221A_1_Review of Semiconductor Physics

8/3/2019 221A_1_Review of Semiconductor Physics

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EE 221A:

PHYSICS OF

SEMICONDUCTOR

DEVICES

Review of Semiconductor Physics

There are two classes of materials (solids): conductors and insulators.

Conductor: The small D.C. conductivity at 0K is non-zero.

Insulator: The small D.C. conductivity at 0K is zero.

All insulators have finite conductivity at T > 0K.

Semiconductors are those insulators with conductivity that i s not too small

at room temperature (300K).

NOTE: The concepts of conductivity, etc. require quantum

mechanics. The solution of the Schrodinger Equation gives the

following results:

The allowed electro nic states are lumped togetherto form bands.

There is a bandgap.

At finite temperature, there are electrons and holes inthe semiconductor.


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)

Band formations:

In a crystalline solid, atoms are arranged in a periodic fashion. Silicon, for

example , is a di amond structure, whereas GaAs is a zinc-blende structure.

The most commonly used technique to solve the electronic states of a

crystal is called single electron approximation. The first treatment was done byF. Block, and the Schrodinger Eqn. is:

22 ( ) ( ) ( )

2k k kV r r E r

m

+ =

(1)

Note that ( )V r is periodic in space for a crystal, ignoring the surfaces, and is called the

crystal potential.

The solution of the above eqn. is called theBlock function, and has the form:

( ) ( , )

jk r

k nr e U k r =

i

(Block Theorem)A great deal of effort has been spent in finding kE and the functional form of ( , )nU k r .

The solution indicates that kE is formed into bands:

Methods used included tight binding, pseudo-potential, K P, orthogonal wave, etc.

The Block Theorem

Now, the Schrodinger Eqn. for a singlee in a rigid lattice is

= +

2 2( ) ( ) ( )2

H r V r r m

(2)

Let RT be an operator that moves the wave function by R, i.e., = +

( ) ( )RT r r R ;

whereR is a lattice constant.

EC

EV

E

k

EC

EV

EC

EV

E

k


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)

Notice that

= + + + + 2 2( ) ( ) ( ) ( )

2RT H r r R V r R r R

m

= + + + 2 2 ( ) ( ) ( )

2r R V r r R

m(3)

since = +

( ) ( )V r V r R i.e., =

( ) ( ( ))R RT H r H T r

Suppose

=

( ) ( )n n nH r E r (solution to the eigenvalue problems)

i.e.

= = +

( ) ( ) ( )R n R n n n n

T H r T E r E r R

=

( ( )) ( ( ))R n n R n H T r E T r (4)

i.e.

( )R nT r is also an eigenfunction of the Hamiltonian. If

( )n r is non-degenerate,

then

= n nR RT C

But lets look at

+= + + =

' ' '

( ) ( ' ) ( )R R R RT T r r R R T r

i.e. + +

= = =

' ' ' ' '( ) ( )R R n n R R R R R R T T C C T r C r , or

+=

' 'R R R R C C C or = ( )m

mR RC C

if

ia is a primitive translation vector, then = 2 i

i

j

aC e and i is a real

number.

In general,

= ii k R

RC e (5)

with = + +

1 2 31 2 3k b b b , and

ib are the reciprocal lattice vectors. That is,

= + =

i

( ) ( ) ( )i k RR n n n T r r R e r


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)

Let =

i

( ) ( ) i k rn nr U r e

and

+=

i

( )( ) ( ( )) i k r R R n R n T U r T r e

+= =

i i

( )( ) ( )i k R i k r R n ne r e U r

( )RU r is periodic in the lattice.

Boundary Condition

If we assume the Born-Von Karman boundary condition:

+ =

11( ) ( )r N a r

+ =

22( ) ( )r N a r (6)

+ = 33( ) ( )r N a r

then, with Blocks Theorem,

+ = =

i

( ) ( ) ( )iiiN k ain i n n r N a e r r

i.e.,= =

i 21ii i iiN k a iN e e

or =i iN m = ( 0, 1, 2,...)m

i.e. =i

i

m

N

Thus the allowablek vectors in a primitive cell of the reciprocal space

= =

i1 2 3( )b b b N number of unit cells in the crystal,

and the volume that a k occupies is

=

i

3 3(2 ) (2 )

N v V

where is the volume of a unit cell.

Consequence of the Block Theorem

Now we see that =

i

( ) ( ) i k rn nr U r e . When we put this into the Schrodinger

Equation, we get:


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22 ( ) ( ) ( )

2n n nV r r E r

m

+ =

2

2, ,( ) ( ) ( ) ( )

2

i k r i k r

n k n n kV r e U r E k e U r m

+ =

i i

2 2 2

, ,( ) ( ) ( ) ( )2 2

n k n n kk V r U r E k U r m m

+ + =

But, , ,( ) ( )n k n kU r R U r + =

, we need only to solve the equation

over a unit cell. Consequently, there is an infinite solution for each k indexed by n.Also,

( ), ,( ) ( )

i k K r

k K n k K n r e U r +

+ +=

i

(K= a reciprocal vector)

,' ( )i k r

k K ne U r+=

i

, ( )i k r

k K ne U r+

i

Thus, the above is also a Block function with a different U.

Note that the differential equation of ,' ( )k K nU r+

is the same as that for , ( )k nU r

the eigenvalues must be the same.with proper choice of the index n, ( )nE k is periodic with K.

Only need to considerk values in the first Brillouin Zone (BZ).Other properties of , ( )n kU r : ( ) ( )n nE k E k=

, ,m k n k mn U U = ; , ' , , 'm k n k m n k k =

NOTE: The band structures are, however, quite complicated. For silicon, the

conduction band minimum is at theXpoints.

There are six equivalent constant energy surfaces.GaAs is, however, a direct bandgap semiconductor with both conduction

band minimum and valence band maximum at .


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Wave Packet

With these block-waves, we next construct a wave packetto study carrier transport. This

is necessary since the Block waves are extended states of the stationary Schrodinger

Equation. In order to treat a wave packet, we require time-dependent perturbation theory.

The results are that, with a wave packet centered at k ,

g

k

m =

, where

2

2

1 1 ( )

ij i j

E k

m k k

=

,

and

1

m

is called the (effective mass)

-1

tensor, and

( )gdk

q E Bdt

= +

However, in reality we have imperfection in the crystal and lattice vibration. Thus

there will be scattering, and the above equation will only be followed between scattering

events. In general, the transport is governed by the Boltzmann transport equation (which

is a semi-classical equation, as it describes the movement of well-constructed wave

packets).

Boltzmann Equation

k rscatt

df fk f f

dt t

= + +

i i (7)

k


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where f is the occupation probability that a particular state in the cell is occupied with

k located in kdV . We will come back and look at the

Boltzmann Equation in greater detail, but let us first considerf .

For an electron (which is what we are interested in), it is afermion, and consequently

it obeys the Pauli Exclusion Principle. A direct result is that its probability of occupancy

of any state with energy E is given by

1( )

1 exp

f EE

kT

=

+

where is the electro-chemical potential, defined as

,E V

S

N T

=

(8)

Sis the entropy, and N is the number of particles. In solid-state, is also called the

Fermi-energy FE . Thus, in a solid with ( )nE k ,

1

( )1 exp

kF

E fE E

kT

= = +

(9)

The number ofe in a band is therefore

2 kk

N f=

Since adjacent k values are very close to each other, we convert the into an integral,

and get

2( )

kk

dVN f

k=

where k is a volume of each k occupied, and3(2 )

V

= ,


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)

i.e.3 . .

2

(2 )k kB Z

V N f dV

=

3 . .

2

(2 )k kB Z

Nn f dV

V = = (10)

Splitting the integration over surfaces of constant energy, and then over all energy in the

band, we have:

( ) ( ) ( )kE k k E k E k k+ = +

i

( )E k= +

dE

Thus, the number ofe states between ( )E k

and dE is:

3( )( ) ( )

4nS EdS

N E dE k k

=

3( ) (4 ) ( )ES E n

dS dEE k

=

i.e.,3( )

( )(4 ) ( )EE S E n

dSn f E dE

E k

=

(11)

( )N E

( ) ( )E

f E N E dE= (12)

density of states


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For a minimum conduction band, the constant energy surface is approximated by an

ellipsoid, and it is very simple to obtain ( )N E :

Let ( )n E = total number of states inside an ellipsoid with constant energy surface of

E

1( )

( )dn E

N E V dE

=

But

2 2 2 21 2 3

1 2 32C

k k kE E

m m m

= + + +

(13)

for an ellipsoid with the proper choice of principal axes, and the principal radii are:

2

2 ( )' i Ci

m E Ek =

i.e.,4

1 2 33' ' '

( )/2

k k kn E

k

=

1

3

(2 )

Vk

=

Thus,3/ 21 2 3

2 3

2 2( ) ( )

3C

m m mVn E E E

=

and1/ 21 2 3

2 3

2( ) ( )C

m m m N E E E

=

(14)

i.e.,( ) ( )CC CEn M f E N E dE

, a familiar formula where

C= number of

equivalent minimum valleys (6 for silicon).

Now,

1/ 21 2 3

2 3 ( )/

2 ( )

1FCC

E E kT E

m m m E En dE

e

=

+


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)

3/ 2 1/ 2

2 3 /

2 ( ) ( )

1FCC dc C

E E kT E

M m E EdE

e

=

+

If 3C FE E kT , then

( )/

/

1

1

E E kT C F

FE E kT e

e

+

and

( )/3/ 2 3/ 2

3/ 2 3

( ) ( )

2

E E kT C FC dc M m kT n e

=

(15)

NC= effective density of states of conduction band

Otherwise,

( )/

3/ 2 3/ 2 1/ 2

2 3 0

2( ) ( )

1E E kT C F

dcC

m kT d n M

e e

=

+

(16)

Defining0

1( )

( 1) 1

j

j f f

dF

j e e

=

+ + = fermi integral of order j,

1/ 2C F

C

E En N F

kT

=

(17)

What about valence bands?

The important thing is that a full band doesNOTconduct, since the velocity of a particle

with k -vectork is negative of that ofk -vector

-k .Summing up, a filled band net current is zero.

Conduction of a current happens when some electron in the valence band is missing.

Thus, we introduce an accounting principle, called a hole an electron state that isempty.

( )/

11

1FP E E kT

f fe

= =+

and


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hh

lhshhh

lhsh

( )VE

P p n E f dE= 3/ 2 3/ 2

*1/ 23/ 2 3

2( ) ( )( )dVV f

m kTM F

=

;

* F Vf

E E

kT

= (18)

where

2

12 21( )dV Em k

=

For silicon, we actually have two valence bands (+ a split-off band):

and

3/ 2 3/ 2 3/ 2

*1/ 23/ 2 3

2 ( )( )

2

lh hh

f

m m kT p F

+ =

*1/ 2( )V fN F = (19)

Note that in this definition of a hole, we have something called hole transformation:

*( ) ( )E k E k *q q *

Pf f

k k 2 2

2 2

E E

k k


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)

Thus, we have

1/ 2 expC F C F

C C

E E E En N F N

kT kT

=

1/ 2 expF V F V

V V

E E E E p N F N

kT kT

=

Now, lets get back to the Boltzmann Equation.

E rscatt

df fk f f

dt t

= + +

i i

It turns out that if we make the following two assumptions:

1.

. .

0

G R

scatt

df f f f

dt t

= + (relaxation time

approximation)

and

2. f is essentially the same as 0f in Maxwells Boltzmann distribution,

Then the Boltzmann Equation is equivalent to the device equations:

1( ) ( )n

nj G n R n

t q

=

(20)

1( ) ( )p

p j G p R p

t q

+ =

(21)

n n n J q n qD n = + (22)

p p p J q p qD p = (23)

Onsager has shown from a very general thermal dynamic argument that, under

isothermal conditions,

n nJ n = +


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)

with

2n

c

TE

m =

2 3/ 20 0

1/ 2

0

( ) (1 )2

3 ( )

C

C

CE

c CE

E E E f f dET

m kT E E f dE

=

(24)

For the purpose of this course, it is sufficient to use Eqs. (20)-(23) except when the

electric field is high, the carrier distributions are no longer Maxwellian, and velocity

overshoot can happen; or whenx

is larger and we must abandon even Boltzmann

transport.

Now, however, that the cm in Eqn. (24) is given by

1

1 2 3

1 1 1 1

3 m m m

+ +

i.e., the arithematic mean. For silicon, it turns out that for conduction minimum, it is a

circular ellipsoid with 1m = 2m =m and 3m = lm . Thus

* 1/ 3( )dc lm m m= (25)

1

* 2 13ccl

mm m

= +

(26)

Now, so far we have discussed pure semi-conductors. In reality, we always try tointroduce a controlled amount of impurities, called donors and acceptors, with

concentrations DN and AN , respectively. These atoms replace some of the host atoms.

E.g., in silicon, the donors are As, P and Sn. They are column V elements 5 valence

e .


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As

B

As

B

D A N P N n

+

+ = + (charge neutrality)

Next, we must address the interaction between the electrons and the holes. For that, we

introduce the concept ofquasi-fermi energies: FnE and FpE . This means that we are

assuming that the electrons in the conduction band are in quasi-equilibrium with an

electro-chemical potential FnE , while the holes in the valence band are in equilibrium

with FpE = (although FnE FpE ).

This concept makes sense when the relaxation time between the electrons and holes, R ,

is much larger than the relaxation times, i , among the electrons and holes themselves.

This is generally true due to scattering (or matrix elements) differences

910 secR ; 135 10 seci

we can write:

1/ 2 expC Fn C Fn

C C

E E E En N F N

kT kT

=

(27)

1/ 2 expFp V Fp V

V V

E E E E p N F N

kT kT

=

(28)

The fifth e is only loosely bound to the

As atom, and has no k value.

At finite T the extra e will moveinto the conduction band.

In case of acceptor, there are only 3 e

around the atom. At finite T, another e will complete the missing bond, leaving

behind a hole, i.e.,


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For non-degenerate material (i.e., the Boltzmann approximation holds), then we have

exp expFn FpC V

C V

E EE Ep n N N

kT kT

=

i

2in

2 expFn Fp

i

E En

kT

=

(29)

Note that2in is the p ni product of non-degenerate semiconductor sample at

equilibrium!

Generation and Recombination of Excess CarriersLet us next consider the relaxation mechanisms between the electrons and the holes (i.e.,

how they get back into equilibrium). Note that one measures the degree of departure

from equilibrium between the holes and the electrons by ( )Fn FpE E ; or, for a non-

degenerate sample, by 2( )inp n .

Recombination Processes

There are a few kinds of recombination processes that can bring about the equilibriumof the electrons and the holes.

1. Band-to-Band (radiative recombination)

If we are not in the regime of simulated emission (e.g. as in the case of LASAR), the

recombination should be proportional to the number of electrons and holes:

0 0( )( )R C n n p p= + + (30)

o

EC

EV


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where the subscript 0 denotes equilibrium values. That is, at equilibrium 0 0R Cn p= , and

must equal the generation rate, 0G . In other words, the net recombination is proportional

to

0 0

U R R R G= =

0 0( )C n p p n n p= + + (31)

If we assume quasi-charge neutrality, n p = , then we have

2

0 0( ) ( )U C n p n C n = + +

0 0( )C n p n +

(32)

for low level injection (i.e. 0 0n n p + ). Since we define excess carrier lifetime to be

nU

=

0 0

1

( )C n p =

+(33)

i.e., as dopant concentration .

2. Auger recombination

This is a process involving three particles; therefore it is also known as the

e e h process.

o

EC

EV

o

EC

EV


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Single trap levelrecombination

process

For the above process, by mass action law, we anticipate the rate by

2R Cn p= (34)

Because of the three particles involved, we expect C to be small. But in heavily doped

material, e.g. heavily doped n-typed material, n is large and R can be important.

3. Shockley-Read-Hall (SRH) recombination (also called trap-assisted recombination)

Now, for indirect gap materials such as silicon, band-to-band recombination is not

possible to the first order.

In order for the above process to occur, both energy and momentum must be conserved;

thus:

GE h= ;C photon k P=

But since 0photonP , then the two conservation laws cannot be met simultaneously for the

above process. Consequently, to first order, such a process does not occur.

Recombination is therefore via trap. The detailed physical mechanisms are unknown,

but we can work out a phenomenological theory from Shockley, Read, and Hall.

o

X

o

CB

VB

Trap


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In the figure above, (a) e capture;

(b) e emission;

(c) hole capture;

(d) hole emission.

We expect process (a) to n , and number of empty traps. That is,

(1 )a n T T R C nN f = (35)

where TN and Tf are the number of traps per cm3, and the occupation probability of a

trap, respectively. Process (b) is number of filled traps:

b n T T R e N f = (36)Similarly, for processes (c) and (d), we get

c p T T R C pN f = (37)

(1 )d p T T R e N f = (38)

i.e., the net rate of recombination of e

is

n a b

nR R R

t

= =

(39)

and the net rate of recombination ofh is

c d

pR R R

t

= =

(40)

In addition, the rate of change of the number ofe

in the traps is:

( )T Ta d b c

N fR R R R

t

= +

( ) ( )a b c d R R R R =

n pR R= (41)


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)

At equilibrium, 0n pR R= = by detailed balance. That is,

(1 )a b n T T n T T R R C nN f e N f = = = (42)

(1 )c d p T T p T T R R C pN f e N f = = = (43)

But in equilibrium, Tf is given by the Fermi-Dirac distribution:

=

+ ( )/1

1 1/ FTT

D

E E kT f

g e(44)

where Dg is the degeneracy of the trap. Also, we have

( )

1/ 2

C FE E

C F kTC C

E En N F N e

kT

=

(45)

Thus, eqn. (42) yields

=

1 Tn

T

nf

C nf

e

( )

( )/1/C F

T F

E E

E E kT kTn C DC N e g e

=

( )

1/C TE E

kTn C DC N g e

= (46)

Let

( )

1

C TE E

C kT

D

Nn e

g

We then get 1n ne C n=

Similarly, using Eqn. (43) and

( )F VE E

kTV p N e

we obtain:

( )T VE E

kT p p V De C N g e

=

1p

1pC p= (47)


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)

we have: 1( (1 ))n n T T T T R C n N f nN f = + (48)

1( (1 )) p p T T T T R C pN f p N f = (49)

All of the above is under pretext of equilibrium. But of course, equations (35)-(38)

still hold in non-equilibrium; we just dont know the functional form of Tf . Also, ne and

pe should not depend on the state of equilibrium. eqs. (48) and (49) still hold.

Now, suppose we have a sample, which is in steady-state, depart from equilibrium:

0T tN f

t

=

n pR R=

(from Eqn. (41) which, in turn, is from Eqns. (48) and (49)).

1 1( (1 ) ) ( (1 ))n T T T T p T T T T C nN f n N f C pN f p N f =

11 1( ) ( )

n p

T

n p

C n C pf

C n n C p p

+=

+ + +(50)

Thus,

1( (1 ) )n p n T T T R R C N n f n f = =

+ += + + +

= + + +

=

+ ++

1 1 1

1 1

1 1

1 1

1 1

1 1

( ) ( )

( ) ( )

( ) ( )

( )

n p n p

n T

n p

p p

n T

n p

T

p n

n C n C p n C n C pC N

C n n C p p

C np C n pC N

C n n C p p

N np n p

n n p p

C C

(51)

Now,

( )

1

C TE E

c kT

D

Nn e

g

=

( )

1

T VE E

kTV Dp N g e

=


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( )

2

1 1

C VE E

kTC V in p N N e n

= = (from (29))

Thus,

2

1 1

( )

1/ ( ) 1/ ( )

T in p

p n

N np n R R

C n n C p p

= =

+ + +

(52)

in steady state.

Very often, we write ,n pC as

, , ,n p n p thn pC =

where thnV is the thermal velocity of the electron, i.e.,

2

*1 32 2

thnCCm kT = ; or *3thnCC

kTm

=

and n is the electron capture cross-section. Notice that

( )

1

C TE E

c kT

D

Nn e

g

=

( )i TE E

i kT

D

ne

g

=

( ) ( )

1

T V T i E E E E

kT kT V D i Dp N g e n g e

= =

where

log log C Vi C Vi i

N NE E kT E kT

n n = +

Fermi energy for an intrinsic semiconductor (i.e., undoped).

With these definitions, (52) gives

2

( )( )

( )

i TT I

T n p thn thp i

n p E EE E

i kT kT n thn p thp i D

D

N np n R R

nn e p n g e

g

= =

+ + +

(53)

(this is also Eqn. (58) in Szes book, chapter 1.)


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Let1( ) p p thp T N

= ;1( )n n thn T N

=

2

1 1

( )

( ) ( )i

n p

p n

np nR R

n n p p

= =

+ + +

Case A: Neutral region recombination

Suppose we have a quasi-neutral region, i.e.,

0 p p p= +

0n n n= +

with n p =

, and assume that T A D N N N

TN is small; we do not need toconsider them for carrier concentration. Now, define recombination lifetime, , as

n n nR

t t

= =

(i.e., /tn Ae = if the excitation is removed). Now,

0 0 0 0

0 1 0 1

( )( )

( ) ( )p n

n n p n n pR

n n n p n p

+ + =

+ + + + +

2

0 0

0 1 0 1

( ) ( )

( ) ( )p n

n p n n

n n n p p n

+ =

+ + + + + (54)

0 1 0 1

0 0 0 0

( ) ( )

( ) ( )n p

p p n n n n

n p n n p n

+ + + + = +

+ + + + (55)

I. ifn-type material and 0n n ( 0 0n p n-type), then0 1 0 1

0 0

( ) ( )n p

p p n n n n

n n

+ + + + = +

If 0 1n n and 0 1n p (i.e., TE is not too close to the band edges), then


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+ + +

= + +

0 1 1

0 0

1n p p p n n n

n n

1p

T thp pN

=

II. Similarly, forp -type material, 0n p :1

n

T thn n N

=

Suppose n is not too small, and TE is around iE . Then, forn-type material,

01

0 0

n p

n np n

n n n n

+ + = +

+ +

+ + 0

n p

n

n n

In the limiting case that 0n n ,

= +n p

This is rather counter-intuitive. It states that under very high-level injection, lifetime is

actually becoming larger than under low level injection! This is caused by the saturation

of the traps and they become less effective.

The above figure represents a variation of with TE for 0 0n n p + for n-type

material.

EC EVEFET

p

EC EVEFET

p


24/36

as 0 0n p ;

+

+

0 1 1

0 0

p n

n n p

n n

These recombination rates (lifetimes) enable us to calculate Equations (20) and (21).

Thus, the only parameters in Eqns. (20)-(23) that we need to discuss are the diffusion

constants nD , D .

Actually, by starting with (24), we can see what nD is, i.e.,

n n Fn J n E=

But 1/ 2Fn C

C

E En N F

kT

=

=

1/ 2Fn C

C

E En N F

kT

1/ 21/2Fn C Fn C

C

E E E EN F

kT kT

=

( )1/ 2

1/ 22Fn C

F nE E

F kT

=

( )1/ 2

1/ 22Fn

F nE q

F kT

= (56)

that is,1/ 2

1/ 2

2Fn

Fn E kT n qn

F

=

= 1/ 2

1/ 2

2n n n

F J kT n q n

F

n nqD n qn = (57)


25/36

1/ 2

1/ 2

2 C Fn

n n

C Fn

E EF

kT kTD

E EqF

kT

=

For a non-degenerate semi-conductor, we have:

n n

kTD

q

Similarly,

p

kTD

q

NOTE: This is therefore always true for minority carriers under low-level injection!

From Eqn. (24), we see that is determined by the scattering lifetime ( )E . In

general, we have two types of carrier scattering:

1.phonon scattering

2. impurity scattering (ionized impurity)

Additionally,

+

11 1

I ph

3/ 2

I

I

T

N

+

(from (24))

and 3/ 2ph T for silicon.

versus IN


26/36


27/36

E

To get a idea of how changes with E, we consider first rate of energy gain from E-field.

2| |dE

e v e edt

= = =

Next, it can be derived that the rate of energy loss to the lattice through inelastic

acoustic phonon scattering is

2 /28 (2 )( )

e

ev m kT dE

T Tdt T

=

where Te is the temperature of the electron gas, v is speed of sound in the

semiconductor and is the mean free path between collision.

Steady-state conditions

Rate of energy gain from E-field = Rate of energy loss to the lattice, we get

( )( )

1/22

28 2 *

ee

v m kT e T T

T

= (59)

Express n as a function of

To express as a function of, we see that at low field is:

* me

m =

mean time between scatterings

*e

m =

But3kT

m= ; 0

3

e

mkT

;

i.e. ifTof electrons is eT ,


28/36

1/2

0e

TT

= . (60)

Actually, accounting for the angle of scattering:

( )0 1/24

3 2 *

e

m kT , (61)

using (59) to solve forTe, and using1T for phonon scattering, we then have

1/2 1/222

00

323

e

e

eT T TT v eeT T T

= . (62)

Finally,

1/ 22

01 31 12 8

eT

T v

= + +

(63)

and

1/ 22 2

0 01 31 12 8 v

= + +

. (64)

Thus, if 0 v ,

2

00

31

64 v

. (65)

If 0 v >>

1/ 4 1/ 232

3ov

Recap of the transport equations:

nF

n n n

E

J q n

n

= +

=


29/36

p p Fp

p E=

n n

nqR q

t

= +

p p J qR q t =

and

T Tn p

dN fR R

dt= .

Also,

( )2 Dq

P n N N

= +

and

cond n pT J J J = +

.

To get the terminal currents, observe the following:

Now

S1 S2

i1

i2

S1 S2

i1

i2


30/36

1

1

Tcond

tot1

S

S

Dnds

t

i J nds

+

=

(66)

Note from Maxwells equation that

( )Dt t t

= =

.

1 11 Tcond

S Si J nds ds

t = +

.

conduction current displacement current

To probe further, lets integrate the continuity equation:

n nV V V

n J dv q R dv q dv

t

= +

Greens theorem

n nS V V

n J nda q R dv q dv

t

= +

Let

1 contact # oneS =

2 contact # twoS =

* remaining area;S =

i.e.,

1 2*

n n n nS S S S nda J nda J nda J nda = + +

, (67)

and similarly forp .


31/36

But

1 1

1 2

1cond

*

n pS S

n p p pS S S V V

i J nda J nda

p J nda J nda J nda q R dv q dv

t

=

= + + + +

(68)

contact1cond *S bdQi i i idt

= + + +

where

2 1contact p nS Si J nda J nda=

= contact recombination.

* * surface recombinationpS S

i J nda= =

(69)

bulk recombinationpb Vi q R dv= = (70)

change in stored chargeVdQ tdt qpdv= = (71)

Simple Examples

Creating excess carriers ( )0nn p p = = at 0x = .

x=0

Injecting surface

x=0

Injecting surface


32/36

First, we want to show that in the bulk,

n n p p p

p p J qD qD

x =

.

Now by quasi-charge neutrality, n nn = .

Also, n n nn

J qn qDx

= + +

p p pp

J qp qDx

=

.

In steady state, 0J = 0n pJ J+ = i.e., n pJ J=

or n n p ppn

qn qD qp qDx x

+ =

( )

;

n p

n p

n p

n p

pqD qD

xqn qp

D D pn p x

+ =

+

+ = +

E

i.e. ( )p p n p pp n

p p p J q D D qD

p n x x

= +

as n p p p J qD x

For this problem, the Continuity Equation gives:

2

20n np

p

P PPD

t = = +

.


33/36

E

Boundary conditions are

( ) ( ) 00 0n n n P x P P = = +

( ) 0n nx P = .

the solution is:

( ) ( )/

0 00 p

x Ln nn n

P x P P P e = +

with diffusion length p p pL D= = .

Now, if the length of the sample is short, and is W is small, (i.e., pW L ), and if( ) 0n n P W P = , the equation will have the solution:

( ) ( )0 exp / exp /n p pn P P A L B Lx x= + + ;

the boundary conditions will be:

( )0(0) 0n nn P P P = +

( ) 0n n P W P =

( ) ( )

( )

( )0sinh /

0sinh /

p

n nn p

W x L P x P P

W L

= +


34/36

x

p

p=pGL

x=0 surface x

p

p=pGL

x=0 surface

Surface recombination

Here we introduce the concept of surface recombination (see Eq. (11)). When the

surface of a semiconductor is notperfect, it will develop recombinations sites. That is,

( ) 00nS nS P P

=

at the surface

( ) 00nts nth N P P

= .

Note that the dimension ofSisLength

time(i.e., that of velocity). calledsurface

recombination velocity.

Consider minority carriers (from our previous argument, we need only consider the case

of diffusion):

In the steady state, the continuity equation gives:

2

20 p L

p

p d p pD G

t x = = +

. (72)


35/36

Butsurface recombination

at 0

p pp

F Dx

U x

= =

= =

.

In other words, ( ) 00

0 p p n nx

p D S P P

x=

=

(73)

Now the general solution for (72) is

( ) exppLp

x

L p x G A

=

Applying (73), we get

( )pp

p pL

D A

LGs =

Therefore, we have

p

p

p

pL

D

LS

S GA

+

=

That is,

( ) exp

p

p

p

p pL

D

L

pLpS

S G x

Lp x G

+

=


36/36

As expected, ifSp0, p(0)=GLp. if pS ,p(0)=0.

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221A_1_Review of Semiconductor Physics

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