EE 5340Semiconductor Device TheoryLecture 10 – Fall 2010

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

L10 22Sep10 2

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/.

L10 22Sep10 3

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

L10 22Sep10 4

Debye length

n

xxn

Nd

0

material. intrinsic for 2npn and type,-p extrinsic for Npn

type,-n extrinsic for N pn :Note

length. transition a , ,

i

a

d

q

kTVpnq

VL tt

D

L10 22Sep10 5

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

L10 22Sep10 6

Schottky diodecapacitance

xn

x

qNd

-Q-Q

Q’d =

qNdxn

x

Ex

-Em

dn

mx qNxE

dxdE

xn

Q’

2aid

d

aidndn

cmCoul VqN2

qNV2qNxqNQ

,

,'

[Fd] xAC and ][Fd/cm xC so

V2qN

dVdQC

nj

2

nj

aid

an

j

,,,'

,''

L10 22Sep10 7

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

Diagrams for ideal metal-semiconductor Schottky diodes. Fig. 3.21 in Ref 4.L10 22Sep10 8

L10 22Sep10

Energy bands forp- and n-type s/c

p-typeEc

Ev

EFi

EFP

qP= kT ln(ni/Na)

Ev

Ec

EFi

EFNqn= kT ln(Nd/ni)

n-type

9

L10 22Sep10

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 EFiEv

q (electron affinity)

qF

q(work function)

10

L10 22Sep10

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

EcEFNEFi

Ev

Ec

EFP

EFi

Ev

0 xnx

-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

11

L10 22Sep10

• 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.)

12

L10 22Sep10

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

13

L10 22Sep10

Depletion approx.charge distribution

xnx

-xp

-xpc xnc

+qNd

-qNa

+Qn’=qNdxn

Qp’=-qNaxp

Due to Charge

neutrality Qp’ + Qn’ = 0, => Naxp =

Ndxn

[Coul/cm2]

[Coul/cm2]14

L10 22Sep10

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

15

L10 22Sep10

Induced E-fieldin the D.R.

xnx

-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

016

L10 22Sep10

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

17

L10 22Sep10

Soln to Poisson’sEq in the D.R.

xn x-xp

-xpc xnc

Ex

-Emax

dx qN

dxdE

ax qN

dxdE

18

L10 22Sep10

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

)VqkT (note ,xNxN2

qdxdVE ,dxEV

nNNlnq

kTthat is D.R. the in P.E. the of solnthe to V of iprelationsh the Now,

t2pa2nd

xx

xxbi2

ida

bi

n

p

19

L10 22Sep10

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

WV2N2qVE then

,WE21V have also must we Since

.NNNNN where ,qN

V2W

then ,xxW let and ,xNxN

bieffbimax

maxbi

dada

effeffbi

pnpand

20

L10 22Sep10

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=0xEx(x=0x) Emax

21

L10 22Sep10

Sample calculations• Vt 25.86 mV at 300K• = ro = 11.7*8.85E-14 Fd/cm

= 1.035E-12 Fd/cm• If Na5E17/cm3, and Nd2E15

/cm3, then for ni1.4E10/cm3, then what is Vbi = 757 mV

22

L10 22Sep10

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

23

L10 22Sep10 24

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.

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