微电子工艺习题解答

. . . zyzl . .

CRYSTAL GROWTH AND EXPITAXY

1．画出一50cm 长的单晶硅锭距离籽晶10cm 、20cm 、30cm 、40cm 、45cm 时砷的掺杂分布。（单晶硅锭从融体中拉出时，初始的掺杂浓度为1017cm -3）

2．硅的晶格常数为5.43?．假设为一硬球模型：

(a)计算硅原子的半径。

(b)确定硅原子的浓度为多少(单位为cm -3)?

(c)利用阿伏伽德罗(Avogadro)常数求出硅的密度。

3．假设有一l0kg 的纯硅融体，当硼掺杂的单晶硅锭生长到一半时，希望得到0.01 Ω·cm 的电阻率，则需要加总量是多少的硼去掺杂?

4．一直径200mm 、厚1mm 的硅晶片，含有5.41mg 的硼均匀分布在替代位置上，求： (a)硼的浓度为多少?

(b)硼原子间的平均距离。

5．用于柴可拉斯基法的籽晶，通常先拉成一小直径(5.5mm)的狭窄颈以作为无位错生长的开始。如果硅的临界屈服强度为2×106g/cm2，试计算此籽晶可以支撑的200mm 直径单晶硅锭的最大长度。

6．在利用柴可拉斯基法所生长的晶体中掺入硼原子，为何在尾端的硼原子浓度会比籽晶端的浓度高?

7．为何晶片中心的杂质浓度会比晶片周围的大?

8．对柴可拉斯基技术，在k 0=0.05时，画出C s /C 0值的曲线。

9．利用悬浮区熔工艺来提纯一含有镓且浓度为5×1016cm -3的单晶硅锭。一次悬浮区熔通过，熔融带长度为2cm ，则在离多远处镓的浓度会低于5×1015cm -3？

10．从式L kx s e k C C /0)1(1/---=，假设k e =0.3，求在x/L=1和2时，C s /C 0的值。

11．如果用如右图所示的硅材料制造

p +-n 突变结二极管，试求用传统的方法

掺杂和用中子辐照硅的击穿电压改变

的百分比。

12．由图10.10，若C m =20％，在T b 时，

还剩下多少比例的液体?

13．用图10.11解释为何砷化镓液体总

会变成含镓比较多?

14．空隙n s 的平衡浓度为

Nexp[-E s /(kT)]，N 为半导体原子的浓

度，而E s 为形成能量。计算硅在27℃、

900℃和1 200℃的n s (假设E s =2.3eV)．

15．假设弗兰克尔缺陷的形成能量(E f )

为1.1eV ，计算在27℃、900℃时的缺陷密度．弗兰克尔缺陷的平衡密度是

. . . zyzl . . ，其中N 为硅原子的浓度(cm -3)，N’为可用的间隙位置浓度(cm -3)，可

表示为N’=1×1027cm -3．

16．在直径为300mm 的晶片上，可以放多少面积为400mm 2的芯片？解释你对芯片形 状和在周围有多少闲置面积的假设．

17．求在300K 时，空气分子的平均速率(空气相对分子质量为29)

．

图 10.10. Phase diagram for the gallium- 图 10.11. Partial pressure of gallium and arsenic arsenic system. over gallium arsenide as a function of temperature.

Also shown is the partial pressure of silicon.

18．淀积腔中蒸发源和晶片的距离为15cm ，估算当此距离为蒸发源分子的平均自由程的10％时系统的气压为多少?

19．求在紧密堆积下(即每个原子和其他六个邻近原子相接)，形成单原子层所需的每单位面积原子数N s ．假设原子直径d 为4.68?．

20．假设一喷射炉几何尺寸为A=5cm 2及L=12cm ．

(a)计算在970℃下装满砷化镓的喷射炉中，镓的到达速率和MBE 的生长速率；

(b)利用同样形状大小且工作在700℃，用锡做的喷射炉来生长，试计算锡在如前述砷化镓生长速率下的掺杂浓度(假设锡会完全进入前述速率生长的砷化镓中，锡的摩尔质量为118.69；在700℃时，锡的压强为2.66×10-6Pa)．

21．求铟原子的最大比例，即生长在砷化镓衬底上而且并无任何错配的位错的Ga x In 1-x As 薄膜的x 值，假定薄膜的厚度是10nm ．

22．薄膜晶格的错配f 定义为，f=[a 0(s)-a 0(f)]/a 0(f)≡△a 0/a 0。a 0(s)和a 0(f)分别为衬底和薄膜

. .

. zyzl . . 在未形变时的晶格常数，求出InAs-GaAs 和Ge-Si 系统的f 值． Solution

1. C 0 = 1017 cm -3

k 0(As in Si) = 0.3

C S = k 0C 0(1 - M /M 0)k 0-1

= 0.3?1017(1- x)-0.7 = 3?1016/(1 - l /50)0.7

2

4

6

8

10

12

14

16

010********

l (cm)

N D (1016 c m -3)

2. (a) The radius of a silicon atom can be expressed as

?175.143.583 so 8

3=?=

=r a r

(b) The numbers of Si atom in its diamond structure are 8. So the density of silicon atoms is 32233atoms/cm 100.5)?43.5(88?===a n (c) The density of Si is 32322

23

cm / g 1002.610509.28/11002.6/???=?=n M ρ = 2.33 g / cm 3.

3. k 0 = 0.8 for boron in silicon

M / M 0 = 0.5

The density of Si is 2.33 g / cm 3.

The acceptor concentration for ρ = 0.01 Ω–cm is 9?1018 cm -3. The doping concentration C S is given by

10

000)1(--=k s M M C k C

. .

. zyzl . . Therefore

3

182.0181000cm 108.9 )5.01(8.0109)1(0---?=-?=-=k s M M k C C

The amount of boron required for a 10 kg charge is

2218102.4108.9338

.2000,10?=?? boron atoms So that

boron g 75.0atoms/mole

1002.6atoms 102.4g/mole 8.102322=???. 4. (a) The molecular weight of boron is 10.81.

The boron concentration can be given as

3

18223

3atoms/cm 1078.9 1

.014.30.101002.6g 81.10/g 1041.5 fer silicon wa of volume atoms boron of number ?=?????==

-b n (b) The average occupied volume of everyone boron atoms in the wafer is

318cm 10

78.911?==b n V We assume the volume is a sphere, so the radius of the sphere ( r ) is the average distance between two boron atoms. Then

cm 109.2437-?==π

V r . 5. The cross-sectional area of the seed is

22cm 24.0255.0=??

? ??π The maximum weight that can be supported by the seed equals the product of the critical yield strength and the seed ’s cross-sectional area:

kg 480g 108.424.0)102(56=?=?? The corresponding weight of a 200-mm-diameter ingot with length l is m. 56.6cm 656 g 48000020.20)g/cm 33.2(2

3==∴=??

? ?? l l π

. zyzl . . 9. The segregation coefficient of Ga in Si is 8 ?10-3

From Eq. 18

L kx s e k C C /0)1(1/---=

We have

cm.

24 ln(1.102)

250 10

5/10511081ln 1082 /11ln 161533-0==???? ????-?-?=???

? ??--=-C C k k L x s 10. We have from Eq.18

00.20.40.60.81

F r a c tion Solidifie d

. .

. zyzl . . ])/exp()1(1[0L x e k e k C s C ---=

So the ratio )]/exp()1(1[0/L x e k e k C s C ---=

= 1/at 52.0)13.0exp()3.01(1==?-?--L x

= 38.0 at x/L = 2.

11. For the conventionally-doped silicon, the resistivity varies from 120 Ω-cm to 155 Ω-cm. The

corresponding doping concentration varies from 2.5?1013 to 4?1013 cm -3. Therefore the range of breakdown voltages of p + - n junctions is given by

V 11600 to 7250/109.2)(10

6.12)103(1005.1)(2171192

5121

2

=?=?????=?----B B B c s B N N N q V E εV 4350725011600=-=?B V

%307250/2±=??

? ???∴B V For the neutron irradiated silicon, ρ = 148 ± 1.5 Ω-cm. The doping concentration is 3?1013 (±1%). The range of breakdown voltage is

. V 9762 to 9570%)1(103/109.2/103.1131717=±??=?=B B N V

V 19295709762=-=?B V %19570/2±=??

? ???∴B

V . 12. We have

l

s C C C C M M m s l m l s =--==b b T at liquid of weight T at G aAs of weight

Therefore, the fraction of liquid remained f can be obtained as following 65.030

1630=+≈+=+=l s l M M M f l s l . 13. From the Fig.11, we find the vapor pressure of As is much higher than that of the Ga.

Therefore, the As content will be lost when the temperature is increased. Thus the composition of liquid GaAs always becomes gallium rich. 14. ??

????-?=?=-=)300/(8.88exp 105)/eV 3.2exp(105)/exp(2222T kT kT E N n s s

. . . zyzl . . = K 300 C 27at 0cm 1023.10316=≈?-- = K 1173 C 900at cm 107.60312=?- = K 1473 C 1200at cm 107.60314=?-.

15. )2/exp(`kT E NN n f f -=

=)300//(7.94242/1.1/8.327221007.7101105T kT eV kT eV e e e ---??=???? = 1027.517-?at 27o C = 300 K

=2.14?1014 at 900o C = 1173 K.

16. 37 ? 4 = 148 chips

In terms of litho-stepper considerations, there are 500 μm space tolerance between the mask boundary of two dice. We pide the wafer into four symmetrical parts for convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the perimeter parts is the worst due to the edge effects.

17. M

kT dv f dv vf v v av πν8 0 0

==??∞

∞ Where ???? ??-??? ??=kT M kT M f 2exp 2422

2/3ννπν

M: Molecular mass

k: Boltzmann constant = 1.38?10-23 J/k

T: The absolute temperature ν: Speed of molecular

So that cm/sec 1068.4m/sec 46810

67.1293001038.122

42723?==?????=--πνav . 18. cm )Pa in (66.0P =

λ Pa 104.4150

66.066

.03-?===∴λP . d

. . . zyzl . . 19. For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle. Each section corresponds to 1/6 of an atom. Therefore d d N s 2321613 triangle the of area triangle in the contained atoms of number ??

==

=282)1068.4(3232-?=d

=214atoms/cm 1027.5?.

20. (a) The pressure at 970?C (=1243K) is 2.9?10-1 Pa for Ga and 13 Pa for As 2. The arrival rate is given by the product of the impringement rate and A/πL 2 :

Arrival rate = 2.64?1020??

? ????? ??2L A MT P π = 2.64?1020??

? ??????? ????-21125124372.69109.2π = 2.9?1015 Ga molecules/cm 2 –s

The growth rate is determined by the Ga arrival rate and is given by

(2.9?1015)?2.8/(6?1014) = 13.5 ?/s = 810 ?/min .

(b) The pressure at 700oC for tin is 2.66?10-6 Pa. The molecular weight is 118.69.

Therefore the arrival rate is

s ??=??? ??????? ?????-2102620

cm molecular/ 1028.212597369.1181066.21064.2π If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an electron concentration of

. cm 1074.121042.4109.21028.23-17221510?=???

? ??????? ????

21. The x value is about 0.25, which is obtained from Fig.

26.

22. The lattice constants for InAs, GaAs, Si and Ge are

6.05, 5.65,5.43, and 5.65 ?, respectively (Appendix

本文来源：https://www.bwwdw.com/article/t5dl.html