EE 5340 Semiconductor Device Theory Lecture 12 – Spring 2011

EE 5340Semiconductor Device TheoryLecture 12 – Spring 2011

Professor Ronald L. [email protected]

http://www.uta.edu/ronc

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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]3

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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

05

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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

8

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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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Soln to Poisson’sEq in the D.R.

xnx

-xp

-xpc xnc

Ex

-Emax(V)

dx qN

dxdE

ax qN

dxdE

-Emax(V-dV)

W(Va)W(Va-dV)

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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

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JunctionC (cont.)

xn

x-xp

-xpc xnc

r+qNd

-qNa

+Qn’=qNdxn

Qp’=-qNaxp

Charge neutrality => Qp’ + Qn’ = 0,

=> Naxp = Ndxn

dQn’=qNddxn

dQp’=-qNadxp

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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

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JunctionC (cont.)• So this definition of the

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

dQ’p(=-dQ’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.

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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

18

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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

,~,~

'

19

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Junction Capacitance

• Estimate CJO• Define y Cj/CJO• Calculate y/(dy/dV) =

{d[ln(y)]/dV}-1

• A plot of r y/(dy/dV) vs. V has

slope = -1/M, andintercept = VJ/M

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dy/dx - Numerical Differentiation

x y dy/ dx (central diff erence)

x(n-1) y(n-1) [y(n) - y(n-2)]/ [x(n) - x(n-2)]

x(n) y(n) [y(n+1) - y(n-1)]/ [x(n+1) - x(n-1)]

x(n+1) y(n+1) [y(n+2) - y(n)]/ [x(n+2) - x(n)]

x(n+2) y(n+2) [y(n+3) - y(n+1)]/ [x(n+3) - x(n+1)]

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Practical Junctions• Junctions are formed by diffusion

or implantation into a uniform concentration wafer. The profile can be approximated by a step or linear function in the region of the junction.

• If a step, then previous models OK.• If linear, let the local charge

density r=qax in the region of the junction.

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Practical Jctns (cont.)

Shallow (steep) implant

N

x (depth)

Box or step junction approx.

N

x (depth)

Na(x)

xj

Linear approx.

NdNd

Na(x)

Uniform wafer con

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Linear gradedjunction• Let the net donor concentration,

N(x) = Nd(x) - Na(x) = ax, so

r =qax, -xp < x < xn = xp

= xo, (chg neu)

xo

-xo

r r = qa x

Q’n=qaxo2/2

Q’p=-qaxo2/2

x

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Linear gradedjunction (cont.)• Let Ex(-xo) = 0, since this is the

edge of the DR (also true at +xo)

2omax

2

omaxx

x

ox-

x

ox-x

x2qa

E

where ,xx

1E)x(E

so ,axdxq

dE Law, Gauss' By

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Linear gradedjunction (cont.)

31

bi

ajj

31

abio

i

otbi

3oabi

VV

10C'dV

'dQC' Letting

.qa2

VV3x so ,

nax

lnV2V

and ,X3qa2

VVV

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Linear gradedjunction, etc.

2m1

abi

1m

j

mj

31

bi

2

oj

VV2mqB

W0'C

,BxN when 'C for formula general

the suggesting ,V12

qax2

0'C

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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.

EE 5340 Semiconductor Device Theory Lecture 12 – Spring 2011

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