1. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/
Semiconductor Device Modeling and Characterization EE5342, Lecture 6-Spring 2010
Professor Ronald L. Carter
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2. Project 1A – Diode parameters to use
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3. Tasks
Using PSpice or any simulator, plot the i-v curve for this diode, assuming Rth = 0, for several temperatures in the range 300 K < TEMP = TAMB < 304 K.
Using this data, determine what the i-v plot would be for Rth = 500 K/W.
Using this data, determine the maximum operating temperature for which the diode conductance is within 1% of the Rth = 0 value at 300 K.
Do the same for a 10% tolerance.
Propose a SPICE macro which would give the Rth = 500 K/W i-v relationship.
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4. Example
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5. Induced E-field in the D.R.Ex
p-contact
N-contact O - O +
p-type CNR
n-type chg neutral reg O - O + O - O +
Exposed Acceptor Ions
Depletion region (DR)
Exposed Donor ions W x
-xpc
-xp xn
xnc
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6. Depletion approx. charge distribution
+Qn’=qNdxn
+qNd
[Coul/cm2]
-xp x
-xpc xn
xnc
-qNa
Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn
Qp’=-qNaxp
[Coul/cm2]
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7. 1-dim soln. of Gauss’ law Ex
-xpc
-xp xn
xnc x
-Emax
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8. 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).
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9. 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-
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10. Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn
Junction C (cont.) r
+Qn’=qNdxn
+qNd
dQn’=qNddxn
-xp x
-xpc xn
xnc
-qNa
Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn
dQp’=-qNadxp
Qp’=-qNaxp
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11. Junction C (cont.)
The C-V relationship simplifies to
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12. Junction C (cont.)
If one plots [C’j]-2 vs. Va Slope = -[(C’j0)2Vbi]-1 vertical axis intercept = [C’j0]-2 horizontal axis intercept = Vbi
C’j-2
C’j0-2 Va
Vbi
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13. Arbitrary doping profile
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 ...
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14. Arbitrary doping profile (cont.)
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15. Arbitrary doping profile (cont.)
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16. Arbitrary doping profile (cont.)
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17. Arbitrary doping profile (cont.)
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18. n x xn Nd
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
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19. 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), 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
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20. Debye length (cont)
So f’ = f’(xn) exp[+(x-xn)/LD]+con. and n = Nd ef’/Vt, x ~ xn, where LD is the “Debye length”
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21. 13% < d < 28% => DA is OK
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
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22. 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
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23. Ideal Junction Theory Assumptions Ex = 0 in the chg neutral reg. (CNR)
MB statistics are applicable
Neglect gen/rec in depl reg (DR)
Low level injections apply so that dnp < ppo for -xpc < x < -xp, and dpn < nno for xn < x < xnc
Steady State conditions
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24. Forward Bias Energy BandsEv Ec
EFi xn
xnc
-xpc
-xp
q(Vbi-Va)
EFP
EFN
qVa x
Imref, EFn
Imref, EFp
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25. Law of the junction (follow the min. carr.)
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26. Law of the junction (cont.)
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27. Law of the junction (cont.)
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28. Injection Conditions
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29. Apply the Continuity Eqn in CNR
Ideal Junction Theory (cont.)
Apply the Continuity Eqn in CNR
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30. Ideal Junction Theory (cont.)
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31. Ideal Junction Theory (cont.)
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32. Excess minority carrier distr fctn
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33. Carrier Injection ln(carrier conc) ln Na ln Nd ln ni ~Va/Vt ~Va/Vt
ln ni2/Nd
ln ni2/Na x
-xpc
-xp
xnc xn
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34. Minority carrier currents
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35. Evaluating the diode current
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36. Special cases for the diode current
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37. Ideal diode equation Assumptions: Current dens, Jx = Js expd(Va/Vt)
low-level injection
Maxwell Boltzman statistics
Depletion approximation
Neglect gen/rec effects in DR
Steady-state solution only
Current dens, Jx = Js expd(Va/Vt)
where expd(x) = [exp(x) -1]
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38. Ideal diode equation (cont.)
Js = Js,p + Js,n = hole curr + ele curr
Js,p = qni2Dp coth(Wn/Lp)/(NdLp) = qni2Dp/(NdWn), Wn << Lp, “short” = qni2Dp/(NdLp), Wn >> Lp, “long”
Js,n = qni2Dn coth(Wp/Ln)/(NaLn) = qni2Dn/(NaWp), Wp << Ln, “short” = qni2Dn/(NaLn), Wp >> Ln, “long”
Js,n << Js,p when Na >> Nd
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39. Diffnt’l, one-sided diode conductance
Static (steady-state) diode I-V characteristic IQ Va VQ
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40. Diffnt’l, one-sided diode cond. (cont.)
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41. Charge distr in a (1- sided) short diode
dpn
Assume Nd << Na
The sinh (see L12) excess minority carrier distribution becomes linear for Wn << Lp
dpn(xn)=pn0expd(Va/Vt)
Total chg = Q’p = Q’p = qdpn(xn)Wn/2
Wn = xnc- xn
dpn(xn)
Q’p x xn
xnc
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42. Charge distr in a 1- sided short diode
dpn
Assume Quasi-static charge distributions
Q’p = Q’p = qdpn(xn)Wn/2
ddpn(xn) = (W/2)* {dpn(xn,Va+dV) - dpn(xn,Va)}
dpn(xn,Va+dV)
dpn(xn,Va)
dQ’p
Q’p x xn
xnc
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43. Cap. of a (1-sided) short diode (cont.)
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44. General time- constant
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45. General time- constant (cont.)
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46. General time- constant (cont.)
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