ENE 311

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.

Basic fabrication steps• -This process is to make a high qu

ality silicon dioxide (SiO2 ) 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, respectively.

• Generally,dr y oxi dat i on i s used t o f or m -thin oxides because of its good Si SiO2 interface characteristics, whi

lewet oxi dat i on i s used f or f or m ingt hi cker l ayer s si nce i t s hi ghe rgr owt h r at e.

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

ot he semi conduct or .• Fordiffusionmethod, t he semi conduct or sur f ace no

t protected by the oxide is expose d to a high concentration of impur

ity. The impurity moves into the c - rystalby sol i d st at e di ff usi on.

• - 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 nt he semi conduct or .

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 rec tification.

• The forward biased voltage is normally less than1 V an d the current increases rapi

dly as the biased voltage in creases.

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

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

Thermal equilibrium condition

• 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.

Thermal equilibrium condition• -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

field 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 Ad dE e N N p ndx 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 Ee n

ln Ap i F

i

kT NE Ee 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)2ln A D

bi n pi

kT N NVe 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

10 100.0259ln9.65 10

0.774 eV

A Dbi

i

bi

N NV kTn

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 en

ergy ion implantations.• The impurity distributio

n varies linearly across the junction.

Abrupt junction • Consider an abrupt junction

as in the figure above, equa tion (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 eN x xdxd eN x xdx

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

Ap

d eN xc

dxd

Edx

eN xE x c

Abrupt junctionApply 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 D

n meN 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

bix 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

NN xxN

(10)

Abrupt junction

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

(12)

2

2

bi Dp

A D A

bi An

A D D

V Nxe N N N

V Nxe N N N

Abrupt junction

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

(13)2 bi A D

p nA D

V N NW x xe 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 10 100.0259ln1.45 10

0.679 eV

A Dbi

i

bi

kT N NVe n

V

Abrupt junction

Soln (b)

14 16 15

19 16 15

From (13)

2

2 8.85 10 11.8 0.679 5 10 101.6 10 5 10 10

0.95 m

bi A D

A D

V N NWe 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 10101.6 10 5 10 10

9.299 10 cm.

( )

1.6 10 10 9.299 108.85 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 Nxe 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 xeN

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 xeN

( ) BmeN xE x E

Abrupt junction

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

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

(16)

Bm

eN WE

( ) 1Bm

eN xE x W x EW

Abrupt junction

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

(17)

( ) 2biV x xxW 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 V9.65 10

2 3.41 10 0.343 m

0.52 10 V/cm

bi

bi

D

Bm

V

VWeNqN 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 e ax xdx dx


• By integrating (18 ) with the boundary conditi - ons that the electric field is zero at W/2 , E(x

) can be found as(19)

• The maximum field Em at x = 0 is

(20)

2 2/ 2( )

2W xeaE x

2

8meaWE


• - The built in potential is given by

(21)

and

(22)

3

12bieaWV

2

/ 2 / 2 2ln ln2bi

i i

aW aWkT kT aWVe 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 8.85 102 10 0.5 102 0.0259 0.645 V

2 2 9.65 10

m

bii

eaWE

kT aWVe n

ENE 311

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