ene 311 lecture 7. p-n junction a p-n junction plays a major role in electronic devices. it is used...

ENE 311 Lecture 7

p-n Junction

• - A p n junction plays a major role in electroni c devices.

• It is used in rectification, switching, and etc.

• It is the simplest semiconductor devices.

• Also, it is a key building block for other elect ronic, microwave, or photonic devices.

Basic fabrication steps

The basic fabrication steps for p-n junction inc lude

• oxidation,

• lithography,

• diffusion or ion implanation,

• and metallization.


• -This process is to make a high qu ality silicon dioxide (SiO

2 ) as an in

sulator in various devices or a bar rier to diffusion or implanation dur ing fabrication process.

• There are two methods to grow SiO2

: dry and wet oxidation, using d ry oxygen and water vapor, respe

ctively.

• Generally, dry oxidation is used t o form thin oxides because of its g

- ood Si SiO2 interface characteristi cs, while wet oxidation is used for

forming thicker layers since its hi gher growth rate.

Oxidation


• This process is calle d photolithography

used to delineate th - e pattern of the p n

junction.

Lithography


• (a) The wafer after the development.

• (b) The wafer after SiO2 removal.

• (c) The final result after a complete lithography process.

Basic fabrication steps• This is used to put the impurity int

o the semiconductor.

• For diffusion method, the semicon ductor surface not protected by th

e oxide is exposed to a high conce ntration of impurity. The impurity

-moves into the crystal by solid sta te diffusion.

• - For the ion implantation method, the impurity is introduced into the

semiconductor by accelerating th - e impurity ions to a high energy le

vel and then implanting the ions i n the semiconductor.

Diffusion & Ion Implantation


• This process is used to form ohmic cont

acts and interconne ctions.

• After this process is - done, the p n juncti

on is ready to use.

Metallization

Thermal equilibrium condition

• The most important charac - teristic of p n junction is re

ctification.

• The forward biased voltag e is normally less than1 V

and the current increases r apidly as the biased voltag

e increases.

• As the reverse bias increas es, the current is still small

until a breakdown voltage i s reached, where the curre nt suddenly increases.


• Assume that both p- and n-type semiconductors are uniformly doped.

• The Fermi level EF is near the valence band edge in the p-type material and near the conduction band edge in the n-type material.


• -Electrons diffuse from n sid - e toward p side and holes di

- -ffuse from p side toward n si de.

• -As electrons leave the n sid e, they leave behind the pos

itive donor ions (ND+ ) near t

he junction.

• In the same way, some of n egative acceptor ions (NA

- ) a re left near the junction as h

-oles move to the n side.


• This forms2 region s called “neutral ” r

egions and “ -spacecharge ” region.

• - The space charge r egion is also called

“ depletion region ” due to the depletio

n of free carriers.

Space-charge region

neutral neutral


• Carrier diffusion induc es an internal electric f

ield in the opposite dir ection to free charge d

iffusion.

• Therefore, the electron diffusion current flows

from left to right, wher eas the electron drift c

urrent flows from right to left.


• At thermal equilibrium, the individual electron and hole current flowin

g across the junction a re identically zero.

• In the other words, the drift current cancels o

ut precisely the diffusi on current. Therefore, the equilibrium is reac hed as EFn = EFp.


• - The space charge density distribution and t he electrostatic potential are given by Poi

sson’s equation as

(1)

• Assume that all donor and acceptor atoms a re ionized.

2

2 D A

d dE eN N p n

dx dx


• Assume NA = 0 and n >> p for n-type neutral region and ND = 0 and p >> n for p-type neutral region.


• The electrostatic potential in of the n- and p-type with respect to the Fermi level can be found with the help of and

as

(2)

(3)

exp /i F in n E E kT

exp /i i Fp n E E kT

ln Dn F i

i

kT NE E

e n

ln Ap i F

i

kT NE E

e n


• The total electrostatic potential difference b - - etween the p side and the n side neutral reg

ion is called the “ - built in potential ” Vbi . It is written as

(5)2

ln A Dbi n p

i

kT N NV

e n

• a ) A -p n junction with a brupt doping changes a

t the metallurgical junction.

• (b ) Energy band diagra m of an abrupt junction at thermal equilibrium.

• (c ) Space charge distribution.

• (d ) Rectangular approxi mation of the space ch

arge distribution.


Ex. - Calculate the built in potential for a silicon- p n junction with NA = 1018cm-3 and ND =

1015 cm-3 at 300 K.

Ex. - Calculate the built in potential for a silicon- p n junction with NA = 1018cm-3 and ND =

1015 cm-3 at 300 K.

Soln


2

18 15

29

ln

10 100.0259ln

9.65 10

0.774 eV

A Dbi

i

bi

N NV kT

n

V

Depletion Region

- The p n junction ma y be classified into t

wo classes dependin g on its impurity dist

ribution:

• the abrupt junction a nd

• the linearly graded j unction.

Depletion Region

• An abrupt junction can b - e seen in a p n junction t

hat is formed by shallow - diffusion or low energy i

on implantation.

• The impurity distribution in this case can be appro

ximated by an abrupt tr ansition of doping conce - ntration between the n

- and the p type regions.

Depletion Region

• In the linearly graded j - unction, the p n junctio

n may be formed by de -ep diffusions or high e

nergy ion implantation s.

• The impurity distributi on varies linearly acros

s the junction.

Abrupt junction • Consider an abrupt junctio

n as in the figure above, eq uation (1 ) can be written as

• The charge conservation is expressed by the condition Q = 0 or

2

2

2

2

for - 0

for 0

Ap

Dn

d eNx x

dx

d eNx x

dx

A p D nN x N x

Abrupt junction

• To solve equation (5), we need to solve it separately for p- and n-type cases.

-p side: Integrate eq.(4) once, we have

We know that

( )

A

A

p

d eN xc

dx

dE

dx

eN xE x c

Abrupt junction

Apply boundary condition: ( ) 0

( )( ) 0

p p

A p

p p

A p

E x x

eN xE x c

eN xc

( )( ) A pp

eN x xE x

(7)

Abrupt junction

-n side:

• Similarly, we can have

(8)

( )( ) D n Dn m

eN x x eN xE x E

Abrupt junction

• Let consider at x = 0

(9)

We may relate this electric field E to the poten tial over the depletion region as

( )(0) (0) A p D np n m

eN x eN xE E E

0

0

( ) ( ) ( )n n

p p

x x

bi

x x n sidep side

V E x dx E x dx E x dx

Abrupt junction

• From (6 ), we have

(11)

2 2

2 2A p D n

bi

eN x eN xV

A pn

D

D np

A

N xx

N

N xx

N

(10)

Abrupt junction

• Substitute (11 ) into (10 ), this yields

(12)

2

2

bi Dp

A D A

bi An

A D D

V Nx

e N N N

V Nx

e N N N

Abrupt junction

• - Hence, the space charge layer width or depl etion layer width can be written as

(13)

2 bi A Dp n

A D

V N NW x x

e N N

Abrupt junction

Ex. - Si p n diode of NA =5 x 1016 cm-3 and ND =1015 cm-3 . Calculate

- (a) built in voltage

(b) depletion layer width

(c) Em

Abrupt junction

Soln (a)

2

16 15

210

ln

5 10 100.0259ln

1.45 10

0.679 eV

A Dbi

i

bi

kT N NV

e n

V

Abrupt junction

Soln (b)

14 16 15

19 16 15

From (13)

2

2 8.85 10 11.8 0.679 5 10 10

1.6 10 5 10 10

0.95 m

bi A D

A D

V N NW

e N N

W

Abrupt junction

Soln (c)

max

14 16

1519 16 15

5

max

19 15 5

14

4max

( )( 0)

2

2 8.85 10 11.8 0.679 5 10

101.6 10 5 10 10

9.299 10 cm.

( )

1.6 10 10 9.299 10

8.85 10 11.8

1.431 10 V/cm

A p D n

bi An

A D D

n

D n

eN x eN xE E x

V Nx

e N N N

x

eN xE

E

Abrupt junction

• If one side has much higher impurity doping concentration than another, i.e. NA >> ND or ND >> NA, then this is called “one-sided junction”.

• Consider case of p+- n ju nction as in the figure (

NA >> ND), 2 bin

D

VW x

eN

Abrupt junction

• Similarly, for n+- p junction of ND >> NA

• - The electric field distribution could be written as

where NB = lightly doped bulk concentration

(i.e., NB = ND for p+- n junction)

2 bip

A

VW x

eN

( ) Bm

eN xE x E

Abrupt junction

• The maximum electric field Em at x =0 can b e found as

• - Therefore, the electric field distribution E(x) - can be re written as

(16)

Bm

eN WE

( ) 1Bm

eN xE x W x E

W

Abrupt junction

• The potential distribution can be found from integrating (16 ) as

(17)

( ) 2biV x xx

W W

Abrupt junction

Ex. - For a silicon one sided abrupt junction wit h NA = 1019 cm-3 and ND = 1016 cm-3 , calculat e the depletion layer width and the maximu m field at zero bias.

Abrupt junction

Soln

19 16

29

5

4

10 100.0259ln 0.895 V

9.65 10

23.41 10 0.343 m

0.52 10 V/cm

bi

bi

D

Bm

V

VW

eN

qN WE

Linearly Graded Junction


• In this case, the Possion equation (1 ) is expr essed by

(18)

where a is the impurity gradient in cm-4 and W is the depletion-layer width

2

2

W W for -

2 2

d dE eax x

dx dx


• By integrating (18 ) with the boundary conditio - ns that the electric field is zero at W/2 , E(x) c an be found as

(19)

• The maximum field Em at x = 0 is

(20)

2 2/ 2( )

2

W xeaE x

2

8m

eaWE


• - The built in potential is given by

(21)

and

(22)

3

12bi

eaWV

2

/ 2 / 2 2ln ln

2bii i

aW aWkT kT aWV

e n e n

Linearly graded junction in thermal equilibrium.

(a) Impurity distribution. (b) Electric-field distribution.

(c) Potential distribution with distance.

(d) Energy band diagram.


Ex. For a silicon linearly graded junction with a n impurity gradient of 1020 cm-4 , the depletio- n layer width is 0.5 μm. Calculate the maxi

- mum field and built in voltage.


Soln

Note: Practically, the Vbi is smaller than that c alculated from (22 ) by about 0.05 to 0.1 V.

219 20 423

14

20 4

9

1.6 10 10 0.5 104.75 10 V/cm

8 8 11.9 8.85 10

2 10 0.5 102 0.0259 0.645 V

2 2 9.65 10

m

bii

eaWE

kT aWV

e n

ene 311 lecture 7. p-n junction a p-n junction plays a major role in electronic devices. it is used...

Documents

pn junction

oxidation slide

lithography slide

metallization slide

basic fabrication steps

fabrication process

n type material

ntype semiconductors