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

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

Professor Ronald L. Carter

[email protected]

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


First assignment
First Assignment EE5342 Lecture 09– Spring 2011

  • 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
Second Assignment EE5342 Lecture 09– Spring 2011

  • 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
Additional University Closure Means More Schedule Changes EE5342 Lecture 09– Spring 2011

  • 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
MT and P1 Assignment on Friday, 2/18/11 EE5342 Lecture 09– Spring 2011

  • 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.


Energy bands for p and n type s c

p-type EE5342 Lecture 09– Spring 2011

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


Making contact in a p n junction

E EE5342 Lecture 09– Spring 2011o

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


Band diagram for p n jctn at v a 0

E EE5342 Lecture 09– Spring 2011fN

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 for p n at v a 0 cont
Band diagram for EE5342 Lecture 09– Spring 2011p+-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


Depletion approximation
Depletion EE5342 Lecture 09– Spring 2011Approximation

  • 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 s equation
Poisson’s EE5342 Lecture 09– Spring 2011Equation

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


Poisson s equation1
Poisson’s EE5342 Lecture 09– Spring 2011Equation

  • 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 fermi energy
Quasi-Fermi EE5342 Lecture 09– Spring 2011Energy


Quasi-Fermi EE5342 Lecture 09– Spring 2011Energy (cont.)


Quasi-Fermi EE5342 Lecture 09– Spring 2011Energy (cont.)


Induced e field in the d r
Induced E-field EE5342 Lecture 09– Spring 2011in 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


Induced e field in the d r1

O EE5342 Lecture 09– Spring 2011

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
Depletion approx. EE5342 Lecture 09– Spring 2011charge 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 of gauss law
1-dim soln. of EE5342 Lecture 09– Spring 2011Gauss’ law

Ex

-xp

xn

xnc

-xpc

x

-Emax


Depletion approxi mation summary
Depletion Approxi- EE5342 Lecture 09– Spring 2011mation (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
One-sided p+n EE5342 Lecture 09– Spring 2011or 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-


Junction EE5342 Lecture 09– Spring 2011C (cont.)

r

+Qn’=qNdxn

+qNd

dQn’=qNddxn

-xp

x

-xpc

xn

xnc

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

-qNa

dQp’=-qNadxp

Qp’=-qNaxp


Junction c cont
Junction EE5342 Lecture 09– Spring 2011C (cont.)

  • The C-V relationship simplifies to


Junction c cont1
Junction EE5342 Lecture 09– Spring 2011C (cont.)

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

C’j-2

C’j0-2

Va

Vbi


Arbitrary doping profile
Arbitrary doping EE5342 Lecture 09– Spring 2011profile

  • 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 doping profile cont
Arbitrary doping EE5342 Lecture 09– Spring 2011profile (cont.)


Arbitrary doping profile cont1
Arbitrary doping EE5342 Lecture 09– Spring 2011profile (cont.)


Arbitrary doping profile cont2
Arbitrary doping EE5342 Lecture 09– Spring 2011profile (cont.)


Arbitrary doping profile cont3
Arbitrary doping EE5342 Lecture 09– Spring 2011profile (cont.)


Debye length

n EE5342 Lecture 09– Spring 2011

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
Debye length EE5342 Lecture 09– Spring 2011 (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), let f = fo + f’, where fo = Vt ln(Nd/ni) so Nd - 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 cont1
Debye length EE5342 Lecture 09– Spring 2011 (cont)

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


Debye length cont2
Debye length EE5342 Lecture 09– Spring 2011 (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
Example EE5342 Lecture 09– Spring 2011

  • 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
References EE5342 Lecture 09– Spring 2011

  • *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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