
2
d2E

2
m
dk
Change
m*
instead of
m* 
- m* is inversely proportional to the curvature
m
2
d 2 E dk 2
This formula is the effective mass of
an electron inside the crystal.
Positive and negative effective mass
Direct-band gap s/c’s (e.g. GaAs, InP, AlGaAs)
m*
E

2
d 2 E dk 2
CB
e-
•
The sign of the effective mass is determined directly
from the sign of the curvature of the E-k curve.
•
The curvature of a graph at a minimum point is a
positive quantity and the curvature of a graph at a
maximum point is a negative quantity.
•
Particles(electrons) sitting near the minimum have a
positive effective mass.
•
Particles(holes) sitting near the valence
maximum have a negative effective mass.
•
A negative effective mass implies that a particle will
go ‘the wrong way’ when an extrernal force is
applied.
k
+
VB
band
Direct an indirect-band gap materials
Direct-band gap s/c’s (e.g. GaAs, InP, AlGaAs)
E
•
For a direct-band gap material, the minimum
of the conduction band and maximum of the
valance band lies at the same momentum, k,
values.
•
When an electron sitting at the bottom of the
CB recombines with a hole sitting at the top
of the VB, there will be no change in
momentum values.
•
Energy is conserved by means of emitting a
photon, such transitions are called as
radiative transitions.
CB
ek
+
VB
Indirect-band gap materials
Indirect-band gap s/c’s (e.g. Si and Ge)
•
E
•
CB
For an indirect-band gap material; the
minimum of the CB and maximum of the
VB lie at different k-values.
When an e- and hole recombine in an
indirect-band gap s/c, phonons must be
involved to conserve momentum.
ePhonon
Eg
k

+

VB

Atoms vibrate about their mean
position at a finite temperature.These
vibrations produce vibrational waves
inside the crystal.
Phonons are the quanta of these
vibrational waves. Phonons travel with
a velocity of sound .
Their wavelength is determined by the
crystal lattice constant. Phonons can
only exist inside the crystal.
Indirect-band gap materials
• The transition that involves phonons without producing
photons are called nonradiative (radiationless) transitions.
• These transitions are observed in an indirect band gap s/c
and result in inefficient photon producing.
• Momentum conservation requires phonon and photon emit
together  Much lower probability.
• So in order to have efficient LED’s and LASER’s, one should
choose materials having direct band gaps such as compound
s/c’s of GaAs, AlGaAs, etc…
Photon vs. Phonon
• For GaAs, calculate a typical (band gap) photon energy and momentum , and
compare this with a typical phonon energy and momentum that might be
expected with this material.
photon
phonon
E(photon) = Eg(GaAs) = 1.43 ev
 = hc / λ
E(photon) = h
s
/λ
= hvs / a0
λ (phonon) ~a0 = lattice constant =5.65x10-10 m
c= 3x108 m/sec
P=h/λ
 = hv
E(phonon) = h
h=6.63x10-34 J-sec
Vs= 5x103 m/sec ( velocity of sound)
λ (photon)= 1.24 / 1.43 = 0.88 μm
E(phonon) = hvs / a0 = 0.037 eV
P(photon) = h / λ = 7.53 x 10-28 kg-m/sec
P(phonon)= h / λ = h / a0 = 1.17x10-24 kg-m/sec
Photon vs. Phonon
•
•
•
•
Photon energy = 1.43 eV
Phonon energy = 37 meV
Photon momentum = 7.53 x 10-28 kg-m/sec
Phonon momentum = 1.17 x 10-24 kg-m/sec
Photons carry large energies but negligible amount of momentum.
On the other hand, phonons carry very little energy but significant
amount of momentum.
Electric conduction
•
•
•
•
•
•
Carrier drift
Carrier mobility
Mobility variation with temperature
A derivation of Ohm’s law
Drift current equations
Semiconductor band diagrams with an electric field
present
• Carrier diffusion
• The Einstein relation
• Total current density
Drift and Diffusion
• As recalls, current is the rate of flow of charge.
• So current depend on the number of charge carriers
and their flowing capabilities.
• There are two current mechanisms which cause
charges to move.
• The two mechanisms are drift and diffusion.
Carrier Drift
• Electron and holes will move under the influence of an
applied electric field since the field exert a force on charge
carriers (electrons and holes).
F  qE
• These movements result a current of
;
Id
I d  nqVd A
I d : drift current
Vd :
A:
drift velocity of charge carrier
n:
number of charge carriers per unit volume
q:
charge of the electron
area of the conductor / semiconductor
Carrier Mobility ,

Vd   E
E:
:
2


    cm 
V  Sec 

applied field
mobility of charge carrier
 is a proportionality factor
Vd 
 
E
 So
is a measure how easily charge carriers move under the influence of
an applied field or
determines how mobile the charge carriers are.

Carrier Mobility ,
Macroscopic understanding
Vd

E
In a perfect Crystal
 0
 
It is a superconductor

Microscopic understanding? (what the carriers
themselves are doing?)
q
 *
m
me*  mh* in general
m ; n  type
*
e
m ; p  type
*
h
Microscopic understanding of mobility?
How long does a carrier move in time before collision ?

The average time taken between collisions is called as relaxation
time, (or mean free time)
How far does a carrier move in space (distance) before a
collision?
The average distance taken between collisions is called as mean
free path, l.
Carrier Mobility ,

• A perfect crystal has a perfect periodicity and
therefore the potential seen by a carrier in a perfect
crystal is completely periodic.
• So the crystal has no resistance to current flow and
behaves as a superconductor. The perfect periodic
potential does not impede the movement of the
charge carriers. However, in a real device or
specimen, the presence of impurities, interstitials,
subtitionals, temperature , etc. creates a resistance
to current flow.
• The presence of all these upsets the periodicity of the
potential seen by a charge carrier.
The mobility two components in s.c.
The mobility has two component
Lattice interaction
component
L
Impurity interaction
component
I
Mobility variation with temperature in s.c.


T
T
Low temperature
High temperature
1
T

1
L

1
ln( )
I
This equation is called as
Mattheisen’s rule.
I
L
ln( T )
Peak depends on
the density of
impurities
Variation of mobility with temperature s.c.
L
At high temperature (as
the lattice warms up)
L
component becomes significant.
decreases when temperature increases.
 L  C1  T

3
2
It is called as a
T

3
2
C1 is a constant.
T 1.5 power law.
Carriers are more likely scattered by the lattice atoms.
Variation of mobility with temperature s.c.
At low temperatures
I
I
component is significant.
decreases when temperature decreases.
 I  C2  T
3
2
C2 is a constant.
Carriers are more likely scattered by ionized impurities.
Thermal velocity
• Assume crystal is at thermodynamic equilibrium (i.e. there is no applied
field). What will be the energy of the electron at a finite temperature?
• The electron will have a thermal energy of kT/2 per degree of freedom. So
, in 3D, electron will have a thermal energy of
3kT
1 * 2 3kT
E
 m Vth 
 Vth 
2
2
2
Vth : therm alvelocity of electron
Vth  T
1
2
 
Vth  m
*
1
2
3kT
m
Random motion  no current
• Since there is no applied field, the movement
of the charge carriers will be completely
random. This randomness result no net
current flow. As a result of thermal energy
there are almost an equal number of carriers
moving right as left, in as out or up as down.
Vdrift Vs. Vth
• Calculate the velocity of an electron in a piece of n-type silicon
due to its thermal energy at RT and due to the application of an
electric field of 1000 V/m across the piece of silicon.
Vth  ?
Vd  ?
V
th

RT  300 K
E  1000 V / m
me*  1.18 m0
  0.15 m 2 /(V  s )
3kT
5
m / sec
10
x
1.08

V

th

m
Vd   E  Vd  150 m / sec
Calculation
Drift velocity=Acceleration x Mean free time
F
Vd 

*
m
Force is due to the applied field, F=qE
Vd
F
qE

 
*
m
m*
Vd 
E



q

m
Calculation- Drift Velocities
• Calculate the mean free time and mean free path for electrons
in a piece of n-type silicon and for holes in a piece of p-type
silicon.
 ?
l ?
me*  1.18 mo
e  0.15 m 2 /(V  s )
e 
e me
q
 10 12 sec
vthelec  1.08 x105 m / s
mh  0.59mo
 h  0.0458 m 2 /(V  s )
h 
 h mh
q
 1.54 x10 13 sec
vthhole  1.052 x105 m / s
le  vthelec  e  (1.08 x105 m / s )(10 12 s )  10 7 m
lh  vthhole h  (1.052 x105 m / s )(1.54 x10 13 sec)  2.34 x10 8 m
Saturated Drift Velocities
• The equation of Vd  .E does not imply that Vd
increases linearly with applied field E.
• Vd increases linearly for low values of E and then it
saturates at some value of Vd which is close Vth at
higher values of E.
• Any further increase in E after saturation point does
not increase Vd instead warms up the crystal.
A Derivation of Ohm’s Law
Vd   E
I d  nqVd A
Id
Jd 
A
q
 
m
J x  nqVd  nq E
nq 
 
m
2
J x   Ex
 nq 2
Jx   
 m

1


 Ex

      m
   1 (  m)
Drift Current Equations
For undoped or intrinsic semiconductor ; n=p=ni
For electron
For hole
J p  pqE  p
J n  nqE n
drift
current for
electrons
number of
free
electrons
per unit
volume
mobility
of
electron
drift
current for
holes
number of
free holes
per unit
volume
mobility
of holes
Drift Current Equations
Total current density
Ji  Je  J h
J i  nqE  n  pqE  P
since
n  p  ni
J i  ni q (  n   p ) E
For a pure
intrinsic
semiconductor
Drift Current Equations
J total  ?
for doped or extrinsic semiconductor
n-type semiconductor;
n  p  J T  nqn E  N D qn E
where ND is the shallow donor concentration
p-type semiconductor;
p  n  JT  pq p E  N Aq p E
where NA is the shallow acceptor concentration
Carrier Diffusion
Current mechanisms
Drift
Diffusion
photons
P  nkT
dP dn
kT

dx dx
dn 1 dP

dx kT dx
Contact with a metal
Einstein Relation
Einstein relation relates the two independent current mechanicms of
mobility with diffusion;
Dn
kT

n
q
and
Dp
kT

p
q
for electrons and holes
Constant value at a fixed temperature
2
cm sec
 volt
2
cm V  sec
kT
 25 mV
q
kT  J / K  K 

 volt
q
C
at room temperature
Total Current Density
When both electric field (gradient of electric potential) and
concentration gradient present, the total current density ;
dn
J n  qn nE  qDn
dx
dp
J p  q p pE  qD p
dx
J total  J n  J p
Semiconductor Band Diagrams with Electric
Field Present
At equilibrium ( with no external field )
EC
All these
energies are
horizontal
Eİ
Pure/undoped semiconductor
EV
How these energies will change with an applied field ?
+
qV
e-
EC
Ef
Eİ
n – type
Electric field
Electron movement
Hole flow
EV
hole
Semiconductor Band Diagrams with Electric
Field Present
• With an applied bias the band energies slope down for the given
semiconductor. Electrons flow from left to right and holes flow from right to left
to have their minimum energies for a p-type semiconductor biased as below.
_
+
e-
EC
qV
Eİ
p – type
Ef
Electric field
Electron movement
Hole flow
EV
hole
Under drift conditions;
•Under drift conditions; holes float and electrons
sink. Since there is an applied voltage, currents are
flowing and this current is called as drift current.
•There is a certain slope in energy diagrams and the
depth of the slope is given by qV, where V is the
battery voltage.
‫תלות בטמפרטורה של הולכה חשמלית במוליכים‬
‫המוליכות ‪ σ‬מוגדרת ע"י ‪:‬‬
‫‪J=σE‬‬
‫‪σ=ne2τ/m‬‬
‫‪ -n‬צפיפות נושאי המטען‪.‬‬
‫‪ - e‬מטען האלקטרון‪.‬‬
‫‪ - m‬המסה האפקטיבית של נושא המטען‪.‬‬
‫‪ - τ‬הוא זמן הרלקסציה‪.‬‬
‫‪ m ,e‬לא תלויים בטמפרטורה ולכן התלות נובעת רק מהשינוי בריכוז נושאי‬
‫המטען ובזמן הרלקסציה‪.‬‬
‫במוליכים ריכוז נושאי המטען גבוה מאוד ואינה משתנה עם הטמפרטורה‪,‬‬
‫ולכן התלות בטמפ' נובעת מהתלות של ‪ τ‬בטמפרטורה‪.‬‬
‫‪ ‬במתכות – גבישים מסודרים – ההתנגשויות בעיקר עם פונונים‪ .‬ההתנגדות עולה‬
‫עם הטמפרטורה באופן מתון – כמו ‪ 1/Tα‬כאשר ‪ α‬מסדר גודל של ‪.1‬‬
‫‪ ‬בסגסוגות – ריבוי פגמים בגביש – ההתנגשויות בעיקר עם פגמים‪ .‬ההתנגדות‬
‫כמעט לא תלויה בטמפרטורה‪ .‬מוליכות גרועה מזו של מתכות מסודרות‪.‬‬
‫תלות בטמפרטורה של הולכה חשמלית במוליכים למחצה‬
‫חלוקה ל ‪ 4‬תחומי טמפרטורה‪:‬‬
‫• תחום הקיפאון (‪ σ=0 T=0 )Freeze out‬המל"מ מבודד‪.‬‬
‫• תחום היינון החלקי (‪kT << Ed, Ea << Eg )Partial ionization‬‬
‫)‪σ  exp(-Ed/kT‬‬
‫)‪σ  exp(-Ea/kT‬‬
‫עבור מוליך למחצה מסוג ‪n‬‬
‫עבור מוליך למחצה מסוג ‪p‬‬
‫• התחום האקסטרינסי (‪Ed, Ea < kT << Eg )Saturation‬‬
‫ריכוז נושאי המטען כמעט קבוע בטמפרטורה‪ ,‬ולכן השינוי בהולכה קטן‬
‫ונובע מהתלות של הנידות בטמפרטורה‪.‬‬
‫• התחום האינטרינסי ‪Ea, Ed < kT < Eg‬‬
‫)‪σ  exp(-Eg/2kT‬‬
‫במל"מ צפיפות נושאי המטען החופשיים ‪ n‬ו ‪ p‬הוא בדרך כלל הגורם דומיננטי בתלות‬
‫המוליכות בטמפרטורה‪ .‬רק בתחום האקסטרינסי‪ ,‬כאשר ריכוז נושאי המטען קבוע‪,‬‬
‫ניתן להבחין בשינוי בזמן הרלקסציה‪.‬‬
Conductivity in solids
Organic (plastic) Semiconductors
‫הולכה חשמלית במוצקים ‪ -‬סיכום‬
‫ההולכה החשמלית במוצקים תלויה בגורמים הבאים‪:‬‬
‫‪ .1‬כמות נושאי המטען שיכולים לנוע כאשר מופעל מתח חשמלי על‬
‫המוצק‪.‬‬
‫‪ .2‬התנגשויות של נושאי המטען עם תנודות השריג‪ ,‬הפונונים‪.‬‬
‫‪ .3‬התנגשויות של נושאי המטען עם פגמים במבנה השריגי (נקעים‬
‫דחיקים וכו')‪.‬‬
‫• תלות ההולכה בטמפרטורה תלויה בהשתנות הגורמים הנ"ל כפונקציה‬
‫של הטמפרטורה‪.‬‬
‫• בחומרים מסוימים ישנו גורם אחד דומיננטי והוא יאפיין את תלות הולכה‬
‫בטמפרטורה‪ .‬לדוגמה‪ :‬מתכת – פונונים‪ ,‬מל"מ – ריכוז נושאי מטען‪,‬‬
‫סגסוגת – התנגשות בפגמים‪.‬‬
p – n junction
Hole
Movement
n-type
+++++
+++++
+++++
+++++
+++++
+++++
+++++
+++++
---------------------------------
p-type
Electron
Movement
++++
++++
++++
Fixed positive
space-charge
Metallurgical
junction
----------
Fixed negative
space-charge
Ohmic
end-contact
p – n junction
p-type
n-type
EC
EC
Ef
Eİ
Eİ
Ef
EV
EV
EC
Eİ
p-type
n-type
EC
Ef
Eİ
Ef
EV
EV
There is a big discontinuity in the fermi level accross the
p-n junction.
p – n junction
 Lots of electrons on the left hand side of the
junction want to diffuse to the right and lots of
holes on the right hand side of the junction want
to move to the left.
 The donors and acceptors fixed,don’t move
(unless you heat up semiconductors, so they can
diffuse) because they are elements (such as
arsenic and boron) which are incorporated to
lattice.
 However, the electrons and holes that come from
them are free to move.
Idealized p-n junction
 Holes diffuse to the left of the metalurgical junction and
combine with the electrons on that side. They leave behind
negatively charged acceptor centres.
 Similarly, electrons diffusing to the right will leave behind
positively charged donor centres. This diffusion process can
not go on forever. Because, the increasing amount of fixed
charge wants to electrostatically attract the carriers that are
trying to diffuse away(donor centres want to keep the
electrons and acceptor centres want to keep the holes).
Equlibrium is reached.
 This fixed charges produce an electric field which slows
down the diffusion process.
 This fixed charge region is known as depletion region or
space charge region which is the region the free carriers
have left.
 It is called as depletion region since it is depleted of free
carriers.
p – n junction
 The drift and diffusion currents are flowing all the
time. But, in thermal equilibrium, the net current
flow is zero since the currents oppose each other.
 Under non-equilibrium condition, one of the
current flow mechanism is going to dominate
over the other, resulting a net current flow.
 The electrons that want to diffuse from the ntype layer to the p-layer have potential barier.
DR
Neutral
p-region
----------
+++
+++
+++
 Current Mechanisms,
Neutral
n-region
 Diffusion of the carriers
cause an electric in DR.
Field Direction
Electron Drift
Hole energuy
Electrıon energy
p – n junction in thermal equilibrium
 Drift current is due to
the presence of electric
Electron Diffusion
field in DR.
EC
Ef
 Diffusion current is due
to the majority carriers.
EV
 Drift current is due to
the minority carriers.
Hole Diffussion
Hole Drift
Appliying bias to p-n junction
+
-
p
n
forward bias
-
+
p
n
reverse bias
 How current flows through the p-n
junction when a bias (voltage) is
applied.
 The current flows all the time whenever
a voltage source is connected to the
diode. But the current flows rapidly in
forward bias, however a very small
constant current flows in reverse bias
case.
Appliying bias to p-n junction
I(current)
Reverse Bias
Vb
I0
Forward Bias
V(voltage)
Vb ; Breakdown voltage
I0 ; Reverse saturation current
 There is no turn-on voltage because current flows in any
case. However , the turn-on voltage can be defined as the
forward bias required to produce a given amount of forward
current.
 If 1 m A is required for the circuit to work, 0.7 volt can be
called as turn-on voltage.
Appliying bias to p-n junction
Zero Bias
p
Forward Bias
+
-
-- ++ n
-- ++
p
Ec
Ec
Ev
n
qVbi  VF 
Vbi
p
---
++
++
n
Ec
Ev
Potential Energy
Ev
qVbi
- +
- +
Reverse Bias
+
q Vbi  Vr 
Vbi  VR
Vbi  VF
Ideal diode equation
J Total
 Dn n po D p pno    qV
 q

 exp 
Lp    kT
 Ln
  qV
 
  1  J o exp 
 
  kT
 
  1
 
multiplying by area ;
  qV
I  I o exp 
  kT
 
  1
 
Ideal diode equation
This equation is valid for both forward and reverse biases; just change the sign of V.
Ideal diode equation reverse bias
•
Change V with –V for reverse bias. When qV > a few kT;
exponential term goes to zero as

 qV
I  I o exp 
 kT

 
  1
 
I   Io
Reverse saturation current
Current
Forward Bias
VB
I0
Voltage
VB ; Breakdown voltage
Reverse Bias
I0 ; Reverse saturation current