phys112 s14 semiconductors - uc berkeley cosmology...

Phys112 (S2014) 9 Semiconductors

Semiconductorscf. Kittel and Kroemer chap. 13

also: S.M. Sze Physics of Semiconductor Devices

States in a semiconductor Bands and gap Impurities Electrons and holes

Position of the Fermi level Intrinsic Doped= Extrinsic

The p-n junction Band bending, depletion region Forward and reverse biasing Current voltage characteristics

Devices Rectification diode The bipolar transistor The MOS FET/Charged coupled devices Optical devices

1


MotivationsPlay with Fermi Dirac distributions !Understand qualitatively transistors

2

ε

E C

B

Emitter base Collector

Change of this potential barrier strongly affects emitter-collector current

Phys 112 (F2014) 7 Fermi Dirac/Bose Einstein B.Sadoulet

Gap from Sze “Physics of Semiconductor devices” p13

Wiley-InterScience 1981

3

E

k

εc

εv

conduction band

valence band

Gap

Ge,Si,GaAs

Phys 112 (F2014) 7 Fermi Dirac/Bose Einstein B.Sadoulet

from Sze “Physics of Semiconductor devices” p850 Wiley-InterScience 1981

4

Example Ge,Si,GaAs

Phys 112 (F2014) 7 Fermi Dirac/Bose Einstein B.Sadoulet5

Intrinsic Semiconductors(no role of impurities)

Statistical distribution Still good approximation to consider free electrons as quantum ideal gas => occupation number !Density of states !!!!!We can then get the total number of electrons

f ε( ) = 1exp ε − µ( ) /τ( ) +1

neT = f ε( )D ε( )dε0

∞

∫ = f ε( )Dh ε( )dε0

εv

∫ + f ε( )De ε( )dεεc

∞

∫

= 1−1

exp µ − ε( ) /τ( ) +1⎛

⎝ ⎜

⎞

⎠ ⎟ Dh ε( )dε

0

ε v

∫ +1

exp ε − µ( )/τ( ) +1De ε( )dε

εc

∞

∫

Dh(ε)dε =

24π 2

2mh*

!2⎛

⎝ ⎜

⎞

⎠ ⎟

32(εv −ε )dε

De(ε)dε =

24π 2

2me*

!2⎛

⎝ ⎜

⎞

⎠ ⎟

32(ε − εc)dεεc

εv

ε ε

Gap

D (ε)

conduction band

valence band

2 spin states

Parabolic at gap edge

Electron state density below the gap


Electrons and HolesThis can be rewritten as !!!!!!Therefore we can describe the electron population by two non

relativistic “gases”: holes and electrons (cf. what we did with metals). The equality of the number of holes and electrons fixes the chemical

potential: Charge neutrality! Fermi level: in the middle of the gap if mh*=me* !!!!!The red expressions apply to the case where the 1 is negligible in front of

exponential (non degenerate semi conductors)

�

neT = nvTtotal invalence bandat zero temperature=total number of elctrons

! − 1exp µ −ε( ) /τ( ) + 1

Dh ε( )dε0

εv

∫holes

" # $ $ $ $ $ % $ $ $ $ $ + 1

exp ε − µ( ) /τ( ) + 1De ε( )dε

εc

∞

∫electrons

" # $ $ $ $ $ % $ $ $ $ $

⇒ 1exp µ −ε( ) /τ( ) + 1

Dh ε( )dε0

ε v

∫holes

" # $ $ $ $ $ % $ $ $ $ $ = 1

exp ε − µ( ) /τ( ) + 1De ε( )dε

εc

∞

∫electrons

" # $ $ $ $ $ % $ $ $ $ $


Chemical potential (Intrinsic)No impurities occupation number !!!!!Yes do intersect at 1/2 but does not fix position! charge neutrality does!

εv εcεµ

f

εv εcµµ

logne(µ)lognh µ( )

1

1/2


Classical limitMeasuring from the edge of the valence and conduction band

respectively !!!!

8

ne =2

4π 22me

*

!2

⎛⎝⎜

⎞⎠⎟

3/21

exp ε '− µ − εc( )( ) /τ( )+1ε ' dε '

0

∞

∫

≈ nQe exp − εc − µτ

⎛⎝⎜

⎞⎠⎟ with nQe ≡ nc = 2 me

*τ2π!2

⎛⎝⎜

⎞⎠⎟

32 in the classical limit: exp( ) >>1

nh =24π 2

2mh*

!2⎛⎝⎜

⎞⎠⎟

3/21

exp ε '− εv − µ( )( ) /τ( )+1 ε ' dε '0

∞

∫

≈ nQh exp − µ − εvτ

⎛⎝⎜

⎞⎠⎟ with nQh = nv = 2

mh*τ

2π!2⎛⎝⎜

⎞⎠⎟

32 in the classical limit: exp( ) >>1


Fermi Level (Intrinsic)Law of mass action

true even if not intrinsic (impurities) !

!!Intrinsic Fermi level (non degenerate)

Imposing neutrality, from previous expressions one can deduce that ! !!!!!very close to middle of the gap!

9

µ =εc + εv2

+τ2log

nvnc

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

εc + εv2

+3τ4log

mh*

me*

⎛

⎝ ⎜

⎞

⎠ ⎟

nenh = ncnv exp −εc − εv

τ⎛ ⎝ ⎜ ⎞

⎠ ⎟ = ni

2

with ni = ncnv exp −εc − εv

2τ⎛ ⎝ ⎜ ⎞

⎠ ⎟

ne = nc exp −εc − µτ

⎛ ⎝

⎞ ⎠ = nh = nv exp −

µ − εvτ

⎛ ⎝

⎞ ⎠


April 30

Evaluation survey !Midterm !Qualitative diode behavior of p-n junction

Heuristic Shockley equation !

Schottky diode A simpler problem Equilibrium Biasing. Shockley equation/

10


Semiconductors: Role of impuritiesLarge role of impurities: localized states (Not band !) in gap

If they are shallow (≈ 40meV (Si) ≈10meV (Ge)), such states can be excited at room temperature. This modifies totally the behavior!

Donors ! Acceptors ! note: 2 A0 state because a bond is missing and the missing

electron can be spin up or down, A- bond established (pair of electrons of antiparallel

spins) : 1 state ⇒The number of free electrons is no more equal to number of holes

Number of electrons can be increased by donors and decreased by acceptors

But we need to keep charge neutrality = method to compute the Fermi level

⇒For large enough impurities concentration, the Fermi level can move close to the edge of the gap

⇒(Thermally generated) conductivity either dominated by • electron like excitation: negative carriers (n type) • hole like excitation: positive carriers (p type)

11

do ↔ d+ + e− nd = nd + + nd o

a− ↔ ao + e− na = na− + nao

k

εc

εv

εDεA


Fermi Level (Extrinsic) General caseNeutrality

at equilibrium !

Non degenerate case: Impose neutrality

!!

very close to energy level of dominant impurity! !cf Kittel & Kroemer Fig. 13.6

12

ne + na− = nh + nd+

common µ

nc exp −εc − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +

na1+ 2exp εa − µ

τ⎛ ⎝ ⎜ ⎞

⎠ ⎟

= nv exp −µ − εvτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +

nd1+ 2exp µ − εd

τ⎛ ⎝ ⎜ ⎞

⎠ ⎟

n (µ )n+

µ

n-

εcεv

nd

na

µ


p-n Junctions: Qualitative propertiesBring together one n type semiconductor

and one p type On n side, majority carriers are

electrons, but on p side, there are hole => n side electrons diffuse to p side and

annihilate with holes/are trapped on acceptors

=> p side holes diffuse to n side and annihilate with electrons/ are trapped on donors !!!!This builds up a dipole layer of charges

(positive on n side, negative on p side) which generates a built-in potential which “bends” the bands and equalizes the Fermi levels on both sides

Note: positive charges =>positive curvature !13

n typep type

∇2ϕ = − ρ∈

but energy εe = −qϕ ⇒∇2εe =ρ∈

where the charge of electron is − q, q > 0

εc

εv

εc

εv

++ ++

-- --


Diode effect

14

Aligning the chemical potentials corresponds to internal voltage Vn −Vp =Vbi > 0

but additional junctions imposed by equilibrium will correspond to no external voltage Vnext =Vpext

εc

++ ++

-- --


Diode effect

This indeed acts as a diode If !, more reverse biasing, we have very

small current current< reverse diffusion current

!For forward biasing, !, current begins to flow !!!celebrated Shockley equation!

15

εc

εv

εc

εv

Vnext >Vp

ext ⇒Vn −Vp >Vbi

Vnext <Vp

ext ⇒Vn −Vp <Vbi

Aligning the chemical potentials corresponds to internal voltage Vn −Vp =Vbi > 0

but additional junctions imposed byequilibrium will correspond to no external voltage Vnext =Vpext

J = J0 expqVτ

⎛⎝⎜

⎞⎠⎟ −1

⎡⎣⎢

⎤⎦⎥

reverse diffusion current

εc

εv

εc

εv


A simpler example: Schottky diode

Junction between semiconductor and metal

At equilibrium Have to align chemical potential = Fermi

level of metal

16

++++++

Metaln type

Assume high but not degenerate doping , e.g, na << ni << nd << ncthe classical approximation gives

ne x( ) = nc exp −εc x( )− µ

τ⎛⎝⎜

⎞⎠⎟= nc exp −

εc x( )− εFτ

⎛⎝⎜

⎞⎠⎟

where εF is the metal Fermi level.

Choosing ϕ = 0, x = 0 at the metal interface εc x( ) = εc − qϕ x( ) where εc is fixed with respect to metal Fermi level by relative work functions.Calling ϕ∞ the potential at x = ∞ where the semiconductor is undisturbed

ne x = ∞( ) = nc exp − εc − qϕ∞ − εFτ

⎛⎝⎜

⎞⎠⎟ = neothe electron concentration at zero field in the n type semiconductor

⇒ nc = neo expεc − qϕ∞ − εF

τ⎛⎝⎜

⎞⎠⎟ and ne x( ) = nc exp −

εc − qϕ x( )− εFτ

⎛⎝⎜

⎞⎠⎟= ne0 exp −

q ϕ∞ −ϕ x( )( )τ

⎛

⎝⎜⎞

⎠⎟

ϕ∞

ϕ x( )

x


Shottky diode at equilibrium

17

Now use Poisson equation

∂2ϕ∂x2

= −q n

d++ nh − ne − na−( )

∈≈ −

q nd+− ne( )∈

≈ −

qnd 1− exp −q ϕ∞ −ϕ x( )( )

τ⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

∈

Multiplying by 2 ∂ϕ∂x

and integrating

∂ϕ x( )∂x

⎛⎝⎜

⎞⎠⎟

2

= Ex2 x( ) = −2 qnd

∈ϕ x( )−ϕ∞ −

τqexp −

q ϕ∞ −ϕ x( )( )τ

⎛

⎝⎜⎞

⎠⎟−1

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟

where we have fixed the constant of integration by imposing that Ex x = ∞( ) = 0Except when ϕ∞ ≈ϕ x( ).

exp −q ϕ∞ −ϕ x( )( )

τ⎛

⎝⎜⎞

⎠⎟is very small and can be neglected. We then have the approximate solution

Using the standard notation ϕ∞ equilibrium( ) =Vbi

ϕ x( ) =qnd∈

xxn −x2

2⎡

⎣⎢

⎤

⎦⎥ for x ≤ xn =

2∈qnd

Vbi≡ϕ∞

! − τq

⎛

⎝⎜

⎞

⎠⎟ = barrier/depletion width

Vbi for x > xn

E x( ) = − ∂ϕ∂x

=qnd∈

x − xn( )for x ≤ xn

0 for x > xnNet current is zero, because we are in equilibrium

Ex

x

xn++++----


From equilibrium to transportEquilibrium Transport

18

at equilibrium, in the non degenerate case (≈classical)

ne x( ) = nc exp −εc x( )− µ

τ⎛⎝⎜

⎞⎠⎟

with εc x( ) = εc − qϕ x( )

or

µ=εc −τ logne x( )nc

⎛⎝⎜

⎞⎠⎟

internal! "### $###

−qϕ x( )external!"# $#

+ Poisson equation

∂2ϕ∂x2

= −q n

d++ nh − ne − na−( )

∈

∂2ϕ∂x2

≈ −q n

d+− ne( )∈

≈ −

qnd 1− exp −q ϕ∞ −ϕ x( )( )

τ⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

∈

Current densities (1D)

Je = q !µeneE + qDe∂ne∂x

Jh = q !µhnhEDrift!"# $# −qDh

∂nh∂x

diffusion! "# $#

Einstein relation

De =τq!µe Dh =

τq!µh

Je = !µe −neq∂ϕ∂x

+τ ∂ne∂x

⎡⎣⎢

⎤⎦⎥

Jh = !µh −nhq∂ϕ∂x

−τ ∂nh∂x

⎡⎣⎢

⎤⎦⎥

+ Poisson equation

∂2ϕ∂x2

= −q n

d++ nh − ne − na−( )

∈

!µe

!µh

are the mobilities:

Drift velocities"we = − !µe

"E

"wh = !µh

"E


Physical Interpretation of Constancy of Chemical Potential

at Equilibrium ***Einstein relation is easily demonstrated in kinetic theory. Let us look at the diffusion equation at equilibrium !!!!!!!!!!

The constancy of the total chemical potential is due to the balance of drift and diffusion at equilibrium

19

Je = !µe −qne∂ϕ∂x

drift!"# $#

+ τ ∂ne∂x

Diffusion!

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

At equilibrium

Je = 0 ⇒ − q ∂ϕ∂x

+ τne

∂ne∂x

= 0

⇒−qϕ +τ log nen0

= constant = µ if we choose n0 = nc


Rate equations

20

Continuity equations = conservation of particles∂ne∂t

+ 1q∂Je∂x

= 0 ∂ne∂t

= Γ0→nh − Γnh→0nenh + Γd0→d++end0− Γ

d++e→d0nd+ne + Γa−→a0+e

na−− Γ

a0+e→a−na0ne

∂nh∂t

+ 1q∂Jh∂x

= 0 ∂nh∂t

= Γ0→nh − Γnh→0nenh + Γa0→a−+hna0− Γ

a−+h→a0na−nh + Γd_→d0+h

nd+− Γ

d0+h→d+nd0nh

Rate equations for donors and acceptors∂nd+∂t

= Γd0→d++e

nd0− Γ

d++e→d0nd+ne + Γd0+h→d+

nd0nh − Γd+→d0+h

nd+ = −∂n

d0

∂t∂n

a−

∂t= Γ

a0→a−+hna0− Γ

a−+h→a0na−nh + Γa0+e→a−

na0ne − Γa−→a0+e

na−= −

∂na0

∂tThe emission terms which only depend on temperature are equal tothe absorption terms at equilibrium. These rates are straighforwardly computed from cross sections

Γ0→nh − Γnh→0nenh = −Γnh→0 ne,nh( ) nenh −ni2

equilibrium!

⎡

⎣⎢⎢

⎤

⎦⎥⎥

In the steady state we set ∂n

d+

∂t= 0

∂na−

∂t= 0 ∂ne

∂t+ 1q∂Je∂x

= 0 ∂nh∂t

+ 1q∂Jh∂x

carriers can be brought along by currents! "##### $#####

= 0


Approximations

21

For analytical results, we make the appropriate approximations given orders of magnitudee.g. , for annililation of minority carriers (e.g. holes on the n-side)

∂ne∂t

= −Γnh→0 ne,nh( ) nenh −ni2

equilibrium!

⎡

⎣⎢⎢

⎤

⎦⎥⎥

≈ −Γnh→0 ne0,nh0( )nh0equilibrium

! "### $###ne −

ni2

nh0=ne0!

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= − ne − ne0

τ e returns to equilibrium density

This usually dominates over other processes!In depletion region the electric field is high and capture is small. Hence in n semiconductornd+≈ nd nd0 ≈ 0 n

a−≈ na << nd


Consequence for Schottky diode

22

In the depletion region in the n side, Poisson equation is still

∂2ϕ∂x2

= −q n

d++ nh − ne − na−( )

∈≈ −

q nd+− ne( )∈

≈ −

qnd 1− exp −q ϕ∞ −ϕ x( )( )

τ⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

∈but now

ϕ∞ =Vbi −V and the approximate solution becomes for x ≤ xn' =2∈qnd

Vbi −V − τq

⎛⎝⎜

⎞⎠⎟

Depletion decreases for forward bias

! "#### $####

ϕ x( ) = qnd∈

xx 'n−x2

2⎡

⎣⎢

⎤

⎦⎥ E x( ) = qnd

∈x − x 'n[ ]

ne x( ) = nd exp −q ϕ∞ −ϕ x( )( )

τ⎛

⎝⎜⎞

⎠⎟

⇒ ne x = 0( ) = nd exp −q VBi −V( )

τ⎛⎝⎜

⎞⎠⎟= ne0 (x = 0)

equilibrium! "# $# exp qV

τ⎛⎝⎜

⎞⎠⎟


Biasing a Shottky diode

Same qualitative behavior as p-n diode

23

Sze

Equilibrium

Forward

Reverse

Sze


Biasing a Shottky diode

Same qualitative behavior as p-n diode

24

At the metal-semiconductor interface

ne = nd exp −q VBi −V( )

τ⎛⎝⎜

⎞⎠⎟= ne0

equilibrium! exp qV

τ⎛⎝⎜

⎞⎠⎟

J = J0 expVτ

⎛⎝⎜

⎞⎠⎟

forward!"# $#

−1reverse

stays the same

!

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

How to determine J0? • Ballistic flux of electrons over the barrier (Bethe) • Slow diffusion up the barrier (Schottky) depends on the semiconductor…


p-n Junctions at Equilibriumn side

ϕ x( ) =qnd∈

xxn −x2

2⎡

⎣⎢

⎤

⎦⎥ for 0 ≤ x ≤ xn =

2∈qnd

ϕ∞n −τq

⎛⎝⎜

⎞⎠⎟

ϕ∞n for x > xn

En x( ) = − ∂ϕ∂x

=qnd∈

x − xn( )for x ≤ xn

0 for x > xnp side

ϕ x( ) = − qna∈

xxp −x2

2⎡

⎣⎢

⎤

⎦⎥ for xp ≤ x ≤ 0 = − 2∈

qna−ϕ∞p −

τq

⎛⎝⎜

⎞⎠⎟

ϕ∞p for x ≤ xp

Ep x( ) = − ∂ϕ∂x

=− qna

∈x − xp( )for xp ≤ x ≤ 0

0 for x ≤ xp

We have to link the two solutionsEp x = 0( ) = En 0( )⇔ xp na = xnnd

⇒Vbi =ϕ∞n −ϕ∞p =εg −τ log

nandni2

⎛⎝⎜

⎞⎠⎟

q=εg −τ log

ncnvnand

⎛⎝⎜

⎞⎠⎟

q⇒W = xn − xp =

2∈q

nandna + nd

⎛⎝⎜

⎞⎠⎟Vbi − 2

τq

⎛⎝⎜

⎞⎠⎟

xxnxp

x

xnxp

nenh

Vp

ϕ∞n

ϕ∞p

Ex

xxn

++++xp

---

-

ϕ

n


Biasing a pn junction

similarly to Schottky diode !!!!How to compute J0?

26

εc

εv

εc

εv

reverse diffusion current

εc

εv

εc

εv

J = J0 expqVτ

⎛⎝⎜

⎞⎠⎟ −1

⎡⎣⎢

⎤⎦⎥.


Biased p-n diode

27

Under the assumption of complete local depletionthe densities at the edge of the depletion region

ne x = xp( ) = nd exp −q VBi −V( )

τ⎛⎝⎜

⎞⎠⎟= ne0 xp( )

equilibrium!"# $#

exp qVτ

⎛⎝⎜

⎞⎠⎟

nh x = xn( ) = na exp −q VBi −V( )

τ⎛⎝⎜

⎞⎠⎟= nh0 xn( )

equilibrium!"# $#

exp qVτ

⎛⎝⎜

⎞⎠⎟

Where do these additional electrons or holes go?Dominant mechanism: annihilation with majority carriers

∂ne xp( )∂t

= −ne xp( )− ne0 xp( )

equilibrium!"# $#

τ e

q∂ne xp( )

∂t= − ∂Je

∂xfrom continuity!"# $#

= − %µe∂∂x

−neq∂ϕ∂x

drift!"# $#

+ τ ∂ne∂x

Diffusion!

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⇒ !µeτ∂2ne∂x2

− qne xp( )− ne0 xp( )

τ e= 0⇒ ne x( ) = ne0 + ne xp( )− ne0⎡⎣ ⎤⎦exp

xDeτ !µeq

!τ e

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟


Biased p-n diode

28

The current generated at xp is Je xp( ) = Deq∂ne∂x

= De

τ eq ne xp( )− ne0

=ne0 expqVτ

⎛⎝⎜

⎞⎠⎟−1

⎡⎣⎢

⎤⎦⎥

! "# $#

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Similarly, the current generated at xn is Jh xn( ) = Dhqτ∂nh∂x

= Dh

τ hq nh xn( )− nh0

=nh0 expqVτ

⎛⎝⎜

⎞⎠⎟−1

⎡⎣⎢

⎤⎦⎥

! "# $#

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

J = Jn xp( )+ Jh xn( ) = De

τ eqne0 +

Dh

τ hqnh0

⎛

⎝⎜⎞

⎠⎟exp qV

τ⎛⎝⎜

⎞⎠⎟ −1

⎡⎣⎢

⎤⎦⎥

of the announced form J = J0 expqVτ

⎛⎝⎜

⎞⎠⎟ −1

⎡⎣⎢

⎤⎦⎥. Shockley equation


Summary

29

Sze


I-V Curve

30

Sze I

V

V

log|I|


Bipolar devices

p-n Diode !!Bipolar transistor

The base potential changes the potential barrier! Ebers-Moll Equation

31

IB = IS expeVBEτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1⎛

⎝ ⎜ ⎞

⎠ ⎟

∂IC∂IB

= hFE = β >>1

ε

E C

B

Emitter base Collector

Change of this potential barrier strongly affects emitter-collector current

n p


MOS devicesBending bands with external potential

Presence of electrons in a metal on other side of an oxide bends the energy levels

Potential on metal can create 2 dimensional pools of carriers of either sign depending on sign of voltage !Easy to fabricate=> MOS + low power! !!!!

MOSFET => CMOS

32

metalOxide

semiconductor


Charge Coupled Devices

Transferring !!!!

!!!!!!!We can play with implantation to bury the channel in order to

prevent trapping at surface!

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Optical Devices

Light emitting diode If the gap is direct (e.g. in GaAs), the hole-electron

recombination can generate photons! !!!!!!!

Laser

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ε

k

hν


Photodiodes

Principle Photon of high enough energy absorbed in depletion region can

promote an electron from the valence band to the conduction band. Depletion field separates electron and hole => current

!!!=> Photodiodes

Solar cells Also: Semiconductor particle detectors

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n p

hν

phys112 s14 semiconductors - uc berkeley cosmology...

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