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EECS 105 Fall 2003, Lecture 6
Lecture 6:
Integrated Circuit Resistors
Prof. Niknejad
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Lecture Outline
Department of EECS
Semiconductors
Si Diamond Structure
Bond Model
Intrinsic Carrier Concentration
Doping by Ion Implantation
Drift
Velocity Saturation
IC Process Flow
Resistor Layout
Diffusion
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Resistivity for a Few Materials
Pure copper, 273K
Pure copper, 373 K
Pure germanium, 273 K
Pure germanium, 500 K
Pure water, 291 K
Seawater
1.56×10-6 ohm-cm
2.24×10-6 ohm-cm
200 ohm-cm
.12 ohm-cm
2.5×107 ohm-cm
25 ohm-cm
What gives rise to this enormous range?
Why are some materials semi-conductive?
Why the strong temp dependence?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Electronic Properties of Silicon
Silicon is in Group IV
–
–
–
Atom electronic structure: 1s22s22p63s23p2
Crystal electronic structure: 1s22s22p63(sp)4
Diamond lattice, with 0.235 nm bond length
Very poor conductor at room temperature:
why?
(1s)2
(2s)2
(2p)6
(3sp)4
Hybridized State
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Periodic Table of Elements
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
The Diamond Structure
3sp tetrahedral bond
2.35 A
5.43 A
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
States of an Atom
Energy
..
.
E3
E2
Allowed
Energy
Levels
Forbidden Band Gap
E1
Atomic Spacing
Lattice Constant
Quantum Mechanics: The allowed energy levels
for an atom are discrete (2 electrons can occupy a
state since with opposite spin)
When atoms are brought into close contact, these
energy levels split
If there are a large number of atoms, the discrete
energy levels form a “continuous” band
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Energy Band Diagram
The gap between the conduction and valence band
determines the conductive properties of the material
Metal
Conduction Band
–
negligible band gap or overlap
Valence Band
Conduction Band
Insulator
–
large band gap, ~ 8 eV
band gap
Valence Band
Semiconductor
–
e-
medium sized gap, ~ 1 eV
Electrons can gain energy from lattice
(phonon) or photon to become “free”
Department of EECS
eUniversity of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Model for Good Conductor
The atoms are all ionized and a “sea” of electrons can
wander about crystal:
The electrons are the “glue” that holds the solid together
Since they are “free”, they respond to applied fields and
give rise to conductions
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
On time scale of electrons, lattice looks stationary…
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Bond Model for Silicon (T=0K)
Silicon Ion (+4 q)
Four Valence Electrons
Contributed by each ion (-4 q)
Department of EECS
2 electrons in each bond
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Bond Model for Silicon (T>0K)
Some bond are broken: free electron
Leave behind a positive ion or trap (a hole)
+
-
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Holes?
-+
Notice that the vacancy (hole) left behind can be filled by a
neighboring electron
It looks like there is a positive charge traveling around!
Treat holes as legitimate particles.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Yes, Holes!
The hole represents the void after a bond is broken
Since it is energetically favorable for nearby
electrons to fill this void, the hole is quickly filled
But this leaves a new void since it is more likely
that a valence band electron fills the void (much
larger density that conduction band electrons)
The net motion of many electrons in the valence
band can be equivalently represented as the motion
of a hole
J vb (q)vi
vb
Filled Band
J vb
(q)v
i
EmptyStates
(q)v
i
EmptyStates
Department of EECS
(q)vi
qv
i
EmptyStates
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
More About Holes
When a conduction band electron encounters a
hole, the process is called recombination
The electron and hole annihilate one another thus
depleting the supply of carriers
In thermal equilibrium, a generation process
counterbalances to produce a steady stream of
carriers
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Thermal Equilibrium (Pure Si)
Balance between generation and recombination
determines no = po
Strong function of temperature: T = 300 oK
G Gth (T ) Gopt
R k (n p)
GR
k (n p) Gth (T )
n p Gth (T ) / k ni (T )
2
ni (T ) 1010 cm 3 at 300 K
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Doping with Group V Elements
P, As (group 5): extra bonding electron … lost
to crystal at room temperature
Immobile
Charge
Left
Behind
Department of EECS
+
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Donor Accounting
Each ionized donor will contribute an extra “free”
electron
The material is charge neutral, so the total charge
concentration must sum to zero:
qn0 qp0 qN d 0
Free
Electrons
Free
Holes
Ions
(Immobile)
By Mass-Action Law: n p ni 2 (T )
ni2
qn0 q qN d 0
n0
qn02 qni2 qN d n0 0
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Donor Accounting (cont)
Solve quadratic: n02 N d n0 ni2 0
N d N d2 4ni2
n0
2
Only positive root is physically valid:
N d N d2 4ni2
n0
2
For most practical situations: N d ni
2
Nd Nd
n0
Department of EECS
ni
1 4
Nd Nd
N
Nd
2
2
2
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Doping with Group III Elements
Boron: 3 bonding electrons one bond is
unsaturated
Only free hole … negative ion is immobile!
-
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Mass Action Law
Balance between generation and recombination:
po no ni
• N-type case:
(T 300 K, ni 1010 cm 3 )
2
d
n0 N N d
• P-type case: p0 N a N a
Department of EECS
ni2
n0
Nd
ni2
p0
Na
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Compensation
Dope with both donors and acceptors:
–
Create free electron and hole!
+-
- +
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Compensation (cont.)
More donors than acceptors: Nd > Na
2
no N d N a ni
po
ni
Nd Na
• More acceptors than donors: Na > Nd
po N a N d ni
Department of EECS
no
n i2
Na Nd
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Thermal Equilibrium
Rapid, random motion of holes and electrons at
“thermal velocity” vth = 107 cm/s with collisions
every c = 10-13 s.
* 2
1
1
2
mn vth 2 kT
Apply an electric field E and charge carriers
accelerate … for c seconds
vth c
zero E field
107 cm / s 10 13 s 10 6 cm
vth
positive E
x
Department of EECS
a c
(hole case)
vth
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Drift Velocity and Mobility
For holes:
Fe
qE
q c
c
c
E
vdr a c
m
m
m
p
p
p
vdr p E
For electrons:
Fe
qE
q c
c
c
E
vdr a c
m
m
m
p
p
p
vdr n E
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Mobility vs. Doping in Silicon at 300 oK
“default” values:
Department of EECS
n 1000
p 400
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Speed Limit: Velocity Saturation
c 3 1010 m / s
Thermal Velocity
V
cm
V
4 V
10
10
1
4
cm
cm 10 m
m
4
The field strength to cause velocity saturation may seem very large
but it’s only a few volts in a modern transistor!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Drift Current Density (Holes)
Hole case: drift velocity is in same direction as E
hole drift
current density
Jpdr
vdp
E
x
The hole drift current density is:
Jp dr = q p p E
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Drift Current Density (Electrons)
Electron case: drift velocity is in opposite direction as E
electron drift
current density
Jndr
vdn
E
J ndr (q)nn E qnn E
x
The electron drift current density is:
Jndr = (-q) n vdn
units: Ccm-2 s-1 = Acm-2
J J pdr J ndr qp p qnn E
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Resistivity
Bulk silicon: uniform doping concentration, away from surfaces
n-type example: in equilibrium, no = Nd
When we apply an electric field,
n = Nd
J n q n nE q n N d E
Conductivity n q n N d ,eff q n ( N d N a )
Resistivity
Department of EECS
1
1
n
n qn N d ,eff
cm
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Ohm’s Law
Current I in terms of Jn
Voltage V in terms of electric field
V IR
I JA JtW
I JA JtW t W E
–
Result for R
L 1
R
W t
Department of EECS
R
L
W t
E V / L
tW
I JA JtW
V
L
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Sheet Resistance (Rs)
IC resistors have a specified thickness – not
under the control of the circuit designer
Eliminate t by absorbing it into a new
parameter: the sheet resistance (Rs)
L L
L
R
Rsq
Wt
t W
W
“Number of Squares”
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Using Sheet Resistance (Rs)
Ion-implanted (or “diffused”) IC resistor
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Idealizations
Why does current density Jn “turn”?
What is the thickness of the resistor?
What is the effect of the contact regions?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Diffusion
Diffusion occurs when there exists a concentration
gradient
In the figure below, imagine that we fill the left
chamber with a gas at temperate T
If we suddenly remove the divider, what happens?
The gas will fill the entire volume of the new
chamber. How does this occur?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Diffusion (cont)
The net motion of gas molecules to the right
chamber was due to the concentration gradient
If each particle moves on average left or right then
eventually half will be in the right chamber
If the molecules were charged (or electrons), then
there would be a net current flow
The diffusion current flows from high
concentration to low concentration:
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Diffusion Equations
Assume that the mean free path is λ
Find flux of carriers crossing x=0 plane
n ( 0)
n ( )
n( )
1
dn
dn
F vth n(0) n(0)
2
dx
dx
1
dn
n( )vth
2
F vth
dx
1
n( )vth
2
Department of EECS
1
F vth n( ) n( )
2
0
J qF qvth
dn
dx
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Einstein Relation
The thermal velocity is given by kT
1
2
mn*vth2 12 kT
vth c
Mean Free Time
kT q c
vth v kT *
mn
q mn*
2
th c
c
kT dn
dn
J qvth
q
n
dx
q
dx
kT
Dn n
q
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 6
Prof. A. Niknejad
Total Current
When both drift and diffusion are present, the total
current is given by the sum:
J J drift J diff
Department of EECS
dn
q n nE qDn
dx
University of California, Berkeley