1
EE143 Midterm Exam #1 Solutions Sp2001
Problem1
(i) The pad oxide is very thin (~100 Å). It can easily be prepared by thermal oxidation.
The silicon nitride is prepared by chemical vapor deposition (CVD)
(ii ) Peak concentration of implant profile = 0.4 φ / ∆Rp. For the shallow channel stop implant, ∆Rp ~
0.1 µm , peak concentration ~ 4E17/cm2
(iii) The gate oxide oxidation step is performed before poly-Si deposition.
(iv) We prefer As for shallow junctions because As has a slower diffusion constant than Phosphorus
[Other secondary reasons: As has a higher solid solubility than P.Therefore, lower sheet resistance and
contact resistance. Also, for the same implant energy , the projected range of As is smaller]
(v) We cannot use thermal oxidation step to replace CVD oxide. The oxide layer is to electrically isolate
the Al metallization from the poly-Si gate and substrate and require sufficient thickness. If too much poly-
Si is consumed by using thermal oxidation, the gate dimensions will be significantly reduced.
(vi) We cannot replace the poly-Si by Al as gate material in this process flow. There is a high temperature
annealing step (>900 oC ) after source/drain implantation. This step will melt the Al.
Problem2
( i) xox2 + Axox = B(t + τ)
For 900°C, O2 : B = 5600 (Å )2/minute, B
A = 2 Å /min ∴ A = 2800 Å
∴ For xox = 1000Å, τ = 0 ⇒ t = 680 min.
( ii) Let xi be the oxide grown during the steam oxidation step such that the final oxide thickness
xox = 5000Å after dry oxidation
∴ (5000)2 + 2800 • (5000) = 5600 (680) + xi2 + 2800 xi ⇒ xi = 4700Å.
( iii) Using a high pressure for oxidation increases the oxidant concentration on the oxide surface ( Henry’s
Law). The oxidation rate will increase according to the Grove Model. Since it takes less time to grow the
0.1 µm gate oxide, the additional oxide grown on the field oxide is less. We need a larger xi initially to
make the final field oxide thickness = 0.5 µm .
( iv) (111) Si surface has more reaction sites than (100) Si. This will increase the surface reaction term (ks).
The oxidation rate will increase according to the Grove Model. Since it takes less time to grow the 0.1 µm
gate oxide, the additional oxide grown on the field oxide is less. We need a larger xi initially to make the
final field oxide thickness = 0.5 µm .
Problem 3
(a) (0.5 - x) • 2.17 = 0.5 ⇒ x = 0.27 µm
(b) (i) Rp = 0.5 µm ∆Rp = 0.09 µm
The boron will need a kinetic energy of 186
keV
(A) For B +ions, we need 186kV accelerating
voltage
(B) For BF2+ions, we need 186 × 11+19+19
11 =
828 kV accelerating voltage
(ii) Np = 1013
2π∆Rp = 4.4×1017/cm3
∴ 4.4×1017e- (xj - Rp)2/2∆Rp = 10 15
x=0
C(x)
x
0.5
µµ
m
x
j1
1E15/cm3
