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EE143 S06
Lecture 3
Electrical Resistance
V
I
+
_
W
t
Material with resistivity ρ
L
Resistance
V
L
R≡ =ρ
I
Wt
(Unit: ohms)
where ρ is the electrical resistivity
Professor N Cheung, U.C. Berkeley
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EE143 S06
Adding parts/billion
to parts/thousand
of “dopants” to pure
Si can change
resistivity by
8 orders of
magnitude !
Resistivity Range of Materials
Lecture 3
Si with
dopants
SiO2, Si3N4
1 Ω-m = 100 Ω-cm
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
The Si Atom
The Si Crystal
“diamond” structure
High-performance semiconductor devices require defect-free crystals
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Carrier Concentrations of Intrinsic (undoped) Si
electron
-
Bottom of
conduction band
Energy gap
=1.12 eV
hole
+
Top of
valence band
n (electron conc)
= p (hole conc)
= ni
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Dopants in Si
By substituting a Si atom with a special impurity atom (Column V
or Column III element), a conduction electron or hole is created.
Donors: P, As, Sb
Professor N Cheung, U.C. Berkeley
Acceptors: B, Al, Ga, In
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EE143 S06
Lecture 3
Semiconductor with both acceptors and donors
has 4 kinds of charge carriers
Hole
Electron
Ionized
Donor
Ionized
Acceptor
Professor N Cheung, U.C. Berkeley
Mobile Charge Carriers
they contribute to current flow
with electric field is applied.
Immobile Charges
they DO NOT
contribute to current flow
with electric field is applied.
However, they affect the
local electric field
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EE143 S06
Charge Neutrality Condition
Lecture 3
Valid for homogeneously doped
semiconductor at thermal equilibrium
Even NA is not equal to ND,
microscopic volume surrounding
any position x has zero net charge
Si atom (neutral)
Ionized
Donor
Ionized
Acceptor
Hole
Electron
Electrons and holes created
by Si atoms with conc ni
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Electron and Hole Concentrations
for homogeneous semiconductor at thermal equilibrium
n: electron concentration (cm-3)
p : hole concentration (cm-3)
ND: donor concentration (cm-3)
NA: acceptor concentration (cm-3)
Assume completely
ionized to form ND+
and NA-
1) Charge neutrality condition:
ND + p = NA + n
2) Law of Mass Action
n• p = ni2
:
Note: Carrier concentrations depend on
NET dopant concentration (ND - NA) !
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
n-type Semiconductor
If ND >> NA (such that ND – NA ≥ 10 ni):
=
+
ND /cm3
n-type
NA /cm3
2
n ≅ ND − N A
and
ni
p≅
ND − N A
Note n >> p
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
p-type Semiconductor
If NA >> ND (such that NA – ND ≥ 10 ni):
=
+
ND /cm3
NA /cm3
p-type
2
p ≅ N A − ND
and
ni
n≅
N A − ND
Note p >> n
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Carrier Drift
• When an electric field is applied to a semiconductor, mobile
carriers will be accelerated by the electrostatic force. This
force superimposes on the random thermal motion of
carriers:
2
3
1
electron
4
3
2
1
electron
4
5
5
E =0
E
E.g. Electrons drift in the direction opposite to the E-field
Æ Current flows
Average drift velocity = |
v|=µE
Carrier mobility
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Carrier Mobility
• Mobile carriers are always in random thermal
motion. If no electric field is applied, the
average current in any direction is zero.
• Mobility is reduced by
1) collisions with the vibrating atoms
“phonon scattering”
Si
-
2) deflection by ionized impurity atoms “Coulombic
scattering”
BAs+
Professor N Cheung, U.C. Berkeley
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EE143 S06
Carrier Mobility µ
Lecture 3
Mobile charge-carrier drift velocity v is proportional to applied E-field:
|v|=µE
µn
Mobility depends
on (ND + NA) !
(Unit: cm2/V•s)
µp
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Electrical Conductivity σ
When an electric field is applied, current flows
due to drift of mobile electrons and holes:
electron current J = ( − q ) nv = qnµ E
n
n
n
density:
hole current
density:
total current
density:
conductivity
Professor N Cheung, U.C. Berkeley
J p = (+ q ) pv p = qpµ p E
J = J n + J p = (qnµ n + qpµ p ) E
J = σE
σ ≡ qnµ n + qpµ p
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EE143 S06
Electrical Resistivity ρ
Lecture 3
1
1
ρ≡ =
σ qnµ n + qpµ p
1
ρ≅
qnµ n
for n-type
1
ρ≅
qpµ p
for p-type
(Unit: ohm-cm)
Professor N Cheung, U.C. Berkeley
Note: This plot does not
apply for compensated
material!
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EE143 S06
Si
Example Calculation 1
Lecture 3
What are n and p values?
What is its electrical resistivity ?
1016
Boron/cm3
Answer:
(NA >> ND Æ p-type)
NA = 1016/cm3 , ND = 0
Æ p ≈ 1016/cm3 and n ≈ 104/cm3
1
1
≅
ρ=
qnµ n + qpµ p qpµ p
[
= (1.6 × 10
−19
16
)(10 )(450)
]
−1
= 1.4 Ω − cm
From µp vs. ( NA + ND ) plot
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Example Calculation 2: Dopant Compensation
Si
1016
Boron/cm3
+
1017
Arsenic/cm3
What are n and p values?
What is its electrical resistivity ?
Answer:
NA = 1016/cm3, ND = 1017/cm3 (ND>>NA Æ n-type)
Æ n ≈ 9x1016/cm3 and p ≈ 1.1x103/cm3
1
1
≅
ρ=
qnµ n + qpµ p qnµ n
[
= (1.6 × 10
−19
)(9 × 10 )(600)
16
From µn vs. ( NA + ND ) plot
]
−1
= 0.12 Ω − cm
* The p-type sample is converted to n-type material by adding more
donors than acceptors, and is said to be “compensated”.
Professor N Cheung, U.C. Berkeley
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EE143 S06
Lecture 3
Summary of Doping Terminology
intrinsic semiconductor: undoped semiconductor
extrinsic semiconductor: doped semiconductor
donor: impurity atom that increases the electron concentration
group V elements (P, As)in Si
acceptor: impurity atom that increases the hole concentration
group III elements (B, In) in Si
n-type material: semiconductor containing more electrons than holes
p-type material: semiconductor containing more holes than electrons
majority carrier: the most abundant mobile carrier in a semiconductor
minority carrier: the least abundant mobile carrier in a semiconductor
mobile carriers: Charge carriers that contribute to current flow when
electric field is applied.
Professor N Cheung, U.C. Berkeley
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Sheet Resistance RS
EE143 S06
I
V
+
_
W
Lecture 3
L
L
= Rs
R=ρ
W
Wt
t
Material with resistivity ρ
L
Rs is the resistance when W = L
ρ
Rs ≡
t
(unit of RS in ohms/square)
if ρ is independent of depth x
• Rs value for a given conductive layer (e.g. doped
Si, metals) in IC or MEMS technology is used
– for design and layout of resistors
– for estimating values of parasitic resistance in a device
or circuit
Professor N Cheung, U.C. Berkeley
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EE143 S06
ρ1, dx
ρ2, dx
ρ3, dx
….
ρn, dx
Lecture 3
RS when ρ(x) is function of depth x
V
I
_
+
W
t
depth x
L
1
dx dx dx
dx
=
+
+
+ .... +
= ( σ1 + σ 2 + ..σ n )dx
R S ρ1 ρ 2 ρ 3
ρn
For a continuous σ(x) function: RS =
1
t
∫ σ ( x )dx
0
=
1
t
∫ [qµ ( x )n( x ) + qµ
n
Professor N Cheung, U.C. Berkeley
0
p
( x ) p( x )]dx
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EE143 S06
Electrical Resistance of Layout Patterns
(Unit of RS: ohms/square)
Metal contact
L=1µm
R = Rs
R ≅ 2.6Rs
1m
R = 3Rs
R = Rs/2
Professor N Cheung, U.C. Berkeley
Top View
1m
W = 1µm
R = Rs
Lecture 3
R = 2Rs
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EE143 S06
Lecture 3
How to measure RS ?
(Typically, s ≈ 1 mm >>t)
• The Four-Point Probe is used to measure Rs
– 4 probes are arranged in-line with equal spacing s
– 2 outer probes used to flow current I through the
sample
– 2 inner probes are used to sense the resultant voltage
drop V with a voltmeter
4.532V
For a thin layer (t ≤ s/2), Rs =
I
If ρ is known, then Rs
measurement can be
used to determine
thickness t
For derivation of expression, see EE143 Lab Manual
http://www-inst.eecs.berkeley.edu/~ee143/fa05/lab/four_point_probe.pdf
Professor N Cheung, U.C. Berkeley
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EE143 S06
Electron mobility vs. T
Professor N Cheung, U.C. Berkeley
For reference only
Lecture 3
Hole mobility vs. T
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