microelectronics devices & circuits
TRANSCRIPT
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Administrativia
Make-up Lecture tomorrow Fr at 3:30pm (streamed)
Another Make-up Lecture Monday at 4pm (streamed)
NO LECTURE ON TUESDAY
Labs start next TU MAKE SURE TO ATTEND
4
Some other reading material
Sedra and Smith, Microelectronic Circuits, Fifth Edition,Oxford University Press
Donald Neamen, Microelectronics Circuit Analysis and
Design, Third Edition, McGraw Hill
R. F. Pierret, Semiconductor Device Fundamentals,Addison Wesley, 1996. (130 Text Book)
R. S. Muller and T. I. Kamins with Mansun Chan, Device
Electronics for Integrated Circuits, 3rd Edition; Wileyand Sons, Publisher.
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Resistivity
Bulk silicon: uniform doping concentration, away from surfaces
n-type example: in equilibrium, no = Nd
When we apply an electric field, n =Nd
ENqnEqJ dnnn ==
Resistivity
Conductivity )(, adneffdnn NNqNq ==
effdnn
nNq ,
11
== cm
6
Ohms Law
( ) R
V
VL
Wt
L
V
WtEWttWJJAI =
=====
W
L
tW
L
tR
==
1
effdnn
nNq ,
11
==with
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Sheet Resistance (Rs)
IC resistors have a specified thickness not underthe control of the circuitdesigner
Eliminate tby absorbing it into a new parameter: the
sheet resistance (Rs)
=
==
W
LR
W
L
tWt
LR sq
Number of Squares
8
Using Sheet Resistance (Rs
)
Ion-implanted (or diffused) IC resistor
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Idealizations
Why does current density Jnturn?
What is the thickness of the resistor?
What is the effect of the contact regions?
10
Diffusion
Diffusion occurs when there exists a concentrationgradient
In the figure below, imagine that we fill the left chamberwith 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?
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Diffusion (cont)
The net motion of gas molecules to the right chamberwas 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 therewould be a net current flow
The diffusion current flows from high concentration to
low concentration:
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Diffusion Equations
Assume that the mean free path is
Find flux of carriers crossing x=0 plane
)(n)0(n
)( n
0
thvn )(2
1 th
vn )(2
1
( ))()(2
1 nnvF th =
+
=
dx
dnn
dx
dnnvF th )0()0(
2
1
dxdnvF th=
dx
dnqvqFJ th==
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Einstein Relation
The thermal velocity is given by kT
kTvm thn 212*
21 =
cthv =Mean Free Time
dx
dnqD
dx
dn
q
kTq
dx
dnqvJ nnth =
==
th
n
n Vq
kTD=
=
**
2
n
c
n
ccthth
m
q
q
kT
mkTvv
===
Mobility
Diffusion Coefficient
Einstein Relation
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Total Current
When both drift and diffusion are present, the totalcurrent is given by the sum:
dx
dnqDnEqJJJ nndiffdrift +=+=
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oxidation
opticalmask
processstep
photoresist coatingphotoresistremoval (ashing)
spin, rinse, dryacid etch
photoresist
stepper exposure
development
Typical operations in a single
photolithographic cycle (from [Fullman]).
IC Fabrication: Photo-Lithographic Process
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IC Fabrication: Si Substrate
Pure Si crystal is starting material (wafer)
The Si wafer is extremely pure (~1 part in a billion
impurities)
Why so pure?
Si density is about 5 1022 atoms/cm3
Desire intentional doping from 1014 1018
Want unintentional dopants to be about 1-2 orders of magnitude
less dense ~ 1012
Si wafers are polished to about 700 m thick (mirrorfinish)
The Si forms the substrate for the IC
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IC Fabrication: Oxide
Si has a native oxide: SiO2
SiO2 (glass) is extremely stable and very convenient for
fabrication
Its an insulator
SiO2 windows are etched using photolithography
These openings allow ion implantation into selectedregions
SiO2 can block ion implantation in other areas
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IC Fabrication: Patterning of SiO2
Si-substrate
Si-substrate Si-substrate
(a) Silicon base material
(b) After oxidation and depositionof negative photoresist
(c) Stepper exposure
Photoresist
SiO2
UV-light
Patternedoptical mask
Exposed resist
SiO2
Si-substrate
Si-substrate
Si-substrate
SiO2
SiO2
(d) After development and etching of resist,chemical or plasma etch of SiO
2
(e) After etching
(f) Final result after removal of resist
Hardened resist
Hardened resist
Chemical or plasmaetch
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Diffusion Resistor
Using ion implantation/diffusion, the thickness anddopant concentration of resistor is set by process
E.g. 100/ (unsilicided), 10/ (silicided)
Shape of the resistor is set by design (layout) Metal contacts are connected to ends of the resistor
Resistor is capacitively isolation from substrate
Reverse-biased PN Junction!
P-type Si Substrate
N-type Diffusion RegionOxide
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Using Sheet Resistance (Rs
)
Ion-implanted (or diffused) IC resistor
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Poly Film Resistor
To lower the capacitive parasitics, we should build theresistor further away from substrate
We can deposit a thin film of poly Si (heavily doped)
material on top of the oxide E.g. 10-100/ (unsilicided), 1/ (silicided) Bad absolute tolerance, very good relative tolerance
Polysilicon Film (N+ or P+ type) Oxide
P-type Si Substrate
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CMOS Process at a Glance
Define active areasEtch and fill trenches
Implant well regions
Deposit and patternpolysilicon layer
Implant source and drainregions and substrate contacts
Create contact and via windowsDeposit and pattern metal layers
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Electrostatics: a Tool for Device Modeling
E( ) =
E =
( )( ) 2 = =
Gausss Law
Potential Def.
Poissons Eqn.
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One-Dimensional Electrostatics
Gausss Law
Potential Def.
Poissons Eqn.
)()(2
2 x
dx
xd=
==
dx
dEE
dx
dE
=
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Electrostatics Review (1)
Electric field go from positive charge to negative charge(by convention)
In words, if the electric field changes magnitude, therehas to be charge involved!
Result: In a charge-free region, the electric field mustbe constant!
+++++++++++++++++++++
= E
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Electrostatics Review (2)
Gauss Law equivalently says that if there is a netelectric field leaving a region, there has to be positive
charge in that region:
+++++++++++++++++++++
Electric Fields are Leaving This Box!
= QdSE
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Electrostatics in 1D
Everything simplifies in 1-D
Consider a uniform charge distribution
==
dx
dEE dxdE
=
')'(
)()(
0
0 dxx
xExE
x
x
+=
)(x
x
xdxx
xE
x
0
0
')'(
)( ==
Zero fieldboundary
condition
1x
0
1x
)(xE
1
0
x
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Electrostatic Potential
The electric field (force) is related to the potential(energy):
Negative sign says that field lines go from highpotential points to lower potential points (negativeslope)
Note: An electron should float to a high potentialpoint:
dx
dE
=
dx
deqEFe
==
1
2
dx
deFe
=
e
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More Potential
Integrating this basic relation, we have that the potentialis the integral of the field:
In 1D, this is a simple integral:
Going the other way, we have Poissons equation in 1D:
= C ldExxr
)()( 0
)(x
)( 0xE
ldr
=x
xdxxExx
0
')'()()( 0
)()(2
2 x
dx
xd=
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Boundary Conditions
Potential must be a continuous function. If not, thefields (forces) would be infinite
Electric fields need not be continuous. We have already
seen that the electric fields diverge on charges. In fact,across an interface we have:
)( 11 E
)( 22 E
=+= insideQSESEdSE 2211 x
00
xinside
Q
02211 =+ SESE
1
2
2
1
=
E
ES
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IC MIM Capacitor
By forming a thin oxide and metal (or polysilicon) plates, acapacitor is formed
Contacts are made to top and bottom plate
Parasitic capacitance exists between bottom plate and substrate
Top PlateBottom Plate Bottom Plate
Contacts
CVQ =
Thin Oxide
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Review of Capacitors
For an ideal metal, all charge must be at surface
Gauss law: Surface integral of electric field overclosed surface equals charge inside volume
+++++++++++++++++++++
+
Vs
= Q
dSE
= Q
dSE
sox VtEdlE == 0ox
s
t
VE =0
== Q
AEdSE 0
QA
t
V
ox
s =
sCVQ =
oxt
AC
=
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Capacitor Q-V Relation
Totalcharge is linearly related to voltage Charge density is a delta function at surface (for
perfect metals)
sCVQ =
sV
Q
y
)(yQ
y+++++++++++++++++++++
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A Non-Linear Capacitor
Well soon meet capacitors that have a non-linear Q-Vrelationship
If plates are not ideal metal, the charge density can penetrate intosurface
)( sVfQ =
sV
Q
y
)(yQ
y+++++++++++++++++++++
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Whats the Capacitance?
For a non-linear capacitor, we have
We cant identify a capacitance
Imagine we apply a small signal on top of a bias
voltage:
The incremental charge is therefore:
ss CVVfQ = )(
s
VV
sss vdV
VdfVfvVfQ
s=
++=)(
)()(
Constant charge
s
VV
s vdV
VdfVfqQQ
s=
++=)(
)(0
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Small Signal Capacitance
Break the equation for total charge into two terms:
s
VV
s vdV
VdfVfqQQ
s=
++=)(
)(0
Constant
Charge
IncrementalCharge
ss
VVvCvdV
Vdfq
s
===
)(
sVVdV
VdfC
=
)(
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Example of Non-Linear Capacitor
Next lecture well see that for a PN junction, the chargeis a function of the reverse bias:
Small signal capacitance:
b
paj
VxqNVQ
= 1)(
ConstantsCharge At N Side of Junction
Voltage Across NPJunction
b
j
b
b
paj
jV
C
V
xqN
dV
dQVC
=
==
11
1
2)(
0