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Electronic Devices 2011-2012 - Problems - Technology

Giovanni GhioneJanuary 21, 2012

1. A Si substrate with (100) orientation is covered by an oxide layer with thickness 0.2 m. How long doesit take to grow at 1000 C in an atmosphere of dry oxygen an oxide layer having the same thickness ontop of the existing one? What would be the time required in the case of wet oxidation (water vapor)?ANSWER We generally describe the oxide thickness vs. time through the equation:

tox = A2 s 1 + t + t 0A2 =4B 1!:

where t 0 is some initial time formally associated to an initial thickness of the oxide and A and B areproper coecients depending on temperature. Let us nd those coecients for the two cases considered.We have 1000=(T + 273) = 1000 =1273 = 0 :785 , thus from the graphs we obtain approximately:

Bdr y = 1 102 m2 =h

Bwet = 5 101 m2 =h

B=Adr y = 5 102 m=h

B=Awet = 1 m=hAdr y = Bdry =(B=Adry ) = 1 10

2 = 5 10 2 = 0 :2 mAwet = Bwet =(B=Awet ) = 5 10 1 =(1) = 0 :5 m

From the basic equation we obtain:

t 2ox + At ox = B (t + t 0 )

and in the two cases of dry and wet oxidation at t = 0 :

t 2ox + At ox = Bt 0 !t 0 ,wet =

t2ox + Awet toxBwet

= 0:22 + 0 :5 0:2

5 10 1 = 0 :28 h

t 0 ,dry = t2ox + Adry tox

Bdry=

0:22 + 0 :2 0:2

1 102 = 8 h

We have now to nd the total time t + t0 needed to grow a thickness of tox,tot = 0 :4 m in the twocases. We have:

twet + t 0 ,wet =t 2ox,tot + Awet tox,tot

Bwet=

0:42 + 0 :5 0:45 10 1

= 0 :72 h ! twet = 0 :72 0:28 = 0 :44 h

tdr y + t 0 ,dry =t 2ox,tot + Adr y tox,tot

Bdry=

0:42 + 0 :2 0:41 10 2

= 24 h ! tdr y = 24 8 = 16 h

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Figure 1: Parabolic ( B ) and linear ( A=B) coecients for oxide growth on Silicon.

2. Consider an oxidation process done in such conditions that B=A = 6 :22 m/h and B = 1 :5 m2 /h.The oxidation is done on a substrate with a pre-existing oxide layer of thickness d0 = 1 m. Evaluatethe total oxide thickness grown in 10 h.ANSWER We have A = B= (B=A) = 1 :5=6:22 = 0 :2412 m. Thus:

d20 + Ad0 = Bt 0 ! t 0 = d20 + Ad0

B =

12 + 0 :2412 11:5

= 0 :823 h

Then for a total oxidation time t = 10 :823 h we have:

d20 + Ad0 = Bt ! d20 + 0 :2412d0 = 1 :5 10:823 ! d0 = 3 : 91 m

3. Evaluate the time needed to grow an oxide thickness d = 0 :5 m on a Si substrate, knowing that theoxidation process is characterized by B=A = 1 :5 m/h and B = 0 :75 m2 /h.ANSWER We have directly:

d2 + Ad = Bt ! t = d20B

+ d0B=A

= 0:52

0:75 +

0:51:5

= 0 :666 h = 40 min

4. A rectangular trench is etched into a Si substrate, the trench width is l = 1 m and the trench depthis d = 3 m. Suppose that on the Si surface a passivation layer is stopping oxidation. The substrateis oxidized with B=A = 3 m/h and B = 0 :5 m2 /h, till the trench is completely lled. Find the nalwidth of the oxide w and the time needed to ll the trench, t .ANSWER We only consider growth on the trench sides. We know that the ratio of the Si thicknessand the nal oxide thickness is 0.44, in other words:

(w d) =2w=2

= w d

w = 0 :44 ! 1

dw

= 0 :44 !dw

= 0 :56 ! w = d0:56

= 1 : 786d = 1 : 786 m

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Since the oxide growth is taking place on the two sides the nal thickness of the oxide is d=2, thus:

d

2

2

+ Ad

2 = Bt

!t =

(d=2)2

B +

d=2

B=A =

(1:786=2)2

0:5 +

1:786=2

3 = 1 :8926 h

5. In a uniformely doped Si substrate with donor concentration N D = 10 15 cm 3 we implant a Borondose S = 10 12 cm 2 at 100 keV. After this we perform a 2 h annealing at 1000 C. Find the depth of the junction after annealing.ANSWER We have to nd the boron projected range and vertical straggle at the given energy. Weuse the two graphs:

Figure 2: Projected range vs. energy.

We nd R p = 0 :35 m and R p = 0 :07 m. Therefore the doping prole is:

N A (x) = S

p 2 R p exp(x R p )2

2 R 2 p =

1012

p 2 0:07 10 4 exp" (x 0:35 104 )2

2 (0:07 10 4 )2 #

After annealing we have:R

0

p = q R 2 p + 2 Dt aFrom the gure we obtain a boron diusivity D = 2 10 14 cm2 /s, thus:

R0

p = q (0:07 10 4 )2 + 2 2 10 14 2 3600 = 0 :18 10 4 cmwith doping prole:

N 0A (x) = S

p 2 R 0 pexp

(x R p )2

2 R 02 p =

1012

p 2 0:18 10 4 exp" (x 0:35 104 )2

2 (0:18 10 4 )2 #

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Figure 3: Straggle vs. energy.

In conclusion we have the two proles:

N A (x) = 5 : 7 1016 exph1: 0 1010 x 3: 5 10 5

2

i cm 3N 0A (x) = 2 : 2 10

16 exph1: 5 109 x 3: 5 10 52

i cm 3The resuling plots are shown in the gure. After annealing there is just one junction around xj =0:8 10 4 cm = 0:8 m.

6. We implant a dopant into Si with a projected range in SiO 2 R p = 300 nm and a straggle R p = 70nm. What is the minimum oxide thickness we need to mask the implant? Assume that masking iseective when the dopant concentration at the Si surface is 10 orders of magnitude less than the peakdopant concentration.ANSWER The condition on the normalized prole is:

exp(x R p )2

2 R 2 p = exp

(x 300)2

2 702 = 10 10

i.e.:

(x 300)2

2 702 = log10 10 = 23

(x 300)2 = 2 702 23 ! x = 775 nm

7. A silicon substrate is doped N D = 5 1015 cm 3 . We make a boron predeposition with a predeposeddose S = 1 1014 cm 2 ; then, a diusion follows at 1100 C. What is the diusion time if we want a junction at x j = 2 m?

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Figure 4: Dopant diusivity in Si vs. temperature.

ANSWER In generale the prole following from the predeposition has a erfc shape; however, weapproximate this with a delta function on the surface:I

N A (x; 0) = S (x):

The prole after a constant dose diusion is a half-gaussian:

N A (x; t ) = S p Dt exp

x2

4Dt :

Assuming D = 2 10 13 cm2 /s we have the equation:

S p Dt exp

x2j4Dt ! = N D !

1 1014

p 2 10 13 t exp2 10 4

2

4 2 10 13 t! = 5 1015 !1: 2616 1020

p t exp50000

t = 5 1015

The solution for t can be done numerically or by trial and error, the result is t 1:89 h.

8. A p type Si substrate has a resistivity = 2 cm. Find the substrate doping assuming for the holemobility h = 500 cm

2 /Vs.ANSWER From the substrate conductivity:

= q h N A ! = 1

q h N A !N A =

1q h

= 1

1:6 10 19 500 2 = 6 : 25 1015 cm 3 :

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10-4

1014

1015

1016

1017

Depth, cm

D o p

i n g ,

c m

- 3

Figure 5: Doping prole before and after annealing.

9. A Si substrate is undergoing wet oxidation (orientation 111) for 5 h, at 1050 C. Evaluate the thicknessof the oxide layer from the graphs directly yielding the oxide thickness and compare with the analyticalexpression with the A and B coecients.ANSWER From the graph:we obtain tox 2 m. Considering the graphs with the parabolic andlinear coecients we obtain:

B=A 4 m/hB 0:8 m

2 /h

A = BB=A

= 0:8

4 = 0 :2 m

i.e.:

tox = A2 s 1 + tA2 =4B 1! = 0:22 s 1 + 50:22 =(4 0:8) 1! = 1 :9 m

roughly consistent with the other result.

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Figure 6: Oxide thickness vs. time.

Figure 7: Parabolic ( B ) and linear ( A=B) coecients for oxide growth on Silicon.

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