Ch16s

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

⎛ ⎞ = =⎜ ⎟⎝ ⎠


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

⎛ ⎞⇒ = =⎜ ⎟⎝ ⎠


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

16.67

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=

Ch16s

Technology

Transcript of Ch16s