Integrated Circuit Devices

Professor Ali Javey Summer 2009 Semiconductor Fundamentals

Evolution of Devices

Yesterday’s Transistor (1947) Today’s Transistor (2006)

Why “Semiconductors”?

• Conductors – e.g Metals • Insulators – e.g. Sand (SiO

2

) • Semiconductors – conductivity between conductors and insulators – Generally crystalline in structure

• In recent years, non-crystalline semiconductors have become commercially very important Polycrystalline amorphous crystalline

What are semiconductors

Elements: Si, Ge, C Binary: GaAs, InSb, SiC, CdSe, etc.

Ternary+: AlGaAs, InGaAs, etc.

Electrons and Holes in Semiconductors

•

Unit cell

of silicon crystal is cubic.

•

Each Si atom has 4 nearest

neighbors.

5.43 Å

Silicon Wafers and Crystal Planes

(011) flat

z

(100)

x

(100) plane

y z

(011)

x y z

(111)

x y

 The standard notation for crystal planes is based on the cubic unit cell. Si (111) plane  Silicon wafers are usually cut along the (100) plane with a flat or notch to help orient the wafer during IC fabrication.

Bond Model of Electrons and Holes (Intrinsic Si)

Si Si Si  Silicon crystal in Si Si Si a two-dimensional representation.

Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si  When an electron breaks loose and becomes a

conduction electron

, a

hole

is also created.

Dopants in Silicon

Si Si Si Si Si Si Si As Si Si B Si Si Si Si Si Si Si N-type Si P-type Si  As (Arsenic), a Group V element, introduces conduction electrons and creates

N-type silicon,

and is called a

donor.

 B (Boron), a Group III element, introduces holes and creates

P-type silicon

, and is called an

acceptor

.

 Donors and acceptors are known as dopants.

Types of charges in semiconductors

Hole Electron

Mobile Charge Carriers

they contribute to current flow with electric field is applied.

Ionized Donor Ionized Acceptor

Immobile Charges

they DO NOT contribute to current flow with electric field is applied. However, they affect the local electric field

EE143 – Vivek Subramanian Slide 1-9

GaAs, III-V Compound Semiconductors, and Their Dopants

Ga As Ga As Ga As Ga As Ga As Ga  GaAs has the same crystal structure as Si.

 GaAs, GaP, GaN are III-V compound semiconductors, important for optoelectronics.

 Which group of elements are candidates for donors? acceptors?

From Atoms to Crystals

conduction band p s Pauli exclusion principle valence band isolated atoms lattice spacing Decreasing atomic separation  Energy states of Si atom (a) expand into energy bands of Si crystal (b).  The lower bands are filled and higher bands are empty in a semiconductor.

 The highest filled band is the  The lowest empty band is the

valence band.

conduction band

.

Energy Band Diagram

Conduction band

E c E g

Band gap

E v

Valence band 

Energy band diagram

shows the bottom edge of conduction band,

E c



E c

, and top edge of valence band,

E v

and

E v

are separated by the .

band gap energy, E

g

.

Measuring the Band Gap Energy by Light Absorption

electron

E c

photons photon energy: h

v

>

E g E g E v

hole •

E g

can be determined from the minimum energy (

h

photons that are absorbed by the semiconductor.

n ) of

E g

Bandgap energies of selected semiconductors

Material (eV) PbTe 0.31

Ge 0.67

Si 1.12

GaAs 1.42

GaP 2.25

Diamond 6.0

Semiconductors, Insulators, and Conductors

E c E c E g

= 9 eV Top of conduction band empty

E g

= 1.1 eV filled

E v E v E c

Si (Semiconductor) SiO 2 (Insulator) Conductor  Totally filled bands and totally empty bands do not allow current flow. (Just as there is no motion of liquid in a .

 Metal conduction band is half-filled.

 Semiconductors have lower

E g

's than insulators and can be doped.

Donor and Acceptor Levels in the Band Model

Conduction Band Donor Level

E c E d

Donor ionization energy Acceptor Level Acceptor ionization energy

E a E v

Valence Band

Ionization energy of selected donors and acceptors in silicon

Dopant Ionization energy

, E c –E d

or

E a –E v

(meV) Sb 39 Donors P 44 As 54 B Acceptors Al In 45 57 160 Hydrogen:

E ion

=

m

0

q 4 8

e 0 2

h

2 = 13.6 eV

Donors n-type Acceptors p-type

Dopants and Free Carriers

Dopant ionization energy ~50meV (very low).

General Effects of Doping on n and p

Charge neutrality:

N a

_

N d

+

n

+

N a

_ -

p

-

N d

+ : number of ionized acceptors /cm3 : number of ionized donors /cm3 = 0 Assuming total ionization of acceptors and donors:

n

+

N a

-

p

-

N d N a

: number of acceptors /cm3 = 0

N d

: number of donors /cm3

E c E v

Density of States

E g c

DE

E

c

g(E) E v g v g c

(

E

)  number D

E

of states in  volume D

E

eV  1 cm 3

g c

(

E

) 

m n

*

g v

(

E

) 

m

*

p

2

m

*

n

 

E

2

h

3 -

E c

 2

m

*

p

 

E v

2

h

3 -

E



Thermal Equilibrium

Thermal Equilibrium

An Analogy for Thermal Equilibrium

Sand particles Dish Vibrating Table  There is a certain probability for the electrons in the conduction band to occupy high-energy states under the agitation of thermal energy (vibrating atoms, etc.)

At

E

=

E F

,

f

(

E

)=1/2

Question

• If

f

(

E

) is the probability of a state being occupied by an electron, what is the probability of a state being occupied by a hole?

N c

is called the

effective density of states (of the conduction band) .

N v

is called the

effective density of states of the valence band.

Intrinsic Semiconductor

• Extremely pure semiconductor sample containing an insignificant amount of impurity atoms.

n = p = n i

E f

lies in the middle of the band gap

Material

E g

(eV)

n i

(1/cm 3 ) Ge

0.67

2 x 10 13

Si

1.12

1 x 10 10

GaAs

1.42

2 x 10 6

Remember: the closer E

f

moves up to E

c

, the larger n is; the closer E

f

For Si, N

c

moves down to E

v

= 2.8

´

10 19 cm -3 and N

v

, the larger p is.

= 1.04

´

10 19 cm -3 .

E c E v E f E c E v E f

Example: The Fermi Level and Carrier Concentrations

Where is E f for n =10 17 cm -3 ? Solution: E c n

-

E f



N c e

(

E c

-

E f



kT

ln 

N c

) /

kT n

  0 .

026 ln  2 .

8  10 19 / 10 17   0 .

14 6 eV 0.146 eV

E c E f E v

The np Product and the Intrinsic Carrier Concentration

Multiply

n



N c e

(

E c

-

E f

) /

kT

and

p



N v e

(

E f

-

E v

) /

kT np



N c N v e

(

E c

-

E v

) /

kT



N c N v e

-

E g

/

kT np



n i

2

n i



N c N v e

-

E g

/ 2

kT

• In an intrinsic (undoped) semiconductor,

n = p = n i .

EXAMPLE: Carrier Concentrations

Question: What is the hole concentration in an N-type semiconductor with 10 15 cm -3 of donors?

Solution: n = 10 15 cm -3 . p



n i

2

n

 10 20 cm -3 10 15 cm 3  10 5 cm 3

After increasing T by 60



C, n remains the same at 10 15 cm -3 increases by about a factor of 2300 because n i

2 

e

-

E g

/

kT .

while p Question: What is n if p = 10 17 cm -3 in a P-type silicon wafer?

Solution:

n



n i

2

p

 10 20 cm -3 10 17 cm 3  10 3 cm 3

General Effects of Doping on n and p

I.

N d

-

N a



n i

(i.e., N-type)

n



N d p



n i

2 -

N a n

If

N d



N a

,

n



N d

and

p



n i

2

N d

II.

N a

-

N d



n i

(i.e., P-type)

p



N a n



n i

2 -

N d p

If

N a



N d

,

p



N a

and

n



n i

2

N a

EXAMPLE: Dopant Compensation

What are n and p in Si with (a) N d and (b) additional 6



10 16 = 6 cm -3 of N a ?



10 16 cm -3 and N a = 2



10 16 cm -3

(a)

n



N d

-

N a

 4  10 16 cm 3

p



n i

2 /

n

 10 20 / 4  10 16  2 .

5  10 3 cm 3 (b)

N a

= 2  10 16 + 6  10 16 = 8  10 16 cm -3 >

N d

!

p



N a

-

N d n



n i

2 /  8  10 16 6  10 16

p

 10 20 / 2  10 16   2  10 16 cm 3 5  10 3 cm 3

n

= 4  10 16 cm -3 + + + + + +

N d

. . . . . .

= 6  10 16 cm -3 + + + + + +

N d

. . . . . .

= 6  10 16 cm -3

N a

= 2  10 16 cm -3 . . . . . . . . . . .

N a

- - - - - - - -

p

= 8  10 16 . . . . . .

= 2  10 16 cm -3 cm -3

Chapter Summary

Energy band diagram. Acceptor. Donor.

m n

,

m p

. Fermi function .

E f

.

n



p



N c e

(

E c

-

E f

) /

kT N v e

(

E f

-

E v

) /

kT n



N d p



N a

-

N a

-

N d np



n i

2

Thermal Motion

• Zig-zag motion is due to collisions or scattering with imperfections in the crystal. • Net thermal velocity is zero.

• Mean time between collisions (mean free time) is 

m ~

0.1ps

Thermal Energy and Thermal Velocity

v th



electron or hole kinetic energy

 3

kT

2  1 2 2

mv th

3

kT m eff

 3  1 .

38  10 23 JK 1  300 K 0 .

26  9 .

1  10 31 kg  2 .

3  10 5 m/s  2 .

3  10 7 cm/s ~8.3 X 10 5 km/hr

Drift

Electron and Hole Mobilities

•

Drift

is the motion caused by an electric field

.

Effective Mass

In an electric field,

E

, an electron or a hole accelerates.

electrons Remember : F=ma=-qE holes

Electron and hole effective masses

m n /m

0

m p /m

0

Si

0.26

0.39

Ge

0.12

0.30

GaAs

0.068

0.50

GaP

0.82

0.60

Remember : F=ma=mV/t = -qE

Electron and Hole Mobilities

m p v



q

E



mp v



q

E



mp m p



p v

  

q



p

E

mp m p



n v

  

n

E

q



mn m n

• 

p

is the hole mobility and 

n

is the electron mobility

Electron and Hole Mobilities

v =



E

;

 has the dimensions of

v/

E

  cm/s V/cm  cm V  s 2   .  

Electron and hole mobilities of selected semiconductors

n

(cm 2 /V∙s)

p

(cm 2 /V∙s)

Si

1400 470

Ge

3900 1900

GaAs

8500 400

InAs

30000 500 Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices?

Drift Velocity, Mean Free Time, Mean Free Path

EXAMPLE: Given

E

= 10 3



p = 470 cm V/cm? What is



mp 2 /V·s, what is the hole drift velocity at and what is the distance traveled between collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units.

Solution:

n

=



p

E

=

470 cm 2 /V·s  10 3 V/cm = 4.7

 10 5 cm/s 

mp =



p m p /q =

470 cm 2 /V ·s  0.39  9.1

 10 -31 kg/1.6

 10 -19 C

=

0.047 m 2 /V ·s  2.2

 10 -12 kg/C = 1  10 -13 s = 0.1 ps

mean free path =



mh

n

th ~

1  10 -13 s  2.2

 10 7 cm/s

=

2.2

 10 -6 cm = 220 Å = 22 nm

This is smaller than the typical dimensions of devices, but getting close.

Mechanisms of Carrier Scattering

There are two main causes of carrier scattering: 1. Phonon Scattering 2. Impurity (Dopant) Ion Scattering

Phonon scattering

mobility decreases when temperature rises

:



phonon

 

phonon



phonon density

1 

carrier thermal velocity



T

1 

T

1 / 2 

T

3 / 2 

= q



/m



T v th



T 1/2

Impurity (Dopant)-Ion Scattering or Coulombic Scattering

Electron + Arsenic Ion There is less change in the direction of travel if the electron zips by the ion at a higher speed.



impurity



N a

3

v th

+

N d



N a

3 / 2

T

+

N d

1600 1400 1200 1000 800 600 400 200 0 1E14 Holes 1E15

Total Mobility

Electrons 1   1 

phonon

+ 1 

impurity

1   1 

phonon

+ 1 

impurity

1E16 1E17 1E18 1E19

N a

+

N d

(cm -3 ) -3 ) 1E20

Temperature Effect on Mobility

10 15

Question

: What

N d d



n

/

dT

will make = 0 at room temperature?

Drift Current and Conductivity E

J p

+

unit area

+

n Current density

EXAMPLE:

J p = qpv

A/cm 2 or C/cm 2 ·sec If

p =

10 15 cm -3 and

J p =

1.6

 10 -19 C  1 .

6 C/s  cm 2  10

v

15

=

= 10 cm -3 4 A/cm cm/s, then  2 10 4 cm/s

Drift Current and Conductivity E

J p

+

unit area

+

n E + -

Remember:

• Holes travel in the direction of the Electric field • Electrons travel in the direction opposite to that of the E-field

Drift Current and Conductivity

J p,drift = qpv = qp



p

E

J n,drift = –qnv = qn



n

E

J drift = J n,drift + J p,drift = (qn



n +qp



p )

E =



E

 

conductivity of a semiconductor is



= qn



n + qp



p resistivity of a semiconductor is

 = 1/ 

Relationship between Resistivity and Dopant Density

N-type P-type

RESISTIVITY (  cm)  = 1/ 

E c and E v vary in the opposite direction from the voltage. That is, E c and E v are higher where the voltage is lower.

E c -E reference = -qV V x E c E E v x

Variation in E c with position is called band bending

Diffusion Current

Particles diffuse from a higher-concentration location to a lower-concentration location.

Diffusion Current

J n

,

diffusion



qD n dn dx J p

,

diffusion

 -

qD p dp dx D

is called the diffusion constant. Signs explained:

n p x x

Total Current – Review of Four Current Components

J TOTAL = J n + J p J n = J n,drift + J n,diff = qn



n

E

+ qD n dn dx J p = J p,drift + J p,diff = qp



p

E

– qD p dp dx

J

TOTAL

= J

n,drift

+ J

n,diff

+ J

p,drift

+ J

p,diff

Chapter Summary

v p v n

 

p

E



-



n

E

J p

,

drift



qp



p

E

J n

,

drift



qn



n

E

J n

,

diffusion



qD n dn dx J p

,

diffusion

 -

qD p dp dx D n



kT



n q D p



kT



p q