Lecture notes-3 of Material Science - SRM UNIVERSITY . FIRST YEAR B.TECH UNIT - 1...
Description
LECTURE 2
SEMICONDUCTORS AND ITS CLASSIFICATION AND
FERMI ENERGY LEVEL DISTRIBUTION IN INTRINSIC SEMICONDUCTORS
VARIATION OF FERMI LEVEL WITH TEMPERATURE IN EXTRINSIC SEMICONDUCTORS 15PY102L UNIT 1 LECTURE 2
Semicondc!o"# In!"odc!ion The
materials are classified on the basis basis of conductiit! and resistiit!" #emiconductors are the materials $hich has conductiit!% resistiit! alue in bet$een conductor and insulator " The resistiit! of semiconductor is in the order of 10 −& to 0"5 'hm(metre" It is not that% the resistiit! alone decides $hether a substance is a semiconductor )or* not % because some allo!s hae resistiit! $hich are in the ran+e of semiconductor,s resistiit!" -ence there are some .ro.erties li/e band +a. $hich distin+uishes the materials as conductors% semiconductors and insulators" 15PY102L UNIT 1 LECTURE 2
semi(conductor is a solid $hich has the ener+! band similar to that of an insulator insulator"" It acts as an insulator at absolute ero and as a conductor at hi+h tem.eratures and in the .resence of im.urities"
#emiconductors are materials $hose electronic .ro.erties are intermediate bet$een those of metals and insulators" These intermediate .ro.erties are determined b! the cr!stal structure% bondin+ characteristics and electronic ener+! bands" The! are a +rou. of materials hain+ conductiities bet$een those of metals and insulators" 15PY102L UNIT 1 LECTURE 2
are of the same /ind )i"e* com.osed of sin+le s.ecies of atoms" )e+* +ermanium and silicon" Compound semiconductor: The! are com.osed of
t$o or more different elements )e+* ) e+* a#% ls ls etc"% 15PY102L UNIT 1 LECTURE 2
C")#!%$ #!"c!"e o& #i$icon %nd 'e"m%nim
The structure of #i and e% $hich are hain+ coalent bondin+" Coalent bondin+s are stereo s.ecific3 i"e" each bond is bet$een a s.ecific .air of atoms" The .air of atoms share a .air of electrons )of o..osite ma+netic s.ins*" 15PY102L UNIT 1 LECTURE 2
Three dimensional re.resentation of the structures #i% and e% $ith the bonds sho$n in belo$ fi+ure% the re+ion of hi+h electron .robabilit! )shaded*"
ll atoms hae coordination number &3 each material has an aera+e of & alence electrons .er atom% and t$o electrons .er bond" Each atom of a material is coordinated $ith its nei+hbours"
The thermal ibrations on one atom influence the ad4acent atoms3 the dis.lacement of one atom b! mechanical forces% or b! an electric field% leads to ad4ustments of the nei+hbourin+ atoms" The number of coordinatin+ nei+hbours that each atom has is im.ortant" Coalent bonds are er! stron+"
In semiconductors and insulators% $hen an eternal electric field is a..lied the a..lied the conduction is not .ossible as .ossible as there is a forbidden +a.% +a.% $hich is absent in metals" In order to conduct% the electrons from the to. of the full alence band hae to moe into the conduction band% b! crossing the forbidden gap " The field that needs to be a..lied to do this $or/ $ill be etremel! lar+e" 15PY102L UNIT 1 LECTURE 2
E+6 #ilicon $here the forbidden +a. is about 1 e7 e7"" The distance bet$een these t$o locations is about 1 8 )10 m*"
−10
field field +radient of a..roimatel! 179 )10 10 m* : 10107m 1 is necessar! to moe an electron from the to. of the alence band to the bottom of the conduction band" −
15PY102L UNIT 1 LECTURE 2
−
The other .ossibilit! b! $hich this transition can be brou+ht about is b! thermal excitation"
t room tem.erature% the thermal ener+! that is aailable can ecite a limited number of electrons across the ener+! +a."" This limited number accounts for semi(conduction" +a. semi(conduction"
;hen the ener+! +a. is lar+e as in diamond% the number of electrons that can be ecited across the +a. is etremel! small"
15PY102L UNIT 1 LECTURE 2
In intrinsic semiconductors% the conduction is due to the intrinsic .rocesses )without the influence of impurities) "
.ure cr!stal of silicon or +ermanium is an intrinsic semiconductor" The electrons that are ecited from the to. of the alence band to the bottom of the conduction band b! thermal ener+! are res.onsible for conduction" The number of electrons ecited across the +a. can be calculated from the
The .robabilit! f )E *of *of an electron occu.!in+ ener+! leel E becomes f )E * : e.)−E g 9 2k BT *"
> The fraction of electrons at ener+! E is e@ual to the .robabilit! f )E *" *" The number n of electrons .romoted .rom oted across the +a.% e.)−E g 9 2k BT * n : N e.) $here N is is the number of electrons aailable for ecitation from the to. of the alence band"
15PY102L UNIT 1 LECTURE 2
The .romotion of some of the electrons across the +a. leaes some acant electron sites in the alence band" These are called holes" n intrinsic semiconductor contains an equal number of holes in the alence band and electrons in the conduction band % that is ne : nh"
Under an eternall! a..lied field% the electrons% $hich are ecited into the conduction band b! thermal means% can accelerate usin+ the acant states aailable in the conduction band" 15PY102L UNIT 1 LECTURE 2
t the same time% the holes in the alence band also moe% but in a direction opposite to that of electrons" The conductiit! of the intrinsic semiconductor de.ends on the concentration of these char+e carriers% ne and nh" In the case of metals% the drift elocit! ac@uired b! the free electrons in an a..lied field" The mobilit! of conduction electrons and holes% µe and µh% as the drift elocit! ac@uired b! them under unit field +radient" 15PY102L UNIT 1 LECTURE 2
The conductiit! σ of an intrinsic semiconductor sem iconductor as σi
: ne e
µe
B nh e
µh
$here e is the electronic char+e% ne and nh are concentrations of electrons and holes .er unit olume"
15PY102L UNIT 1 LECTURE 2
Fermi level
The number of free electrons .er unit olume in an intrinsic semiconductor is n = 2
h
3 / 2
* 2π me kT 2
E F − E c ex p kT
The number of holes .er unit olume in an intrinsic semiconductor is p :
2m ∗π k T 2 h 2 h
3
2
E − E KT
. exp
V
F
#ince n : . in intrinsic semiconductors" 15PY102L UNIT 1 LECTURE 2
2
3
(m ) ∗
e
3
2
exp
h
( E − E ) F
C
= m∗
h
kT
3
2 E F − E c 2π mh∗k T 2 Ev − E F ex p exp = 2 2 kT kT h
* 2π me k T 2
3
2
Ev − E exp KT F
or
e
kT
3
m 2 E + E = * exp kT m ∗
2 E F
h
v
e
Ta/in+ lo+ on both sides% 2 E F kT
mh∗ E + E c log e exp v = log e * + kT 2 m e
2 E F kT
3
mh∗ E v + E c = log e * + 2 kT m e 3
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