EE 5340Semiconductor Device TheoryLecture 8 - Fall 2009

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

L 08 Sept 18 2

Test 1 – Sept. 24, 2009• 8 AM Room 108 Nedderman Hall• 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 previous semesters.

L 08 Sept 17 3

Si and Al and model (approx. to scale)

qm,Al ~ 4.1 eV

Eo

EF

mEFp

EFn

Eo

Ec

Ev

EFi

qs,n

qsi~ 4.05 eV

Eo

Ec

Ev

EFi

qs,p

metal n-type s/c p-type s/c

qsi~ 4.05 eV

L 08 Sept 17 4

Making contact be-tween metal & s/c• Equate the EF in the

metal and s/c materials far from the junction

• Eo(the free level), must be continuous across the jctn.

N.B.: q = 4.05 eV (Si),and q = qEc - EF

Eo

EcEF EFiEv

q (electron affinity)

qF

q(work function)

L 08 Sept 17 5

Equilibrium Boundary Conditions w/ contact• No discontinuity in the free level, Eo at

the metal/semiconductor interface.• EF,metal = EF,semiconductor to bring the

electron populations in the metal and semiconductor to thermal equilibrium.

• Eo - EC = qsemiconductor in all of the s/c.• Eo - EF,metal = qmetal throughout metal.

L 08 Sept 17 6

Ideal metal to n-typebarrier diode (m>s,Va=0)

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qBnqi

q’n

No disc in Eo

Ex=0 in metal ==> Eoflat

Bn=m- s = elec mtl to s/c barr

i=Bn-n= m-s elect s/c to mtl barr Depl reg

L 08 Sept 17 7

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 region

Acc reg

qi

L 08 Sept 17 8

Ideal metal to p-typebarrier diode (m<s)

EFp

Eo

Ec

Ev

EFi

qs,p

qs

p-type s/c

qm

EF

m

metal

qBn

qi

qp<0

No disc in Eo

Ex=0 in metal ==> Eoflat

Bn= m- s = elec mtl to s/c barr

Bp= m- s + Eg = hole m to si = Bp-s,p =

hole s/c to mtl barr

Depl regqBp

qi

L 08 Sept 17 9

Metal to p-typenon-rect cont (m>s)

No disc in Eo

Ex=0 in metal ==> Eo flat

B,n=m- s,n = elec mtl to s/c barr

Bp= m- s + Eg = hole m to s

Accumulation region

EFi

Eo

Ec

Ev

EfP

qs,n

qs

n-type s/c

qm

EF

m

metal

qBnq(i)

qpAccum reg

qBpqi

L 08 Sept 17 10

Metal/semiconductorsystem typesn-type semiconductor• Schottky diode - blocking for m > s

• contact - conducting for m < s

p-type semiconductor• contact - conducting for m > s

• Schottky diode - blocking for m < s

L 08 Sept 17 11

Real Schottkyband structure1

• Barrier transistion region,

• Interface statesabove o acc, p neutrlbelow o dnr, n neutrl

Dit -> oo, qBn= Eg- oFermi level “pinned”

Dit -> 0, qBn= m - Goes to “ideal” case

L 08 Sept 17 12

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

L 08 Sept 17 13silicon for 711

andFd/cm, ,14E858with , ypermitivit the is

xEE where, ,E

r

o

ro

x

..

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

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

L 08 Sept 17 14

Poisson’s Equation• n = no + n, and p = po + p, in non-

equil• For n-type material, N = (Nd - Na) > 0,

no = N, and (Nd-Na+p-n)=-n +p +ni2/N

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

2/N• So neglecting ni

2/N

0n or p with material type-pand type-n for ,npqE

L 08 Sept 17 15

Ideal metal to n-typebarrier diode (m>s,Va=0)

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qBnqbi

q’n

No disc in Eo

Ex=0 in metal ==> Eoflat

Bn=m- s = elec mtl to s/c barr

bi=Bn-n= m-s elect s/c to mtl barr

Depl reg

0 xn xnc

L 08 Sept 17 16

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

so = q(Nd-Na+p-n) = qNd

• For xn < x < xnc, assume n = no = Nd, so = q(Nd-Na+p-n) = 0

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

L 08 Sept 17 17

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

xn

x

qNd

Q’d = qNdxn

x Ex

-Em

dnmx qN

xE

dxdE

xn

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

dctsmnBni

ix

0xdin

NNV

dxE- , qN2x n

/ln

L 08 Sept 17 18

Debye length

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

• In the region of xn, Poisson’s eq is E = / --> dEx/dx = q(Nd -

n), and since Ex = -d/dx, we have-d2/dx2 = q(Nd - n)/ to be solved

n

xxn

Nd

0

L 08 Sept 17 19

Debye length (cont)• Since the level EFi is a reference for equil, we

set = Vt ln(n/ni)• In the region of xn, n = ni exp(/Vt), so

d2/dx2 = -q(Nd - ni e/Vt), let = o + ’, where o = Vt ln(Nd/ni) so Nd - ni e/Vt = Nd[1 - e/Vt-o/Vt], for - o = ’ << o, the DE becomes d2’/dx2 = (q2Nd/kT)’, ’ << o

L 08 Sept 17 20

Debye length (cont)• So ’ = ’(xn) exp[+(x-xn)/LD]+con. and n

= Nd e’/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 ,qkTV ,pnq

VL

iad

ttD

L 08 Sept 17 21

Debye length (cont)• LD estimates the transition length

of a step-junction DR. Thus, i

t

0V

dD2V

WNLa

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

assumption OK

L 08 Sept 17 22

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

L 08 Sept 17 23

dimax

d

in

xa

aix

0x

NVa2qE

and ,qNVa2x

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 17 24

Ideal metal to n-typeSchottky (Va > 0)

qVa = Efn - Efm

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

qBn the sameDR decr

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qBnq(i-Va)

q’nDepl reg

L 08 Sept 18 25

Schottky diodecapacitance

xn

x

qNd

-Q-Q

Q’d = qNdxn

x

Ex

-Em

dnmx qN

xE

dxdE

xn

Q’

VQ

VQC

VVVQQQ

area jctn.A where AQQ

j

aiai

nn

'''

,'

L 08 Sept 18 26

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

qNV2qNxqNQ

,

,'

L 08 Sept 18 27

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] xAC and ][Fd/cm xC so

V2qN

dVdQC

nj

2

nj

aid

an

j

,,,'

,''

L 08 Sept 18 28

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 29

Schottky Capacitance(continued)• The C-V relationship simplifies to

][Fd/cm 2qNAC herew

equation model a V1CC

2

id

0j

21

ia

0jj

,

,

L 08 Sept 18 30

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

iVa

Cj0-2

L 08 Sept 17 31

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 model.

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

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