Recombination and Generation
Carrier generation and recombination are processes in which
electrons and holes are created and eliminated
RecombinationGeneration Processes Include
Bandtoband recombination
generally results in light emission
Bandtoba
MOS Capacitors
MOSFET structure
G (gate)
S (source)
D (drain)
L
W
Metal (or polysilicon)
Oxide
Semiconductor
Field Effect Transistor
L
channel
W: gate width
L: gate length
WL: gate area
Cross sections of MOSFETs
PMOS
NMOS
Source
Gate
n+
Drain
Source
Drain
MOSFETs (NMOS shown)
G (gate)
D (drain)
S (source)
L
n+
L
W
Metal (or polysilicon)
Oxide
Semiconductor
Field Effect Transistor
n+
p
channel
W: gate width
L: gate length
WL: gate area
MOSFETs are symmetric
MOSFETs contain:
2 pn junctions (in TE or RB)
A MO
PN Junction not in Thermal
Equilibrium
ID
+ VD
ID
p
n
VD
Previously T.E.
PN junction Applying a voltage
+ VA
ID
PNjunction in Thermal Equilibrium
p
Builtin Potential
         
+
+
+
 +
+ 
Applied Voltage
 +
n
Vbi
Va
 Va
The convention is
Introduction to Semiconductors
Atoms Contain
e
nucleus
Electrons
eNucleus
Protons
Neutrons
Valence Electrons:
The number of
electrons in the
outermost shell
Semiconductors: an introduction
Classification of Materials in terms of electrical resistivity:
Einsteins Relationship
Relating the Diffusion Coefficient and Mobility
Thermal Equilibrium
Material at a constant temperature with no applied external forces (light,
current, voltage)
In Thermal Equilibrium
np = ni2
dE F
0
dx
There is no NET current
N
Band Bending
Setting the stage nonuniform doping
P
P
Si
P
Si
Si
P
P
Si
P
Si
Si
P
Si
Si
Si
P
P
P+
Si
Si
Si
P
P
P
P
P
P
P
P
P
P
Electron diffuses
You can have an Efield in a
material without applying a
voltage!

Jndiff
Creates Efield
E
Leading to drift
Metal Semiconductor (MS)
Junctions/Contacts
Two Kinds
p
n
Schottky Diodes
Ohmic contacts
Ohmic contacts
Metalsemiconductor (MS) junctions
Many of the properties of pn junctions can be realized by forming a
metalsemiconductor rectifying contact (Schott
PN Junction in Thermal Equilibrium
ID
+ VD
ID
p
n
VD
T.E.
PN junction fabrication with notation for
doping
p
n
Metallurgical junction
n
N D , N A
N A N A N D
p
n
N D
ND
Step junction assumes an abrupt transition
between the p and the nsides
PN junction
Carrier Properties
Charge
Hole charge: q = 1.6 x 1019 Coul. Electron charge = q
Effective Mass
Concentration
Carrier property: Effective mass
Mass, like charge, is a very basic property of electrons and holes. The mass of
an electron in a semiconduct
Three types of carrier action
Drift
Diffusion
Recombination Generation
Frequently current density is used instead of current
I
J
area
2
(units : amps / cm )
Drift is the motion of a charged particle
in response to an applied electric field.
What is an
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PN Junction in Thermal Equilibrium
ID
+ VD
ID
p
n
VD
T.E.
PN junction fabrication with notation for
doping
p
n
Metallurgical junction
n
N D , N A
N A N A N D
p
n
N D
ND
Lecture 8
1
Step junction assumes an abrupt transition
between the p and the nsides
Introduction to Semiconductors
Atoms Contain
e
nucleus
Electrons
e
Valence Electrons:
The number of
electrons in the
outermost shell
Nucleus
Protons
Neutrons
Semiconductors: an introduction
Classification of Materials in terms of electrical resistivity:
Deriving electron and hole
concentrations (n and p)
Game Plan:
Density of states function
gC(E)
Distribution of allowable
energy states for a given
material
 material specific
X
Fermi Function
f(E)
Probability that an electron
could occupy a given
energy
A few words about SPICE
Simulation Program with Integrated Circuit Emphasis
Developed in 1970s at Berkeley
Many commercial versions are available
HSPICE is a robust industry standard for MOSFETS
There are different kinds of SPICE simulations
Bias Point
BJT operation: Forward Active
C
IC Is eqVBE/KT
B
n
IB
p
I E = IB + I C IC
n+
E
Physical structure of BJT
Emitter doping is larger than
base and collector doping
E
n+
BE junction is
forward biased
Base is very narrow (not
drawn to scale)
p
B
n
CB junction
B. OLSON
ECE 330
Homework 3
Unless told otherwise, use the parameters below:
KT (T = 300K) = 26meV
K = 8.617 105 eV /K
q = 1.6 1019 coul
1. A sample of innitely long ptype silicon (with NA = 1015 ) is continuously injected with electrons at the position
B. OLSON
ECE 330
Homework 2  SOLUTIONS
1. At 300K: ND ni therefore:
n
= ND = 1013 cm3 and p =
n2i
n
=
1020
1013
= 107 cm3 .
The corresponding resistivity is:
=
=
1
1
(since p p n n)
=
q(n n + p p)
q p p
1
= 1302 cm
19
1.6 10
480 1013
R=
= 19.2cm
A
2.
B. OLSON
ECE 330
Homework 4  SOLUTIONS
1) Calculate the builtin potential Barrier Vbi for Si, and Ge pnjunctions if they have the following
dopant concentrations (T = 300K)
a)
b)
KT
Vbi
q
Si: Nd = 3x1014 cm3
Ge: Nd = 1014 cm3
Na = 1017 cm3
Na = 1
Study Questions
1. What is a diamond lattice? The unit cell arrangement of the lattice (crystal)
structure of Si and Ge or the crystal structure of Si and Ge. crystal structure which
belongs to the facecentered cubic class of crystals
2. What is a zinc b
B. OLSON
ECE 330
Quiz 5Aa
Equations that you might wish to use:
dnp
d2 np
d2 pn
np
dpn
pn
R);
R)
= Dn
= Dp
+ (G
+ (G
2
dt
dx
n
dt
dx2
p
In Thermal Equilibrium: ni = np, Generally: Ln = Dn n , Lp = Dp p
1. A sample of ptype silicon (with NA = 1014 ) is
B. OLSON
ECE 330
Homework 6
Parameters to use:
For Silicon: S = KS O = (11.8)(8.85 1014 ) F/cm
For SiO2 : OX = KOX O = (3.9)(8.85 1014 ) F/cm
1. A MOSC is maintained at T = 300K. The oxide thickness, tox, is:
tox = 100 angstroms (1010 m.) and the Si dopi
B. OLSON
ECE 330
Homework 1  SOLUTIONS
1. First we determine KT in eV:
KT (eV ) = 500K 8.617 105 eV /K = 43 103 eV.
Then we determine it in Joules:
KT (J) = KT (eV ) 1.6 1019 = 6.88 1021 J.
2.
m 2mn (E Ec)
gc (E) = n
2 h3
Number of states/cm2 =
mn 2mn E
Study Questions
1. In a nondegenerate semiconductor, indicate the two dominant scattering
mechanisms
2. What is the significance of a critical Efield (Ec)?
3. Know how to compute the drift velocity given the Efield
4. Know how to draw the cross section
B. OLSON
ECE 330
Homework 2  SOLUTIONS
1. At 300K: ND ni therefore:
n
= ND = 1013 cm3 and p =
n2i
n
=
1020
1013
= 107 cm3 .
The corresponding resistivity is:
=
=
1
1
(since p p n n)
=
q(n n + p p)
q p p
1
= 1302 cm
19
1.6 10
480 1013
R=
= 3.84cm
A
2.
B. OLSON
ECE 330
Quiz 5Ba
Equations that you might wish to use:
d2 np
d2 pn
np
dpn
pn
dnp
R);
R)
= Dn
=
D
+
(
G
+ (G
p
2
2
dt
dx
n
dt
dx
p
In Thermal Equilibrium: ni = np, Generally: Ln = Dn n , Lp = Dp p
1. A sample of innitely long ntype silicon (wit
Parameters to use for the problems:
Silicon:
Eg = 1.12eV
Si(T = 300K):
ni (T = 300K)= 1010/cm3
s(Silicon) = Ks o = (11.8)(8.85x1014 F/cm)
ox(SiO2) = Kox o = (3.9)(8.85x1014 F/cm)
3) Below is a band diagram of a Silicon based MOS Capacitor at room temper
B. OLSON
ECE 330
Homework 1
Parameters to use:
q = 1.6 1019 coulombs
K = 8.617 105 eV /K
KT (T = 300K) = 26meV
For Silicon at room temperature(T = 300K):
EG = 1.12eV
ni = 1 1010 cm3
1. Determine KT at 500K. Express your answer in eV and in Joules.
2. Deve