EE 5340Semiconductor Device TheoryLecture 11 – Spring 2011

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

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Metal/semiconductorsystem types

n-type semiconductor• Schottky diode - blocking for fm >

fs

• contact - conducting for fm < fs

p-type semiconductor• contact - conducting for fm > fs

• Schottky diode - blocking for fm < fs

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Real Schottkyband structure1

• Barrier transistion region, d• Interface states

above fo acc, p neutrl

below fo dnr, n neutrl

Ditd -> oo, qfBn = Eg- fo

Fermi level “pinned”

Ditd -> 0, qfBn = fm - cGoes to “ideal” case

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Fig 8.41 (a) Image charge and electric field at a metal-dielectric interface (b) Distortion of potential barrier at E=0 and (c) E0

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silicon for 711

andFd/cm, ,14E858

with , ypermitivit the is

xE

E where, ,E

r

o

ro

x

.

.

Poisson’s Equation• The electric field at (x,y,z) is

related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation:

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Poisson’s Equation• n = no + dn, and p = po + dp, in

non-equil• For n-type material, N = (Nd - Na) >

0, no = N, and (Nd-Na+p-n)=-dn +dp +ni

2/N

• For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = dp-dn-ni

2/N

• So neglecting ni2/N

0n or p with material type-p

and type-n for ,npq

E

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Ideal metal to n-typebarrier diode (fm>fs,Va=0)

EFn

Eo

Ec

Ev

EFi

qfs,n

qcs

n-type s/c

qfm

EF

m

metal

qfBn

qfbi

qf’n

No disc in Eo

Ex=0 in metal ==> Eoflat

fBn=fm- cs = elec mtl to s/c barr

fbi=fBn-fn= fm-fs elect s/c to mtl barr

Depl reg

0 xn xnc

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DepletionApproximation• For 0 < x < xn, assume n << no =

Nd, so r = q(Nd-Na+p-n) =

qNd

• For xn < x < xnc, assume n = no =

Nd, so r = q(Nd-Na+p-n) = 0

• For x = 0-, there is a pulse of charge balancing the qNdxn in 0 <

x < xn

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

xn

x

qNd

d

Q’d =

qNdxn

x

r Ex

-Em

d

n

mx qNxE

dxdE

xn

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

dctsmnBni

i

x

0xdin

NNV

dxE- , qN2xn

/ln

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Debye length

• The DA assumes n changes from Nd to 0 discontinuously at xn.

• In the region of xn, Poisson’s eq is

E = r/e --> dEx/dx =

q(Nd - n), and since Ex = -df/dx, we

have -d2f/dx2 = q(Nd -

n)/e to be solved

n

xxn

Nd

0

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Debye length (cont)• Since the level EFi is a reference

for equil, we set f = Vt ln(n/ni)

• In the region of xn, n = ni exp(f/Vt),

so d2f/dx2 = -q(Nd - ni ef/Vt), let

f = fo + f’, where fo = Vt

ln(Nd/ni) so Nd - ni ef/Vt = Nd[1 -

ef/Vt-fo/Vt], for f - fo = f’ << fo, the

DE becomes d2f’/dx2 =

(q2Nd/ekT)f’, f’ << fo

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Debye length (cont)• So f’ = f’(xn) exp[+(x-xn)/LD]

+con. and n = Nd ef’/Vt, x ~ xn,

where LD is the “Debye

length”

material. intrinsic for 2n and type-p

for N type,-n for N pn :Note

length. transition a ,q

kTV ,

pnqV

L

i

ad

tt

D

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Debye length (cont)

• LD estimates the transition length of a step-junction DR. Thus,

i

t

0V

dD

2V

W

NL

a

• For Va = 0, i ~ 1V, Vt ~ 25 mV d < 11% DA

assumption OK

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Effect of V 0• Define an external voltage source,

Va, with the +term at the metal contact and the -term at the n-type contact

• For Va > 0, the Va induced field

tends to oppose Ex caused by the DR

• For Va < 0, the Va induced field

tends to aid Ex due to DR

• Will consider Va < 0 now

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dimax

d

in

xa

ai

x

0x

NVa2qE

and ,qN

Va2x

are Solutions .E reduce to tends V to

due field the since ,VdxE

that is now change only Then

Effect of V 0

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Ideal metal to n-typeSchottky (Va > 0)

qVa = Efn - Efm

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

qfBn the same

DR decr

EFn

Eo

Ec

Ev

EFi

qfs,n

qcs

n-type s/c

qfm

EF

m

metal

qfBn

q(fi-Va)

qf’nDepl reg

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Schottky diodecapacitance

xn

x

qNd

-Q-dQ

Q’d =

qNdxn

x

r

Ex

-Em

d

n

mx qNxE

dxdE

xn

dQ’

VQ

VQ

C

VVV

QQQ

area jctn.A

where AQQ

j

aiai

nn

'''

,'

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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

,

,'

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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

,,,'

,'

'

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Schottky Capacitance(continued)• So this definition of the

capacitance gives a parallel plate capacitor with charges dQ’n and

dQ’p(=-dQ’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.

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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

,

,

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

-2 vs. Va

Slope = -[(Cj0)2Vbi]-1

vertical axis intercept

= [Cj0]-2 horizontal axis

intercept = fi

Cj-2

fiVa

Cj0-2

Diagrams for ideal metal-semiconductor Schottky diodes. Fig. 3.21 in Ref 4.

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Energy bands forp- and n-type s/c

p-typeEc

Ev

EFi

EFP

qfP= kT ln(ni/Na)

Ev

Ec

EFi

EFNqfn= kT ln(Nd/ni)

n-type

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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.: qc = 4.05 eV (Si),

and qf = qc + Ec - EF

Eo

EcEF EFi

Ev

qc (electron affinity)

qfF

qf(work function)

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

Ec

EFNEFi

Ev

Ec

EFP

EFi

Ev

0 xn

x-xp

-xpc xnc

qfp < 0

qfn > 0

qVbi = q(fn - fp)

*Na > Nd -> |fp| > fn

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

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

2)

is necessary to set EFp = Efn

• For -xp < x < 0, Efi - EFP < -qfp, = |qfp|

so p < Na = po, (depleted of maj. carr.)

• For 0 < x < xn, EFN - EFi < qfn,

so n < Nd = no, (depleted of maj. carr.)

-xp < x < xn is the Depletion Region

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

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

0, so r = q(Nd-Na+p-n) = -qNa, -xp

< x < 0, and p = po = Na for -xpc <

x < -xp, so r = q(Nd-Na+p-n) =

0, -xpc < x < -xp

• Assume n << no = Nd for 0 < x <

xn, so r = q(Nd-Na+p-n) = qNd, 0 <

x < xn, and n = no = Nd for xn < x

< xnc, so r = q(Nd-Na+p-n) =

0, xn < x < xnc

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Depletion approx.charge distribution

xn

x-xp

-xpc xnc

r+qNd

-qNa

+Qn’=qNdxn

Qp’=-qNaxp

Due to Charge

neutrality Qp’ + Qn’ =

0, => Naxp =

Ndxn

[Coul/cm2]

[Coul/cm2]29

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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

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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

031

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

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

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

+qNd, 0 < x < xn, with

Ex = 0 for the remaining range

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Soln to Poisson’sEq in the D.R.

xnx

-xp

-xpc xnc

Ex

-Emax

dx qN

dxdE

ax qN

dxdE

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Soln to Poisson’sEq in the D.R. (cont.)

)Vq

kT (note ,xNxN

2q

dxdV

E ,dxEVn

NNln

qkT

that is D.R. the in P.E. the of soln

the to V of iprelationsh the Now,

t2pa

2nd

x

x

xxbi2

i

da

bi

n

p

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Soln to Poisson’sEq in the D.R. (cont.)

WV2N2qV

E then

,WE21

V have also must we Since

.NN

NNN where ,

qNV2

W

then ,xxW let and ,xNxN

bieffbimax

maxbi

da

daeff

eff

bi

pnpand

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Comments on theEx and Vbi

• Vbi is not measurable externally since Ex is zero at both contacts

• The effect of Ex does not extend beyond the depletion region

• The lever rule [Naxp=Ndxn] was obtained assuming charge neutrality. It could also be obtained by requiring Ex(x=0-dx) = Ex(x=0+dx) = Emax

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Sample calculations• Vt = 25.86 mV at 300K

• e = ereo = 11.7*8.85E-14 Fd/cm= 1.035E-12 Fd/cm

• If Na=5E17/cm3, and Nd=2E15

/cm3, then for ni=1.4E10/cm3, then

what is Vbi = 757 mV

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Sample calculations• What are Neff, W ?

Neff, = 1.97E15/cm3

W = 0.707 micron• What is xn ?

= 0.704 micron• What is Emax ? 2.14E4 V/cm

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References1Device Electronics for Integrated Circuits, 2 ed.,

by Muller and Kamins, Wiley, New York, 1986. See Semiconductor Device Fundamentals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model.

2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

3Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997.

4Device Electronics for Integrated Circuits, 3/E by Richard S. Muller and Theodore I. Kamins. © 2003 John Wiley & Sons. Inc., New York.

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References1 and M&KDevice Electronics for Integrated

Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. See Semiconductor Device Fundamentals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model.

2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

3 and **Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997.

Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.