Lecture 1: Economic importance; device trends

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Electrical Engineering 2


Microelectronics 2

Dr. Peter Ewen



(Room G08, SMC; email
-

pjse)

ELECTRICAL ENGINEERING 2


Microelectronics 2


Dr. P.J.S. Ewen

LECTURES: Mondays 12.10
-
13.00


Swann 7.20



Fridays 10.00
-
10.50
JCMB 5327



TUTORIALS:
Mondays
11.10
-
12.00

Eng. CR 4



Tuesdays
11.10
-
12.00

Eng. CR 4


N.B. Tutorials run in weeks
3, 5
, 7, 9, 11

Microelectronics 2 Lecture Topics




LECTURES 1
-
6 Semiconductor theory


Basic semiconductor properties:


e.g. energy bands, carrier


concentrations, mobility, conductivity



LECTURES 7
-
12


Fabrication & p
-
n junctions




LECTURES 13
-
17


Field Effect Transistors



LECTURES 18
-
20


Bipolar Junction Transistors



Structure and
fabrication



Principles of
operation



Properties



Applications

The
handout

contains additional
explanatory notes as well as the
Powerpoint

slides


this additional
information generally appears in a
pane headed
Notes


to
the right of
the corresponding slide.



Recommended text



T.F. Bogart “Electronic

Devices and Circuits”



Formula sheet



Issued with exam paper

Resources also

include
lecture

examples
,
tutorial

solutions
, etc.



Powerpoint slides



Slides available on web


follow link on
“Microelectronics 2” LEARN page:



Microelectronics 2

LECTURE 1




Importance of semiconductor devices




Device miniaturization




Revision of basic concepts:



Resistance and resistivity


Conductance and conductivity


Electric field


Energy of a charged particle


Electronics is America’s Largest Industry

Jobs

Industry shipments

Electronics

Motor

vehicles

and parts

Aerospace

Steel mill

products

3.0


2.5


2.0


1.5


1.0


0.5


0

300


250


200


150


100


50


0

Jobs (millions)

Shipments ($billions)

Fig. 1a

Fig. 1b

How the different
electronics industries
contribute to the
overall totals

The semiconductor
industry underpins
all the other
electronics industries

1980 1990 2000 2010

Fig. 2

Gross world product (GWP) and sales
volumes of various manufacturing industries from
1980 to 2000 and projected to 2010.

Year

Global sales ($ Billions)

10
5




10
4




10
3




10
2




10
1

GWP

Steel

Electronics

Automobiles

Semi
-

conductors

Source: “Fundamentals of semiconductor

Fabrication” by G.S. May and S.M. Sze

The UK Electronics Industry



Annual sales


£23 billion




Employs 250,000 people




8000+ electronics companies




5
th

largest in the world in terms of production




Accounts for 6% of UK manufacturing output




& 2.5% of UK’s GDP




40% of Europe’s semiconductor design revenue


comes from UK




Electronics overtook whisky as
Scotland’s No.1 export in the early 1980’s

Sources: IET Holyrood Briefings Dec. 2008


Innovation UK 2008


National Microelectronics Institute


sector information

The UK Semiconductor Industry



Employs 4,700




£4.5 billion revenue




13 semiconductor manufacturing plants




19.6% of Europe’s semiconductor market share




Semiconductor fabrication companies in


Scotland include:















Sources: ICAF Industry study 2007


Hansard 2008

The Scottish Microelectronics Centre (SMC)



Part of the School of Engineering




Located at west end of the KB site




Contains >£10M of equipment dedicated to making


IC’s





The Scottish Microelectronics Centre

Lithography area

Access corridor to the 8

cleanroom bays


Wet etching bench

Jack Kilby wins 2000 Nobel Prize for Physics



Jack Kilby received his B.Sc. in
Electrical Engineering from the
University of Illinois in 1947.





Won the 2000 Nobel Prize in
Physics for his part in the
invention and development of
the integrated circuit.





Kilby invented the IC in 1958
while at Texas Instruments.





Jack Kilby wins 2000 Nobel Prize for Physics



The Nobel citation says:


“…
Kilby’s invention enabled the
microelectronics field to grow to
become the basis of all modern
technology.”



Spot the difference


Larry, Curly and Mo


the Three Stooges


a
well
-
known comedy trio
of the 1950’s.

Walter Brattain, William
Shockley and John
Bardeen


inventors of the
transistor, for which they
were awarded the Nobel
Prize for Physics in 1956.

On 16
th

December 1947
the first transistor was
demonstrated


it was a
germanium point
-
contact
device.

Ge (base)

Au (emitter)

Au (collector)

1 cm

Fig. 3

1
-

30,000

2
-

300,000

3
-

3 million

4
-

30 million

5
-

300 million

6
-

3 billion

How many transistors (approximately) are
there in the latest Intel microprocessor chip?

Latest Intel processor “Ivy Bridge”


May 2011

How big, approximately, is 22nm?


(1nm = 10
-
9
m)


1
-

Size of a red blood cell

2
-

Width of a human hair

3
-

Size of a staphylococcus bacterium

4
-

Width of a DNA molecule

5
-

90 silicon atoms side by side

6
-

Size of a cold virus

22 nm

90,000nm

500nm

5nm


20 atoms

20nm

6,000nm

600nm

2nm

staphylococcus

bacterium

DNA molecule

cold virus

red blood cell

hair

atoms in a silicon crystal



source

gate

drain

n
-
type Si

n
-
type Si

p
-
type Si wafer

d

L

T

MOS
Transistor

Fig. 4


L


T


d

Gordon E. Moore,

co
-
founder of Intel.

“Moore’s Law”


The number of transistors per chip
(die) doubles approximately every 18 months.

Fig. 5

World’s Smallest

Practical Transistor

source

drain

gate

Fig. 6

REASONS FOR CONTINUING
SIZE REDUCTION

1.

Higher productivity


2.

Faster operation


3.

Lower power consumption


4.

Higher yield


5.

New products

(a) 52 possible IC’s

but only 40 are

defect
-
free


Yield = 40/52


=
77%

Defects

(b) 240 possible IC’s

but only 224 are

defect
-
free


Yield = 224/240


=
93%

By making the IC’s

smaller, more of

them manage to

avoid the

defect sites.

Fig. 7

Resistance and Resistivity

a

b

r

l

A

Cross
-
sectional


area A


A = ab



A =

r
2
R =
ρ
l
/A




A = ?


ρ

is
Resistivity



depends only on
the material

R is
Resistance



depends
on the material and
geometry of the object

Fig. 8: The resistance

of an object with uniform
cross
-
sectional area A.

l

l




CONDUCTANCE G = 1/R




CONDUCTIVITY
σ

= 1/
ρ






Quantity

Unit

Symbol

Resistance, R

ohm

Ω

Resistivity,
ρ

ohm
-
metre

Ω
m

Conductance, G

siemens

S

Conductivity,
σ


siemens/metre

Sm
-
1

Lecture Examples


1. Resistance


The figure shows a composite block consisting of a
copper cylinder embedded in a stainless steel bar of
square cross
-
section. Calculate the resistance
between the opposite ends of the block if the
conductivities are:



σ
Cu
= 5.7x10
7

Sm
-
1



σ
s.s.

= 10
6
Sm
-
1



The copper cylinder can be represented

by a resistor R
1
...

1. Resistance

R
1

and R
2

are in
parallel, so the total
resistance between the
ends will be given by:

R
1

R
2

σ
Cu
= 5.7x10
7

Sm
-
1

σ
s.s.

= 10
6
Sm
-
1


σ

= 1/
ρ

steel surrounding it

by a resistor R
2
:

and the stainless

+




-

E



F

F

+

F

= q
E

E

is
Electric Field




F

is
Force

on
Charge

q


Direction of the Electric Field is defined by the
direction of the force on a
positive

charge


i.e. if
q is positive,
F

and
E

are in the same direction.


for V = kx + C

V i
Potential

Units for E: Vm
-
1

Fig. 9

ELECTRIC FIELD

Δ
V

E


Δ
x

-
ve

+ve

Δ
x

Δ
V

E


+ve

-
ve

Δ
V

Δ
x

|E| =

Fig. 10

Fig. 11

2. Electric field


A Cu wire (
ρ
Cu
= 1.7x10
-
8
Ω
m
) is joined to
an Fe wire (
ρ
Fe
= 1x10
-
7
Ω
m
). If both wires
are 10 m long and have the same radius,
calculate the field in each wire if the
voltage across the ends is 100 V.

2. Electric Field

The two wires can be regarded as two resistors in series, R
Cu

and R
FE ,

with
voltage differences ΔV
Cu

and ΔV
Fe

across them.


100 V


R
Cu


R
Fe

ΔV
Cu

ΔV
Fe

Cu

Fe


The two resistors in series, R
Cu

and R
FE
, form a voltage divider, so:


ρ
Cu
= 1.7x10
-
8
Ω
m

ρ
Fe
= 1x10
-
7
Ω
m

Δ
V

Δ
x

|E| =

10m

10m

R =
ρ
l
/A

2. Electric Field

100 V


R
Cu


R
Fe

ΔV
Cu

ΔV
Fe

Cu

Fe

ρ
Cu
= 1.7x10
-
8
Ω
m

ρ
Fe
= 1x10
-
7
Ω
m

14.5V

Δ
V

Δ
x

|E| =

Energy acquired by a charge, q, moving
through a potential difference,
Δ
V

V
1


Δ
V = V
2



V
1


V
2

q

Δ
x


Energy (Work) = Force
×

Distance



= qE
×

Δ
x




= q
×

Δ
x




= q
Δ
V

Δ
V

Δ
x

If q and
Δ
V are both positive, the change
in the energy of the particle will be
positive, i.e. its energy will
increase
.

Fig. 12

SUMMARY




Importance of semiconductor devices






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and the number of people employed.





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† ††



Challenge of semiconductor devices






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㔮 乥N produ捴s




Revision of basic concepts:





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ρ
l/A




䍯ndu捴慮捥 慮d 捯ndu捴楶楴礠
σ

= 1/
ρ




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-

dV⽤x




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Δ
V