EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2003

L 08 Sept 18 1

EE 5340Semiconductor Device TheoryLecture 8 - Fall 2003

Professor Ronald L. [email protected]

http://www.uta.edu/ronc

L 08 Sept 18 2

Ideal n-type Schottky depletion width (Va=0)

xn

x

qNd

Q’d =

qNdxd

x

Ex

-Em

d

n

mx qNxE

dxdE

xn

(Sheet of negative charge on metal)= -Q’d

dctsmnBni

i

x

0xdin

NNV

dxE- , qN2xn

/ln

L 08 Sept 18 3

Ideal metal to n-typeSchottky (Va > 0)

qVa = Efn - Efm

Barrier for electrons from sc to m reduced to q(bi-Va)

qBn the same (data - p.166)

DR smaller

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qBn

q(i-Va)

q’nDepl reg

L 08 Sept 18 4

daimax

d

ain

xa

ai

x

0x

NV2qE

and ,qN

V2x

are Solutions .E reduce to tends V to

due field the since ,VdxE

that is now change only Then

Effect of V 0

L 08 Sept 18 5

Schottky diodecapacitance

xn

x

qNd

-Q-Q

Q’d =

qNdxn

x

Ex

-Em

d

n

mx qNxE

dxdE

xn

Q’

VQ

VQ

C

VVV

QQQ

area jctn.A

where AQQ

j

aiai

nn

'''

,'

L 08 Sept 18 6

Schottky Capacitance(continued)• The junction has +Q’n=qNdxn (exposed

donors), and Q’n = - Q’metal (Coul/cm2),

forming a parallel sheet charge capacitor.

2aid

d

aidndn

cmCoul

VqN2

qNV2

qNxqNQ

,

,'

L 08 Sept 18 7

Schottky Capacitance(continued)• This Q ~ (i-Va)

1/2 is clearly non-linear, and Q is not zero at Va = 0.

• Redefining the capacitance,

[Fd] xA

C and ][Fd/cm x

C so

V2qN

dVdQ

C

nj

2

nj

ai

d

a

nj

,,,'

,'

'

L 08 Sept 18 8

Schottky Capacitance(continued)• So this definition of the capacitance

gives a parallel plate capacitor with charges Q’n and Q’p(=-Q’n), separated by, L (=xn), with an area A and the capacitance is then the ideal parallel plate capacitance.

• Still non-linear and Q is not zero at Va=0.

L 08 Sept 18 9

Schottky Capacitance(continued)

• The C-V relationship simplifies to

][Fd/cm 2qN

AC herew

equation model a V

1CC

2

i

d0j

21

i

a0jj

,

,

L 08 Sept 18 10

Schottky Capacitance(continued)• If one plots [Cj]

-2 vs. Va

Slope = -[(Cj0)2Vbi]

-1 vertical

axis intercept = [Cj0]-2 horizontal

axis intercept = i

Cj-2

i

Va

Cj0-2

L 08 Sept 18 11

Profiling dopantsin a Schottky diode

xn

x

qNd

-Q-Q

Q’d =

qNdxn

x

Ex

-Em

d

n

mx qNxE

dxdE

xn

Q’

VQ

VQ

C

VVV

QQQ

area jctn.A

where AQQ

j

aiai

nn

'''

,'

L 08 Sept 18 12

Arbitrary dopingprofile• If the net donor conc, N = N(x), then at xd,

the extra charge put into the DR when Va->Va+Va is Q’=-qN(xn)x

• The increase in field, Ex =-(qN/)x, by Gauss’ Law (at xn, but also const).

• So Va=-xnEx= (W/) Q’

• Further, since qN(xn)x = -Q’metal, we have the dC/dx as ...

L 08 Sept 18 13

Arbitrary dopingprofile (cont.)

j

n2

3j

n

jj

2j

2nn

j

CA

x and ,

dVdC

qA

C

A1

dCVd

qC

dxCd

xN

so , dVCd

dCxd

qNdV

xdqN

dVdQ'

C' further

ACC ,C

xxdxd

dx

dC

'''

L 08 Sept 18 14

Ideal metal to n-typeSchottky (Va > 0)

qVa = Efn - Efm

Barrier for electrons from sc to m reduced to q(bi-Va)

qBn the same

DR decr

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qBn

q(i-Va)

q’nDepl reg

L 08 Sept 18 15

Ideal m to n s/c Schottky diode curr

t0B

2sT

tasmssm

taiDsa

sntiDs

mmmssnssma

mmmssnssm

VTAJ

1VVJJJJ so

,VVNn ,0V

constv ,VNn and

,qvnJqvnJ ,0V

qvnJ ,qvnJ

/exp*

/exp

/exp

/exp ,

,

,

L 08 Sept 18 16

Metal to n-typenon-rect cont (m<s)

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qB,n

qn

No disc in Eo

Ex=0 in metal ==> Eo flat

B,n=m - s =

elec mtl to s/c barr

i= Bn-n< 0

Accumulation regionAcc reg

qi

L 08 Sept 18 17

Metal to n-typeaccum region (m<s)

junctiontorsemiconduc to metal

the at ionconcentrat electron the is

kTNkTEENn

and ,qnV

L ,L 2x1

qn

by given is region

accum the in density charge local The

idFsccs

s

tD

D

s

/exp/exp

/

L 08 Sept 18 18

Metal to s/c contact resistance

0x

NN for ,

N

m2R

VTAV

R

1VV

VTAJ

cm-ohm , VJ

R

n

cd

d

Bnntunnelc

t

Bn2

tSchottkyc

t

a

t

Bn2n

2

0V

1n

c

*

,

*,

exp

exp

expexp*

L 08 Sept 18 19

Energy bands forp- and n-type s/c

p-type

Ec

Ev

EFi

EFP

qP= kT ln(ni/Na)

Ev

Ec

EFi

EFNqn= kT ln(Nd/ni)

n-type

L 08 Sept 18 20

Making contactin a p-n junction• Equate the EF in

the p- and n-type materials far from the junction

• Eo(the free level), Ec, Efi and Ev must be continuous

N.B.: q = 4.05 eV (Si),

and q = qEc - EF

Eo

EcEF EFi

Ev

q (electron affinity)

qF

q(work function)

L 08 Sept 18 21

Band diagram forp+-n jctn* at Va = 0

Ec

EFNEFi

Ev

Ec

EFP

EFi

Ev

0 xn

x-xp

-xpc xnc

qp < 0

qn > 0

qVbi = q(n - p)

*Na > Nd -> |p| > n

p-type for x<0 n-type for x>0

L 08 Sept 18 22

• A total band bending of qVbi = q(n-p) = kT ln(NdNa/ni

2) is necessary to set EFp = Efn

• For -xp < x < 0, Efi - EFP < -qp, = |qp| so p < Na = po, (depleted of maj. carr.)

• For 0 < x < xn, EFN - EFi < qn, so n < Nd = no, (depleted of maj. carr.)

-xp < x < xn is the Depletion Region

Band diagram forp+-n at Va=0 (cont.)

L 08 Sept 18 23

DepletionApproximation• Assume p << po = Na for -xp < x < 0, so

= q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Na for -xpc < x < -xp, so = q(Nd-Na+p-n) = 0, -xpc < x < -xp

• Assume n << no = Nd for 0 < x < xn, so = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Nd for xn < x < xnc, so = q(Nd-Na+p-n) = 0, xn < x < xnc

L 08 Sept 18 24

Depletion approx.charge distribution

xn

x-xp

-xpc xnc

+qNd

-qNa

+Qn’=qNdxn

Qp’=-qNaxp

Charge neutrality => Qp’ + Qn’ = 0,

=> Naxp = Ndxn

[Coul/cm2]

[Coul/cm2]

L 08 Sept 18 25

Induced E-fieldin the D.R.• The sheet dipole of charge, due to

Qp’ and Qn’ induces an electric field which must satisfy the conditions

• Charge neutrality and Gauss’ Law* require that Ex = 0 for -xpc < x < -xp and Ex = 0 for -xn < x < xnc QQAdxEAdVdSE '

p'n

xx

xxx

VS

n

p

0

L 08 Sept 18 26

Induced E-fieldin the D.R.

xn

x-xp-xpc xnc

O-O-O-

O+O+

O+

Depletion region (DR)

p-type CNR

Ex

Exposed Donor ions

Exposed Acceptor Ions

n-type chg neutral reg

p-contact N-contact

W

0

L 08 Sept 18 27

Induced E-fieldin the D.R. (cont.)• Poisson’s Equation E = /, has

the one-dimensional form, dEx/dx = /,

which must be satisfied for = -qNa, -xp < x < 0,

and = +qNd, 0 < x < xn, with Ex = 0 for the remaining range

L 08 Sept 18 28

Soln to Poisson’sEq in the D.R.

xnx

-xp

-xpc xnc

Ex

-Emax

dx qN

dxdE

ax qN

dxdE

L 08 Sept 18 29

Test 1 - 25Sept03

• 8 AM Room 206 Activities Building• Open book - 1 legal text or ref., only.• You may write notes in your book.• Calculator allowed• A cover sheet will be included with

full instructions. See http://www.uta.edu/ronc/5340/tests/ for examples from Fall 2002.

EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2003

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