EE 5340Semiconductor Device TheoryLecture 9 – Fall 2010

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

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Test 1 – W 29Sep10• 11 AM Room 108 Nedderman Hall• Covering Lectures 1 through 10• 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. For examples see http://www.uta.edu/ronc/5340/tests/.

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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 =q(Nd-Na-p-n) by the Poisson Equation:

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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 +ni

2/N

• For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = p-n-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 (m>s,Va=0)

EFn

Eo

Ec

Ev

EFi

qs,n

qs

n-type s/c

qm

EF

m

metal

qBn

qbi

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

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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

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

xn

x

qNd

Q’d =

qNdxn

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

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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 = / --> dEx/dx = q(Nd - n),

and since Ex = -d/dx, we have

-d2/dx2 = q(Nd - n)/ 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 = 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) soNd - ni e/Vt

= Nd[1 - e/Vt-o/Vt], for - o = ’ << o, the

DE becomes d2’/dx2 = (q2Nd/kT)’,

’ << o

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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 ,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 < 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(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

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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

'''

,'

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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 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.

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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 = i

Cj-2

i

Va

Cj0-2

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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 model.

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

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