Transcript Week 10 - Help-A-Bull

EEE 3394
Electronic Materials
Chris Ferekides
Fall 2014
Week 10++
Equilibrium Carrier Concentrations
REM:
 gC(E)d(E)
 f(E)
represents the number of available states (cm-3) in the energy interval E+dE.
is the probability a state is occupied by an electron; (1-f(E) holes);
 gC(E)f(E)d(E) gives the number of electrons (cm-3) in the interval E+dE;
Therefore the TOTAL number of electrons n (and holes p) in the conduction band (and in the valence
band) can be obtained by integrating the relationships:
no 
ETO P
g
C
(E)f(E)dE
EC
gc (E) 
po 
EV
g
V
(E)[1 f(E)]dE
EBotom
m
*
n
2m (E  Ec )
*
n
2 3
 
gv (E) 
mp* 2mp* (Ev  E)
 2 3
Equilibrium Carrier Concentrations
no =
let
m
*
n
2m
p2
E - E C dE
* E top
n
3
ò
EC
1+ e
(E - E C )
h=
kT
Þ no =
m
*
n
(E-E F )
(E F - E C )
hC =
kT
and
*
n
2m (kT)
π2 3
3
kT
2 ¥
1
h 2dh
ò 1+ eh h
-
C
0
The integral is called the Fermi-Dirac integral of order ½ (F1/2(ηC)) and can be found in tables.
Equilibrium Carrier Concentrations
Solving the above integrals yield:
no = NC
2
p
F1 (hC )
2
po = N V
2
p
F1 (hV )
2
…where NC and NV are the effective densities of states in the conduction and valence bands
respectively.
Note:
Effective Density of States: if all electron (hole) states were located at EC (EV).
2 ×π × m*n kT 3 2
NC = 2[
]
2
h
N V = 2[
2π × m*p kT
h2
]
3
2
When does n=p
g(E) (E Ec)1/2
Ec+c
E
E
E
[1 f(E)]
CB
Area = nE (E )dE  n
For
electrons
Ec
Ec
nE(E)
Ev
pE(E)
EF
EF
Ev
For holes
Area = p
VB
0
g(E)
f(E)
nE(E) or pE(E)
Equilibrium Carrier Concentrations
Note:
Previous relationships are GENERAL!
Yet inconvenient to calculate every time.
HOWEVER, if EF is not “close” to the conduction band or valence band i.e.
E V + 3kT £ E F £ EC - 3kT
…then the Equilibrium Electron and Hole Concentrations can be expressed as:
n o = NCe-(EC -EF ) kT
po = N Ve-(EF -EV ) kT
When the Fermi Level is too close to the Conduction (or Valence) Band, or lies within the Conduction
(or Valence) Band, the semiconductor is said to be Degenerate; i.e. Heavily Doped, with the electron
(or hole) concentration being very high.
Intrinsic Fermi Level
Note:
Ei is the Fermi Level (energy) in an intrinsic semiconductor.
In intrinsic semiconductors the electron and hole concentrations are equal.
n = p = ni
…then
n i = NC e
-(EC -E i ) kT
pi = N Ve
Ec
…solving for NV and NC
NC = n i e
(EC -E i ) kT
Ec
N V = n ie
(E i -E V ) kT
…finally..
n = n ie(EF -Ei ) kT
CB
-(E i -E V ) kT
p = nie(Ei -EF ) kT
EFp
Ev
EFi
Ec
EFn
Ev
Ev
VB
np product
What is ni2 and the product np equal to???
pi = N Ve-(Ei -E V ) kT
n i = NCe-(EC -Ei ) kT
n 2i = NCN Ve-(EC -EF ) kT = NCN Ve-EG kT
Þ n i = NC N V e-EG
…also ..
2kT
n o = n ie(EF -Ei ) kT
po = n ie(Ei -EF ) kT
Þ n o po = n
2
i
REMEMBER: ALL OF THE ABOVE WERE CALCULATED ASSUMING EQUILIBRIUM CONDITIONS
AND NONDEGENERATE SEMICONDUCTORS.
Charge Neutrality
 A uniformly doped semiconductor under equilibrium conditions is charge-neutral; i.e. the NET
charge is equal to zero. (WHY?)
 Are there any charges inside the semiconductor?
Electrons, holes, ionized acceptors, ionized donors.
…therefore …
ep - en + eN+D - eN-A = 0
Note:



Number of ionized Donors:
ND+
Number of ionized Acceptors: NAAt room temperature complete ionization will be assumed.
  ND+ = ND
  N A- = N A
Calculating n & p
np = n 2i
p - n + ND - NA = 0
p = n 2i n

n 2 - n(ND - NA ) - n 2i = 0
…therefore …
ù
N D - N A éæ N D - N A ö
2
n=
+ êç
+ ni ú
÷
ø
2
2
ëè
û
2
1
2
ù
N A - N D éæ N A - N D ö
2
2
p = ni n =
+ êç
+ ni ú
÷
ø
2
2
ëè
û
2
1
2
The above equations are general expressions and can be simplified for most practical
applications.
Calculating n & p
(1)
Intrinsic semiconductors
ND=NA=0
n=p=ni
(2)
Doped Semiconductors:
ND-NA>>ni or NA-ND>>ni
The above is usually true since the controlled doping of semiconductors yields more
donors than acceptors, and since typical intentional dopants are at least 1014 (rem: ni=1010
for Si).
For
N D ññN A
and
N D ññn i

n » ND
p » n 2i ND
For
N A ññN D
and
N A ññn i

p » NA
n » n 2i N A
Calculating n & p
(3)
Doped with ni>>ND (or ni>>NA)
The semiconductor is considered intrinsic.
At high enough temperatures all semiconductors are intrinsic.
(4)
Compensated Semiconductor:
When ND or NA are comparable or equal; keep both concentrations and use general
expressions.
ù
N D - N A éæ N D - N A ö
2
n=
+ êç
+ ni ú
÷
ø
2
2
ëè
û
2
1
2
ù
N A - N D éæ N A - N D ö
2
2
p = ni n =
+ êç
+ ni ú
÷
ø
2
2
ëè
û
2
1
2
General Expressions for n and p
ù
N D - N A éæ N D - N A ö
2
n=
+ êç
+ ni ú
÷
ø
2
2
ëè
û
2
1
2
ù
N A - N D éæ N A - N D ö
2
2
p = ni n =
+ êç
+ ni ú
÷
ø
2
2
ëè
û
2
(1)
What happens when ND>>NA & ND>>ni??
(2)
What happens when NA>>ND & NA>>ni??
(3)
What happens when ni >> NA & ND??
1
2
Temperature Dependence of n (or p)
T < Ts
Ts < T < Ti
T > Ti
CB
EF
As+ As
As
Eg
As
+
+
+
+
EF As As As As
EF
As+ As+ As+ As+
VB
(a)T=T1
(b)T=T2
(c)T=T3
(a) Below Ts, the electron concentration is controlled by the ionization of
the donors. (b) Between Ts and Ti, the electron concentration is equal to
the concentration of donors since they would all have ionized. (c) At high
temperatures, thermally generated electrons from the VB exceed the
number of electrons from ionized donors and the semiconductor behaves
as if intrinsic.
From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (©
McGraw-Hill, 2005)
Temperature dependence of n and p
Þ n i = NC N V e-EG
2kT
Fermi Level (Energy)
Note:
Derivations will be based on equations derived previously that relate n, p, and EF; given
one of these three variables one can calculate the others.
(1)
Intrinsic semiconductors
ND=NA=0
n=p=ni
NCe(Ei -EC ) kT = N Ve(EV -Ei ) kT
Ei =
E C + E V kT æ N V ö
+
ln ç
2
2
è N C ÷ø
Note:
The Fermi Level for intrinsic semiconductors is located “near” mid gap!
….. also …
æ NV ö æ m ö
çè N ÷ø = çè m ÷ø
C
*
p
*
n
Note:
3
2
æ m*p ö
EC + EV 3
Ei =
+ kTln ç * ÷
2
4
è mn ø
The fermi level is exactly at mid-gap when the effective masses for electrons and holes are
equal or when the temperature is zero!
Fermi Level (Energy)
(2)
Using …
Doped Semiconductors: (assuming complete ionization and non-degenerate
semiconductors)
n = n ie(EF -Ei ) kT
and
p = nie(Ei -EF ) kT
EF - Ei = kTln(n/ni ) = -kTln(p/ni )
E F - Ei = kTln(ND /n i )
N D ññN A
N D ññn i
For n-type
E i - E F = kTln(NA /n i )
N A ññN D
N A ññn i
For p-type
Drift
Drift:
charged particle motion under an applied E-field.
Inside a semiconductor: Under an applied electric field both the electrons and holes experience a
force (qE) and move in opposite directions. The carrier motion is interrupted due to
collisions with lattice atoms and ionized impurities (scattering).
Note: the carriers are always moving - even in the absence of an electric field!: thermal motion;
however, thermal motion is completely random and it “averages out” to zero. An
electron does not necessarily return to its original position, but examining a group of
electrons the net effect of thermal motion is zero; on a macroscopic scale thermal motion
can be neglected.
Drift Velocity: On a macroscopic scale the average carrier motion can be described by a constant
velocity: vd


vd  μE
where μ is the mobility of the carrier
Drift
MOBILITY: is a constant that describes the ease by which carriers can move within a material. It is a
property of semiconductors often used to describe the quality of the materials.
Typical RT values:
for Si with NA=ND=1014/cm3 μn=1360 cm2/V-sec μp=460 cm2/V-sec
for uncompensated (high purity) GaAs with ND or NA<1015/cm3
μn=8000 cm2/V-sec; μp=320 cm2/V-sec
Notes:
electrons have (typically) higher mobility than holes
the motion of electrons is “impeded” by collisions
mobility is the measure of the ease of electron (carrier) motion
therefore the degree of scattering must influence the carrier mobility
Drift Current Density
Electron current density due to drift:
Je drift = qme nE
Hole current density due to drift:
Jp drift = qmh pE
Total current density:
Jdrift = q(nme + pmh )E = s E
CONDUCTIVITY
s = q(nme + pmh )
Scattering Mechanisms
 Lattice Scattering: collisions with the thermally agitated lattice atoms;
 Ionized impurity scattering: due to donor or acceptor site collisions.
 For doping concentrations below 1015/cm3 (Si) the mobility is essentially independent of the
doping concentration; and ionized impurity scattering is negligible.
 For doping concentrations above 1015/cm3 the mobility decreases with increasing doping
concentration due to increased ionized impurity scattering.
 Lattice Scattering increases with temperature; i.e. mobility decreases.
The temperature dependence is T-3/2
 Impurity Scattering (dominates at the lower temperatures) increases with decreasing
temperature.
The temperature dependence is T3/2
Scattering Mechanisms
Electron Drift Mobility(cm2 V-1s-1)
50000
LT -1.5
10000
Nd =1014
Ge
Nd =1013
Nd =1016
Nd =1017
1000
Nd =1018
100
Nd =1019
Si
 T1.5
10
70
100
Temperature (K)
800
Log-log plot of drift mobility vs temperature for n-type Ge and n-type
Si samples. Various donor concentrations for Si are shown. Nd are in
cm-3. The upper right inset is the simple theory for lattice limited
mobility whereas the lower left inset is the simple theory for impurity
scattering limited mobility.
Mobility as a function of Temperature
Effect of Doping on Mobility
Conductivity, Mobility & Temperature
600oC 400oC
L L
Semiconductor
Metal
log(n)
T
EXTRINSIC
Lattice
IONIZATION
scattering
T
log( )
-3/2
T
200oC
3/2
Impurity
scattering
L
L
2.41013 cm-3
1015
Ge
1012
1.451010 cm-3
109
Si
106
2.1106 cm-3
GaAs
1/T
High Temperature
27oC 0oC
1018
Intrinsic Concentration (cm-3)
INTRINSIC
R esistivity
LO G A R IT H M IC S C A LE
log( )
Low Temperature
Temperature dependence of electrical conductivity for a doped (ntype) semiconductor.
103
1
1.5
2.5
3.5
3
4
1000/T (1/K)
The temperature dependence of the intrinsic concentration.
2
Diffusion Current Density
DIFFUSION:
 A process where particles (not necessarily charged) redistribute as a result of thermal
motion, migrating from regions of high concentration to regions of low concentration…….
Eventually the diffusion process will lead to a uniform distribution.
In semiconductors the mobile particles are electrons and holes (charged!!). i.e. any
diffusion-related motion would lead to current flow…. i.e. DIFFUSION CURRENTS…..
J P DIFF =-qDpÑp
J N DIFF =qDnÑn
d
d
d
Ñ = x̂ + ŷ + ẑ
dx
dy
dz
J(p)
+++++++
ND,A(x)

---------
d/dx
x [length]
Semiconductor
J(n)
Optical Absorption
 Photons with energy higher than the semiconductor band gap will be absorbed by the
semiconductor, resulting in the generation of excess carriers.
hn > EG
 Semiconductors are “transparent” to photons with energy less than their bandgap!
hn < EG
 The absorption (transmission) of light by a semiconductor is given by the relationship below:
where α is the absorption coefficient [cm-1], and IO is the
intensity of the incident light beam; x is the distance
inside the semiconductor.
I(x)
I(x) = IOe-ax
The photon energy wavelength relationship is:
hc
1240
E[eV] =
=
l l[nm]
Thickness (x)
Optical Absorption
 If sample k micrometers thick then…..
It
IO
It
-ak
=
e
IO
k
i.e. above will give us the % of light transmitted through the sample: Optical transmission!


So far all energy band diagrams are drawn
with EC and EV constant….
When an Electric Field is present inside a
semiconductor EC,V become a function of
space i.e. EC,V(x)
Energy band diagram
 Energies represent the total electron
energy……
 Conduction band electrons can be described
as “motionless”
 If energy in excess of EG is added, excess
energy i.e. E-EC is kinetic energy
 Total Energy = KE + PE
 The potential energy of an electron is equal to
-qV where V is the electrostatic potential
Total Electron Energy
Band Bending
EC
-
-
KE
E-Field
EV
Energy Reference level
x
Total Electron Energy
Band Bending
1
V = - (E C - E REF )
q
dV
E = -ÑV = x̂
dx
EC
-
-
KE
E-Field
EV
Energy Reference level
x
1 dE C
1 dE V
1 dE i
E=
x̂ =
x̂ =
x̂
q dx
q dx
q dx

Electric Field = Band Bending
Donor Concentration, ND
Concentration Gradient
If complete
ionization
 n.
Semiconductor
x
EC
EF
EV
Ei
Einstein Relationship




therefore…
dEF/dx= dEF/dy= dEF/dz=0
If ND is a function of space i.e. ND(x)
Therefore … band bending
Therefore … E-field
What happens as a result of the concentration
gradient?
In what direction will the electrons diffuse?
Donor Concentration, ND
Under equilibrium Conditions the Fermi Level is Constant as a Function of
position.
If complete
ionization
 n.
Semiconductor
x
EC
EF
EV
Ei
Donor Concentration, ND
Einstein Relationship
Consider the total electron current:
dn(x)
Jn = qmn n(x)E + qDn
dx

What do the two terms represent?
If complete
ionization
 n.
Semiconductor
x
EC
n(x) = nie(EF -Ei (x)) kT
EF
dn 1
q
(E F -E i (x)) kT dE i (x)
=
nie
(
) = - nE
dx kT
dx
kT
Ei
EV
Note: under equilibrium the total current inside the semiconductor is zero!!
q
qmn nE + qDn ( nE) = 0
kT

Dn
kT
=
mn
q
Dp
kT
=
mp
q
Generation – Recombination
 Generation – electrons and holes are created
 Recombination – electrons and holes are “destroyed”
 The thermal creation and recombination of carriers (an
on going process – i.e. at all times) is typically dominated
by indirect R-G.
 R-G centers (or traps) introduce allowed energy levels
near the center of the band gap; ET in the figure.
 Under equilibrium conditions the recombination rate is
equal to the generation rate and therefore there is not
net change in the carrier concentrations.
g=r
 If the equilibrium is “disturbed” (perturbation) the thermal R-G rates change favoring return to
equilibrium conditions.
 Photo-generation always results in excess carriers.
Excess Carriers
Excess carries:
concentrations of electrons or holes above their equilibrium
values, created by an external “force”/excitation (for
example light).
Generation:
process by which electrons and holes are created
(generated).
Recombination:
process by which electrons and holes are annihilated
(destroyed).
Direct recombination:
electrons from the conduction band recombining with holes
in the valence band.
Indirect recombination:
recombination of (conduction band) electrons and (valence
band) holes via recombination centers (R-G centers; traps)
located in the energy gap.
Excess Carriers
Notation
 no, po: carrier concentrations under equilibrium conditions;
 n, p:
carrier concentrations under arbitrary conditions (anytime);
 Δn=n-no at time zero: excess electron concentration at t=0;
 Δp=p-po at time zero: excess hole concentration at time t=0;
 δn(t), δp(t): instantaneous excess carrier concentrations;
dn(t)
= a r n 2i - a r n(t)p(t)
dt
Assume that at time t=0 an excess of EHP is created (light pulse):
Dn = Dp
Note:
the excess electrons and holes are generated by a pulse of light!!. i.e.
the excitation source is removed!
Excess Carriers
d
d n(t) = a r n 2i - a r [n o + d n(t)][po + d p(t)]
dt
= -a r [(n o + p o )× d n(t) + d n 2 (t)]
Note:
Above equation is difficult to solve!.... So we will simplify!
We’ll assume “Low Level Injection”:
REM: Majority vs. Minority carriers
n-type material:
Δn=Δp<<no
p-type material:
Δp=Δn<<po


n≅no
p≅po
(Δp>>po
(Δn>>no
p≅ Δp)
n≅ Δn)
d
d n(t) = -a r pod n(t)
dt
d n(t) = Dne
- a r po t
= Dne
-t t n
τ: recombination lifetime!
1
tn =
a r po
Generation - Recombination
Example:
Si at RT
ND=1014/cm3
Perturbation (i.e. disturbance/an external force – could be light):
Δp= Δn=109/cm3 (same number of excess electrons and holes; they
in pairs EHP)
Is this a LLI case??
Initially under equilibrium conditions:
no=ND=1014/cm3
po=ni2/no=106/cm3
After excitation:
n=no+ Δn ≅ no and Δn << no
p=po+ Δp ≅ Δp and Δp >> po
are generated
We’ll assume “Low Level Injection”:
n-type material: Δp<<no n≅no
p-type material: Δn<<po p≅po
τ: recombination lifetime!
under low level injection the majority carriers are essentially unchanged but
carriers increase by several orders of magnitude.
the minority
Minority Carrier Diff Eq
 No time to derive … will go over derivation next semester …
¶d n
¶ dn dn
= Dn
- + GL
2
¶t
¶x
tn
2
¶d p
¶ dp dp
= Dp
+ GL
2
¶t
¶x
tp
2
Minority Carrier Diffusion Equations
 How do we use the MCDEs?
δn
 δn δn
 Dn

 GL
2
t
x
τn
2
 Under steady state conditions……
 No concentration gradient……
δp
 δp δp
 Dp

 GL
2
t
x
τp
2

 0
t

 0
x
 No thermal R-G……
δn
 0
τn
 No illumination
GL  0
Example
A silicon sample is doped uniformly with ND=1015 cm-3. The sample is illuminated
at t=0. Assuming that the EHP generation rate is 1017 cm-3s-1 (throughout the
semiconductor) and that p=10-6 sec find p(t).
Setting up and understanding the problem……
1. equilibrium prior to light turned on;
2. uniform doping and illumination;
3. assume (implied) room temperature;
4. what happens??? For time t<0 the sample is at equilibrium:
ND=1015 cm-3 >> ni;
therefore n=1015 cm-3; p=ni2/n=105cm-3
Example
5. when the light comes on it is absorbed and EHP are being generated; i.e. n and
p begin to increase; since the equilibrium is disturbed by increasing the
generation rate, the recombination rate (thermal) must increase in order to
balance the “excess” generation; eventually the recombination rate “catches up”
with generation rate and steady state conditions are reached.
When steady state conditions are reached:
generation=recombination; GL=Δp/τ.
6. to find δp(t) we must solve the minority carrier diffusion equation (are all
assumptions met???? YES - verify)
δp
 2 δp δp
 Dn

 GL
2
t
x
τn
δp δp

 GL
t
τp
δp(t)  GL τP  Ae
 t/τP
Example
 How do we evaluate A?
δp(t)  GL τP  Ae
 t/τP
 We need initial Conditions!
 i.e. what is δp @ t=0?..... δp(0)=0
δp(t)  GL τP  GL τP e
 Graph this!
 t/τP
G and pn(t)
Gph
Illumination
0
pno+pn()
pn(t') = pn(0)exp(-t'/h)
hGph
pno
0
Time, t
toff
t'
Illumination is switched on at time t = 0 and then off at t = toff. The excess
minority carrier concentration, pn(t) rises exponentially to its steady
state value with a time constant h. From toff, the excess minority carrier
concentration decays exponentially to its equilibrium value.
From Principles of Electronic Materials and Devices, Third Edition, S.O. Kasap (© McGraw-Hill, 2005)
Example
A semi-infinite bar of Si is doped with ND=1015 cm-3. At the x=0 end of the bar
is illuminated so that pn=1010 cm-3 at x=0. The light does not penetrate
beyond x=0. Find pn(x).
First – set up the problem…
1. What type of Si? – n-type
2. What are the equilibrium carrier concentrations? - nno ND=1015 cm-3; pno105
cm-3
3. What does semi-infinite bar imply?
4. Is the temperature important for this problem? – Assume room temp.
5. What does “does not penetrate beyond x=0” mean?
pno
???
X=0
X=
Example
6. Solve the MCDE to find pn(x,t)…
p ( x, t )
 2p p
 Dp

 GL
2
t
x
p
7. Simplify ………..
 2p p
0  Dp
  GL
2
x
p
8. Solution ………
p( x )  Ae
 x / Lp
 Be
x / Lp
….. where
Lp 
Dp p
Example
6. Evaluate constants A and B
7. What are the boundary conditions (could be initial i.e. t=0 conditions)..…
δpn( 0 )  pn
pn (  )  0
Therefore ………
1.2E+10
1.0E+10
The excess EHP are generated at x=0; due
to the concentration gradient generated
they will diffuse into the material.
“On average” they recombine after they
travel a distance LP known as the hole
diffusion length
8.0E+09
δpn (x)
pn ( x)  pne
 x / Lp
6.0E+09
4.0E+09
2.0E+09
0.0E+00
0.0E+00 5.0E-05 1.0E-04 1.5E-04 2.0E-04
distance
Minority Carrier Diffusion length: The average distance minority carriers will
diffuse (travel) before recombining.
Quasi Fermi Levels
 Under non-equilibrium conditions the np=ni2 is not valid any more;
 If we have a piece of Si doped with ND=1015 then the electron and hole
concentrations are: no=1015 and po=105;
 Let’s assume the sample is illuminated and an excess of EHP are generated
n=p=1012 …. What is the new np product??
 The relationships derived previously that relate the position of the fermi level
to n and p are not valid any more!
 A new variable is defined Quasi Fermi level which relates n and p under steady
state conditions when excess carriers are present
no  ni e(EF  Ei ) kT
po  ni e(Ei  EF ) kT
n  ni e
p  ni e
(Fn  Ei ) kT
(Ei  Fp ) kT
 Under equilibrium conditions
 Under non- equilibrium
conditions
Quasi Fermi Levels
Example ………….ND=1014 cm-3 & ni=1.15x1010
g=1013 EHP cm-3μs-1 & τn=τp=2μs
no  10
14
po 
ni2
no
 2.25  106
g·τn=Δn … why ??? … 2x1013 cm-3
n  no  n  1.2  1014  ni e(Fn Ei ) kT
EC
EFFn
Ei
Fp
EV
Fn  Ei  0.233eV
p  ni e
(Ei  Fp ) kT
 2  1013
Ei  Fp  0.187eV
 The number of electrons is not affected significantly therefore Fn EF
 However the number of holes increases by several orders of magnitude
therefore the position of FP shifts by a significant amount.
Generation - Recombination
CB
Ec
Er
Ev
VB
Recombination
center
Er
Er
Phonons
Recombination
CB
Ec
Ev
VB
Et
Et
Trapping
center
Trapping
Et