EE 5340Semiconductor Device TheoryLecture 11 - Fall 2010

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

L11 27Sep10

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

2

L11 27Sep10

Depletion approx.charge distribution

xn

x-xp

-xpc xnc

+qNd

-qNa

+Qn’=qNdxn

Qp’=-qNaxp

Due to Charge

neutrality Qp’ + Qn’ =

0, => Naxp =

Ndxn

[Coul/cm2]

[Coul/cm2]3

L11 27Sep10

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

04

L11 27Sep10

Soln to Poisson’sEq in the D.R.

xnx

-xp

-xpc xnc

Ex

-Emax

dx qN

dxdE

ax qN

dxdE

5

L11 27Sep10

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

6

L11 27Sep10

Effect of V 0

• Define an external voltage source, Va, with the +term at the p-type contact and the -term at the n-type contact

• For Va > 0, the Va induced field tends to oppose Ex due to DR

• For Va < 0, the Va induced field tends to add to Ex due to DR

• Will consider Va < 0 now7

L11 27Sep10 8

Band diagram forp+-n jctn* at Va 0

EcEFN

EFi

Ev

Ec

EFP

EFi

Ev

0 xn

x-xp

-xpc xnc

qp < 0

qn > 0

q(Vbi-Va)

*Na > Nd -> |p| > n

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

q(Va)

L11 27Sep10 9

Soln to Poisson’sEq in the D.R.

xnx

-xp

-xpc xnc

Ex

-Emax(V)

dx qN

dxdE

ax qN

dxdE

-Emax(V-V)

W(Va)W(Va-V)

L11 27Sep10

effbimax

eff

bi

xa

abinx

pxx

NVaV2qE

and ,qN

VaV2W

are Solutions .E reduce to tends V to

due field the since ,VVdxE

that is now change only The

Effect of V 0

10

L11 27Sep10

Effect of V 0• Lever rule, Naxp = Ndxn, still applies

• Vbi = Vt ln(NaNd/ni2), still applies

• W = xp + xn, still applies

• Neff = NaNd/(Na + Nd), still applies

• Q’n = qNdxn = -Q’p = qNaxp, still applies

• For Va < 0, W increases and Emax increases

11

L11 27Sep10

One-sided p+n or n+p jctns• If p+n, then Na >> Nd, and

NaNd/(Na + Nd) = Neff --> Nd, and W --> xn, DR is all on lightly d. side

• If n+p, then Nd >> Na, and NaNd/(Na + Nd) = Neff --> Na, and W --> xp, DR is all on lightly d. side

• The net effect is that Neff --> N-, (- = lightly doped side) and W --> x-

12

L11 27Sep10

Depletion Approxi-mation (Summary)• For the step junction defined by

doping Na (p-type) for x < 0 and Nd, (n-type) for x > 0, the depletion width

W = {2(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni

2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn,

xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd).

13

L11 27Sep10

Debye length• The DA assumes n changes from Nd to

0 discontinuously at xn, likewise, p changes from Na to 0 discontinuously at -xp.

• 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

14

L11 27Sep10

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

15

L11 27Sep10

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

16

L11 27Sep10

Debye length (cont)• LD estimates the transition length of a step-

junction DR (concentrations Na and Nd with Neff =

NaNd/(Na +Nd)). Thus,

bi

efft

da0V

dDaDV2

NV

N1

N1

W

NLNL

a

• For Va=0, & 1E13 < Na,Nd < 1E19

cm-3

13% < < 28% => DA is OK17

L11 27Sep10

JunctionC (cont.)

xn

x-xp

-xpc xnc

+qNd

-qNa

+Qn’=qNdxn

Qp’=-qNaxp

Charge neutrality => Qp’ + Qn’ = 0,

=> Naxp = Ndxn

Qn’=qNdxn

Qp’=-qNaxp

18

L11 27Sep10

JunctionCapacitance• The junction has +Q’n=qNdxn (exposed

donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor.

2da

daabi

da

daabi

a

d

dndn

cm

Coul ,

NN

NNVVq2

,NqN

NNVV2

N

N1

qNxqN'Q

19

L11 27Sep10

JunctionC (cont.)• So this definition of the capacitance

gives a parallel plate capacitor with charges Q’n and Q’p(=-Q’n), separated by, L (=W), 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.

20

L11 27Sep10

JunctionC (cont.)• This Q ~ (Vbi-Va)

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

• Redefining the capacitance,

[Fd] W

C and ][Fd/cm W

C so

NNVVNqN

dVdQ

C

Aj

2j

daabi

da

a

nj

,,,'

,'

2

21

L11 27Sep10

JunctionC (cont.)• The C-V relationship simplifies to

][Fd/cm ,NNV2

NqN'C herew

equation model a ,VV

1'C'C

2

dabi

da0j

21

bi

a0jj

22

L11 27Sep10

JunctionC (cont.)• If one plots [Cj]

-2 vs. Va

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

vertical axis intercept = [Cj0]-2 horizontal axis intercept = Vbi

Cj-2

Vbi

Va

Cj0-2

1M31

VVJ C0CJ

VJV

10CJACC

:Equation Model

bi0j

M

jj

,~,~

'

23