advsemi_lec3 2013-03-11

Advanced Semiconductor Devices

Lecture 3

Advanced Semiconductor Devices

Lecture 3

2© V. Ariel 2012 Advanced Semiconductor Devices Lecture 3

Lecture outline

Non-equilibrium semiconductors: Quasi-Fermi levels

Excess carrier generation-recombination

Carrier lifetime and diffusion length

PN – junction in equilibrium

PN – junction under bias

PN – junction current


Non-equilibrium: Quasi-Fermi levels

In non-equilibrium, can not describe semiconductor properties with

a single constant Fermi level

Introduce quasi-Fermi levels EFN

and EFP

for electrons and holes

FN C V FPE E E E

kT kTC Vn N e p N e

- -

= × = ×

CE

GE FNEFPE

VE

0

0

n n n

p p p

= + D

= + D

Excess carrier concentrations: Δn and Δp


Non-equilibrium total current

In general, for electrons in 3-dimensions

CE

GE FNEFPE

VE

Similarly, for holes

Total current

dr diffn n n n n n FNqn qD n n Em m= + = + Ñ = ÑJ J J E

dr diffp p p p p p FPqp qD p p Em m= + = - Ñ = ÑJ J J E

n p n FN p FPn E p Em m= + = Ñ + ÑJ J J


Generation - recombination of carriers

Main physical processes are

– Direct band generation and recombination

– R-G center generation-recombination via impurities and traps

– Surface recombination


Generation -recombination of carriers

Assume low-level injection

Minority carrier recombination rate can be approximated as

0 0,p p n nD D= =

,n p

n pR R

t tD D; ;

Band-to-band R-G in indirect band semiconductors is complex and

requires involvement of phonons for momentum conservation


Continuity equation

Δx

( )nJ x ( )

nJ x x+ D

nG

nR

Consider electron current in a semiconductor bar

Assume electron generation and recombination

Let Δx approach zero to derive the continuity equation for electrons

( )( ) ( )n nn n

J x J x xdnx G R x

dt q q

+ DD = - + - D

- -

Similarly for holes

( )1 nn n

dJdnG R

dt q dx= + -

( )1 pp p

dJdpG R

dt q dx= - + -


Minority carrier diffusion equationNo electric field

Low level injection in p-type semiconductor

The equilibrium minority carriers

Approximate minority carrier current

Obtain minority carrier diffusion equation

0=E

n n n n

dn dnJ qn qD qD

dx dxm= + ;E

2

2

1 pnn

d ndJD

q dx dx

D; ( )1p n

n n

d n dJG R

dt q dx

D= + -

0p p pn n n= + D 0p pn pD =

0 const( )pn x=

( )2

2

p pn n n

d n d nD G R

dt dx

D D= + -


Diffusion length

whSi x

A p-type Si bar is illuminated with light resulting in Δn0 minority

carriers on the surface (no generation in the body of Si)

Find the distribution of minority carriers in the body of Si

Assume low-level injection condition

We are only interested in minority carrier behavior

0nD


Diffusion length

And assuming steady-state condition

Write minority carrier diffusion equation

General exponential solution

0pdn

dt=

2

20p p

nn

d n nD

dx tD D

- =

1 2( ) n n n n

x x

D Dpn x C e C et t

-

D = × + ×

Using minority carrier recombination rate

pn

n

nR

tD

=


Diffusion length Assume that excess minority carriers approach zero far from the

illuminated surface. Use the following boundary conditions

Obtain solution

Where we define the diffusion length and calculate it using

Einstein relationship

0

, 0

0,

p

p

x n

x n n

®¥ D ®

® D ®D

0 0( ) n n N

x xD L

pn x n e n et- -

D = D × = D ×

N n n n n

kTL D

qt m tº =


Poisson equationRelationship between semiconductor charge density and electric

fields

12 F1 10cmse

-@ ×

2

2

( )

s

d d x

dx dx

y re

= - = -E

( )( ) D Ax q p n N Nr = - + -

Define total semiconductor charge density (assuming complete

ionization of impurities)

,D D A AN N N N+ -; ;

Semiconductor permittivity


Consider two bars of semiconductor material of types n and p respectively

CE

VE

iE

CE

VE

iE

Dn N;Ap N;

Abrupt pn - junction in equilibrium

F CE E

kTCn N e

-

= ×V FE E

kTVp N e

-

= ×

F iE E

kTin n e

-

= ×i FE E

kTip n e

-

= ×

FE

FE


Ap N;Dn N;

DN+

AN-

px- nx

Three different regions present:

– Space-charge depletion region

– Neutral p-type region

– Neutral n-type region

SCp-type n-type



CE

VE

iECE

VE

iE


FEFE

For pn - junction under thermal equilibrium const (x)FE =

( ) ln ln Ai F p

i i

p NE E kT kT

n n

æ ö æ ö- = × ×ç ÷ ç ÷

è ø è ø; ( ) ln ln D

F i ni i

n NE E kT kT

n n

æ ö æ ö- = × ×ç ÷ ç ÷

è ø è ø;


Band bending in equilibriumBand bending maintains constant Fermi level in equilibrium

Introduce the built-in voltage VBI

CE

VE

CE

VE

BIqVFE

px-

nx

niE

piE

2ln A D

BIi

kT N NV

q n

æ ö= × ç ÷

è ø

( ) ( )( )BI i F F ip nqV E E E E= - + -


Carrier distribution in equilibrium

nx

x

ln( )n

20 /n i Dp n N»

0n Dn N»0p Ap N»

20 /p i An n N»

in

CE

VE

iECE

VE

iE

C FPE E- BIqV

FEFE

px- nx

px-


Solution for electric fieldFirst, solve for the electric field

●

( )

s

d x

dx

re

=E

maxA D

p ns s

qN qNx x

e e= - = -E

A p D nN x N x=

nx

xpx-

maxE

( )xE

( ) ( )

, 0 , 0

( ) , ( )

A p D n

A Dp n

s s

qN x x qN x x

qN qNx x x x x x

r r

e e

= - - < < = < <

= - + = -E E


Solution for electric potentialSecond, solve for electrostatic potential

d

dx

y= -E

( )( )( )( )

2 2

22

( ) , 02

( ) , 02

Ap p p

s

Dn n n

s

qNx x x x x x

qNx x x x x x

ye

ye

= + - - < <

= - - - < <

2 2( ) , ( )2 2

A Dp p n n

s s

qN qNx x x xy y

e e- = - =

nx

xpx-

( )xy

( )2 2( ) ( )2BI n p n D p A

s

qV x x x N x Ny y

e= - - = +


Depletion region widthFinally, solve for the depletion region width

( )2 2

2BI n D p As

qV x N x N

e= + D n

A p D n pA

N xN x N x x

N= Þ =

( )2 s A

n BID A D

Nx V

q N N N

e= ×

+( )2 s D

p BIA A D

Nx V

q N N N

e= ×

+

2 1 1sp n BI

A D

W x x Vq N N

e æ ö= + = × +ç ÷

è ø


Electric field and junction charge

maxA D

p ns s

qN qNx x

e e= - = -E

max

2 A DBI

s D A

q N NV

N Ne= ×

+E

Calculate maximum electric field

Define junction charge

J A p D nQ qN x qN x= =

2 A Ds BI

D A

N NQ q V

N Ne= ×

+

22CMOS Digital Camera © Viktor Ariel 2012 (6)

Energy bands under reverse bias

CE

VE

iE

CE

VE

iE

( )BI Aq V V-FPE

FNE

px-

CE

VE

iECE

VE

iE

BIqV

FEFE

px- nx

nx

E

AqV

Most of the applied voltage drops in the depletion region because of the high resistivity

Far from the junction, material remains quasi-neutral

Energy band bending is increased under reverse bias by qV

A

The distance between quasi-Fermi levels in the depletion region: qV

A

The depletion region width is increased


nx

x

ln( )n

20 /n i Dp n N»

0n Dn N»0p Ap N»

20 /p i An n N»

in

Carrier distribution under revers biasCE

VE

iE

CE

VE

iE

( )BI Aq V V-

FPE

FNE

px- nx

px-

E

AqV


Depletion region width under reverse biasThe depletion region width increases due to applied reverse voltage

2 1 1sp n BI A

A D

W x x V Vq N N

e æ ö= + = × + -ç ÷

è ø

max

2 A DBI A

s D A

q N NV V

N Ne= × -

+E

2 A Ds BI A

D A

N NQ q V V

N Ne= × -

+

Define depletion region capacitance

1 12

2sA D

sA D A BI A

dQ N NC q

dV N N WV V

ee= = × =+ -


Energy bands under forward bias

CE

VE

iECE

VE

iE

BIqV

FEFE

px- nx

CE

VE

iE CE

iEFPE FNE

px- nx

( )BI Aq V V-

AqV

Depletion region width decreases under forward bias

Energy band bending is decreased under forward bias by qV

A

The minority carrier concentration in the quasi-equilibrium region is higher than the equilibrium concentration


nx

x

ln( )n

20 /n i Dp n N»

0n Dn N»0p Ap N»

20 /p i An n N»

in

Carrier distribution under forward bias

px-

CE

VE

iE CE

iEFPE FNE

px-

nx

AqV

E

( )BI Aq V V-


Charge density and applied voltageCharge concentration in terms of the quasi-Fermi levels

lnFN ii

nE E kT

n

æ ö- = × ç ÷

è ølni FP

i

pE E T

n

æ ö- = × ç ÷

è ø

Within the depletion (space charge region)

FN FP AE E qV- =

On the border of the depletion region px x= -

( )

0 0

FN FP Aq E E qV

kT kTp p pn n e n e

-

= × = ×

Similarly when nx x=

( )

0 0

FN FP Aq E E qV

kT kTn n np p e p e

-

= × = ×


Junction current

In the neutral n- type region the minority carriers density is pn

Since there is no electric field, the continuity equation is

Ap N;Dn N;

DNAN

px-nx

SCp-type n-type

( )20 0

20n n n n

p p

d p p p p

d x D t- -

- =

Boundary condition 0( )n np x p= ¥ =

( ) ( )/ ( )/0 0 0 1

A

n p n p

n

qVx x L x x LkT

n n n n nx xp p p p e p e e- - - -

=

æ ö- = - × = -ç ÷

è øp p pL D tº

nppn


Junction current

Assume the diffusion current only on the border of the neutral

region0 1

A

n

qVp nn kT

p px x p

qD pdpJ qD e

dx L=

æ ö= - = -ç ÷

è øSimilarly for the p-type region

0 1A

p

qVp n p kT

n nnx x

dn qD nJ qD e

dx L=-

æ ö= = -ç ÷

è ø

Since current is constant at every place along the device, we

calculate current at the boundary between the depletion and

neutral regions


Junction current

Charge density and current under external bias


Junction current The total current results in Shockley equation

0

0 00

1AqV

kTn p

n p p n

n p

J J J J e

qD n qD pJ

L L

æ ö= + = -ç ÷

è ø

= +

J

0JAV


Junction current Current in practical Si diode

Reverse current is much higher than theoretical

due to generation-recombination and surface effects

advsemi_lec3 2013-03-11

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