Chapter 16 Problem Solutions 16.1 (a)
( )( )( )( )
148
8
1/ 219 14 15 8
22 2
(3.9)(8.85 10 ) 7.67 10450 10
2 2 1.6 10 11.7 8.85 10 8 10 5.15 10
s aTN fp SB fp
ax
axax
ax
s a
e NV V
C
Ct
e N
φ φ
−−
−
− − −
∈ ⎡ ⎤Δ = + −⎣ ⎦
∈ ×= = = ××
⎡ ⎤∈ = × × × = ×⎣ ⎦
Then 8
8
5.15 10 2(0.343) 2(0.343)7.67 10TN SBV V
−
−
× ⎡ ⎤Δ = ⋅ + −⎣ ⎦×
For 1 :SBV V=
0.671 1.686 0.686 0.316TN TNV V V⎡ ⎤Δ = − ⇒ Δ =⎣ ⎦
For 1 :SBV V=
0.671 2.686 0.686 0.544TN TNV V V⎡ ⎤Δ = − ⇒ Δ =⎣ ⎦
(b) For 2.5 V, 5 V,GS DSV V= = transistor biased in the saturation region.
[ ]( )
[ ]( )
2
2
2
2
( )For 0,
0.2(2.5 0.8) 0.578 mAFor 1,
0.2 2.5 0.8 0.316 0.383 mA
For 2,
0.2 2.5 0.8 0.544 0.267 mA
D n GS TN
SB
D
SB
D
SB
D
I K V VVI
V
I
V
I
= −== − ==
= − + =
=
= − + =
16.2 (a)
( )( ) ( )
2
23
54 2
2( )
5 (0.1) 2 5 0.8 0.1 0.140 10
8 10or 1.476 10 /2
DD OD n GS TN O O
D
n
n
V vI K V V v v
R
K
WK A VL
−−
− ⎡ ⎤= = − −⎣ ⎦
− ⎡ ⎤= − −⎣ ⎦×× ⎛ ⎞= × = ⎜ ⎟
⎝ ⎠
So that 3.69WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
b. From Equation (16.10). [ ] [ ]
[ ] [ ]
[ ]
[ ]
2
2
2
0
(0.1476)(40) 0.8 0.8 5 0
1 (1) 4(0.1476)(40)(5)or 0.8
2(0.1476)(40)or 0.8 0.839
n D It TN It TN DD
It It
It
It
K R V V V V V
V V
V
V
− + − − =
− + − − =
− ± +− =
− =
So that 1.64 VItV =
(max)D DDP I V= ⋅
and 5 (0.1)(max) 0.1225 mA40DI −= =
or 0.6125 mWP = 16.3 a. From Equation (16.10), the transistor point is found from
2
2
2
0
( ) ( ) 050 / , 20 , 0.8
(0.05)(20)( ) ( ) 5 0
1 1 4(0.05)(20)(5)1.79 V So 2.59 V
2(0.05)(20)1.79 V
n D It TN It TN DD
n D TN
It TN It TN
It TN It
t
K R V V V V VK A V R k V V
V V V V
V V V
V
μ− + − − =
= = Ω =− + − − =
− ± +− = = =
=
Output voltage for 5 VIv = is determined from Equation (16.12): 2
0 0 0
20 0
2
0
5 (0.05)(20) 2(5 0.8)
9.4 5 0
9.4 (9.4) 4(1)(5)So 0.566 V
2(1)
v v v
v v
v
⎡ ⎤= − − −⎣ ⎦− + =
± −= =
b. For 200 k ,DR = Ω
( )
( )
( ) ( )( )( )
0
20 0 0
20 0
2
0
1 1 4(0.05)(200)(5)0.659 V So 1.46 V
2(0.05)(200)0.659 V
5 (0.05)(200) 2 5 0.8
or 10 85 5 0
85 85 4 10 50.0592 V
2 10
It TN It
t
V V V
V
v v v
v v
v
− ± +− = = =
=
⎡ ⎤= − − −⎣ ⎦− + =
± −= =
16.4 (a) P IV=
( )
( )
2
2
0.25 (33) 75.76 μA3.3 0.15 41.6 K0.07576
28075.76 3.3 0.8 0.3032
nGS TN
I I
R R
k WI V VL
W WL L
= ⇒ =−= ⇒ =
′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
(b)
( )
( )( ) ( )
( )
2
2
2
2
(sat)(sat)
3.3 0.80.08 0.303 1.6 0.642 41.6
0.504 1.6 0.64 4.1
0.504 0.1936 3.777 0
0.1936 0.03748 4(0.504)(3.777)2(0.504)
2.55V
DS GS TN
DD DSD n GS TN
GSGS GS
GS GS GS
GS GS
GS
GS
V V VV VI K V V
RV
V V
V V V
V V
V
V
= −−
= − =
− −⎛ ⎞ − + =⎜ ⎟⎝ ⎠
− + = −
+ − =
− ± +=
=
For 0.8 2.55VGSV≤ ≤ Transistor biased in saturation region 16.5 (a)
0.25 (3.3) 75.76 μADDP I V
I I= ⋅
= ⇒ =
For Sat Load
( )
( )( ) ( )( ) ( )
2
2 2
8075.76 3.3 0.15 0.8 0.3432
80 800.343 3.3 0.15 0.8 75.76 2 2.5 0.8 0.15 0.152 2
3.89
L L
D
D
W WIL L
WIL
WL
⎛ ⎞⎛ ⎞ ⎛ ⎞= = − − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎡ ⎤− − = = = − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
Eq 16.21 3.893.3 0.8 0.8 10.343 5.994
4.36773.8910.343
0.8 1.372 V
It
GS
V
V
⎛ ⎞− + +⎜ ⎟⎜ ⎟
⎝ ⎠= =+
≤ ≤
16.6 (a) From Equation (16.23)
( )( ) ( ) ( )2 22 3 0.5 0.25 0.25 3 0.25 0.5 4.26D D
L L
K KK K
⎡ ⎤− − = − − ⇒ =⎣ ⎦
(b) ( )( ) ( ) ( )2 22 2.5 0.5 0.25 0.25 3 0.25 0.5 5.4D D
L L
K KK K
⎡ ⎤− − = − − ⇒ =⎣ ⎦
(c) ( )2 2
2
( )0.080 (1)(3 0.25 0.5) 0.203
2(0.203)(3) 0.608
D L GSL TNL L DD O TNL
D
D DD
i K V V K V v V
i mA
P i V P mW
= − = − −
⎛ ⎞= − − ⇒ =⎜ ⎟⎝ ⎠
= ⋅ = ⇒ =
for both parts (a) and (b). 16.7
( ) ( )
2
2
0.4 (3) 0.1333
( )0.0800.1333 3 0.1 0.5 0.2304
2
D DD D D
D L DD O TNL
L L
P mW i V i i mA
i K V v VW WL L
= = ⋅ = ⇒ =
= − −
⎛ ⎞⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
So 0.579L
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
( )( ) ( ) ( )2 22 2.5 0.5 0.1 0.1 3 0.1 0.5D
L
KK
⎡ ⎤− − = − −⎣ ⎦
14.8D
L
KK
⇒ = so that 8.55D
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
( )3 0.5 0.5 1 14.8
1 14.8ItV− + +
=+
or 1.02 , 0.52It OtV V V V= =
16.8 We have
( ) ( )
( )( ) ( )( ) ( ) ( )
( )( ) ( )( )( ) ( )
( )( ) [ ] ( )
( )
22
2 2
22 2
2
/2 0.08 0.08 0.08
/
/2 2 0.2 0.08 0.0064 0.92 0.2 0.5184
/
/ /0.096 0.5184 5.4
/ /
DI TND O O DD O TNL
L
DDD TN TN DD DD DD DD TN
L
DDD DD DD DD DD DD
L
D D
L L
K v V v v V v VKW L
V V V V V V V VW L
W LV V V V V V
W L
W L W LW L W L
⎡ ⎤− − = − −⎣ ⎦
⎡ ⎤− − − = − −⎣ ⎦
⎡ ⎤− − = − =⎡ ⎤⎣ ⎦⎣ ⎦
= ⇒ =
16.9
Logic 1OH B TNV V V= − = So (a) 4 3B OHV V V V= ⇒ = (b) 5 4B OHV V V V= ⇒ = (c) 6 5B OHV V V V= ⇒ = (d) 7 5 ,since 0B OH DSV V V V V= ⇒ = = For I OHv V=
( ) [ ]222D I T O O L B O TK v V v v K V v V⎡ ⎤− − = − −⎣ ⎦ Then (a) ( ) ( ) ( )[ ]221 2 3 1 0.4 4 1 0.657OL OL OL OLV V V V V⎡ ⎤− − = − − ⇒ =⎣ ⎦
(b) ( ) ( ) ( )[ ]221 2 4 1 0.4 5 1 0.791OL OL OL OLV V V V V⎡ ⎤− − = − − ⇒ =⎣ ⎦
(c) ( ) ( ) ( )[ ]221 2 5 1 0.4 6 1 0.935OL OL OL OLV V V V V⎡ ⎤− − = − − ⇒ =⎣ ⎦ (d) Load in non-sat region
( ) ( ) ( )( ) ( )22(1) 2 5 1 0.4 2 7 1 5 5DD OL
OL OL OL OL OL
i i
V V V V V
=
⎡ ⎤⎡ ⎤− − = − − − − −⎣ ⎦ ⎣ ⎦
( ) ( )( ) ( )( ) ( )( )
2 2
2 2
2 2
8 0.4 2 6 5 25 10
0.4 2 30 11 25 10
0.4 60 22 2 25 10
OL OL OL OL OL OL
OL OL OL OL
OL OL OL OL
V V V V V V
V V V V
V V V V
⎡ ⎤− = − − − − +⎣ ⎦⎡ ⎤= − + − + −⎣ ⎦⎡ ⎤= − + − + −⎣ ⎦
( )( )( )
2 2
2
8 14 4.8 0.4
1.4 12.8 14 0
12.8 163.84 4 1.4 142 1.4
1.27
OL OL OL OL
OL OL
OL
OL
V V V V
V V
V
V V
− = − +
− + =
± −=
=
For load ( )sat 7 1.27 1 4.73
5 1.27 3.73 non-satDS
DS
V VV
= − − == − =
16.10 a. For load 5 2 3Ot DD TNLV V V V= + = − =
( )
( ) ( )500 0.8 2100
1.69Load
3
DIt TND TNL
L
It
It
Ot
K V V VK
V
V VV V
⋅ − = −
− = − −
⇒ = ⎫⎬= ⎭
0
Driver: 1.69 0.8 0.891.69 V
Driver0.89 V
Ot It TND
It
t
V V V VVV
= − = − == ⎫
⎬= ⎭
b. From Equation (16.29(b)):
( )
( ) ( )( )( )
220 0
20 0
2
0 0
500 2(5 0.8) 21005 42 4 0
42 42 4 5 40.0963 V
2 5
v v
v v
v v
⎡ ⎤⋅ − − = − −⎡ ⎤⎣ ⎦⎣ ⎦
− + =
± −= ⇒ =
c. ( ) ( ) 22 100 2 400D L TNL Di K V i Aμ= − = − − ⇒ =⎡ ⎤⎣ ⎦
16.11
( )( ) ( ) ( )2 2500 2 3 0.5 0.1 0.150 TNLV⎛ ⎞ ⎡ ⎤− − = −⎜ ⎟ ⎣ ⎦⎝ ⎠
So ( )2 4.9 2.21TNL TNLV V V− = ⇒ = − 16.12 (a)
( )
( )
( )( ) ( ) ( )
( )( )
2
2
22
150 3 50
( )8050 1 1.252
2 3 0.5 0.1 0.1 1
/2.04 2.55
/
D DD
D D
D L TNL
L L
D
L
D D
DL L
P i Vi i A
i K VW WL L
KK
W LK WK W L L
μ= ⋅
= ⋅ ⇒ =
= −
⎛ ⎞⎛ ⎞ ⎛ ⎞= − − ⇒ =⎡ ⎤⎜ ⎟⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠⎝ ⎠ ⎝ ⎠
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦⎣ ⎦
⎛ ⎞= = ⇒ =⎜ ⎟⎝ ⎠
For the Load:
( ) ( )3 1 2
2.04 0.5 1 1.20
Ot DD TNL Ot
It It
V V V V V
V V V
= + = − ⇒ =
− = − − ⇒ =⎡ ⎤⎣ ⎦
For the Driver: 1.20 0.5 0.70
1.20Ot It TND Ot
It
V V V V V
V V
= − = − ⇒ =
=
(b) L IL OLU
H OHU IH
NM V VNM V V
= −= −
( )( )( )
( )( )
10.5 0.902
2.04 1 2.04
2 10.5 1.31
3 2.04
IL
IH
V V
V V
− −⎡ ⎤⎣ ⎦= + =+
− −⎡ ⎤⎣ ⎦= + =
( ) ( )( )Then 3 1 2.04 0.902 0.5 2.82OHUV V= − + − =
( )1.31 0.50.405
20.902 0.405 0.497
2.82 1.31 1.51
OLU
L L
H H
V V
NM NM V
NM NM V
−= =
= − ⇒ =
= − ⇒ =
16.13 a. From Equation (16.29(b)):
( )( ) ( ) ( )2 22 2.5 0.5 0.05 0.05 [ 1 ]
1
D L
L
W WL L
WL
⎛ ⎞ ⎛ ⎞⎡ ⎤− − = − −⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
Then 5.06D
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
b. ( ) ( ) 280 1 12Di
⎛ ⎞= − −⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠
or 40 ADi μ=
( )( )40 2.5 100 WD DDP i V P μ= ⋅ = ⇒ = 16.14 a. i. 0.5 V 0 0I Dv i P= ⇒ = ⇒ = ii. 5 V,Iv = From Equation (16.12),
( )( ) ( )
( ) ( )( )( )
( )( )
20 0 0
20 0
2
0 0
5 0.1 20 2 5 1.5
2 15 5 0
15 15 4 2 50.35 V
2 25 0.35 0.2325 mA
200.2325 5 1.16 mW
D
D DD
v v v
v v
v v
i
P i V P
⎡ ⎤= − − −⎣ ⎦− + =
± −= ⇒ =
−= =
= ⋅ = ⇒ =
b. i. 0.25 V 0 0I Dv i P= ⇒ = ⇒ = ii. 4.3 V,Iv = From Equation (16.23),
( ) [ ]220 0 0
2 20 0 0 0
100 2 4.3 0.7 10 5 0.7
10 7.2 18.49 8.6
v v v
v v v v
⎡ ⎤− − = − −⎣ ⎦⎡ ⎤− = − +⎣ ⎦
Then
( ) ( )( )( )
20 0
2
0 0
11 80.6 18.49 0
80.6 80.6 4 11 18.490.237 V
2 11
v v
v v
− + =
± −= ⇒ =
Then [ ]
( )( )
210 5 0.237 0.7 165 A
165 5 825 WD
D DD
i
P i V P
μμ
= − − =
= ⋅ = ⇒ =
c. i. 0.03 V 0 0I Dv i P= ⇒ = ⇒ = ii.
( ) ( ) ( )( )( )
22
5 V
10 2 40
40 5 200 W
I
D L TNL
D DD
v
i K V A
P i V P
μ
μ
=
= − = − − =⎡ ⎤⎣ ⎦= ⋅ = ⇒ =
16.15
1 3.8Vov = Load & Driver in Sat, region, ML1, MD1
( ) ( )
( ) ( )( )( )
2 2
2 21
2
(1) 5 0.8 10 0.8
0.16 10 0.80.9265V
DL DD
GSL TNL GSD TNDL D
o I
I
I
i iW Wv V v VL L
v v
vv
=
⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− − = −
= −=
Now MD2: Non Sat and ML2: Sat
( ) ( )
( )( ) ( ) ( )( )
2 21 2 2
2 22 2 2
2 22 2 2
2 22 2 2 2
22 2
2
1 5 0.8 10 2 3.8 0.8
4.2 10 6
17.64 8.4 60 10
11 68.4 17.64 0
DL DD
GSL TNL o TND o oL D
o o o
o o o
o o o o
o o
i iW Wv V v V v vL L
v v v
v v v
v v v v
v v
=
⎛ ⎞ ⎛ ⎞ ⎡ ⎤− = − −⎜ ⎟ ⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠
⎡ ⎤− − = − −⎣ ⎦
⎡ ⎤− = −⎣ ⎦− + = −
− + =
( )( )( )2
2
68.4 4678.56 4 11 17.642 11
0.270 V
o
o
v
v
± −=
=
16.16 a. From Equation (16.41),
( )( ) 01
2 20.8 1.95 V
3 4IH IHV V v
− −⎡ ⎤⎣ ⎦= + ⇒ = =
2DM in non-saturation and 2LM in saturation.
( ) ( )( ) ( ) ( )
( ) ( )( )( )
2201 02 02
2202 02
202 02
2
02 02
2
4 2 1.95 0.8 1 2
4 9.2 4 0
9.2 9.2 4 4 40.582 V
2 4
D TND L TNLK v V v v K V
v v
v v
v v
⎡ ⎤− − = −⎣ ⎦
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦⎣ ⎦− + =
± −= ⇒ =
Both 1DM and 1LM in saturation region. From Equation (16.28(b)).
( ) ( )4 0.8 2Iv⋅ − = − − or 1.8 VIv =
b. ( )( ) 01
20.8 1.25 V
4 1 4ILV v
+= + = =
+
2DM in saturation, 2LM in non-saturation
[ ] ( )( ) ( )
( ) ( )( ) ( )( ) ( )
( ) ( )( )( )
2 21 2 2
2 202 02
202 02
2
02
2 5 5
4 1.25 0.8 2 2 5 5
5 4 5 0.81 0
4 4 4 1 0.815 0.214 V
2 1
D O TND L TNL O OK v V K V v v
v v
v v
v
⎡ ⎤− = − − − −⎣ ⎦
− = − − −
− − − + =
± −− = =
so 02 4.786 Vv =
To find :Iv
( ) ( ) ( )( )2201
01
01
4 0.8 1 2
0.8 11.8 V
v
vv
− = − −
− =− =
c. 1.95 V, 1.25 VIH ILV V= = 16.17 a. i. Neglecting the body effect,
0 DD TNv V V= − Assume 5 V,DDV = then 0 4.2 Vv = ii. Taking the body effect into account: From Problem 16.1.
0 0.671 0.686 0.686TN TN SBV V V⎡ ⎤= + + −⎣ ⎦
and 0SBV v= Then
( )
( )
( ) ( )
0 0
0 0
0 0
20 0 0
20 0
2
0 0
5 0.8 0.671 0.686 0.686
4.756 0.671 0.686
0.671 0.686 4.756
0.450 0.686 22.62 9.51
9.96 22.3 0
9.96 9.96 4 22.33.40 V
2
v v
v v
v v
v v v
v v
v v
⎡ ⎤= − + + −⎣ ⎦
= − +
+ = −
+ = − +
− + =
± −= ⇒ =
b. PSpice results similar to Figure 16.13(a). 16.18 Results similar to Figure 16.13(b). 16.19 a. XM on, YM cutoff. From Equation (16.29(b)):
( )( ) ( ) ( ) 222 5 0.8 0.2 0.2 2D
L
KK
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦⎣ ⎦
or 2.44D
L
KK
=
b. For .5 VX Yv v= =
( ) ( ) ( )
( ) ( )( )( )
220 0
20 0
2
0
2 2.44 2 5 0.8 2
4.88 41.0 4 0
41 41 4 4.88 42 4.88
v v
v v
v
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦⎣ ⎦− + =
± −=
or 0 0.0987 Vv =
c.
( ) ( )
( )( )
280 1 2 160 A2
160 5 800 W
Di
P P
μ
μ
⎛ ⎞= − − =⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠= ⇒ =
for both parts (a) and (b). 16.20 (a) Maximum value of Ov in low state- when only one input is high, then,
( )( ) ( ) ( ) 222 3 0.5 0.1 0.1 1
2.04
D
L
D
L
KKKK
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦⎣ ⎦
=
(b)
( )
( )
2
2
0.1 (3) 33.3
28033.3 1 0.83252
D DD
D D
nD TNL
L
L L
P i Vi i Ak Wi V
LW WL L
μ= ⋅= ⇒ =
′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= − − ⇒ =⎡ ⎤⎜ ⎟⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠⎝ ⎠ ⎝ ⎠
Then 1.70D
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
(c) ( ) ( ) ( ) 223 2.04 2 3 0.5 1 0.0329O O Ov v v V⎡ ⎤− − = − − ⇒ =⎡ ⎤⎣ ⎦⎣ ⎦ 16.21 (a) One driver in non-sat,
( ) ( )
( ) ( )( ) ( )
2 2
2 2
2
1 2 3.3 0.5 0.1 0.1
1.82
D L TNL D GS TND DSD DSD
D
L
D
L
I K V K V V V V
KK
KK
⎡ ⎤= − = − −⎣ ⎦
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦ ⎣ ⎦
=
(b)
( )
( ) 2
0.1 3.3 30.3 A
8030.3 1 0.75752
1.38
DD
L L
D
P VI
W WIL L
WL
μ= ±= ± ⇒ =
⎛ ⎞⎛ ⎞ ⎛ ⎞= = − − ⇒ =⎡ ⎤⎜ ⎟⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
(c)
(i) Two inputs High, 0.1 0.05 V2ov ≈ =
(ii) Three inputs High, 0.1 0.0333 V3ov ≈ =
(iii) Four inputs High, 0.1 0.025 V4ov ≈ =
16.22 a.
( )
[ ]
( )
'2
11
2
1
250 5 50
2
6050 22
D DD
D D
nD TNL
ML
ML
P i Vi i
k Wi VL
WL
μ= ⋅
= ⇒ = Α
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞= − −⎡ ⎤⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠
So that 1
0.417ML
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
( ) [ ]
( )( ) ( ) ( )
22
22
2
2 5 0.8 0.15 0.15 2
DI TND O O TNL
L
D
L
K v V v v VKKK
⎡ ⎤− − = −⎣ ⎦
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦⎣ ⎦
or 1
3.23 1.35D
MDL
K WK L
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
b. For 010 5X Yv v v= = ⇒ = and 03 4.2v = Then
( ) ( ) [ ]
( ) ( ) ( ) ( )
22 22 1 2 2 3 3 2 2 2 2
2 3 222 2
02 02 02 02
2 202 02 02 02
2 2
8, 8, 1
8 2 5 0.8 8 2 4.2 0.8 1 2
67.2 8 54.4 8 4
D O TND O O D O TND O O L TNL
D D L
K v V v v K v V v v K V
K K K
v v v v
v v v v
⎡ ⎤ ⎡ ⎤− − + − − = −⎣ ⎦ ⎣ ⎦∝ ∝ ∝
⎡ ⎤ ⎡ ⎤− − + − − = − −⎡ ⎤⎣ ⎦⎣ ⎦ ⎣ ⎦− + − =
Then
( ) ( )( )( )
20 0
2
02
16 121.6 4 0
121.6 121.6 4 16 42 16
v v
v
− + =
± −=
So 02 0.0330 Vv = 16.23
a. We can write ( ) ( ) [ ]22 22 2x X TN DSX DSX y Y DSX TN DSY DSY L DD O TNK v V v v K v v V v v K V v V⎡ ⎤ ⎡ ⎤− − = − − − = − −⎣ ⎦ ⎣ ⎦
where 0 DSX DSYv v v= + We have
9.2 , 10 , 0.8X Y DD TNv v V V V V V= = = = As a good first approximation, neglect the 2
DSXv and 2DSYv terms. Let 0 2 DSXv v≈ . Then from the first and
third terms in the above equation. ( ) ( )( )
( )
29 2 9.2 0.8 1 10 2 0.8
151.2 84.64 36.8DSX DSX
DSX DSX
v v
v v
− ≅ − −⎡ ⎤⎣ ⎦≅ −
So that 0.450 VDSXv = From the first and second terms of the above equation.
( ) ( )9 2 9.2 0.8 9 2 9.2 0.8DSX DSX DSYv v v− ≅ − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ or ( )( ) ( )16.8 0.45 2 9.2 0.45 0.8 DSYv= − − which yields 0.475 VDSYv =
Then 0 0.450 0.475DSX DSYv v v= + = + or 0 0.925 Vv =
We have 9.2 VGSXv =
and 9.2 9.2 0.45GSY DSXv v= − = − or 8.75 VGSYv =
b. Since 0v is close to ground potential, the body effect will have minimal effect on the results. From a PSpice analysis: For part (a):
00.462 V, 0.491 V, 0.9536 V, 9.2 V, and 8.738 VDSX DSY GSX GSYv v v v v= = = = = For part (b):
0.441 V,DSXv = 00.475 V, 0.9154 V, 9.2 V, and 8.759 VDSY GSX GSYv v v v= = = = 16.24 a. We can write
( ) ( ) [ ]22 22 2x X TNX DSX DSX y Y DSX TNY DSY DSY L TNLK v V v v K v v V v v K V⎡ ⎤ ⎡ ⎤− − = − − − = −⎣ ⎦ ⎣ ⎦
From the first and third terms, (neglect 2DSXv ),
( ) ( ) ( ) 24 2 5 0.8 1 1.5DSXv− = − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
or 0.067 VDSXv =
From the second and third terms, (neglect 2 ),DSYv
( ) ( ) ( ) 24 2 5 0.067 0.8 1 1.5DSYv− − = − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
or 0.068 VDSYv = Now
5, 5 0.067 4.933 VGSX GSY GSYv v v= = − ⇒ = and 0 0 0.135 VDSX DSYv v v v= + ⇒ =
Since 0v is close to ground potential, the body-effect has little effect on the results. 16.25
(a) We have DSV of each driver 0.2 0.05 V4
≈ =
[ ] ( )
( ) ( )( ) ( )
2 2
2 2
2
1 2 3.3 0.4 0.05 0.05
3.478
L TNL D GSD TN DSD DSD
D
L
D
L
K V K V V V V
KK
KK
⎡ ⎤− = − −⎣ ⎦
⎡ ⎤− − = − −⎡ ⎤⎣ ⎦ ⎣ ⎦
=
(b)
( )
( ) 2
0.15 3.3 45.45 A
8045.45 1 1.142
3.95
DD
L L
D
P I VI I
W WL L
WL
μ=
= ⇒ =
⎛ ⎞⎛ ⎞ ⎛ ⎞= − − ⇒ =⎡ ⎤⎜ ⎟⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.26 Complement of (B AND C) OR ( )A B C A⇒ ⋅ + 16.27 Considering a truth table, we find
A B Y 0 0 0 0 1 1 1 0 1 1 1 0
which shows that the circuit performs the exclusive-OR function. 16.28 ( )( )A B C D+ + 16.29 (a) Carry-out ( )A B C B C= • + + •
(b) For 1 0.2Ov Low V= =
( )( ) ( ) ( ) 222 5 0.8 0.2 0.2 1.5D
L
KK
⎡ ⎤− − = − − ⇒⎡ ⎤⎣ ⎦⎣ ⎦
For 1,L
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
then 1.37D
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
So, for 66
: 1.37WML
⎛ ⎞ =⎜ ⎟⎝ ⎠
To achieve the required composite conduction parameter,
For 1 51 5
: 2.74WM ML −
⎛ ⎞− =⎜ ⎟⎝ ⎠
16.30
AE: 0.075DSV ≈
( ) ( )( ) ( )2 2(1) 1 2 3.3 0.4 0.075 0.075
2.33 4.66
D
A E B C D
WL
W W W W WL L L L L
⎛ ⎞ ⎡ ⎤− − ≅ − −⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = ⇒ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
16.31 Not given
16.32 a. From Equation (16.43),
( )0 0
0 0
5 0.8 0.8 2.5 V1 1
channel, 2.5 0.8 3.3 V
channel, 2.5 0.8 1.7 V
I It It
Pt Pt
Nt Nt
v V V
p V V
n V V
− += = = =+
− = − − ⇒ =
− = − ⇒ =
c For 2 V,Iv = NMOS in saturation and PMOS in nonsaturation. From Equation (16.49),
( ) ( )( ) ( )2 20 0
20 0
2 0.8 2 5 2 0.8 5 5
1.44 4.4(5 ) (5 )
v v
v v
⎡ ⎤− = − − − − −⎣ ⎦= − − −
So ( ) ( )20 05 4.4 5 1.44 0v v− − − + =
( ) ( ) ( )( )2
0
4.4 4.4 4 1 1.445
2v
± −− =
or 0 05 0.356 4.64 Vv v− = ⇒ =
By symmetry, for 03 V, 0.356 VIv v= = 16.33
(a) ( )
( )
2
2
80 2 80 /240 4 80 /2
n
p
K A V
K A V
μ
μ
⎛ ⎞= =⎜ ⎟⎝ ⎠⎛ ⎞= =⎜ ⎟⎝ ⎠
(i) 3.3 0.4 (1)(0.4)1 1
1
1.65
nDD TP TN
pIt
n
p
It
KV V VK
VKK
V V
+ + ⋅− += =
++
=
PMOS: ( )1.65 0.4 2.05Ot It TP OtV V V V V= − = − − ⇒ =
NMOS: ( )1.65 0.4 1.25Ot It TN OtV V V V V= − = − ⇒ =
(iii) For 0.4 :Ov V= NMOS: Non-sat: PMOS:Sat
( ) [ ]( )( ) ( ) ( )
22
2 2
2
2 0.4 0.4 0.4 3.3 0.4 1.89
n GSN TN DS DS p SGP TP
I I I
K V V V V K V V
v v v V
⎡ ⎤− − = +⎣ ⎦
− − = − − ⇒ =
For 2.9 ,Ov V= By symmetry
( )1.65 1.89 1.65 1.41I Iv v V= − − ⇒ =
(b) ( )
( )
2
2
80 2 80240 2 402
n
p
K A V
K A V
μ
μ
⎛ ⎞= = /⎜ ⎟⎝ ⎠⎛ ⎞= = /⎜ ⎟⎝ ⎠
(i) ( )803.3 0.4 0.4
40 1.4480140
It ItV V V− + ⋅
= ⇒ =+
PMOS: ( )1.44 0.4 1.84Ot OtV V V= − − ⇒ =
NMOS: 1.44 0.4 1.04Ot OtV V V= − ⇒ =
(iii) For 0.4 Ov V=
( ) ( )( ) ( ) ( )[ ]2280 2 0.4 0.4 0.4 40 3.3 0.4 1.62I I Iv v v V⎡ ⎤− − = − − ⇒ =⎣ ⎦
For 2.9 :Ov V= NMOS: Sat, PMOS: Non-sat
( )[ ] ( ) ( )( ) ( )2 280 0.4 40 2 3.3 0.4 0.4 0.4 1.18I I Iv v v V⎡ ⎤− = − − − ⇒ =⎣ ⎦
16.34 (a) From Eq. (16.43), switching voltage (i)
( ) ( ) ( )
( )
2 43.3 0.4 0.4 3.226612 1.776 V
1.81652 41 112
nDD TP TN
pI Itt
n
p
KV V VK
v vKK
+ + ⋅ + − += = = ⇒ =
+ +
(ii) 0 3.1 Vv = , PMOS, non-sat; NMOS, sat
( ) ( )
( ) ( )( ) ( ) ( )[ ]
[ ]
( )
22
22
2
2
22 2
40 8012 2 3.3 0.4 3.3 3.1 3.3 3.1 4 0.42 2
12 1.16 0.4 0.04 8 0.8 0.16
8 1.6 12.16 0
1.6 2.56 4 8 12.1
p nSG TP SD SD GS TN
p n
I I
I I I
I I
I
k kW WV V V V V VL L
v v
v v v
v v
v
′ ′⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎡ ⎤+ − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞⎡ ⎤− − − − − = −⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠⎡ ⎤− − = − +⎣ ⎦
− − =
± +=
( )( )
61.337 V
2 8 Iv⇒ =
(iii) 0 0.2 Vv = PMOS: sat, NMOS, non-sat.
( )[ ] ( ) ( )( ) ( )
[ ]
( )( )( )
2 2
2
2
40 8012 3.3 0.4 4 2 0.4 0.2 0.22 2
12 8.41 5.8 8 0.4 0.2
12 72.8 102.52 0
72.8 5299.84 4 12 102.522 12
2.222 V
I I
I I I
I I
I
I
v v
v v v
v v
v
v
⎛ ⎞ ⎛ ⎞ ⎡ ⎤− − = − −⎜ ⎟ ⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎡ ⎤− + = −⎣ ⎦
− + =
± −=
=
(b)
(i) ( ) ( ) ( )
( )
2 63.3 0.4 0.4 3.59284
2.7322 61
41.315 V
It
It
v
v
+ − += =
+
=
(ii) From (a), (ii) [ ]
( )( )( )
2
2
4 1.16 0.4 0.04 12 0.8 0.16
12 8 2.56 0
8 64 4 12 2.560.903 V
2 12
I I I
I I
I I
v v v
v v
v v
⎡ ⎤− − = − +⎣ ⎦− − =
± += ⇒ =
(iii) From (a), (iii) [ ]
( )( )( )
2
2
4 8.41 5.8 12 0.4 0.2
4 28 36.04 0
28 784 4 4 36.041.70 V
2 4
I I I
I I
I I
v v v
v v
v v
⎡ ⎤− + = −⎣ ⎦− + =
± −= ⇒ =
16.35 a. For 1 20.6 5O TN Ov V v V= < ⇒ =
1N in nonsaturation and 1P in saturation. From Equation (16.45),
( )( ) ( ) [ ]22
2
2 0.8 0.6 0.6 5 0.8
1.2 1.32 17.64 8.4
I I
I I I
v v
v v v
⎡ ⎤− − = − −⎣ ⎦− = − +
or
( ) ( )( )
2
2
9.6 18.96 0
9.6 9.6 4 1 18.962
I I
I
v v
v
− + =
± −=
or 2.78 VIv =
b. 0 02 0Nt PtV v V≤ ≤ From symmetry, 2.5 VItV =
0 2.5 0.8 3.3 VPtV = + = and 0 2.5 0.8 1.7 VNtV = − = So 021.7 3.3 Vv≤ ≤ 16.36 a. 0 01 0Nt PtV v V≤ ≤ By symmetry, 2.5 VItV =
0 2.5 0.8 3.3 VPtV = + = and 0 2.5 0.8 1.7 VNtV = − = So 011.7 3.3 Vv≤ ≤
b. For 2 30.6 5O TN Ov V v V= < ⇒ =
2N in nonsaturation and 2P in saturation. From Equation (16.57),
( )( ) ( ) [ ]222 2
22 2 2
2 0.8 0.6 0.6 5 0.8
1.2 1.32 17.64 8.4
I I
I I I
v v
v v v
⎡ ⎤− − = − −⎣ ⎦− = − +
or 22 29.6 18.96 0I Iv v− + =
So 2 01 2.78 VIv v= =
For 01 1 12.78, both andv N P= in saturation. Then 2.5 VIv =
16.37 a.
( )( ) ( )1/ 20.1 2.5 0.8 0.538 mA
Peak n I TN
Peak
i K v V
i
= −
= ⋅ − =
b. ( ) ( )1/ 20.1 1.65 0.8 0.269 mAPeaki = ⋅ − =
16.38
(a) ( )
( )
( ) ( )
2
2
2 2, 1
50 2 50 /225 4 50 /2
50 2.5 0.8
n
p
D peak n TN
K A V
K A V
I K v V
μ
μ
⎛ ⎞= =⎜ ⎟⎝ ⎠⎛ ⎞= =⎜ ⎟⎝ ⎠
= − = −
or , 144.5D peakI Aμ=
(b) 2 250 / , 25n pK A V K A Vμ μ= = / From Equation (16.55),
505 0.8 (0.8)25 2.2150125
Itv V− +
= =+
Then ( )22
, ( ) 50 2.21 0.8D peak n It TNI K V V= − = − or , 99.4D peakI Aμ= 16.39
(a) Switching Voltage, Eq. (16.43) ( ) ( )
( )
( )( )
It
2peak , peak
2 43.3 0.4 0.4
8 1.65 V2 4
18
80 4 1.65 0.4 250 μA2
It
D, D
v v
i i =
− += = =
+
⎛ ⎞= − ⇒⎜ ⎟⎝ ⎠
(b)
( ) ( )
( )
( )( )2, peak , peak
2 43.3 0.4 0.4
4 1.436 V2 4
14
80 4 1.436 0.4 172 μA2
It It
D D
v v
i i
− += = =
+
⎛ ⎞= − ⇒ =⎜ ⎟⎝ ⎠
(c)
( ) ( )
( )
( )( )2, peak , peak
2 43.3 0.4 0.4
12 1.776 V2 4
112
80 4 1.776 0.4 303 μA2
It It
D D
v v
i i
− += ⇒ =
+
⎛ ⎞= − ⇒ =⎜ ⎟⎝ ⎠
16.40 a. 2
L DDP fC V= For 6 12 25 V, (10 10 )(0.2 10 )(5)DDV P −= = × × or 50 WP μ=
For 6 12 215 V, (10 10 )(0.2 10 )(15)DDV P −= = × × or 450 WP μ=
b. For 6 12 25 V, (10 10 )(0.2 10 )(5)DDV P −= = × × or 50 WP μ= 16.41 (a)
( )( )( )22 6 12 3150 10 0.4 10 5 1.5 10 W / inverterL DDP fC V − −= = × × = ×
Total power: ( )( )6 32 10 1.5 10 3000 !!!!T TP P W−= × × ⇒ = (b) For 300f MHz=
( )( )3 6 12 21.5 10 300 10 0.4 10 3.54DD DDV V V− −× = × × ⇒ = 16.42
(a) 77
3 3 10 W10
P −= = ×
(b) 22L DD L
DD
PP fC V CfV
= ⇒ =
(i) ( )( )
7
26
3 10 0.0024pF5 10 5
L LC C−×= ⇒ =
×
(ii) ( )( )
7
26
3 10 0.00551pF5 10 3.3
L LC C−×= ⇒ =
×
(iii) ( )( )
7
26
3 10 0.0267 pF5 10 1.5
L LC C−×= ⇒ =
×
16.43
(a) 66
10 2 10 W5 10
P −= = ××
(b) 2LDD
PCfV
=
(i) ( )( )
6
26
2 10 0.01 pF8 10 5
L LC C−×= ⇒ =
×
(ii) ( )( )
6
26
2 10 0.023 pF8 10 3.3
L LC C−×= ⇒ =
×
(iii) ( )( )
6
26
2 10 0.111 pF8 10 1.5
L LC C−×= ⇒ =
×
16.44 (a) For ,I DDv V≅ NMOS in nonsaturation
( ) 22D n I TN DS DSi K v V v v⎡ ⎤= − −⎣ ⎦ and 0DSv ≅
So ( )1 2Dn DD TN
ds DS
di K V Vr dv
= ≅ −⎡ ⎤⎣ ⎦
Or
( )1
22
dsn
DD TNn
rk W V V
L
=′⎛ ⎞⎛ ⎞ ⋅ −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
or
( )1
ds
n DD TNn
rWk V VL
=⎛ ⎞′ ⋅ −⎜ ⎟⎝ ⎠
For 0,Iv ≅ PMOS in nonsaturation
( ) 22D p DD I TP SD SDi K V v V v v⎡ ⎤= − + −⎣ ⎦
and 0SDv ≅ for 0.Iv ≅ So
( )1 22
pDDD TP
psd SD
kdi W V Vr dv L
′⎛ ⎞⎛ ⎞= ≅ ⋅ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
or
( )
1
1sd
DD TPp p
rW V V
k L
=⎛ ⎞
⋅ +⎜ ⎟⎜ ⎟′ ⎝ ⎠
(b) For 2, 4n p
W WL L
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( )( )( )
( )( )( )
1 2.3850 2 5 0.8
1 2.3825 4 5 0.8
ds ds
sd sd
r r k
r r k
= ⇒ = Ω−
= ⇒ = Ω−
For 2,.p
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
( )( )( )1 4.76
25 2 5 0.8sd sdr r k= ⇒ = Ω−
Now, for NMOS:
ds d dsv i r= or 0.5 0.212.38
dsd d
ds
vi i mAr
= = ⇒ =
For PMOS: For 2.38 ,sdr k= Ω
0.5 0.212.38
sdd d
sd
vi i mAr
= = ⇒ =
For 4.76sdr k= Ω , 0.5 0.1054.76
sdd d
sd
vi i mAr
= = ⇒ =
16.45 From Equation (16.63)
( )31.5 10 1.5 1.5 4.125 V8IL ILV V= + ⋅ − − ⇒ =
and Equation (16.62)
( )01 2 4.125 10 1.5 1.52HUV = ⋅ + − +⎡ ⎤⎣ ⎦
or 0 9.125 VHUV = From Equation (16.69)
( )51.5 10 1.5 1.5 5.875 V8IH IHV V= + ⋅ − − ⇒ =
and Equation (16.68)
( )01 2 5.875 10 1.5 1.52LUV = ⋅ − − +⎡ ⎤⎣ ⎦
or 0 0.875 VLUV = Now
0 4.125 0.875 3.25 VL IL LU LNM V V NM= − = − ⇒ =
0 9.125 5.875 3.25 VH HU TH HNM V V NM= − = − ⇒ = 16.46 From Equation (16.71)
( ) ( )100
10 1.5 1.5 501.5 2 1 1.5 7 2 0.632 1100100 315050
ILV
⎡ ⎤⎢ ⎥− −⎢ ⎥= + − = + −⎡ ⎤⎣ ⎦⎛ ⎞ ⎢ ⎥+−⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦
or 3.348 VILV =
From Equation (16.70)
( ) ( )01 100 1001 3.348 10 1.5 1.52 50 50HUV ⎧ ⎫⎛ ⎞ ⎛ ⎞= ⋅ + + − +⎨ ⎬⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎩ ⎭
or 0 9.272 VHUV = From Equation (16.77)
( ) [ ]100210 1.5 1.5 501.5 1 1.5 7 1.51 1
100 1001 3 150 50
IHV
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟− − ⎝ ⎠⎢ ⎥= + − = + −⎢ ⎥⎛ ⎞ ⎛ ⎞− ⎢ ⎥+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎢ ⎥⎝ ⎠⎣ ⎦
or 5.07 VIHV =
From Equation (16.76)
( ) ( )0
100 1005.07 1 10 1.5 1.550 50
100250
LUV
⎛ ⎞ ⎛ ⎞+ − − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=
⎛ ⎞⎜ ⎟⎝ ⎠
or 0 0.9275 VLUV =
Now 0 3.348 0.9275L IL LUNM V V= − = − or 2.42 VLNM =
0 9.272 5.07H HU IHNM V V= − = − or 4.20 VHNM = 16.47 (a)
( )
( )
( ){ }
( )
( ){ }
3830.4 3.3 0.4 0.4 1.3375 V8
1 2 1.3375 3.3 0.4 0.422.9875 V
50.4 3.3 0.4 0.4 1.9625 V8
1 2 1.9625 3.3 0.4 0.420.3125 V
2.9875 1.9625 1.02
n P
IL TN DD TP TN
IL
OHu
OHu
IH IH
OLu
OLu
H OHu IH H
K K
V V V V V
V
V
V
V V
V
VNM V V NM
=
= + + −
= + − − ⇒ =
= + − +
=
= + − − ⇒ =
= − − +
== − = − ⇒ = 5 V
1.3375 0.3125 1.025 VL IL OLu LNM V V NM= − = − ⇒ =
(b)
( )( )( )
( )
( ) ( ) ( )
( )( ) ( ) ( ) ( ) { }
2 43.3 0.4 0.4 2.5120.4 2 1 0.4 0.147
2 4 0.3332 4 311212
1.505 V
2 4 2 41 11 1.505 3.3 0.4 0.4 2.5083 3.3 0.2667 0.42 12 12 2
2.9708 V
3.3 0.4 0.0.4
IL
IL
OHu
OHu
IH
V
V
V
V
V
⎡ ⎤⎢ ⎥− − ⎢ ⎥= + − = + −⎡ ⎤⎣ ⎦⎢ ⎥ −⎛ ⎞ +− ⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠
=
⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪= + + − + = + − +⎜ ⎟ ⎜ ⎟⎨ ⎬⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭
=
− −= +
( )( )
( )
( )( ) ( ) [ ]
( ) ( ) ( ) ( )
( )
2 42
124 2.51 0.4 0.2302 2.1282 V0.3332 4 2 41 3 112 12
2 4 2 42.1282 1 3.3 0.4 0.4
12 12 3.547 3.3 0.2667 0.4 0.2853 V1.3332 4
212
2.9708 2.128
IH
OLu OLu
H OHu IH
V
V V
NM V V
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥− = + − ⇒ =⎢ ⎥ −⎛ ⎞⎢ ⎥− +⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦
⎛ ⎞ ⎛ ⎞+ − − +⎜ ⎟ ⎜ ⎟
− − +⎝ ⎠ ⎝ ⎠= = ⇒ =⎛ ⎞⎜ ⎟⎝ ⎠
= − = − 2 0.8426 V
1.505 0.2853 1.22 VH
L IL OLu L
NM
NM V V NM
⇒ =
= − = − ⇒ =
16.48 a. 5 VA Bv v= =
1N and 2N on, so 1 2 0 VDS DSv v≈ ≈
1P and 2P off So we have a 3 3P N− CMOS inverter. By symmetry, 2.5 VCv = (Transition Point). b. For A B C Iv v v v= = ≡ Want , ,n eff p effK K=
32 3 2n P
n P
k kW WL L
′ ′⎛ ⎞ ⎛ ⎞⋅ = ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
With 2 ,n Pk k′ ′= then 2 1 1 32 3 2n P
W WL L
⎛ ⎞ ⎛ ⎞⋅ ⋅ = ⋅ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Or 92n P
W WL L
⎛ ⎞ ⎛ ⎞= ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
c. We have 2 9
2 2 2
2
pnn
n p
pp
p
kk W WKL L
k WKL
′′ ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Then from Equation (16.55)
( ) ( )5 0.8 0.8
1
n
pIt
n
p
KK
VKK
+ − + ⋅=
+
Now
( ) 92 92
n
p
KK
⎛ ⎞= =⎜ ⎟⎝ ⎠
Then ( ) ( )5 0.8 3 0.8
1.651 3It ItV V V
+ − += ⇒ =
+
16.49 By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V. State 1N 2N 3N 4N 5N 0v 1 off on off on on 0 2 off off on on off 0 3 on on off off on 5 4 on on off on on 0 Logic function ( ) ( )OR ANDX Y X Zv v v v⊗
Exclusive OR of ( )ORX Yv v with ( )ANDX Zv v 16.50
NMOS in Parallel 2n
WL
⎛ ⎞⇒ =⎜ ⎟⎝ ⎠
4-PMOS in series ( )4 4 16p
WL
⎛ ⎞⇒ = =⎜ ⎟⎝ ⎠
(b) LC doubles⇒ current must double to maintain switching speed.
4
32
n
p
WL
WL
⎛ ⎞⇒ =⎜ ⎟⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.51
4-NMOS in series ( )4 2 8n
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
4-PMOS in parallel 4p
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
(b) 16
8
n
p
WL
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.52
(a) NMOS in parallel 2n
WL
⎛ ⎞⇒ =⎜ ⎟⎝ ⎠
3-PMOS in series ( )3 4 12P
WL
⎛ ⎞⇒ = =⎜ ⎟⎝ ⎠
(b) 4
24
n
p
WL
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.53
(a) 3-NMOS in series ( )3 2 6n
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
3-PMOS in parallel 4p
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
(b) 12
8
n
p
WL
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.54 (a) ( )( )Y A B C D E= + + (b)
(c) For NMOS in pull down mode, 3 in series ( )3 2 6n
WL
⎛ ⎞⇒ = =⎜ ⎟⎝ ⎠
For PMOS
( )
,
, , , ,
4
2 4 8
P A
P B C D E
WL
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
⎛ ⎞ = =⎜ ⎟⎝ ⎠
16.55 (a) ( )Y A BD CE= + (b)
(c) NMOS: 3 transistors in series for pull down mode.
For twice the speed: ( )( )2 3 2 12n
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
PMOS: ( ),
2 4 8P A
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
( )( ), , , ,
2 2 4 16P B C D E
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
16.56 (a) Y A BC DE= + + (b)
(c) NMOS:,
2n A
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
, , , ,
4n B C D E
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
PMOS: 3 transistors in series for the pull-up mode
( )3 4 12p
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
16.57 (a) ( )( )Y A B D C E= + + + (b)
(c) NMOS: ( ),
2 2 4n A
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
( )( ), , , ,
2 4 8n B C D E
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
PMOS:
( ) ( )2 3 4 24p
WL
⎛ ⎞ = =⎜ ⎟⎝ ⎠
16.58 (a) A classic design is shown:
, ,A B C signals supplied through inverters.
(b) For Inverters, 1n
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
and 2p
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
For PMOS in Logic function, let 1p
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
, then for NMOS in Logic function, 2.25n
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.59 (a) A classic design is shown:
(b) , ,
, ,
1, 2
8, 4
ND NA NB NC
PA PB PC PD
W WL L
W WL L
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
16.60 ( ) OR AND A B C 16.61
5-NMOS in series ( )5 2 10n
WL
⎛ ⎞⇒ = =⎜ ⎟⎝ ⎠
5-PMOS in parallel 4p
WL
⎛ ⎞⇒ =⎜ ⎟⎝ ⎠
16.62 By definition: NMOS off if gate voltage 0= NMOS on if gate voltage 5 V= PMOS off if gate voltage 5 V= PMOS on if gate voltage 0= State 1N 1P AN BN CN 01v 2N 2P 02v 1 off on off off off 5 on off 0 2 on off on off off 5 on off 0
3 off on off off off 5 on off 0 4 on off off off on 5 on off 0 5 off on off off off 5 on off 0 6 on off off on on 0 off on 5 Logic function is
( )02 OR ANDA B Cv v v v= 16.63 State 01v 02v 03v
1 5 5 0 2 0 0 5 3 5 5 0 4 5 0 5 5 5 5 0 6 0 5 0 Logic function:
( )03 OR ANDX Z Yv v v v= 16.64
16.65
16.66
16.67
2 CdVI C
dt= −
So
( )1 2CV I tC
Δ = − ⋅
For 0.5CV VΔ = −
( )12
15
2 2 100.5 3.125
25 10
x tt ms
x
−
−
⋅− = − ⇒ =
16.68 (a) (i) 0Ov = (ii) 4.2Ov V= (iii) 2.5Ov V= (b) (i) 0Ov = (ii) 3.2Ov V= (iii) 2.5Ov V= 16.69 (a) (i) 0ov = (ii) 2.9 Vov = (iii) 2.4 Vov = (b) (i) 0ov = (ii) 2.0 Vov = (iii) 2.0 Vov = 16.70 Neglect the body effect. a. 01v (logic 1) 4.2 V= , 02v (logic 1) 5 V= b. 15 V 4.2 VI GSv v= ⇒ =
1M in nonsaturation and 2M in saturation. From Equation (16.23)
( ) ( )
( )( ) ( ) ( )[ ]
221 1 11
22
2
2 4.2 0.8 0.1 0.1 1 5 0.1 0.8
GS TND O O DD O TNLD L
D
W Wv V v v V v VL L
WL
⎛ ⎞ ⎛ ⎞⎡ ⎤− − = − −⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎡ ⎤− − = − −⎜ ⎟ ⎣ ⎦⎝ ⎠
Or
( )0.67 16.81 25.1D D
W WL L
⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Now 01 34.2 V 4.2 VGSv v= ⇒ =
3M in nonsaturation and 4M in saturation. From Equation (16.29(b)).
( ) [ ]
( )( ) ( ) ( ) ( )
223 2 2
22
2
2 4.2 0.8 0.1 0.1 2 1.5
(0.67) 2.25
GS TND O O TNLD L
D
D
W Wv V v v VL L
WL
WL
⎛ ⎞ ⎛ ⎞⎡ ⎤− − = −⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎡ ⎤− − = − −⎡ ⎤⎜ ⎟ ⎣ ⎦⎣ ⎦⎝ ⎠
⎛ ⎞ =⎜ ⎟⎝ ⎠
Or 3.36D
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.71
A B Y 0 0 1 0 1 0 1 0 0 1 1 0.1 ⇒ indeterminate
Without the top transistor, the circuit performs the exclusive-NOR function. 16.72
A A B B Y Z 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0
Y A AB A B= + = + Z Y= or Z AB= 16.73
16.74 For 1, 0,φ φ= = then .Y B= And for 0, 1,φ φ= = then Y A= . A multiplexer. 16.75 Y AC BC= + 16.76 Y AB AB A B= + = ⊗ 16.77
A B Y 0 0 0 1 0 1 0 1 1 1 1 0 Exclusive-OR function. 16.78 This circuit is referred to as a two-phase ratioed circuit. The same width-to-length ratios between the driver and load transistors must be maintained as discussed previously with the enhancement load inverter. When 1φ is high, 01v becomes the complement of Iv . When 2φ goes high, then 0v becomes the complement of 01v or is the same as Iv . The circuit is a shift register. 16.79 Let 0Q = and 1Q = ; as S increases, Q decreases. When Q reaches the transition point of the 5 6M M− inverter, the flip-flop with change state. From Equation (16.28(b)),
( )LIt TNL TND
D
KV V VK
= ⋅ − +
where 6LK K= and 5.DK K= Then
( )30 2 1 2.095100It ItV V Q V= ⋅ − − + ⇒ = =⎡ ⎤⎣ ⎦
This is the region where both 1M and 3M are biased in the saturation region. Then
( ) ( )3
1
30 2 1200TNL TND
KS V V
K= ⋅ − + = ⋅ − − +⎡ ⎤⎣ ⎦
or 1.77S V=
This analysis neglects the effect of 2M starting to turn on at the same time. 16.80 Let ,Yv R= ,Xv S= 02 ,v Q= and 01 .v Q= Assume 0.5 VThNV = and 0.5 V.ThPV = − For 0,S = we have the following:
If we want the switching to occur for 2.5 V,R = then because of the nonsymmetry between the two
circuits, we cannot have Q and Q both equal to 2.5 V.
Set 2.5 VR Q= = and assume Q goes low. For the 1 5M M− inverter, 1M in nonsaturation and 5M in saturation. Then
( ) [ ]2 22 2.5 0.5 2.5 0.5n pK Q Q K⎡ ⎤− − = −⎢ ⎥⎣ ⎦
Or 2
4 4 p
n
KQ Q
K⎛ ⎞
− = ⎜ ⎟⎝ ⎠
For the other circuit, 2 4M M− in saturation and 6M in nonsaturation. Then
( ) ( )( ) ( )2 2 22.5 0.5 ( 0.5) 2 5 0.5 2.5 2.5n n pK K Q K Q⎡ ⎤− + − = − − −⎣ ⎦
Combining these equations and neglecting the 3Q term, we find
1.4Q V= and 0.9p
n
Kk
=
16.81
( )3.3 0.4 0.51.7 V
1 1Itv+ − +
= =+
1.5 VIv = NMOS Sat; PMOS Non Sat
( ) ( )( ) ( )2 21 1 1
1
0.5 2 3.3 0.4 3.3 3.3 2.88 V
1.6 V 2.693 VI I o o o
I o
v v v v v
v v
⎡ ⎤− = − − − − − ⇒ =⎣ ⎦= =
11.7 V variable (switching region)I ov v= = 1.8 V NMOS Non Sat; PMOS SatIv =
( ) ( )2 21 1 13.3 0.4 2 0.5 0.607 VI I o o oV v v v v⎡ ⎤− − = − − ⇒ =⎣ ⎦
Now 1
1
1.5 V, 2.88 V 0V1.6 V, 2.693 V
I o o
I o
v v vv v
= = ⇒ ≈= =
NMOS Non Sat; PMOS Sat ( ) ( )2 2
1 13.3 0.4 2 0.5
0.00979 Vo o o o
o
v v v v
v
⎡ ⎤− − = − −⎣ ⎦=
11.7 V, vI ov = = Switching Mode 0v⇒ = Switching Mode.
11.8 V, 0.607 VI ov v= = NMOS Sat; PMOS Non Sat
( ) ( )( ) ( )2 201 01 0 0 00.5 2 3.3 0.4 3.3 3.3 3.298 Vv v v v v⎡ ⎤− = − − − − − ⇒ =⎣ ⎦
16.82 For DDR Vφ= = and 0 0, 1S Q Q= ⇒ = = For DDS Vφ= = and 0 1, 1R Q Q= ⇒ = = The signal φ is a clock signal. For 0,φ = The output signals will remain in their previous state. 16.83 a. Positive edge triggered flip-flop when CLK 1,= output of first inverter is D and then Q D D= = .
b. For example, put a CMOS transmission gate between the output and the gate of 1M driven by a
CLK pulse. 16.84 For 1,J = 0,K = and CLK 1;= this makes 1Q = and 0Q = . For 0,J = 1,K = and CLK 1= , and if 1,Q = then the circuit is driven so that 0Q = and 1.Q = If initially, 0,Q = then the circuit is driven so that there is no change and 0Q = and 1.Q =
1,J = 1,K = and CLK 1,= and if 1,Q = then the circuit is driven so that 0.Q = If initially, 0Q = , then the circuit is driven so that 1.Q = So if 1,J K= = the output changes state. 16.85 For 1, 0,X YJ v K v= = = = and CLK 1,Zv= = then 0 0.v = For 0, 1,X YJ v K v= = = = and CLK 1,Zv= = then 0 1.v = Now consider CLK 1.J K= = = With 1,X Zv v= = the output is always 0 0,v = So the output does not change state when CLK 1.J K= = = This is not actually a J K− flip-flop. 16.86 64 65,536K ⇒ transistors arranged in a 256 256× array. (a) Each column and row decoder required 8 inputs. (b) (i) Address 01011110= so input 7 6 5 4 3 2 1 0a a a a a a a a= (ii) Address 11101111= so input 7 6 5 4 3 2 1 0a a a a a a a a= (c) (i) Address 00100111= so input 7 6 5 4 3 2 1 0a a a a a a a a= (ii) Address 01111011= so input 7 6 5 4 3 2 1 0a a a a a a a a= 16.87 (a) 1-Megabit memory
1,048,576 1024 1024⇒
= ⇒ ×
Nuclear & input row and column decodes lines necessary 10= (b) 250K 4× bits 262,144 4⇒ × bits 512 512⇒ × For 512 lines ⇒ 9 row and column decoder lines necessary. 16.88 Put 128 words in a 8 16× array, which means 8 row (or column) address lines and 16 column (or row) address lines. 16.89 Assume the address line is initially uncharged, then
CdVI C
dt= or 1
CIV Idt t
C C= = ⋅∫
Then ( ) ( )12
6
2.7 5.8 10
250 10CV C
tI
−
−
×⋅= = ⇒
×
86.26 10 62.6t s ns−= × ⇒ 16.90
(a) ( )( ) ( )25 0.1 35 2 5 0.7 0.1 0.11 2
WL
− ⎛ ⎞⎛ ⎞ ⎡ ⎤= − −⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠
or 0.329WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
(b) 16 16,384K ⇒ cells 2 21Di Aμ≅ =
Power per cell (2 )(2 ) 4A V Wμ μ= = Total Power (4 )(16,384) 65.5T TP W P mWμ= = ⇒ =
Standby current (2 )(16,384) 32.8TA I mAμ= ⇒ = 16.91 16 16,384 cells
200200 Power per cell 12.216,384
12.2 2.54.88 0.5122.5
T
DDD
DD
K
P mW W
VPi A R MV R R
μ
μ
⇒
= ⇒ = ⇒
= = = ≅ = ⇒ = Ω
If we want 0.1Ov V= for a logic 0, then
( )
( )( ) ( )
2
2
22
354.88 2 2.5 0.7 0.1 0.12
nD DD TN O O
k Wi V V v vL
WL
′⎛ ⎞⎛ ⎞ ⎡ ⎤= − −⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎡ ⎤= − −⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠
So 0.797WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
16.92
0, 1Q Q= =
So 1 5D Logic V= =
A very short time after the row has been addressed, D remains charged at 5 .DDV V= Then 3, , p AM M and
1NM begin to conduct and D decreases. In steady-state, all three transistors are biased in the nonsaturation region. Then
( ) ( ) ( )2 2 23 3 3 3 3 1 1 1 1 12 2 2p SG TP SD SD nA GSA TNA DSA DSA n GS TN DS DSK V V V V K V V V V K V V V V⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ − = − − = − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Or ( )( ) ( ) ( )( ) ( )
( )
2 23 3
21 1
2 2
2 (1)
p DD TP DD DD nA DD TNA
n DD TN
K V V V D V D K V Q V D Q D Q
K V V Q Q
⎡ ⎤ ⎡ ⎤+ − − − = − − − − −⎣ ⎦ ⎣ ⎦⎡ ⎤= − −⎣ ⎦
Equating the first and third terms:
( ) ( )( ) ( ) ( ) ( )2 220 401 2 5 0.8 5 5 2 2 5 0.8 (2)2 2
D D Q Q⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤− − − − = − −⎜ ⎟ ⎜ ⎟ ⎣ ⎦⎣ ⎦⎝ ⎠ ⎝ ⎠
As a first approximation, neglect the ( )25 D− and 2Q terms. We find 1.25 0.25Q D= − (3)
Then, equating the first and second terms of Equation (1):
( ) ( )( ) ( ) ( ) ( )( ) ( )2 220 401 2 5 0.8 5 5 1 2 5 0.82 2
D D Q D Q D Q⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤− − − − = − − − − −⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠
Substituting Equation (3), we find as a first approximation: 2.14D V= Substituting this value of D into equation (2), we find
( ) ( )2 28.4 5 2.14 5 2.14 4 8.4Q Q⎡ ⎤− − − = −⎣ ⎦
We find 0.50Q V= Using this value of ,Q we can find a second approximation for D by equating the second and third terms of equation (1). We have
( )( ) ( ) ( )2 220 2 4.2 40 2 4.2Q D Q D Q Q Q⎡ ⎤ ⎡ ⎤− − − − = −⎣ ⎦⎣ ⎦
Using 0.50 ,Q V= we find 1.79D V= 16.93 Initially 1NM and AM turn on.
1,NM Nonsat; AM , sat.
[ ] ( )
( )[ ] ( ) ( )
2 21 1
2 2
2
40 401 5 0.8 2 2 5 0.82 2
nA DD TN n DD TNK V Q V K V V Q Q
Q Q Q
⎡ ⎤− − = − −⎣ ⎦⎛ ⎞ ⎛ ⎞ ⎡ ⎤− − = − −⎜ ⎟ ⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠
which yields 0.771Q V=
Initially 2PM and BM turn on Both biased in nonsaturation reagion
( )( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )
2 2
2 3
2 2
2 2
20 404 2 5 0.8 5 5 1 2 5 0.82 2
P DD TP DD DD nB DD TNBK V V V Q V Q K V V Q Q
Q Q Q Q
⎡ ⎤ ⎡ ⎤+ − − − = − −⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤− − − − = − −⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦⎝ ⎠ ⎝ ⎠
which yields 3.78Q V=
Note: ( / )W L ratios do not satisfy Equation (16.86) 16.94 For Logic 1, 1:v ( )( ) ( )( ) ( ) 1 1
2
2 2
5 0.05 4 1 1 0.05 4.0476
:(5)(0.025) (4)(1) (1 1.025) 4.0244
v v V
vv v V
+ = + ⇒ =
+ = + ⇒ =
For Logic 0, 1:v
1 1
2
2 2
(0)(0.05) (4)(1) (1 0.05) 3.8095
:(0)(0.025) (4)(1) (1 0.025) 3.9024
v v V
vv v V
+ = + ⇒ =
+ = + ⇒ =
16.95 Not given 16.96 Not given 16.97 Not given
16.98
(a) Quantization error 1 1% 0.05 V2
LSB= ≤ ≤
Or 0.10 VLSB ≤
For a 56- , 0.078125 V64
bit word LSB = =
(b) 51 0.078125 V64
LSB− = =
(c) 3.5424 64 45.34 455
n× = ⇒ =
Digital Output 101101= 45 5 3.515625
6413.5424 3.515625 0.026775 .2
LSB
× =
Δ = − = <.
16.99
(a) Quantization error 1 0.5% 0.05 V2
LSB= ≤ ≤
1 0.10 VLSB− =
For a 7-beit word, 10 0.078125 V128
LSB = =
(b) 1 0.078125 VLSB− =
(c) 3.5424 128 45.34272 4510
n× = ⇒ =
Digital output 0101101=
Now 45 10 3.515625128
× =
13.5424 3.515625 0.026775 V2
LSBΔ = − = <
16.100
(a) ( )0 1 0 1 52 4 8 16
1.5625 V
o
o
v
v
⎛ ⎞= + + +⎜ ⎟⎝ ⎠
=
(b) ( )1 0 1 0 52 4 8 16
3.125 V
o
o
v
v
⎛ ⎞= + + +⎜ ⎟⎝ ⎠
=
16.101
(a) ( )1 5 0.3125 V16
1 0.15625 V2
LSB
LSB
⎛ ⎞= =⎜ ⎟⎝ ⎠
=
Now 1
10 (5)20ov
R⎛ ⎞
= ⎜ ⎟+ Δ⎝ ⎠
For 2.5 0.15625 2.65625 Vov = + = ( )( )
1 1
10 520 1.176 K
2.65625R R+ Δ = ⇒ Δ = −
For 2.5 0.15625 2.34375 Vov = − =
( )( )1
1
10 520
2.34375 V1.333 K
R
R
+ Δ =
Δ = +
For 1 11.176 K 5.88%R RΔ = ⇒ Δ =
(b) For ( )44
10: 5160oR v
R⎛ ⎞
= ⎜ ⎟+ Δ⎝ ⎠
( )( )
( )( )
4 4
4 4
0.3125 0.15625 0.46875 V10 5
160 53.33K0.46875
Or 0.3125 0.15625 0.15625 V10 5
160 160 K0.15625
o
o
v
R R
v
R R
= + =
+ Δ = ⇒ Δ = −
= − =
+ Δ = ⇒ Δ =
For 4 453.33K 33.33%R RΔ = ⇒ Δ = 16.102 (a) 5
6
7
8
320 k640 k1280 k2560 k
RRRR
= Ω= Ω= Ω= Ω
(b) ( )10 5 0.01953125 V2560ov ⎛ ⎞= =⎜ ⎟
⎝ ⎠
16.103 (a)
1 1
12
23
34
45
56
5 0.50 mA2 10
0.25 mA2
0.125 mA2
0.0625 mA2
0.03125 mA2
0.015625 mA2
REFVI IR
II
II
II
II
II
−= = ⇒ = −
= = −
= = −
= = −
= = −
= = −
(b) ( )( )6 0.015625 50.078125 V
o F
o
v I Rv
Δ = =Δ =
(c) [ ] [ ]( )2 5 6 0.25 0.03125 0.015625 51.484375 V
o F
o
v I I I Rv
= − + + = + +=
(d) For 101010; ( )( )0.50 0.125 0.03125 5 3.28125 Vov = + + =
For 010101; ( )( )0.25 0.0625 0.015625 5 1.640625 Vov = + + = 1.640625 VovΔ =
16.104 1 5 0.3125 V2 8 2 16 16
REF REFV VRLSBR
⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Ideal
Av for 011 ( )3 3 1.875 V8 8
REFREF
V R VR
⎛ ⎞⇒ = =⎜ ⎟⎝ ⎠
Range of 11.8752Av LSB= ±
or 1.5625 2.1875 VAv≤ ≤ 16.105 6-bits 62 64⇒ = resistors
62 1 63− = comparators 16.106 (a) 10- bit output 1024⇒ clock periods
1 clock period 6
1 1 1 μS10f
= = =
May conversion time 1024 μS 1.024 mS= = (b)
( )
1 1 5 0.002441406 V2 2 1024
5128 16 2 0.712890625 V1024A
LSB
v
⎛ ⎞= =⎜ ⎟⎝ ⎠
⎛ ⎞′ = + + =⎜ ⎟⎝ ⎠
So range of 12A Av v LSB′= ±
0.710449219 0.715332031 VAv≤ ≤ (c) 0100100100 256 32 4 292 clock pulses⇒ + + = 16.107
(a) 53.125 640 512 1281024N N×= ⇒ = ⇒ +
Output 1010000000= (b)
51.8613 381.19 381 256 64 32 16 8 4 11024N N N×= ⇒ = ⇒ = ⇒ + + + + + +
Output 0101111101=