L 29 Dec 4 1

EE 5340Semiconductor Device TheoryLecture 29 - Fall 2003

Professor Ronald L. Carterronc@uta.edu

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.

L 29 Dec 4 6

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,

L 29 Dec 4 20

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

L 29 Dec 4 21

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

L 29 Dec 4 24

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

L 29 Dec 4 25

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