CHAPTER 3 SOLUTIONS
5
2.9 The threshold is increased by applying a negative body voltage so V
sb
> 0.
2.11 The nMOS will be OFF and will see V
ds
= V
DD
, so its leakage is
2.13 Assume V
DD
= 1.8 V. For a single transistor with n = 1.4,
For two transistors in series, the intermediate voltage x and leakage current are
found as:
In summary, accounting for DIBL leads to more overall leakage in both cases.
However, the leakage through series transistors is much less than half of that
through a single transistor because the bottom transistor sees a small Vds and much
less DIBL. This is called the stack effect.
For n = 1.0, the leakage currents through a single transistor and pair of transistors
are 13.5 pA and 0.9 pA, respectively.
2.15 V
IL
= 0.3; V
IH
= 1.05; V
OL
= 0.15; V
OH
= 1.2; NM
H
= 0.15; NM
L
= 0.15
2.17 Either take the grungy derivative for the unity gain point or solve numerically for
V
IL
= 0.46 V, V
IH
= 0.54 V, V
OL
= 0.04 V, V
OH
= 0.96 V, NM
H
= NM
L
= 0.42 V.
2.19 Take derivatives or solve numerically for the unity gain points: V
IL
= 0.43 V, V
IH
=
0.50 V, V
OL
= 0.04 V, V
OH
= 0.97 V, NM
H
= 0.39, NM
L
= 0.47 V.
2.21 (a) 0; (b) 0.6; (c) 0.8; (d) 0.8
Chapter 3
3.1 First, the cost per wafer for each step and scan. 248nm – number of wafers for four
II vee pA
leak dsn T
V
nv
t
T
== =
−
β
218
69
.
499
Ivee pA
)
()
21.8 21.8
1
1
69 mV; 69 pA
DD t
t
TT T
DD t
t
TT T
VxVx
Vx
x
nv v nv
leak T T
VxVx
Vx
x
nv v nv
leak
I vee e vee
eee
xI
η
η
η
η
ββ
−−
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−
−−−
−+
−
⎛⎞
=−=
⎜⎟
⎜⎟
⎝⎠
⎛⎞
−=
⎜⎟
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⎝⎠
==
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