Semiconductor device modeling and characterization ee5342 lecture 09 spring 2011
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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc/. First Assignment. e-mail to [email protected] In the body of the message include subscribe EE5342

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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011

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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011

Professor Ronald L. Carter

[email protected]

http://www.uta.edu/ronc/


First Assignment

  • e-mail to [email protected]

    • In the body of the message include subscribe EE5342

  • This will subscribe you to the EE5342 list. Will receive all EE5342 messages

  • If you have any questions, send to [email protected], with EE5342 in subject line.


Second Assignment

  • Submit a signed copy of the document that is posted at

    www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf


Additional University Closure Means More Schedule Changes

  • Plan to meet until noon some days in the next few weeks. This way we will make up for the lost time. The first extended class will be Monday, 2/14.

  • The MT changed to Friday 2/18

  • The P1 test changed to Friday 3/11.

  • The P2 test is still Wednesday 4/13

  • The Final is still Wednesday 5/11.


MT and P1 Assignment on Friday, 2/18/11

  • Quizzes and tests are open book

    • must have a legally obtained copy-no Xerox copies.

    • OR one handwritten page of notes.

    • Calculator allowed.

  • A cover sheet will be published by Wednesday, 2/16/11.


p-type

Ec

Ec

Ev

EFn

qfn= kT ln(Nd/ni)

EFi

Ev

Energy bands forp- and n-type s/c

n-type

EFi

qfp= kT ln(ni/Na)

EFp


Eo

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.: qc = 4.05 eV (Si),

    and qf = qc + Ec - EF

qc(electron affinity)

qf

(work function)

Ec

Ef

Efi

qfF

Ev


EfN

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

Ec

qVbi = q(fn -fp)

qfp < 0

Ec

Efi

EfP

Ev

Efi

qfn > 0

*Na > Nd -> |fp|> fn

Ev

p-type for x<0

n-type for x>0

x

-xpc

xn

0

-xp

xnc


Band diagram forp+-n at Va=0 (cont.)

  • A total band bending of qVbi = q(fn-fp) = kT ln(NdNa/ni2)is necessary to set EfP = EfN

  • For -xp < x < 0, Efi - EfP < -qfp, = |qfp| so p < Na = po, (depleted of maj. carr.)

  • For 0 < x < xn, EfN - Efi < qfn, so n < Nd = no, (depleted of maj. carr.)

    -xp < x < xn is the Depletion Region


DepletionApproximation

  • Assume p << po = Nafor -xp < x < 0, so r = q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Nafor -xpc < x < -xp, so r = q(Nd-Na+p-n) = 0, -xpc < x < -xp

  • Assume n << no = Ndfor 0 < x < xn, so r = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Ndfor xn < x < xnc, so r = q(Nd-Na+p-n) = 0, xn < x < xnc


Poisson’sEquation

  • The electric field at (x,y,z) is related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation:


Poisson’sEquation

  • For n-type material, N = (Nd - Na) > 0, no = N, and (Nd-Na+p-n)=-dn +dp +ni2/N

  • For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = dp-dn-ni2/N

  • So neglecting ni2/N, [r=(Nd-Na+p-n)]


Quasi-FermiEnergy


Quasi-FermiEnergy (cont.)


Quasi-FermiEnergy (cont.)


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 thatEx = 0 for -xpc < x < -xp and Ex = 0 for -xn < x < xnc

h0


O

O

O

O

O

O

+

+

+

-

-

-

Induced E-fieldin the D.R.

Ex

N-contact

p-contact

p-type CNR

n-type chg neutral reg

Depletion region (DR)

Exposed Donor ions

Exposed Acceptor Ions

W

x

-xpc

-xp

xn

xnc

0


Depletion approx.charge distribution

r

+Qn’=qNdxn

+qNd

[Coul/cm2]

-xp

x

-xpc

xn

xnc

Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn

-qNa

Qp’=-qNaxp

[Coul/cm2]


1-dim soln. ofGauss’ law

Ex

-xp

xn

xnc

-xpc

x

-Emax


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 = {2e(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn, xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd).


One-sided p+n or n+p jctns

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

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

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


JunctionC (cont.)

r

+Qn’=qNdxn

+qNd

dQn’=qNddxn

-xp

x

-xpc

xn

xnc

Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn

-qNa

dQp’=-qNadxp

Qp’=-qNaxp


JunctionC (cont.)

  • The C-V relationship simplifies to


JunctionC (cont.)

  • If one plots [C’j]-2vs. VaSlope = -[(C’j0)2Vbi]-1vertical axis intercept = [C’j0]-2 horizontal axis intercept = Vbi

C’j-2

C’j0-2

Va

Vbi


Arbitrary dopingprofile

  • If the net donor conc, N = N(x), then at xn, the extra charge put into the DR when Va->Va+dVa is dQ’=-qN(xn)dxn

  • The increase in field, dEx =-(qN/e)dxn, by Gauss’ Law (at xn, but also const).

  • So dVa=-(xn+xp)dEx= (W/e) dQ’

  • Further, since N(xn)dxn = N(xp)dxp gives, the dC/dxn as ...


Arbitrary dopingprofile (cont.)


Arbitrary dopingprofile (cont.)


Arbitrary dopingprofile (cont.)


Arbitrary dopingprofile (cont.)


n

Nd

0

xn

x

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, the 1-dim Poisson equation is dEx/dx = q(Nd - n), and since Ex = -df/dx, the potential is the solution to -d2f/dx2 = q(Nd - n)/e


Debye length (cont)

  • Since the level EFi is a reference for equil, we set f = Vt ln(n/ni)

  • In the region of xn, n = ni exp(f/Vt), so d2f/dx2 = -q(Nd - ni ef/Vt), letf = fo + f’, where fo = Vt ln(Nd/ni) soNd - ni ef/Vt = Nd[1 - ef/Vt-fo/Vt], for f - fo = f’ << fo, the DE becomes d2f’/dx2 = (q2Nd/ekT)f’, f’ << fo


Debye length (cont)

  • So f’= f’(xn) exp[+(x-xn)/LD]+con. and n = Nd ef’/Vt, x ~ xn, whereLD is the “Debye length”


Debye length (cont)

  • LD estimates the transition length of a step-junction DR (concentrations Na and Nd with Neff = NaNd/(Na +Nd)). Thus,

  • For Va=0, & 1E13 <Na,Nd< 1E19 cm-3

  • 13% <d< 28% => DA is OK


Example

  • An assymetrical p+ n junction has a lightly doped concentration of 1E16 and with p+ = 1E18. What is W(V=0)?

    Vbi=0.816 V, Neff=9.9E15, W=0.33mm

  • What is C’j? = 31.9 nFd/cm2

  • What is LD? = 0.04 mm


References

  • *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.

  • **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.

  • M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.

  • 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.

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

  • 3 Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990.


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