EE 5340 Semiconductor Device Theory Lecture 24 – Spring 2011 Professor Ronald L. Carter
EE 5340 Semiconductor Device Theory Lecture 29 - Fall 2003
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Transcript of EE 5340 Semiconductor Device Theory Lecture 29 - Fall 2003
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EE 5340Semiconductor Device TheoryLecture 29 - Fall 2003
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
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Notes in Progress• This is a draft of L29. A final copy
will appear soon• Bring a copy
of-http://www.uta.edu/ronc/5340/project/40ProjectSolution03.pdf to class.
• Final copy of L29 will include “Problems of interest” from Chs. 8, 9, and 10 in Mueller and Kamins, 3rd edition.
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Part A DataFirst, the range for fitting the data was chosen as there was considerable “scat-ter” in the data. This was noted by plotting C^-1/M vs. Va for M = ½ and 1/3. This is shown in Figure A1.1. Figure A1.1. C^-1/M vs. Va
for M = ½ and 1/3 for the data given
0
5
10
15
20
25
30
-10 -8 -6 -4 -2 0 2Va (Volts)
(Cj0
/Cj)^
1/M M = 1/2
M = 1/3
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Part A Data re-plottedThe data “scatter” seemed to be most pronounced for the small values of C corresponding to the most negative values of V. The same data was plotted in the range -1 V < V < 0.4 V for the same M values. This is shown in Figure A1.2. Figure A1.2. C^-1/M vs. Va for M = ½ and 1/3 for the
data given in Table 1 over the range-1 V < V < 0.4 V.
y = -1.7632x + 1.0161R2 = 0.9995
y = -1.0671x + 0.9592R2 = 0.9907
0
1
2
3
-1.0 -0.5 0.0 0.5
Va (Volts)
(Cj0
/Cj)^
1/M
M = 1/2M = 1/3
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Determining ValuesThe M value of 1/3 seems to fit the data better, as demonstrated by the R^2 value of 0.9995. The value of M, etc., will be determined using the approach outlined in Lectures 11 and 14. In order to apply the theory developed in Lectures 11 and 14, it was decided to use the range-1 V ≤ V ≤ 0.4 V to calculate the central and forward derivatives. The range should be chosen by determining the minimum (1 - R2) value (as in Figure A1.2). One could try to use 5 points (rather than the 6 chosen here) or 4 (using the highest voltages only), omit the highest and use the next 4 highest, and so on.
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Estimating Junction Capacitance Parameters• Notation different than L14• If CJ = CJO {1 – Va/VJ}-M
• Define y {d[ln(CJ)]/dV}-1
• A plot ofy = yi vs. Va = vi has
slope = -1/M, andintercept = VJ/M
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Derivatives DefinedThe central derivative is defined as (following Lecture 14 and 11)
yi,Central = (vi+1 – vi-1)/(lnCi+1 – lnCi-1), with vi = (vi+1 + vi-1)/2 Equation A1.1
The Forward derivative (as applied to the theory in L11 and L14) is defined in this case as
yi,Forward = (vi+1 – vi)/(lnCi+1 – lnCi), with vi,eff = (vi+1 + vi-1)/2 Equation A1.2
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Data calculationsTable A1.1. Calculations of yi and vi for the Central and Forward derivatives for the data in Table 1. The yi and vi are defined in Equations A1.1 and A1.2.
Va (V) Cj (Fd) viCentral Forward
0.40 2.51E-12 derivative derivative0.35 0.585
0.30 2.11E-12 0.30 0.8160.25 1.347
0.20 1.96E-12 0.20 1.1520.15 1.007
0.10 1.78E-12 0.10 1.3250.05 1.938
0.00 1.69E-12 -0.20 2.231-0.25 2.300
-0.50 1.36E-12 -0.50 2.946-0.75 4.096
-1.00 1.20E-12 -1.00 4.132
yi = (dlnc/dv)^-1
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y = -2.551x + 1.6326R2 = 0.9977, Central
y = -2.9965x + 1.7788R2 = 0.9517, Forward
0
1
2
3
4
5
-1.0 -0.5 0.0 0.5 1.0
Vi (Volts)
yi (V
olts
^-1)
Central
Forward
Linear (Central)
Linear (Forward)
y vs. Va plotsFigure A1.3. The yi and vi values from the theory in L11 and L14 with associa-ted trend lines and slope, intercept and R^2 values.
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Comments on thedata interpretationIt is clear the Central derivative gives the more reliable data as the R^2 value is larger. M is the reciprocal of the magnitude of the slope obtained by a least squares fit (linear) plot of yi vs. ViVJ is the horizontal axis intercept (computed as the vertical axis intercept divided by the slope)Cj0 is the vertical axis intercept of a least squares fit of Cj-1/M vs. V (must use the value of V for which the Cj was computed). The computations will be shown later.The results of plotting Cj-1/M vs. V for the M value quoted below are shown in Figure A1.4
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Calculating theparametersM = 1/2.551 = 0.392
(the data were generated using M = 0.389, thus we have a 0.77% error).
VJ = yi(vi=0)/slope =1.6326/2.551 = 0.640
(the data were generated using fi = 0.648, thus we have a 1.24% error).
Cj0 = 1.539E30^-.392 = 1.467 pF (the data were generated using Cj0 =
1.68 pF, thus we have a 12.6% error)
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Linearized C-V plotFigure A1.4. A plot of the data for Cj^-1/M vs. Va using the M value determined for this data (M = 0.392).
y = -1.539E+30x + 1.058E+30R2 = 9.976E-01
0.00E+00
1.00E+30
2.00E+30
3.00E+30
-1.0 -0.5 0.0 0.5 1.0Va (Volts)
Cj^
-1/M
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Doping ProfileThe data were equal-ly spaced (V=0.1V), the central differ-ence was used, for -7.4V ≤ V ≤ 0.4V, which for Cj = /x, corresponds to a range of 2.81E-5 cm ≤ x ≤ 8.99E-5 cm. These data are shown. The trend line is also shown for a linear fit. Since R^2 = 1.000, a linear N(x) relationship can be assumed.
y = 1.888E+20x + 1.861E+15R2 = 1.000E+00
6.0E+15
8.0E+15
1.0E+16
1.2E+16
1.4E+16
1.6E+16
1.8E+16
2.0E+16
2.0E-05 4.0E-05 6.0E-05 8.0E-05 1.0E-04
Depletion depth, x (cm)
Dop
ing
Con
cent
ratio
n (c
m̂-3
)
dVCdqA
xN n )(2)( 2
2
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Parameter values• B2. Using the same procedure, as for
part A (central derivatives), one finds M = 0.3448. The data were generated using M = 0.345, thus we have less than 0.1% error.
• B3. Since we found a linear fit in B1, and M ~ 0.333 (the value for linear doping), the two observations are in agreement that the doping profile is linear.
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Part CUse the depletion approximation for a spherical diode to find the breakdown voltage as a function N- and junction radius, rj. Place the heavily doped side of the junction (Na = N+) in the range r+c < r < rj and the lightly doped side of the junction (Nd = N-) in the range rj < r < r-c (r+c and r-c are the respective contacts). Use the Gauss Law form of the Poisson Equation to solve the E(r) relationship and then solve for Emax = E(rj) as a function of Va = V(r+) - V(r-). In this (one-dimensional) case, the Gauss Law form of the Poisson Equation is
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Reverse biasjunction breakdown
8/3
4/3
0
4/3
2/3
20
161/
1.1/ 120 so
,161/
1.1/ 60 gives *,***
usually , 2
D.A. theand diode sided-one a Assuming
EN
EqNVE
ENEV
BVCasey
BVqNEBV
g
Sicrit
B
g
icritSi
i
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Ecrit for reverse breakdown (M&K**)
Taken from p. 198, M&K**
Casey Model for Ecrit
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rpcrprj rnrnc
Gauss’ Law
Surface r
rErdSE0
Surfacein Enclosed2 Q)(4
2
3a
max
33a2
3qN so
,3
4qN 4
j
pjr
Surfacepr
rr
rEE
rrErdSE
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Spherical DiodeFields calculations
2
3d
2
2
max 3qN
rr
rrr
EE jjr
Setting Er = 0 at r = rn, we get
3
d
max
qN 31
jjn r
Err
Note that the equivalent of the lever law for this spherical diode is
33d
33a NN jnpj rrrr
For rj < ro ≤ rn,
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Spherical DiodeFields calculations
Assume Na >> Nd, so rn – rj d >> rj – rp. Further, setting the usual definition for the potential difference, and evaluating the potential difference at breakdown, we havePHIi – Va = BV and Emax = Em = Ecrit = Ec. We also define = 3eEm/qNd[cm].
njj
njjnj rr
rrr
rrr 11E11E2E BV 2
c3c22c
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Showing therj ∞ limit
C1. Solve for rn – rj = as a function of Emax and solve for the value of in the limit of rj . The solution for rn is given below.
theorem.binomial apply the limit, thegwhen takin
11 so
,qNE3
, 1
1/3
,0d
crit
1/3
jjjn
Sirj
jn
rrrr
rrr
.
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Solving for theBreakdown (BV)
Solve for BV = [i – Va]Emax = Ecrit, and solve for the value of BV in the limit of rj . The solution for BV is given below.
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Spherical diodeBreakdown Voltage
1.0
10.0
100.0
1.00E+14 1.00E+15 1.00E+16 1.00E+17
Substrate Concentration (cm^-3)
Bre
akdo
wn
Volta
ge (V
olt)
rj = 0.1 micronrj = 0.2 micronrj = 0.5 micronrj = 1.0 micron
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Final Exam• For BOTH sections, 051 and 001
– 8:00 to 10:30 AM– Tuesday, December 9 in – 206 ACT– Cover sheet on web page at
http://www.uta.edu/ronc/5340/tests/• The Final is comprehensive
– 20% to 25% on Test 1 material– 20% to 25% on Test 2 material– Balance of final on material since Test 2
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Emphasis• Chapters 1 through 7
– Lecture notes– Problems assigned– Previous tests
• Chapters 8 through 10– As above– Some key problems in 9 and 10
are P9:1,3,5,7 and P10:8