6
6.1
(a CHAPTER ) Pavg = 1W 10-5W / gate = 10 W / gate (b) I = = 4 A / gate 105 gates 2.5V
6.2
(a) Pavg = 100 5x10-6W / gate = 5 W / gate (b) I = = 2 A/ gate 2.5V 2x10 7 gates
(c) I total = 2
(2x10 gates)= 40 A gate
7
A
6.3
2.5 - 0 5 (a ) VH = 2.5 V | VL = 0 V | P H = I R = 0 mW | P L = 10 = 62.5 W V V 5 10
2 2
3.3 - 0 5 (b) VH = 3.3 V | VL = 0 V | P H = I R = 0 mW | P L = 10 = 109 W V V 5 10
2 2
6.4
vO VH (3.3 V)
vI V (0V)
L
1.1 V (V REF)
3.3V V+
6.5
v V H (3.3 V)
O
V L (0V) 1.1 V (V REF) 3.3V V+
vI
Z= A =A
()
6-1
6.6
V REF vI
AV
6.7
V H = 3 V | VL = 0 V | VIH = 2 V | VIL = 1 V | AV =
dvO -3V = = -3 dv I 1V
6.8
V (3 V)
H
v
O
Slope = +9 1.5 V
v V (0V)
L
I
1.33 V
1.67 V
1.5 V
3V V
+
6.9 VOH = 5 V VOL = 0 V
VIH = VREF = 2 V VIL = VREF = 2 V
NM H = 5 - 2 = 3 V NM L = 2 - 0 = 2 V 6.10 We would like to achieve the highest possible noise margins for both states and have them be symmetrical. Therefore VREF = 3.3/2=1.65 V. 6.11 V H = 3.3 V | VL = 0 V | VIH = 1.8 V | VOL 0.25 V | VIL = 1.5 V | VIH 3.0 V NM H = 3.0 -1.8 = 1.2 V | NM L = 1.5 - 0.25 = 1.25 V 6.12 VH = 2.5 V | VL = 0.20 V
6-2
6.13 V H = -0.80 V | VL = -1.35 V 6.14 VIH = VOH - NM H = -0.8 - 0.5 = -1.3 V | VIL = NM L + VOL = 0.5 + (-2) = -1.5 V 6.15 -13 -4 -9 P = PDP/P = 10 J/10 W = 10 s = 1 ns 6.16
4 x10-6W / gate 1W = 4 W / gate (b) I = = 1.60 A / gate 2.5V 2.5 x105 gates (c) PDP = 2ns (4 W ) = 8 fJ (a ) Pavg =
6.17
1W / gate 100W = 1 W / gate (b) I = = 0.4 A / gate 8 2.5V 10 gates (c) PDP = 1ns (1W ) = 1 fJ (a) Pavg =
6.18
PDP 250 100 Slope = 2
10 Slope = 1 1 P 1 10 50 100
6-3
6.19
dv (t ) dvc (t ) | v(t ) = RC c + vC (t ) | v(t ) = 1 for t 0 dt dt t t v(t ) = 1 - exp - | 0.9 = 1 - exp - 90% t90% = - RC ln (0.1) RC RC t 0.1 = 1 - exp - 10% t10% = - RC ln (0.9 ) | tr = t90% - t10% = RC ln (9) = 2.20 RC RC t t (b) v(t ) = 0 vC (0 ) = 1 v(t ) = exp - | 0.9 = exp - 90% t90% = - RC ln (0.9) RC RC t 0.1 = exp - 10% t10% = - RC ln (0.1) | t f = t10% - t90% = RC ln (9 ) = 2.20 RC RC (a ) v(t ) = i (t )R + vC (t ) | i (t ) = C
6.20 (a) VH = 2.5V | VL = 0.20V
(b) V10% = VL + 0.1V = 0.20 + 0.23 = 0.43V t10% 23 ns for vO V90% = VL + 0.9V = 0.20 + 2.07 = 2.27V t90% 33 ns for vO t r = 33 - 23 = 10 ns For fall time : t10% 2.5 ns for vO t90% 0.8 ns for vO t f = 1.7 ns For v I , t10% 0 ns t90% 1 ns t r = 1 ns | t f 1 ns
(c)
PHL
1.5ns - 0.5ns = 1 ns | PLH 26ns - 21ns = 5 ns (d ) P =
1+ 5 ns = 3 ns 2
6.21 (a) VH = -0.78V | VL = -1.36V
V90% = VL + 0.9V = -1.36 + 0.9(0.58) = -0.84V t90% 42 ns for vO tr = 42 - 32.5 = 9.5 ns For fall time : t10% 11.5 ns for vO t90% 2 ns for vO t f = 9.5 ns For vI , t10% 0 ns t90% 1 ns tr = 1 ns | t f 1 ns
(b) V10% = VL + 0.1V = -1.36 + 0.1(0.58) = -1.30V t10% 32.5 ns for vO
(c ) V50% = - 0.78 - 1.36 = -1.07V | PHL 4 ns | PLH
2
4 ns (d ) P =
4+4 ns = 4 ns 2
6-4
6.22 (A + B)(A + C)
AA + AC + BA + BC A + AC + BA + BC A(1 + C) + AB + BC A + AB + BC A(1+ B) + BC A + BC
6.23 Z = ABC + ABC + ABC
Z = ABC + + ABC + ABC + ABC Z = AB C + C + A + A BC Z = AB(1) + (1)BC Z = AB + BC
(
) (
)
6.24 A B C Z
0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 1
Z = AB+BC
6-5
6.25 Z = ABC + ABC + ABC + ABC
Z = C AB + AB + AB + AB
( ) Z = C (AB + AB + AB + AB) Z = C (A(B + B)+ A(B + B) ) Z = C (A(1) + A(1)) Z = C (A + A)
Z = C (1) Z =C
6.26 A B C 0 0 0
0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 0 1 0
Z 0 1 0 1 0 1 0 1
1 1 1 Z =C
6-6
6.27 A B C D Z 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 6.28 A B C Z1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 1 6.29 (a) Fanout = 2
Z = AB + CD Z = ABCD Z = ABCD
Z2 1 1 0 1 0 1 1 1
Z1 = AB = AB Z 2 = AB + C
(b) Fanout = 1
6-7
6.30
i(t) 0.132 A
For each line : i = C i = 40 x10-12 F
1 ns 0 t
dv dt
3.3V = 132mA = 0.132 A 10- 9 s For all 64 lines, I = 64(0.132 A) = 8.45 A!
6.31
i(t) 1.32 A
For each line : i = C i = 40 x10-12 F
t 0
dv dt
0.1 ns
3.3V = 1.32 A 10-10 s For all 64 lines, I = 64(1.32 A) = 84.5 A!
6.32
F 7.5mm 0.1cm 3.9 8.854 x10 -14 (1.5m ) ox A 3.9 o LW cm 2 mm = 0.583 pF C = 3 =3 t =3 t 1m ox ox
6.33
" CV KCoxWLV | Let W* = W and L* = L T = = 1 I " W nCox (VGS - VTN )2 2 L
C *V * K (W )(L )(V ) T = = = T * 1 W I 2 n (VGS - VTN ) 2 L
*
V W 2 (VGS - VTN ) = n ox 2 Tox L V W 2 2 P* = n ox (VGS - VTN ) = P 2 Tox L P = VI = V " W nCox 2 L P P P* 2P P = | *= = A WL A W (L ) A
2 (VGS - VTN )
PDP* = P*T * = (T ) 2 P = 3 PT = 3 PDP
Power density =
6-8
6.34
1 1 W " W nCox (VGS - VTN )2 = n ox (VGS - VTN )2 2 2 Tox L L W 1 ox 2 * (V - VTN )2 = 2 I D I D = n Tox L GS 2 2 2 (b) P* = V (2 I ) = 2VI = 2 P - The power has increased by a factor of two. W L CG " * (c) CG = CoxWL = ox WL | CG = ox = Tox 2 2 Tox 2 2 The capacitance has decreased by a factor of two. (a) I D = W L K (V ) T C V 2 2 = (d ) T * = = * 4 I W 1 2 2 n (VGS - VTN ) 2 L 2
* *
6.35
( a ) Pavg = 0.5 2 x10 gates
(
1W
6
)
= 1 W / gate ( b ) I =
1W / gate = 0.556 A / gate 1.8V
6.36
(a) Pavg =
20W 1.5W / gate = 1.5 W / gate (b) I = = 0.833 A/ gate 2 1.8V 6 20x10 gates 3
6.37
V 2.5 - 0.2 50 W = 20 A | Let VL = TN = 0.2V | R = = 115 k 2x10-5 2.5V 3 0.926 1 0.2 W W = M S is in the triode region : 20x10-6 = 60x10-6 2.5 - 0.6 - 0.2 = 1 1.08 2 L S L S I DD =
6-9
6.38 (a ) For MS off, I D = 0 and VH = 2.5V.
3 A 2.5 - VL V A = Kn VH - VTN - L VL | Kn = 60 2 = 180 2 200k 2 V 1 V V A 2.5 - VL = 2x105 180 2 2.5 - 0.6 - L VL 36VL2 -138.8VL + 5 = 0 2 V 2.5 - 0.0364 = 12.3 A | P = 2.5V ( 12.3 A) = 30.8 W VL = 0.0364 V | I D = 200k 0.0364 A Checking : I D = 180 2 2.5 - 0.6 - 0.0364 = 12.3 A 2 V (b) For MS off, I D = 0 and VH = 2.5V. For VL , I D =
(
)
6 A 2.5 - VL V A = Kn VH - VTN - L VL | Kn = 60 2 = 360 2 400k 2 V 1 V V A 2.5 - VL = 4x105 360 2 2.5 - 0.6 - L VL 144VL2 - 549.2VL + 5 = 0 2 V 2.5 - 0.00913 = 6.23 A | P = 2.5V (6.23 A)= 15.6 W VL = 9.13 mV | I D = 400k 0.00913 A Checking : I D = 360 2 2.5 - 0.6 - 0.00913 = 6.21 A 2 V For VL , I D =
(
)
6.39 (a ) For MS off, I D = 0 and VH = 2.5V.
3 A 2.5 - VL V A = Kn VH - VTN - L VL | Kn = 60 2 = 180 2 200k 2 V 1 V V A 2.5 - VL = 2x105 180 2 2.5 - 0.8 - L VL 36VL2 -124.4VL + 5 = 0 2 V 2.5 - 0.0407 = 12.3 A | P = 2.5V ( 12.3 A) = 30.7 W VL = 0.0407 V | I D = 200k 0.0407 A Checking : I D = 180 2 2.5 - 0.8 - 0.0407 = 12.3 A 2 V For VL , I D =
(
)
(b) 2.5 - V = (2x10 )180 V
5 L
A
VL 2 2.5 - 0.4 - VL 36VL -153.2VL + 5 = 0 2 2
VL = 0.0329 V | I D =
2.5 - 0.0329 = 12.3 A | P = 2.5V ( 12.3 A) = 30.8 W 200k 0.0329 A Checking : I D = 180 2 2.5 - 0.4 - 0.0329 = 12.3 A 2 V
6-10
6.40
(a) V
IL
= VTN +
1 1 1 = 0.6V + = 0.6 + = 0.627 V Kn R 36 3 A 60 (200k) 1 V 2
VOH = VDD - VIH = VTN -
1 1 2VDD 5 = 2.5 - = 2.49V | VOL = = = 0.215V 3Kn R 2Kn R 72 108 1 V 1 2.5 + 1.63 DD = 0.6 - + 1.63 = 1.00V Kn R 36 36 Kn R
NM L = 0.627 - 0.215 = 0.412 V | NM H = 2.49 -1.00 = 1.49 V 1 1 1 = 0.6 + = 0.607 V (b) VIL = VTN + K R = 0.6V + 6 A 144 n 60 (400k) 1 V2 VOH = VDD - VIH = VTN - 1 1 2VDD 5 = 2.5 - = 2.50V | VOL = = = 0.108V 3Kn R 2Kn R 288 432 1 V 1 2.5 + 1.63 DD = 0.6 - + 1.63 = 0.807V Kn R 144 144 Kn R
NM L = 0.607 - 0.108 = 0.499 V | NM H = 2.50 - 0.807 = 1.69 V
6.41 (a ) For MS off, I D = 0 and VH = 2.5V.
6 A 2.5 - VL V A = Kn VH - VTN - L VL | Kn = 60 2 = 360 2 400k 2 V 1 V V A 2.5 - VL = 4x105 360 2 2.5 - 0.6 - L VL 144VL2 - 549.2VL + 5 = 0 2 V 2.5 - 0.00913 = 6.23 A | P = 2.5V (6.23 A)= 15.6 W VL = 9.13 mV | I D = 400k 0.00913 A Checking : I D = 360 2 2.5 - 0.6 - 0.00913 = 6.23 A 2 V V (b) 2.5 - VL = 1442.5 - 0.5 - 2L VL 144VL2 - 578VL + 5 = 0 2.5 - 0.00867 = 6.33 A | P = 2.5V (6.33 A)= 15.8 W VL = 8.67 mV | I D = 400k A 0.00867 Checking : I D = 360 2 2.5 - 0.5 - 0.00867 = 6.23 A 2 V For VL , I D =
(
)
6-11
(c) 2.5 - V
L
V = 1442.5 - 0.7 - L VL 144VL2 - 520.4VL + 5 = 0 2
VL = 9.63 mV | I D =
2.5 - 0.00963 = 6.23 A | P = 2.5V (6.23 A)= 15.6 W 400k A 0.00963 Checking : I D = 360 2 2.5 - 0.7 - 0.00963 = 6.22 A 2 V
In this design, we see that VL is not sensitive to VTN.
6.42
(a) V
IL
= VTN +
1 1 1 = 0.6V + = 0.6 + = 0.607V Kn R 144 6 A 60 (400k) 1 V2
VOH = VDD - VIH = VTN -
1 1 2VDD 5 = 2.5 - = 2.50V | VOL = = = 0.108V 2Kn R 288 432 3Kn R 1 V 1 2.5 + 1.63 DD = 0.6 - + 1.63 = 0.807V Kn R 144 144 Kn R
NM L = 0.607 - 0.108 = 0.499 V | NM H = 2.50 - 0.807 = 1.69 V
(b) V
6.43
IL
= 0.6 +
1 5 | VOL = | NM L = VIL-VOL = 0 Kn R 3Kn R
Solving for Kn R yields no solution. Zero noise margin will not occur.
(a) I
D
=
P 0.25mW V - VL 2.5 - 0.5 = = 100A | R = DD = = 20.0 k ID VDD 2.5V 1x10-4
Using the values corresponding to Fig. 6.12, K 'p = 100A/V 2 W W 1.21 0.5 100A = 100x10-6 2.5 - 0.6 - 0.5 = 2 1 L S L S
(
)
(b) V
IL
= VTN +
1 1 1 = 0.6V + = 0.6 + = 1.01 V Kn R 2.42 A 1.21x100 2 (20k) V
VOH = VDD - VIH = VTN -
1 1 2VDD 5 = 2.5 - = 2.29V | VOL = = = 0.830V 2Kn R 4.84 7.26 3Kn R 1 V 1 2.5 + 1.63 DD = 0.6 - + 1.63 = 1.84V Kn R 2.42 2.42 Kn R
NM L = 1.01- 0.83 = 0.18 V | NM H = 2.29 -1.84 = 0.450 V
6-12
6.44
P 0.25mW V - VL 3.3 - 0.2 = = 75.76A | R = DD = = 40.9 k ID VDD 3.3V 75.76x10-6 W W 1.52 0.2 75.76x10-6 = 100x10-6 3.3 - 0.7 - 0.2 = 2 1 L S L S
(a) I
D
=
(
)
(b) V
IL
= VTN +
1 = 0.7V + Kn R
1 1 = 0.7 + = 0.861 V 6.22 A (1.52)100 V 2 (40.9k)
VOH = VDD - VIH = VTN -
1 1 2VDD 6.6 = 3.3 - = 3.22 | VOL = = = 0.594V 3Kn R 2Kn R 12.4 18.7 1 V 1 3.3 + 1.63 DD = 0.7 - + 1.63 = 1.73V Kn R Kn R 6.22 6.22
NM L = 0.861- 0.594 = 0.267 V | NM H = 3.22 -1.73 = 1.49 V
6.45
VDD - VL 3 - 0.25 = = 83.3 k ID 33x10-6 W W 1.04 0.25 33x10-6 = 60x10-6 3 - 0.75 - 0.25 = 2 1 L S L S
(a) R =
(
)
(b) SPICE yields VL = 0.249 V with ID = 33.0 A.
6.46
VDD - VL 2 - 0.15 = = 185 k ID 10x10-6 W W 0.15 1 10-5 = 75x10-6 2 - 0.6 - 0.15 = 2 L S L S 1.49
(a) R =
(
)
(b) SPICE yields VL = 0.15 V with ID = 10 A.
6-13
6.47
1 = 417 10 ' W -6 V Kn ( GS - VTN ) 60x10 (5 -1) L 1 1 1 = = 1000 (b) Ron = ' W -6 10 V K p ( SG + VTP ) 25x10 (5 -1) L 1 (c) A resistive connection exists between the source and drain. 1 1 20 W = ' = = (d ) -6 L Kn ( GS - VTN )Ron 60x10 (3 -1)(417) 1 V (a) Ron = = W 1 1 20 = ' = = -6 L K p ( SG + VTP )Ron 25x10 (3 -1)( V 1000) 1
6.48
1
VH = VDD - VTO +
(
(V
SB
+ 2 F - 2 F
)) V
H
= 3.3 - 0.75 + 0.75 VH + 0.7 - 0.7
(
(
))
(V
H
2 - 4.88) = 0.5625( H + 0.7) VH - 6.918VH + 9.706 = 0 V 2
VH = 4.962V , 1.956V VH = 1.96 V
Checking : VTN = 0.75 + 0.75 1.956 + 0.75 - 0.75 = 1.345V | VH = 3.3 -1.345 = 1.96V
(
)
6.49
VH = VDD - VTO +
(
2
(V
SB
+ 2 F - 2 F
)) V
H
= 3.3 - 0.6 + 0.6 VH + 0.6 - 0.6
(
(
))
(V
H
2 - 3.165) = 0.36( H + 0.6) VH - 6.69VH + 9.80 = 0 V
VH = 2.166V , 4.524V VH = 2.17 V Checking : VTN = 0.5 + 0.6
( 2.166 + 0.6 -
0.6 = 1.133V | VH = 3.3 -1.133 = 2.167V
)
6.50
VH = VDD - VTO +
(
(V
SB
+ 2 F - 2 F
)) V
H
= 2.5 - 0.5 + 0.85 VH + 0.6 - 0.6
(
(
))
(V
H
2 - 2.659) = 0.7225( H + 0.6) VH - 6.04VH + 6.634 = 0 V 2
VH = 1.444V , 4.596V VH = 1.44 V
Checking : VTN = 0.5 + 0.85 1.444 + 0.6 - 0.6 = 1.057V | VH = 2.5 -1.057 = 1.443V
(
)
6-14
6.51 For = 0, VH = VDD - VTN = 3.3 - 0.6 = 2.7V | For VL : I DL = I DS ' 2 Kn 1 VL ' 4 2 (3.3 - VL - 0.6) = Kn 2.7 - 0.6 + - VL 9VL - 39VL + 7.29 = 0 2 2 2 1 2 60x10-6 1 VL = 0.1958V | I DD = (3.3 - 0.1958 - 0.6) = 94.1A 2 2
P = (3.3V )(94.08A)= 0.311 mW
4 0.1958 Checking : I DD = 60x10-6 2.7 - 0.6 - 0.1958 = 94.1A 2 1
6.52 (a ) For = 0, VH = VDD - VTN = 3.3 - 0.8 = 2.5V | For VL : I DL = I DS
' 4 2 Kn 1 V (3.3 - VL - 0.8) = Kn' 1 2.5 - 0.8 - 2L VL 9VL2 - 32.2VL + 6.25 = 0 | VL = 0.206V 2 2 2 60x10-6 1 I DD = (3.3 - 0.206 - .8) = 78.9A | P = 3.3V(78.9A)= 0.260 mW 2 2 4 0.206 Checking : I DD = 60x10-6 2.5 - 0.8 - 0.206 = 79.0A 2 1 (b) For = 0, VH = VDD - VTN = 3.3 - 0.4 = 2.9V | For VL : I DL = I DS ' 4 2 Kn 1 V (3.3 - VL - 0.4) = Kn' 1 2.9 - 0.4 - 2L VL 9VL2 - 45.8VL + 8.41 = 0 | VL = 0.191 V 2 2 2 60x10-6 1 (3.3 - 0.191- 0.4) = 110A | P = 3.3V(6.55A)= 0.363 mW 2 2 4 0.191 Checking : I DD = 60x10-6 2.9 - 0.4 - 0.191 = 110A 2 1
I DD =
6.53 VIL = VTNS = 0.6 V | VOH = VH = VDD - VTNL = 3.3 - 0.6 = 2.7V
At VIH (See Eq. 6.29 Second Edition) VOL =
2
VDD - VTNL 3.3 - 0.6 1+ 3KR 4 1+ 3 0.5
= 0.540V
VIH = VTNS
(VDD - VOL - VTNL ) = 0.6 + 0.54 + 0.5 1 3.3 - 0.54 - 0.6 2 = 1.41V V + OL + ( ) 2 2 2KRVOL 2(4) 0.54
NM H = 2.7 -1.41 = 1.29 V | NM L = 0.60 - 0.54 = 0.06 V These values are readily confirmed with SPICE.
6-15
4.0
2.0
0
-2.0
6.54 (a ) For = 0, VH = VDD - VTN = 3.3 - 0.6 = 2.7V | For VL : I DL = I DS
' 8 2 Kn 1 V (3.3 - VL - 0.6) = Kn' 1 2.7 - 0.6 - 2L VL 9VL2 - 39VL + 7.29 = 0 2 1 -6 2 60x10 1 VL = 0.1958V | I DD = (3.3 - 0.1958 - 0.6) = 188A 2 1
188 P = (3.3V )( A)= 0.621 mW
8 0.1958 Checking : I DD = 60x10-6 2.7 - 0.6 - 0.1958 = 188A - check is ok 2 1 (b) VIL = VTNS = 0.6 V | VOH = VH = VDD - VTNL = 3.3 - 0.6 = 2.7V
At VIH (See Eq. 6.29 Second Edition) VOL =
2
VDD - VTNL 3.3 - 0.6 1+ 3KR 4 1+ 3 0.5
= 0.540V
VIH = VTNS
(VDD - VOL - VTNL ) = 0.6 + 0.54 + 0.5 1 3.3 - 0.54 - 0.6 2 = 1.41V V + OL + ( ) 2 2KRVOL 2 2(4) 0.54
NM H = 2.7 -1.41 = 1.29 V | NM L = 0.60 - 0.54 = 0.06 V These values are easily checked with SPICE. See Prob. 6.53, (c) For = 0, VH = VDD - VTN = 3.3 - 0.7 = 2.6V | For VL : I DL = I DS
' 8 2 Kn 1 V (3.3 - VL - 0.7) = Kn' 1 2.6 - 0.7 - 2L VL 9VL2 - 32.2VL + 6.25 = 0 2 1
VL = 0.200V | I DD =
2 60x10-6 1 (3.3 - 0.200 - 0.7) = 173A 1 2 173 P = (3.3V )( A)= 570 W
8 0.200 Checking : I DD = 60x10-6 2.6 - 0.7 - 0.200 = 173A - check is ok 2 1
6-16
6.55
(a) VH = VDD - VTO +
(
(V
SB
+ 2 F - 2 F
)) V
H
= 3.3 - 0.7 + 0.5 VH + 0.6 - 0.6
(
(
))
(V
H
2 - 2.987) = 0.25( H + 0.6) VH - 6.225VH + 8.772 = 0 VH = 2.156 V V 2
VL = 0.20V | I D =
W 0.25mW 0.20 = 75.76A | 75.76 = 100 2.156 - 0.7 - 0.20 3.3V 2 L S
W 2.79 | VTNL = 0.7 + 0.5 0.2 + 0.6 - 0.6 = 0.760V = 1 L S W 2 1 100 W 75.76 = (3.3 - 0.20 - 0.760) = 2 L L L L 3.61 (b) VIL = VTNS = 0.70V | VOH = VH = 2.16V
(
)
Finding VIH (See Eq 6.29 in 2nd Ed.) : VOL =
VDD - VTNL 1+ 3
(W / L) (W / L)
=
S L
1 + 3(2.79)(3.61)
3.3 - VTNL
=
3.3 - VTNL 5.587
VTNL = 0.7 + 0.5 VOL + 0.6 - 0.6
(
)|
5.587VOL = 3.3 - 0.7 + 0.5 VOL + 0.6 - 0.6
(
(
))
Using the quadratic equation : VOL = 0.4432V VTNL = 0.8234V VIH = VTNS + VIH = 0.7 +
4.0
W 2 VOL ( / L)L 1 + (VDD - VOL - VTNL ) 2 (W / L) 2VOL
S
2 0.443 1 1 1 + (3.3 - 0.443 - 0.823) | VIH = 1.39V 2 10.07 2 0.443
NM H = 2.16 - 1.39 = 0.77 V | NM L = 0.7 - 0.443 = 0.26 V
2.0
0
-2.0
6-17
6.56
0.4mW = 160A | VTNL = 0.6 + 0.5 0.3 + 0.6 - 0.6 = 0.687V 2.5V W 1.40 2 100x10-6 W 160x10-6 = (2.5 - 0.3 - 0.687) = 2 1 L L L L W W 0.3 6.67 160x10-6 = 100x10-6 1.55 - 0.6 - 0.3 = 2 1 L S L S I DD =
6.57 (a) VDD = 3.3 V VTN =1 V ID = 75 A VL = 0.2 V VH = VDD - VTN = 3.3 V - 0.6V = 2.7 V
+3.3 V
(
)
+
M
L
V
+
V = 3.1 V
GS
DSL
= 3.1 V
ID
VL = 0.2V
+
M
S
V
H
= 2.7 V
V
DSS
= 0.2 V
-
I DS = I DL = 75A V ' W I DS = Kn VGSS - VTHS - DSS VDSS 2 L S
2 K ' W I DL = n ( GSL - VTNL ) V 2 L L
75A = 100 100
W 1.88 0.2 2.7 - 0.6 - 0.2 = 2 1 V L S L S
2
A W
A
75A =
V2 2
W W 2 1 (3.3 - 0.2 - 0.6) = L L L L 4.17
H
(b) VH = VDD - VTO +
(
(V
SB
+ 2 F - 2 F
)) V
= 3.3 - 0.6 + 0.5 VH + 0.6 - 0.6
(
(
))
(V
H
2 - 3.087) = 0.25( H + 0.6) VH - 6.424VH + 9.381 = 0 VH = 2.245 V 2
75A = 100
W 0.2 2.43 2.245 - 0.6 - 0.2 = 2 2 1 V L S L S 2 K ' W I DL = n ( GSL - VTNL ) | VTNL = 0.6 + 0.5 0.2 + 0.6 - 0.6 = 0.660V V 2 L L
A W
(
)
75A =
100
A
V2 2
W W 2 1 (3.3 - 0.2 - 0.66) = L L L L 3.97
6-18
6.58
(a ) For = 0, V
H
= VDD - VTO = 2 - 0.6 = 1.4V
(b) For = 0.6, V
W W V 0.15 2.30 ' W I DS = Kn VH - VTNS - L VL | 25x10-6 = 100x10-6 1.4 - 0.6 - 0.15 = 2 1 2 L S L S L S W 2 2 K ' W 100x10-6 W 1 I DL = n ( GSL - VTNL ) | 25x10-6 = V (2 - 0.15 - 0.6) = 2 L L 2 L L L L 3.13
H
= VDD - VTNL = 2 - 0.6 + 0.6 VH + 0.6 - 0.6 VH = 1.09V
[
(
)]
W W V 0.15 4.02 ' W I DS = Kn VH - VTNS - L VL | 25x10-6 = 100x10-6 1.09 - 0.6 - 0.15 = 2 1 2 L S L S L S
For vO = VL = 0.15V , VTN = 0.6 + 0.6 I DL =
( 0.15 + 0.6 -
0.6 = 0.655V
)
' W 2 2 Kn W 100x10-6 W 1 VGSL - VTNL ) | 25x10-6 = ( (2 - 0.15 - 0.655) = 2 L L 2 L L L L 2.86 (c) Using LEVEL=1 KP=100U VTO=0.6 GAMMA=0, the values of ID and VL agree with our hand calculations. The results also agree for GAMMA=0.6.
6.59
' W 2 VDSL Kn W I DS = I DL | K VGSS - VTNS - V VDSL = ( GSL - VTNL ) 2 2 L L L S ' 2 Kn 1 VO ' 4.71 Kn 2.5 - 0.6 - VO = (2.5 - VO - VTNL ) 2 2 1.68 1 ' n
VTNL = 0.6 + 0.5 VO + 0.6 - 0.6
6.60
(
)
An iterative solution yields VO = 0.1061 V
2 V K ' W ' W I DS = I DL | Kn VH - VTNS - L VL = n (2.5 - VL - VTNL ) 2 2 L L L S
which is independent of K'n . Ratioed logic maintains VL and VH independent
' of K'n . So VH = 1.55V and VL = 0.20V. However, I DS = I DL Kn :
So, I D = 80A
80 A V 2 = 64.0 A P = 2.5V (64A)= 0.160 mW 100 A V 2 A 4.71 0.2 Checking : I DS = 80 2 1.55 - 0.6 - 0.2 = 64.1A 2 V 1
6-19
6.61
2 V K ' W ' W I DS = I DL | Kn VH - VTNS - L VL = n (2.5 - VL - VTNL ) 2 2 L L L S
which is independent of K'n . Ratioed logic maintains VL and VH independent
' of K'n . So VH = 1.55V and VL = 0.20V. However, I DS = I DL Kn :
So, I D = 80A
120 A V 2 = 96.0 A P = 2.5V (96A)= 0.240 mW 100 A V 2 0.2 A 4.71 Checking : I DS = 120 2 1.55 - 0.6 - 0.2 = 96.1A 2 V 1
6.62
Noise Margins vs. KR
2.5
2
1.5 NMH NML 1
0.5
0 0 2 4 6 8 10 12
-0.5 KR
6.63
(a) VH = VDD VTNL does not depend upon . However, VL is dependent upon . (b) SPICE yields VL = 0.20 V, 0.207 V, 0.217 V, and 0.232 V for = 0, 0.02/V, 0.05/V, and 0.1/V respectively. The current also increases: IDD = 80.1, 82.8, 86.9 and 93.3 A, respectively.
6.64 VTNL = 0.6 + 0.5 0.20 + 0.6 - 0.6 = 0.660V
(
)
VGSL - VTNL = 4 - 0.2 - 0.66 = 3.14V | VDSL = 2.5 - 0.2 = 2.30V Triode region W 2.3 1 A W 80A = 100 2 4 - 0.2 - 0.66 - 2.3 = 2 V L L L L 5.72 W 2.22 0.2 A W 80A = 100 2 2.5 - 0.6 - 0.2 = 2 1 V L S L L
6-20
6.65
For linear operation at vo = VL : VTNL = 0.8 + 0.5 0.2 + 0.6 - 0.6 = 0.860V VGSL - VTNL VDSL : VGG - 0.20 - 0.860 2.5 - 0.2 VGG 3.36V We also require : VGG 2.5 + VTNL = 2.5 + 0.8 + 0.5 2.5 + 0.6 - 0.6 = 3.79V so VGG 3.79V
6.66
(
)
(
)
If VH = 3.3 V , VTNL = 0.6 + 0.5 3.3 + 0.6 - 0.6 = 1.2V 5 -1.2 = 3.8V > 3.3V so VH = 3.3 V is correct. ' 2 VL Kn 1 ' 5 I DS = I DL | Kn 3.3 - 0.6 - VL = (3.3 - VL - VTNL ) 2 2 2 1 VTNL = 0.6 + 0.5 VL + 0.6 - 0.6 I DS =
(
)
(
) An interative solution gives V = 0.1222 V , V
L
TNL
= 0.6376 V
2 100A 1 161 (3.3 - 0.1222 - 0.6376) = 161 A | P = 3.3V ( A) = 0.532 mW 2 2
6.67 We require VGG VDD + VTNL so VH = VDD
VTNL = VTO +
(V
SB
+ 0.6 - 0.6 = 0.6 + 0.6
)
( 3.3 + 0.6 -
0.6 = 1.32V
)
VGG 3.3 + 1.32 = 4.62V
6.68 We require VGG VDD + VTNL so VH = VDD
VTNL = VTO +
(V
SB
+ 0.6 - 0.6 = 0.6 + 0.6 3.3 + 0.6 - 0.6 = 1.32V
)
(
)
VGG 3.3 + 1.32 = 4.62V | Design decision - Choose VGG = 5 V I DD = 300W = 90.9A 3.3V
W W 1.75 0.2 For MS : 90.9A = 100A 3.3 - 0.6 - 0.2 = 2 1 L S L S For ML : VTNL = 0.6 + 0.6 0.2 + 0.6 - 0.6 = 0.672V W W 3.3 - 0.2 1 90.9A = 100A 5 - .2 - 0.672 - (3.3 - 0.2) = 2 L L L L 8.79
6.69 We require VTNL 0 : -1+
(
)
( 2.5 + 0.6 -
0.6 0 1.014
)
6-21
6.70
(a) V
H
2 V K ' W ' W = VDD | I DS = I DL | Kn VDD - VTNS - L VL = n ( TNL ) V 2 2 L L L S
For ratioed logic, both VH and VLare independent of K'n . VH = 2.5 V | VL = 0.2 V 80 ' However, I D Kn | I DS = 80A = 64A | P = 2.5V (64A) = 0.160 mW 100 120 (b) VH = 2.5 V VL = 0.2 V I DS = 80A100 = 96A | P = 2.5V (96A)= 0.240 mW
6.71
0.20mW = 60.1A VTNL = -1+ 0.5 3.3 + 0.6 - 0.6 = -0.400V VH = 3.3V 3.3V W W 1.16 0.20 | For VO = VL = 0.2V , 60.1A = 100A 3.3 - 0.6 - 0.20 = 2 1 L S L S W 1.36 2 100A W VTNL = -1+ 0.5 0.20 + 0.6 - 0.6 = -0.940V | 60.1A = (-0.940) | = 1 2 L L L L I DD =
(
)
(
)
6.72
Assume VH = VDD = 3.3V | Checking : VTNL = -2 + 0.5 3.3 + 0.6 - 0.6 = -1.40 VTNL < 0, so our assumption is correct. | I DD = P 250W = = 75.8A VDD 3.3V W W 1.46 0.2 For MS in the triode region, 75.8A = 100A 3.3 - 0.6 - 0.2 = 2 1 L S L S For ML in the saturation region, VTNL = -2 + 0.5 0.2 + 0.6 - 0.6 = -1.94V and 75.8A =
2 100A W (0 - VTNL ) 2 L L
(
)
(
)
W 1 = L L 2.48
6.73 (a) No, VH does not depend upon . (b) As increases, IDD increases in ML, and VL increases.
0 0.02/V 0.05/V 0.1/v
IDD 78.2 A 81.4 A 86.0 A 93.6 A
VL 195 mV 203 mV 214 mV 231 mV
6-22
6.74 (a) The PMOS load is still saturated, so I DD remains the same : I DD = 80A. W W 1.80 0.25 Also, VH = 2.5 V. 80x10-6 = 100x10-6 2.5 - 0.6 - 0.25 = 2 1 L n L n 1.80( ) 100 V +V K 2.5 - 0.6 = 1.02 V (b) VIL = VTNS + DD2 TP KR = KS = 1.11 40 = 4.05 VIL = 0.6 + ( ) 4.052 + 4.05 K R + KR L KR 4.05 = 2.5 - (2.5 - 0.6) 1- = 2.30V VOH = VDD - ( DD + VTP ) 1- V 5.05 KR + 1 V + VTP 2.5 - 0.6 = = 0.545V VIH = VTNS + 2VOL = 0.6 + 2(0.545) = 1.69V VOL = DD 3KR 3(4.05)
NM H = VOH - VIH = 2.30 -1.69 = 0.610 V
NM L = VIL - VOL = 1.02 - 0.545 = 0.475 V
6.75 (a) The PMOS load is still saturated, so I DD remains the same : I DD = 80A. 2.22 VL Also, VH = 2.5 V. 80x10-6 = 100x10-6 2.5 - 0.5 - VL VL = 0.189 V 2 1 n V +V K 2.22 100 2.5 - 0.6 (b) VIL = VTNS + DD2 TP KR = KS = 1.11 40 = 5.00 VIL = 0.5 + 2 = 0.847 V 5 +5 KR + KR L KR 5 = 2.5 - (2.5 - 0.6) 1- = 2.33V VOH = VDD - ( DD + VTP ) 1- V 5 + 1 KR + 1 V + VTP 2.5 - 0.6 = = 0.491V VIH = VTNS + 2VOL = 0.5 + 2(0.491) = 1.48V VOL = DD 3KR 3(5)
NM H = VOH - VIH = 2.33 -1.48 = 0.849 V NM L = VIL - VOL = 0.847 - 0.491 = 0.356 V
(c) The PMOS load is still saturated, so I DD remains the same : I DD = 80A. 2.22 VL Also, VH = 2.5 V. 80x10-6 = 100x10-6 2.5 - 0.7 - VL VL = 0.213 V 2 1 n V +V K 2.22 100 2.5 - 0.6 (d) VIL = VTNS + DD2 TP KR = KS = 1.11 40 = 5.00 VIL = 0.7 + 2 = 1.05 V 5 +5 KR + KR L KR = 5 2.5 - (2.5 - 0.6) 1- = 2.33V VOH = VDD - ( DD + VTP ) 1- V 5 + 1 KR + 1 V + VTP 2.5 - 0.6 = = 0.490V VIH = VTNS + 2VOL = 0.7 + 2(0.490)= 1.68V VOL = DD 3KR 3(5)
NM H = VOH - VIH = 2.33 -1.68 = 0.650 V NM L = VIL - VOL = 1.05 - 0.490 = 0.560 V
6-23
6.76 (a) The PMOS load is still saturated, so I DD remains the same : I DD = 80A. 2.22 VL Also, VH = 2.5 V. 80x10-6 = 120x10-6 2.5 - 0.6 - VL VL = 0.165 V 2 1 n V +V K 2.22 120 2.5 - 0.6 (b) VIL = VTN + DD2 TP KR = KS = 1.11 40 = 6 VIL = 0.6 + 2 = 0.893V 6 +6 KR + KR L
VOH = VDD - ( DD + VTP ) 1- V VDD + VTP 3KR = 2.5 - 0.6 3(6) KR 6 = 2.5 - (2.5 - 0.6) 1- = 2.36V 6 + 1 KR + 1 = 0.448V VIH = VTN + 2VOL = 0.6 + 2(0.448)= 1.50V
VOL =
NM H = VOH - VIH = 2.36 -1.50 = 0.860 V NM L = VIL - VOL = 0.893 - 0.448 = 0.445 V (c) The PMOS load is still saturated, so I DD remains the same : I DD = 80A. 2.22 VL Also, VH = 2.5 V. 80x10-6 = 80x10-6 2.5 - 0.6 - VL VL = 0.254 V 2 1 n V +V K 2.22 80 2.5 - 0.6 (d) VIL = VTN + DD2 TP KR = KS = 1.11 40 = 4 VIL = 0.6 + 2 = 1.03V 4 +4 KR + KR L VOH = VDD - ( DD + VTP ) 1- V VOL = VDD + VTP 3KR = 2.5 - 0.6 3(4) KR 4 = 2.5 - (2.5 - 0.6) 1- = 2.30V 4 + 1 KR + 1 = 0.549V VIH = VTN + 2VOL = 0.6 + 2(0.549)= 1.70V NM L = VIL - VOL = 1.03 - 0.549 = 0.481 V
NM H = VOH - VIH = 2.30 -1.70 = 0.600 V
6-24
6.77 (a) The PMOS load is still saturated, and VH = 2.5 V. 2 40x10-6 1.11 I DD = (2.5 - 0.5) I DD = 88.8 A 2 1 n 2.22 VL For MS : 88.8x10-6 = 100x10-6 2.5 - 0.6 - VL VL = 0.224 V 2 1 n V +V K 2.22 100 2.5 - 0.5 (b) VIL = VTN + DD2 TP KR = KS = 1.11 40 = 5 VIL = 0.6 + 2 = 0.965V 5 +5 KR + KR L
VOH = VDD - ( DD + VTP ) 1- V VOL = VDD + VTP 3KR = 2.5 - 0.5 3(5) KR 5 = 2.5 - (2.5 - 0.5) 1- = 2.33V 5 + 1 KR + 1 = 0.516V VIH = VTN + 2VOL = 0.6 + 2(0.516) = 1.63V
NM H = VOH - VIH = 2.33 -1.63 = 0.700 V NM L = VIL - VOL = 0.965 - 0.516 = 0.449 V (c) The PMOS load is still saturated, and VH = 2.5 V. 2 40x10-6 1.11 I DD = (2.5 - 0.7) I DD = 71.9 A 2 1 n 2.22 VL For the NMOS device : 71.9x10-6 = 100x10-6 2.5 - 0.6 - VL VL = 0.179 V 2 1 n V +V K 2.22 100 2.5 - 0.7 (d) VIL = VTN + DD2 TP KR = KS = 1.11 40 = 5 VIL = 0.6 + 2 = 0.929V 5 +5 KR + KR L VOH = VDD - ( DD + VTP ) 1- V VOL = VDD + VTP 3KR = 2.5 - 0.7 3(5) KR 5 = 2.5 - (2.5 - 0.7) 1- = 2.34V 5 + 1 KR + 1 = 0.465V VIH = VTN + 2VOL = 0.6 + 2(0.465) = 1.53V NM L = VIL - VOL = 0.929 - 0.465 = 0.464 V
NM H = VOH - VIH = 2.34 -1.53 = 0.810 V
6-25
6.78 (a) The PMOS load is still saturated, and VH = 2.5 V. 2 50x10-6 1.11 I DD = (2.5 - 0.6) I DD = 100 A 2 1 n 2.22 VL For MS : 100x10-6 = 100x10-6 2.5 - 0.6 - VL VL = 0.254 V 2 1 n V +V K 2.22 100 2.5 - 0.6 (b) VIL = VTN + DD2 TP KR = KS = 1.11 50 = 4 VIL = 0.6 + 2 = 1.03V 4 +4 KR + KR L
VOH = VDD - ( DD + VTP ) 1- V VOL = VDD + VTP 3KR = 2.5 - 0.6 3(4) KR 4 = 2.5 - (2.5 - 0.6) 1- = 2.30V 4 + 1 KR + 1 = 0.549V VIH = VTN + 2VOL = 0.6 + 2(0.549)= 1.70V
NM H = VOH - VIH = 2.30 -1.70 = 0.600 V NM L = VIL - VOL = 1.03 - 0.549 = 0.481 V (c) The PMOS load is still saturated, and VH = 2.5 V. 2 30x10-6 1.11 I DD = (2.5 - 0.6) I DD = 60.0 A 2 1 n 2.22 VL For the NMOS device : 60x10-6 = 100x10-6 2.5 - 0.6 - VL VL = 0.148 V 2 1 n V +V K 2.22 100 2.5 - 0.6 = 0.866V (d) VIL = VTN + DD2 TP KR = KS = 1.11 30 = 6.67 VIL = 0.6 + 2 6.67 + 6.67 KR + KR L VOH = VDD - ( DD + VTP ) 1- V VOL = VDD + VTP 3KR = 2.5 - 0.6 3(6.67) KR 6.67 = 2.5 - (2.5 - 0.6) 1- = 2.37V 6.67 + 1 KR + 1 = 0.425V VIH = VTN + 2VOL = 0.6 + 2(0.425)= 1.45V NM L = VIL - VOL = 0.866 - 0.425 = 0.441 V
NM H = VOH - VIH = 2.37 -1.45 = 0.920 V
6.79
(a) I DD =
P 100W = = 55.6A VDD 1.8V W 2 25x10-6 W 2.63 1.8 ( - 0.5) = 2 L p 1 L p
For the saturated PMOS load : 55.6x10-6 =
W W 3.86 0.2 For the linear NMOS switch : 55.6x10-6 = 60x10-6 1.8 - 0.5 - 0.2 = 2 1 L n L n
6-26
(b) V
VOH
K + KR = VDD - ( DD + VTP ) 1- V
2 R
IL
= VTN +
VDD + VTP
KR =
KS 3.86 60 = = 3.52 KL 2.63 25
VIL = 0.5 +
1.8 - 0.5 3.522 + 3.52
= 0.826V
KR 3.52 = 1.8 - ( - 0.5) 1- = 1.65V 1.8 KR + 1 3.52 + 1 = 0.400V VIH = VTN + 2VOL = 0.5 + 2(0.400) = 1.30V NM L = VIL - VOL = 0.826 - 0.400 = 0.426 V
VOL =
VDD + VTP 3KR
=
1.8 - 0.5 3(3.52)
NM H = VOH - VIH = 1.60 -1.30 = 0.300 V
6.80
(a) I D =
P 200W = = 66.7A VDD 3V W 2 25x10-6 W 0.926 1 = (3 - 0.6) = 2 L p 1 1.08 L p
For the saturated PMOS load : 66.7x10-6 =
W W 1.65 0.3 For the linear NMOS switch : 66.7x10-6 = 60x10-6 3 - 0.6 - 0.3 = 2 1 L n L n 60 V +V K 3 - 0.6 = 1.11V (b) VIL = VTN + DD2 TP KR = KS = 1.65(1.08) 25 = 4.28 VIL = 0.6 + 4.282 + 4.28 KR + KR L VOH = VDD - ( DD + VTP ) 1- V VDD + VTP 3KR = 3 - 0.6 3(4.28) KR 4.28 = 3 - (3 - 0.6) 1- = 2.76V 4.28 + 1 KR + 1 = 0.670V VIH = VTN + 2VOL = 0.6 + 2(0.670)= 1.94V NM L = VIL - VOL = 1.11- 0.670 = 0.440 V
VOL =
NM H = VOH - VIH = 2.76 -1.94 = 0.821 V
6.81 With A = 1 = B, the circuit is equivalent to a single 4.44/1 switching device. 4.44 2 VL 100A 1.81 V 100A 2.5 - 0.6 - VL = ( TNL ) | VTNL = -1+ 0.5 VL + 0.6 - 0.6 2 2 1 1 2 100A 1.81 Solving iteratively VL = 0.1033V | VTNL = -0.968V (b) I DD = (0.968) = 84.8 A 2 1
(
)
6-27
6.82
W W 0.1 4.32 80A = 100A 2.5 - 0.6 - 0.1 = 2 1 L A L A
VTNB = 0.6 + 0.5 0.1+ 0.6 - 0.6 = 0.631 W W 0.1 4.65 80A = 100A 2.5 - 0.1- 0.631- 0.1 = 2 1 L B L A
6.83
(
)
We require
Ron R R + on = on and the total area AT (WL)A + (WL)B W W K L A L B
Setting L = 1,
1 1 1 KW B KW B W B2 + = WA = AT + WB = WB - K WB - K WB - K WA WB K d W B2 W B2 - 2KW B Finding the minimum : = 0 W B = 2K & WA = 2K. = dWB W B - K (W B - K )2
6.84
+2.5 V ML 1.81 1 Y M B MC C 2.22 1 D MD 2.22 1
MA A 2.22 1
B
2.22 1
6-28
6.85
+2.5 V ML 1.81 1
Y D MD 8.88 1 8.88 1 8.88 1 8.88 1
C
MC
B
MB
A
MA
6.86
+2.5 V ML 1.11 1 Y MB B 2.22 1 C MC 2.22 1
MA A 2.22 1
(a ) With A = B = C = 1, the circuit is equivalent to a single 6.66/1 switching device.
6.66 2 40A 1.81 VL 100A 2.5 - 0.6 - VL = (0.6) VL = 0.1033V 2 2 1 1 2 40A 1.81 (b) I DD = 2 1 (0.6) = 13.0 A
6-29
6.87
+2.5 V (a) ML 1.11 1 Y C MC 6.66 1 6.66 1 6.66 1
B
MB
A
MA
(b) The PMOS device remains saturated with I
DD
= 80A.
6.66 VDSA For MA : 80A = 100A 2.5 - 0.6 - VDSA VDSA = 0.0643V 2 1 For MB :VTNB = 0.6 + 0.5 0.0643 + 0.6 - 0.6 = 0.620 6.66 VDSB 80A = 100A 2.5 - .0643 - 0.620 - VDSB VDSB = 0.0674V 2 1 For MC :VTNC = 0.6 + 0.5 0.0643 + 0.674 + 0.6 - 0.6 = 0.640 6.66 VDSC 80A = 100A 2.5 - .0674 - .0643 - 0.640 - VDSC VDSC = 0.0709V 2 1 VL = VDSA + VDSB + VDSC = 0.203 V
(
)
(
)
(c) Assume equal values of
0.2 = 0.0667V 3 W W 0.0667 6.43 For MA : 80A = 100A 2.5 - 0.6 - 0.0667 = 2 1 L A L A VDS =
For MB :VTNB = 0.6 + 0.5 0.0667 + 0.6 - 0.6 = 0.621 W W 0.0667 6.74 80A = 100A 2.5 - 0.0667 - 0.621- 0.0667 = 2 1 L B L B For MC :VTNC = 0.6 + 0.5 0.1334 + 0.6 - 0.6 = 0.641 W W 0.0667 7.09 80A = 100A 2.5 - 0.1334 - 0.641- 0.0667 = 2 1 L B L B
(
)
(
)
6-30
6.88
VDD 1/1.7 vO
A
2/14.7/1
4.7/1 2/1
B
Ground
6.89
VDD 1/1.7 A 2/14.7/1 4.7/1 2/1 4.7/1 2/1 vO C
Ground
6.90
B
W 2.22 6.66 | =3 = 1 L A-F 1
W 1.81 Y = (A + B)(C + D)(E + F ) | = 1 L L
6.91 (a) The only change to the schematic is to connect the gate of load transistor ML to its drain instead of its source. (b) There is no change to the logic function Y = (A + B)(C + D)(E + F ) W W 4.71 14.1 1 | =3 = (c) = 1 L L 1.68 L ABCDEF 1
6-31
6.92
W 1.11 Y = (A + B)(C + D)E | = 1 L L
W 2.22 6.66 | =3 = 1 L A- E 1
6.93 (a) In the new circuit schematic, the PMOS transistor is replaced with a saturated NMOS load device as in Fig. 6.29(b). (b) The logic function is unchanged: Y = (A + B)(C + D)E W W 4.71 14.1 1 (c) = | =3 = 1 L L 1.68 L ABCDE 1 6.94
W 1.11 3.33 = | ACDF path contains 4 devices =3 1 1 L L
(a) Y = ACE + ACDF + BF + BDE (b)
2.22 26.6 W W 1 1 1 1 17.8 = 3 4 | + + = = = W 26.6 W 2.22 1 1 L A, C , D , F L B , E 1 3 1 L B 1 L E
6.95 (a) In the new circuit schematic, the PMOS transistor is replaced with a saturated NMOS load device as in Fig. 6.29(b). (b) There is no change to the logic function Y = ACDF + ACE + BDE + BF W W 4.71 18.8 1 (c) L = 1.68 | L = 4 1 = 1 L ACDF
RoB + RonD + RonE setting RoB = RonE : W 12.6 1 1 1 2 1 1 + + = + = = W W 18.8 W 18.8 4.71 L B 1 L B L E L B Checking : RoB + RonF = 1 1 1 1 + = so path BF is ok. 18.8 12.6 7.42 4.71
6-32
6.96
+2.5 V
Y D E
B C
A
W 1.81 = 1 L L DCA and ECA paths contain three devices W 2.22 6.66 = 3 = 1 L A,C , D, E 1 1 1 1 + = W W 2.22 L A L B 1 W 1 1 1 3.33 + = = 6.66 W 2.22 L 1 B 1 A 1 B 1
6-33
6.97
+2.5 V
C
E
B D
A
W 1 1.11 1 = = L L 2 1 1.80 CBA and EDA paths contain three devices W 1 2.22 3.33 = (3) = 1 L A- E 2 1
6-34
6.98
+2.5 V
Y
C E
B D
A
W 1 = L L 1.68
CBA and EDA paths contain three devices W 4.71 14.1 = 3 = 1 L A- E 1
6-35
6.99
+2.5 V ML Y
D B C
E
A
W 1.11 = 1 L L
W 2.22 = 1 L E
DCA path contains three devices W 2.22 6.66 = 3 = 1 L A,C , D 1 W 1 1 1 3.33 + = = 6.66 W 2.22 L 1 B 1 A 1 B 1
6.100
Y = A(B + D)(C + E )+ (C + E )G + F = (C + E ) A(B + D)+ G + F 2.22 13.3 W = 2 3 = 1 L A- E 1
[
]
W 1.81 3.62 = | =2 1 1 L L W 2.22 4.44 = 2 = 1 L F 1
|
W 1 1 1 6.67 + = = W 13.3 2.22 L G 1 2 1 L G 1
6-36
6.101 (a)
+2.5 V ML
Y
D B C
E
A
W 1.81 = 1 L L
W 2.22 = 1 L E
DCA path contains three devices W 2.22 6.66 = 3 = 1 L A,C , D 1 W 1 1 1 3.33 + = = 6.66 W 2.22 L 1 B 1 A 1 B 1 (b) Device E remains the same.
0.20 VL 0.20 V = = 0.0667V | B : VDS = 2 L = 2 = 0.133V 3 3 3 3 W W 0.0667 6.43 100A 2.5 - .6 - 0.0667 = 80A = 2 1 L A L A
A, C , D : VDS = VTNB = VTNC = 0.6 + 0.5 0.0667 + 0.6 - 0.6 = 0.621V W W 0.133 3.45 100A 2.5 - 0.0667 - 0.621 - 0.133 = 80A = 2 1 L B L B W W 0.0667 6.74 100A 2.5 - 0.0667 - 0.621 - 0.0667 = 80A = 2 1 L C L C VTND = 0.6 + 0.5 0.133 + 0.6 - 0.6 = 0.641V W W 0.0667 7.09 100A 2.5 - 0.133 - 0.641 - 0.0667 = 80A = 2 1 L D L D
(
)
(
)
6-37
6.102
W 2.22 Device A remains the same. = 1 L A B,C, D : VDS =
VL 0.20 = = 0.100V 2 2 W W 0.100 4.32 100A 2.5 - 0.6 - 0.100 = 80A = 2 1 L C, D L C , D VTNB = 0.6 + 0.5 0.100 + 0.6 - 0.6 = 0.631V W W 0.100 4.65 100A 2.5 - 0.100 - 0.631- 0.100 = 80A = 2 1 L B L B
(
)
6.103 The load device remains the same.
VL 0.20 V 0.20 = = 0.0667V | A : VDS = 2 L = 2 = 0.133V 3 3 3 3 W W 0.0667 6.43 100A 2.5 - .6 - 0.0667 = 80A = 2 1 L B L B B,C, D : VDS = VTNA = VTND = 0.6 + 0.5 0.0667 + 0.6 - 0.6 = 0.621V W W 0.133 3.45 100A 2.5 - 0.0667 - 0.621- 0.133 = 80A = 2 1 L A L A W W 0.0667 6.74 100A 2.5 - 0.0667 - 0.621- 0.0667 = 80A = 2 1 L D L D VTNC = 0.6 + 0.5 0.133 + 0.6 - 0.6 = 0.641V W W 0.0667 7.09 100A 2.5 - 0.133 - 0.641- 0.0667 = 80A = 2 1 L C L C
(
)
(
)
6-38
6.104 The load device remains the same.
VL 0.20 V 0.20 = = 0.10V | C, D : VDS = L = = 0.050V 2 4 2 4 W W 0.10 4.32 100A 2.5 - .6 - 0.10 = 80A = 2 1 L B L B A, B : VDS = VTNA = VTND = 0.6 + 0.5 0.1+ 0.6 - 0.6 = 0.631V W W 0.10 4.65 100A 2.5 - 0.10 - 0.631- 0.10 = 80A = 2 1 L A L A W W 0.05 9.17 100A 2.5 - 0.10 - 0.631- 0.05 = 80A = 2 1 L D L D VTNC = 0.6 + 0.5 0.15 + 0.6 - 0.6 = 0.646V W W 0.05 9.53 100A 2.5 - 0.15 - 0.646 - 0.05 = 80A = 2 1 L C L C
6.105 Device E and the load device remain the same. In the worst case, for paths BCD or ADE
(
)
(
)
VL 0.20 = = 0.0667V 3 3 W W 0.0667 6.43 100A 2.5 - .6 - 0.0667 = 80A = 2 1 L B , E L B , E A, B,C, D, E : VDS = VTND = 0.6 + 0.5 0.0667 + 0.6 - 0.6 = 0.621V W W 0.0667 6.74 100A 2.5 - 0.0667 - 0.621- 0.0667 = 80A = 2 1 L D L D VTNA = VTNC = 0.6 + 0.5 0.133 + 0.6 - 0.6 = 0.641V W W 0.0667 7.09 100A 2.5 - 0.133 - 0.641- 0.0667 = 80A = 2 1 L A, C L A,C
(
)
(
)
6-39
6.106 A 0 (a) 0 1 1
B Y 0 1 1 0 0 1 0 1
(b) Y = AB + AB = A B
(c) Assuming equal voltage drops (0.10V) across MP and MS : MP must carry one unit of load current with one - half the drain - source W 4.44 voltage (VDS = 0.10V ) of the switching transistor in Fig.6.29(d). = 1 L P MS must carry two units of load current with one - half the drain - source W 8.88 voltage (VDS = 0.10V ) of the switching transistor in Fig.6.29(d). = 1 L S (d) MS will not change. MP will need to be somewhat larger. (e) Coincidence gate (Exclusive NOR)
6.107 Original design 0.20 mW - 1 mW requires 5 times larger current. W 28.8k 2.22 11.1 (a) R = = 5.76k = 5 = 5 1 1 L S
W 1 2.98 (b) = 5 = 1.68 1 L L W 1 1 (c) = 5 = 5.72 1.14 L L W 1.81 9.05 (d) = 5 = 1 1 L L
W 4.71 23.6 = =5 1 1 L S W 2.22 11.1 = =5 1 1 L S W 2.22 11.1 = =5 1 1 L S W 1.11 5.55 W 2.22 11.1 (e) = 5 = = =5 1 1 1 1 L L L S
6.108 W 1.81 7.24 = | =4 1 1 L L W 2.22 8.88 = 4 = 1 L F 1
2.22 26.6 W = 4 3 = 1 L A- E 1 W 13.3 1 1 1 | + = = W 26.6 2.22 1 L G 4 1 L G 1
6-40
6.109 W 1 1.81 1 = = L L 4 1 2.21
W 1 2.22 1.67 | = 3 = 1 L A- F 4 1
6.110 W 1.81 5.43 W 6.66 20.0 W 3.33 9.99 (a) L = 3 1 = 1 | L = 3 1 = 1 | L = 3 1 = 1 L BCD A W 1 1.81 W W 1 1 1 6.66 1.33 1 3.33 (b) L = 5 1 = 2.76 | L = 5 1 = 1 | L = 5 1 = 1.50 L BCD A 6.111 W W W 1 1 1 1 1 1.81 1 4.44 (a) L = 10 1 = 5.53 | L = 10 1 = 2.25 | L = 2 2.25 = 1.13 L AB CD W 2.5 1.81 4.53 W W 11.1 22.2 2.5 4.44 11.1 (b) L = 1 1 = 1 | L = 1 1 = 1 | L = 2 1 = 1 L AB CD 6.112 W 1.11 4.44 W 6.66 26.6 (a) L = 4 1 = 1 | L = 4 1 = 1 L ABCDE W 1 1.11 W 1 1 6.66 2.22 (b) L = 3 1 = 2.70 | L = 3 1 = 1 L ABCDE 6.113
2 2 1 1 W " W V V (a) I D = nCox ( GS - VTN ) = n ox ( GS - VTN ) 2 2 Tox L L W 2 1 I* * I D = n ox 2 ( GS - VTN ) = 2I D | D = 2 V ID 2 Tox L 2 2
* (b) PD = V (2I )= 2VI = 2PD - Power dissipation has increased by a factor of two.
6-41
6.114
For each line : i = C
dv | Assume the transition occurs in T seconds generating dt 2.5V a current pulse with constant amplitude I = 10x10-12 F . T 2.5x10-11 T Then I avg = = 500A and P = 64(2.5V )I avg = 64(2.5)(0.50mA)= 80 mW 50ns T 3.3 2 2 (b) P V so P = 80mW = 139 mW 2.5
6.115
C C PHL and PLH KS KL C C" WL L2 ox | For either case, PHL = = n KS " W nCox L
6.116
P =
6.117
PDP 100 fJ 10-13 J = = -4 = 1 ns PD 100W 10 W 2.5 + 0.20 = 1.35V 2 = 0.25 + 0.23 = 0.48V
VH = 2.5V | VL = 0.20V | V50% = V90% = 2.5 - 0.23 = 2.27V | V10% vI : t f = 62 - 55 = 7 ns
(a) vI : t r = 22.5 -1.5 = 21 ns | vO : t r = 81- 58 = 23 ns | vO : t r = 12.5 - 6 = 6.5 ns 2.5 + 7 = 4.8 ns 2 (b) PHL = 2.5 ns | PLH = 7 ns (c) P =
6.118
(a) T = 301( PHL + PLH ) = 602
P = 602(0.1ns) = 60.2 ns 2 (b) An even number of inverters has a potential steady state and may not oscillate.
( PHL + PLH ) = 602
6-42
6.119 t r = 2.2RC = 2.2(28.8k)(0.5 pF )= 31.7 ns
t f 3.7RonS C =
3.7(0.5 pF ) 3.7C = = 4.39 ns V Kn ( GS - VTN ) 2.22 10-4 (2.5 - 0.6)
PLH = 0.69RC = 0.69(28.8k)(0.5 pF )= 9.94 ns PHL 1.2RonS C =
( )
1.2(0.5 pF ) 1.2C = = 1.78 ns V Kn ( GS - VTN ) 2.22 10-4 (2.5 - 0.6)
( )
P =
9.94 + 1.78 = 5.86 ns 2
6.120 t r = 2.2RC = 2.2(28.8k)(0.5 pF )= 31.7 ns
t f 3.7RonS C =
3.7(0.5 pF ) 3.7C = = 3.09 ns V Kn ( GS - VTN ) 2.22 10-4 (3.3 - 0.6)
PLH = 0.69RC = 0.69(28.8k)(0.5 pF )= 9.94 ns PHL 1.2RonS C =
6.121
( )
1.2(0.5 pF ) 1.2C = = 1.00 ns V Kn ( GS - VTN ) 2.22 10-4 (3.3 - 0.6)
( )
P =
9.94 + 1.00 = 5.47 ns 2
Resistive Load : P =
PLH + PHL
2
VH = 2.5V
VL = 0.20V
VTNS = 0.6V
PLH = 0.69RC and PHL = 1.2RonS C for RonS =
C C 0.526C = = KS KS ( H - VTNS ) KS (2.5 - 0.6) V
V 2.5 - 0.20 2.5 - VL = KS VGS - VTN - L VL KS R = = 6.39 R 2 0.20 2.5 - 0.6 - 0.20 2 6.39 0.526 W 1 987 16.5 A 2.5ns = ( 1pF ) 0.69 = and R = 6.47 k KS = 987 2 | = + KS 2 1 V L S 60 KS I DDL =
(2.5 - 0.20)V = 356A
6.47k
| P=
2.5(356A) 2
= 0.444 mW
6.122 Resistive load inverter has very little effect on the results:
= 0 : t r = 3.8ns t f = 1.3ns PLH = 10.0ns PHL = 1.6ns = 0.04/V : t r = 31.6ns t f = 3.6ns PLH = 9.9ns PHL = 1.5ns
6.123
6-43
Ignore body effect for simplicity. Equate drain currents to find VL : ' 2 V 1 Kn (2.5 - VL - 0.6) = 4Kn' 2.5 - 0.6 - 0.6 - 2L VL VL = 0.156V 2 2 1 1 RonL = = = 19.1k KL ( GS - VTN ) 0.5 60x10-6 (2.5 - 0.156 - 0.6) V
(
)
RonS =
KS ( GS - VTN ) 4 60x10-6 (2.5 - 0.6 - 0.6) V
1
=
PLH 3.0RonLC = 3.0(0.5 pF )( 19.1k)= 28.7 ns PHL 1.2RonS C = 1.2(0.5 pF )(3.21k)= 1.93 ns P =
28.7 + 1.93 = 15.3 ns 2
t f 3.7RonS C = 3.7(0.5 pF )(3.21k) = 5.94 ns
t r 11.9RonLC = 11.9(0.5 pF )( 19.1k) = 114 ns
(
)
1
= 3.21k
6.124 Ignore body effect for simplicity. Equate drain currents to find VL :
VTN = 0.6 VH = 3.3 - 0.6 = 2.7V
-6 2 4 VL 1 60x10 -6 60x10 2.7 - 0.6 - VL = (3.3 - VL - 0.6) 2 2 1 2
(
)
(
)
9VL2 - 39.0VL + 7.29 = 0 VL = 0.196V RonL = RonS = 0.5 60x10 4 60x10
(
-6
)
(3.3 - 0.196 - 0.6)
= 1.98k
1
= 13.3k
t r = 11.9RonLC = 11.9( 13.3k)(0.3 pF )= 47.5 ns 1.98k)(0.3 pF )= 2.20 ns t f = 3.7RonS C = 3.7( 12.0 + 0.713 = 6.36 ns 2
(
1
-6
)(2.7 - 0.6)
PLH = 3.0RonLC = 3.0( 13.3k)(0.3 pF ) = 12.0 ns PHL = 1.2RonS C = 1.2( 1.98k)(0.3 pF )= 0.713 ns P =
6-44
6.125
2 2x10-9 s 3.0RonL + 1.2RonS P = = C 3.0RonL + 1.2RonS = = 4000 2 2 10-12 F 2 V K KS VGSS - VTNS - L VL = L ( GSL - VTNL ) Ignore body effect for simplicity. V 2 2 2 0.25 KL KS 2.5 - 0.6 - 0.6 - 0.25 = (2.5 - 0.25 - 0.6) KS = 4.63KL 2 2
PLH + PHL
(
)
3.0 1.2 + = 4000 KL = 5.04x10-4 A/V 2 KL (2.5 - .25 - 0.6) 4.63KL (2.5 - 0.6 - 0.6) W 5.04x10-4 8.41 = = -5 1 L L 6.0x10 W 8.41 = 38.9 = 4.63 1 L S
6.126 VH = 2.5V VL = 0.2V
RonL = RonS = V KL ( GS - VTN ) 1 1 = =
VTNL = 0.6 + 0.5 0.2 + 0.6 - 0.6 = 0.66V
(
)
(
6x10
-5
)
(4 - 0.2 - 0.66)
1
-5
5.72
= 30.4k = 3.95k
V KS ( GS - VTN ) 2.22 6x10
t r 3.7RonLC = 3.7(0.7 pF )(30.4k) = 78.7 ns
(
)
(2.5 - 0.6)
PLH 0.69RonLC = 0.69(0.7 pF )(30.4k)= 14.7 ns PHL 1.2RonS C = 1.2(0.7 pF )(3.95k)= 3.32 ns P =
14.7 + 3.32 = 9.00 ns 2
t f 3.7RonS C = 3.7(0.7 pF )(3.95k) = 10.2 ns
6-45
6.127
3.0V
2.0V
1.0V
0V 3.0V
2.0V
1.0V
0V
Results: tf = 1.0 ns, tr = 22.3 ns, PHL = 0.47 ns, PLH = 4.0 ns, P = 4.2 ns
6.128
2 3x10-9 s 3.6RonL + 1.2RonS P = = C 3.6RonL + 1.2RonS = = 6000 2 2 10-12 F 2 V K KS VGSS - VTNS - L VL = L (-VTNL ) Ignore body effect for simplicity. 2 2 0.25 KL 2 KS 3 - 0.6 - 0.25 = (3) KS = 7.91KL 2 2 3.6 1.2 + = 6000 KL = 2.11x10-4 A/V 2 KL (3) 7.91KL (3 - 0.6)
PLH + PHL
(
)
W 2.11x10-4 3.52 = = -5 1 L L 6.0x10 t r = 8.1RonL
( ) = 12.8 ns C= 3.52( 6x10 ) 3) (
8.1 10-12
-5
W 3.52 = 27.8 = 7.91 1 L S t f = 3.7RonS C =
( ) = 0.924 ns 27.8( 6x10 ) 3 - 0.6) (
3.7 10-12
-5
6-46
6.129
2 10-9 s 3.6RonL + 1.2RonS P = = C 3.6RonL + 1.2RonS = = 10.0k 2 2 0.2x10-12 F 2 V K KS VGSS - VTNS - L VL = L (-VTNL ) Ignore body effect for simplicity. 2 2 KL 2 0.20 KS 3.3 - 0.75 - 0.20 = (2) KS = 4.08KL 2 2
PLH + PHL
(
)
1.2 3.6 + = 10k KL = 1.92x10-4 A/V 2 KL (2) 4.08KL (3.3 - 0.75) W 1.92x10-4 3.19 = = -5 1 L L 6.0x10 I DD = W 3.19 = 13.0 = 4.08 1 L S
3.3(384A) 2 KL 1.92x10-4 2 -VTNL ) = ( (2) = 384A P = 2 = 0.634 mW 2 2
6.130 (a) VTN = 0.6 + 0.5 0.20 + 0.6 - 0.6 = 0.660V
(
)
80x10-6 =
W 2 100x10-6 W 1 (2.5 - 0.20 - 0.66) = 2 L L L L 1.68
-6
(b) 80x10 (c) 80x10
=
W 2 100x10-6 W 1 (2.5 - 0.20 - 0.6) = 2 L L L L 1.81
W W 2.3 1 = 100x10-6 4 - 0.20 - 0.66 - 2.3 = 2 L L L L 5.72 W W 2.3 1 (d ) 80x10-6 = 100x10-6 L 4 - 0.20 - 0.6 - 2 2.3 L = 5.89 L L
-6
(e) V
TN
= -1+ 0.5 0.20 + 0.6 - 0.6 = -0.940V
-6
(
)
W 1.81 2 100x10 W (-0.940) = 2 1 L L L L W 1.60 2 100x10-6 W (f ) 80x10-6 = 2 L (-1) L = 1 L L 80x10-6 =
6-47
6.131 For VDD = -2.5 V, we have VH = -0.20 V with a power dissipation of 0.20 mW. Since these gates are all ratioed logic design, the ratio of the W/L ratios of the load and switching transistors does not change. We only need to scale both equally to achieve the power level.
W 100 2.22 5.55 = (a) RL = 28.8k | = 1 L S 40 1 W 100 1 W 100 4.71 11.8 1.49 (b) = = | = = 1 1 L L 60 1.68 L S 60 1 W 100 1 W 100 1 5.55 (c) = = | = 2.22 = 1 L L 60 5.72 2.29 L S 60 W 100 1.81 4.53 W 100 2.22 5.55 (d ) = = | = = 1 1 L L 60 1 L S 60 1 W 100 1.11 2.78 W 100 2.22 5.55 (e) = = | = = 1 1 L L 60 1 L S 60 1
6.132
2 2 -V 1 25x10-6 VL = -2.5 + 0.6 = -1.9 V | 25x10-6 -1.9 - (-0.6)- H (-VH )= -2.5 - VH - (-0.6) 2 4 2 1
(
)
(
)
2 9VH - 24.6VH + 3.61 = 0 VH = -0.156 V
6.133 Pretend this is an NMOS gate with VDD = 3.3V and VL = 0.33V
VH = 3.3 - 0.6 + 0.75 VH + 0.7 - 0.7 VH = 2.08V VTNL = 0.6 + 0.75 0.33 + 0.7 - 0.7 = 0.734 | I DD = 0.1mW = 30.3A 3.3V W 0.303 2 40A W 1 30.3A = = (3.3 - 0.33 - 0.734) = 1 3.30 2 L L L L W W 1.75 0.33 30.3A = 40A 2.08 - 0.60 - 0.33 = 2 1 L S L S
[
(
)]
(
)
6-48
6.134
VL = -VTPL | VL = - -0.6 - 0.75 2.5 - VL + 0.7 - 0.7 VL = 1.07V W K 'p W 2 VDS K VGS - VTPS - V VDS = ( GSL - VTPL ) 2 2 L L L S 2 VH - 2.5 1 1 3 V V 1.07 - 2.5 - (-0.6)- ( H - 2.5) = ( H + VTPL ) 2 3 1 2
' p
[
(
)]
VTPL = -0.6 - 0.75 VH + 0.7 - 0.7 Solving the last two equations iteratively : VH = 2.30 V
6.135 Y is low only when both A and B are high : Y = AB or Y = AB.
(
)
Alternatively, Y is high when either A or B is low : Y = A + B = AB
6.136 Y is high only when both A and B are low : Y = AB or Y = A + B 6.137
0.0V -0.5V
-1.0V
-1.5V
VL = -1.90 and VH = -0.156 agree with the hand calculations in Prob. 6.132
6.138
-0.0V
2 0V
-1.0V
-2.0V
3 0V
VH = -0.33 and VL -2.08 agree with the design values in Prob. 6.133.
6-49
6.139
0.0V -0.5V
-1.0V
-1.5V
tr = 16.8 ns, tf = 560 s, PLH = 11.7 ns, PHL = 60 ns, P = 35.9 ns
6.140
-0.0V
2 0V
-1.0V
-2.0V
-3.0V 0s
0 2us
0 4us
0 6us
0 8us
1 0us
1 2us
1 4us
1 6us
1 8us
2 0us
tr = 46 ns, tf = 1.1
s, PLH = 21 ns, PHL = 122 ns, P = 72 ns
6-50