Chapter 16 Exercise Solutions EX16.1

( )( )

( )( )( )( )

( )

147 2

8

19 14 15 12

7

3.9 8.85 101.726 10 F / cm

200 10

2 1.6 10 11.7 8.85 10 100.1055 V

1.726 100.576 0.576 0.1055 0.576 5 0.576

0.169 V

OX

TN SB

TN

C

V r V

V

γ

−−

−

− −

−

×= = ×

×

× ×= =

×⎡ ⎤ ⎡ ⎤Δ = + − = + +⎣ ⎦ ⎣ ⎦

Δ =

EX16.2 (a)

( )

( ) ( )( ) ( )

2

2

3 2 3 0.52

0.10.060.1 3 5 5 0.1 0.1

20.1 3 0.0735

39.5 K

o DD D D

no o o D

o

D

D

D

v V I Rk Wv v v R

Lv

R

RR

= −′⎛ ⎞⎛ ⎞ ⎡ ⎤= − − −⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠

=

⎛ ⎞ ⎡ ⎤= − −⎜ ⎟ ⎣ ⎦⎝ ⎠= −=

(b)

( )( )( ) ( )

( ) ( )

( ) ( )( )( )

2

2

0.06 5 39.5 0.5 0.5 3 02

5.925 0.5 0.5 3 0

1 1 4 5.925 30.5

2 5.9250.632 V

1.132 V

It It

It It

It Ot

Ot

It

V V

V V

V V

VV

⎛ ⎞ − + − − =⎜ ⎟⎝ ⎠

− + − − =

− ± +− = =

==

EX16.3 (a) (i)

3 0.42.6 V

o DD TNL

o

v V Vv

= − = −=

(ii)

( ) [ ]

( ) [ ]

( )( )( )

22

22

2 2

2

2 0.4 0.4

16 2 2.6 0.4 2 3 0.4

35.2 8 6.76 5.29 40.4 6.76 0

40.4 1632.16 4 9 6.762 9

0.174 V

I o o DD oD L

o o o

o o o o

o o

o

o

W Wv v v V vL L

v v v

v v v vv v

v

v

⎛ ⎞ ⎛ ⎞⎡ ⎤− − = − −⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠

⎡ ⎤− − = − −⎣ ⎦− = − +

− + =

± −=

=

(b)

[ ]260 (2) 3 0.174 0.42

353.1 A1.06 mW

D

D

D DD

i

iP i V

μ

⎛ ⎞= − −⎜ ⎟⎝ ⎠

== ⋅ =

EX16.4 (a)

( ) ( )( )

( ) ( )

( )( )

22

2

2

2 0.4 0.8

6 2 3 0.4 2 0.64

6 31.2 1.28 0

31.2 973.44 4 6 1.282(6)

41.4 mV

I o oD L

o o

o o

o

o

W Wv v vL L

v v

v v

v

v

⎛ ⎞ ⎛ ⎞⎡ ⎤− − = − −⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠

⎡ ⎤− − =⎣ ⎦− + =

± −=

=

(b)

( ) ( )

( ) ( )

2 2

2

0.4 ( 0.8)

6 0.4 2 0.64

0.862 VDriver

0.862 0.4 0.462 V

0.862 VLoad

3 0.8 2.2 V

ItD L

It

It

Ot

It

Ot DD TNL

W WvL L

v

vv

vv V V

⎛ ⎞ ⎛ ⎞− = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− =

⇒ = ⎫⎬= − = ⎭

= ⎫⎬= + = − = ⎭

(c)

( )( )260 (2) 0.8 38.4 A2

115.2 W

D

D DD

i

P i V

μ

μ

⎛ ⎞= − − =⎜ ⎟⎝ ⎠

= ⋅ =

EX16.5 We have

{ }{ }

2 2

5 0.8 0.35 0.73 0.73

4.499 0.35 0.73

OH DD TNLO fP SB fP

OH OH

OH OH

V V V r V

V V

V V

φ φ⎡ ⎤= − + + −⎣ ⎦

⎡ ⎤= − + + −⎣ ⎦

− = − +

Squaring both sides 2

2

8.998 20.241 0.1225(0.73 )9.1205 20.15 0

9.1205 83.1835 4(20.15)2

3.76 V

OH OH OH

OH OH

OH

OH

V V VV V

V

V

− + = +− + =

± −=

=

EX16.6 a. i. logic 1 10 V, logic 0A B= = = “A” driver in nonsaturation. “B” driver off

( ) ( )

( ) ( ) ( )( )

( ) ( )( )

2 2

2 20 0

20 0

20 0

2

0 0

22 2

2 3 10 2 10 1.5

9 5 17

5 85 9 0

85 85 4 5 90.107 V

2(5)

nTNL I TND OL OL

L D

L L

L L

L L

L L

k W k WV v v V VL L

V V

V V

V V

V V

′ ′⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎡ ⎤− = − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

⎡ ⎤= − −⎣ ⎦

= −

− + =

± −= ⇒ =

ii. A = B = logic 1

( ) ( )

( ) ( )( ) ( )( )

( ) ( )( )( )

2 2

2 20 0

20 0

20 0

2

0 0

2 22 2

2 3 2 10 2 10 1.5

9 10 17

10 170 9 0

170 170 4 10 90.0531 V

2 10

nTNL I TND OL OL

L D

L L

L L

L L

L L

k W k WV v v V VL L

V V

V V

V V

V V

′ ′⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎡ ⎤− = − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

⎡ ⎤= − −⎣ ⎦

= −

− + =

± −= ⇒ =

b. Both cases.

( ) ( )235 2 3 315 A 3.15 mW2D D DDi P i V Pμ= ⋅ = ⇒ = ⋅ ⇒ =

EX16.7

( )

( )( ) ( )

2

2

800 160 A5

35160 1.4 34.3 4.662

35160 2 5 0.8 0.12 0.12 9.202

D DD D

DL L L

DD D

P i V i

W W WiL L L

W WiL L

μ= ⋅ ⇒ = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = ⋅ = ⇒ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎡ ⎤= = ⋅ − − ⇒ =⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠

EX16.8

(a) 2.1 1.05 V2 2

1.05 ( 0.4) 1.45 V1.05 0.4 0.65 V

DDIt

OPt It TD

ONt It TN

VV

V V VV V V

= = =

= − = − − == − = − =

(b) 2.1 ( 0.4) 0.5(0.4) 1.16 V1 0.5

1.16 0.4 1.56 V1.16 0.4 0.76 V

It

OPt

ONt

V

VV

+ − += =+

= + == − =

(c) 2.1 ( 0.4) 2(0.4) 0.938 V1 2

0.938 0.4 1.338 V0.538 V

It

OPt

ONt

V

VV

+ − += =+

= + ==

EX16.9

( ) ( )( )

2

26 12

4

0.10 10 0.5 10 3

2.22 10 Hz 22.2 kHz

L DDP f C V

f

f f

− −

= ⋅ ⋅

× = ×

= × ⇒ =

EX16.10 a.

0

0

10 2 2.5(2)/ 200 / 80 2.5 4.32 V1 2.5

6.32 V2.32 V

n p It It

Pt

Nt

K K V V

VV

− += = ⇒ = ⇒ =+

==

b.

{ }0

0

0

0

10 2 2 2.52 2 1 3.39 V2.5 1 2.5 3

1 (1 2.5)(3.39) 10 (2.5)(2) 229.43 V

10 2 2 2(2.5)2 1 4.86 V2.5 1 3(2.5) 1

(4.86)(1 2.5) 10 (2.5)(2) 22(2.5)

0.802 V

IL IL

HU

HU

IH IH

LU

LU

V V

V

V

V V

V

V

⎡ ⎤− −= + ⋅ − ⇒ =⎢ ⎥− +⎣ ⎦

= + + − +

=

⎡ ⎤− −= + ⋅ − ⇒ =⎢ ⎥− +⎢ ⎥⎣ ⎦+ − − +=

=

c. 0

0

3.39 0.802 2.59 V

9.43 4.86 4.57 VL IL LU L

H HU IH H

NM V V NM

NM V V NM

= − = − ⇒ =

= − = − ⇒ =

EX16.11 3 PMOS in series and 3 NMOS in parallel. Worst Case: Only one NMOS is ON in Pull-down mode ⇒ same as the CMOS inverter .nW W⇒ = All 3 PMOS are on during pull-up mode 3(2 ) 6 .pW W W⇒ = = EX16.12 NMOS: Worst Case, ,NAM NBM on, 2( )nW W= or ,NCM NDM or NCM , NEM on 2( ).nW W⇒ = PMOS: PAM and PCM on or PAM and PBM on 2(2 ) 4WPW W⇒ = = If PDM and PEM on, need 2(4 ) 8WPW W= = EX16.13 a. 05 V 4 VIv vφ= = ⇒ =

b. 03 V, 5 V 3 VIv vφ= = ⇒ =

c. 4.2 V,Iv = 05 V 4 Vvφ = ⇒ =

d. 5 V,Iv = 03 V 2 Vvφ = ⇒ = EX16.14 (a) 8 ,Iv V= 10 8GSDV v Vφ = ⇒ =

DM in nonsaturation

[ ]

( )( ) ( ) [ ]

2

2

22

2( )

2 8 2 0.5 0.5 10 0.5 2 9.78

D GSD TND O O

L DD O TNL

D D

L L

K v V v v

K V v VK KK K

⎡ ⎤− −⎣ ⎦

− −

⎡ ⎤− − = − − ⇒ =⎣ ⎦

(b)

( )( ) ( ) [ ]22

8 6

2 6 2 0.5 0.5 10 0.5 2 15

I GSD

D D

L L

v V v VK KK K

φ= = ⇒ =

⎡ ⎤− − = − − ⇒ =⎣ ⎦

EX16.15 16 16384K ⇒ cells Total Power 125 (2.5) 50 T TmW I I mA= = ⇒ =

Then, for each cell, 50 3.05 16384

mAI I Aμ= ⇒ =

Now, DDVIR

≅ or 2.5 0.82 3.05

DDVR R MI

= = ⇒ = Ω

TYU16.1

2

2

2

750 150 A5

35150 (5 0.2 0.8)2

150 280 0.536

2( )235150 2(4.2 0.8)(0.2) (0.2)2

150 23.1 6.49

D DD D

L

L L

nD I TND O O

D

D

D D

P i V i

WLW WL L

k Wi v V v vL

WL

W WL L

μ= ⋅ ⇒ = =

⎛ ⎞= − −⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′⎛ ⎞⎛ ⎞ ⎡ ⎤= − −⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠⎛ ⎞ ⎡ ⎤= − −⎜ ⎟ ⎣ ⎦⎝ ⎠

⎛ ⎞ ⎛ ⎞= ⋅ ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

TYU16.2

( )

( )

( )( ) ( )

2

2

2

350 70 A5

23570 2 12

35 2 5 0.8 0.05 0.052

70 7.31 9.58

D DD D

nD TNL

L

L L

DD

D D

P i V I

k Wi VL

W WL L

WiLW WL L

μ= ⋅ ⇒ = =

′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎡ ⎤= ⋅ − −⎜ ⎟ ⎣ ⎦⎝ ⎠

⎛ ⎞ ⎛ ⎞= ⋅ ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

TYU16.3

( )

( )( ) ( )

2

2

800 160 A5

35160 1.4 4.662

35 1160 A 2 5 0.8 0.12 0.12 27.62 3

D DD D

DL L

DD D

P i V i

W WiL L

W WiL L

μ

μ

= ⋅ ⇒ = =

⎛ ⎞ ⎛ ⎞= = ⋅ ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎡ ⎤= = ⋅ ⋅ − − ⇒ =⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠

TYU16.4 a. From the load transistor:

( ) ( )( )2 235 0.5 5 0.15 0.72 2n

DL GSL TNLL

k WI V VL

′⎛ ⎞⎛ ⎞= − = − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

or 150.7 ADLI μ=

Maximum 0v occurs when either A or B is high and C is high. For the two NMOS is series, the effective

Nk is cut in half, so

( ) 21 22 2

nDL GSD TND DS DS

D

k WI V V V VL

′⎡ ⎤⎛ ⎞⎛ ⎞ ⎡ ⎤= − −⎢ ⎥⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠⎣ ⎦

or

( )( ) ( )21 35150.7 2 5 0.7 0.15 0.152 2 D

WL

⎡ ⎤⎛ ⎞ ⎡ ⎤= ⋅ − −⎢ ⎥⎜ ⎟ ⎣ ⎦⎝ ⎠⎣ ⎦

which yields

13.6D

WL

⎛ ⎞ =⎜ ⎟⎝ ⎠

b. ( )( )150.7 5 753 WD DDP i V P μ= ⋅ = ⇒ = TYU16.5 a. 0 (max)v occurs when 1A B= = and 0C D= = or 0A B= = and 1C D= =

2 2

2 2

1( ) 2( )2

1(0.5)(1.2) 2(5 0.7)(0.15) (0.15)2

0.72 (0.634) 1.14

TNL I TND O OL D

D

D D

W WV v V v vL L

WL

W WL L

⎛ ⎞ ⎛ ⎞ ⎡ ⎤− = ⋅ − −⎜ ⎟ ⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎡ ⎤= ⋅ − −⎜ ⎟ ⎣ ⎦⎝ ⎠

⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

b.

[ ]22 35( ) (0.5) ( 1.2)2 2

12.6 A(12.6)(5) 63 W

nD TNL

L

D

D DD

k Wi VL

iP i V P

μμ

′⎛ ⎞⎛ ⎞ ⎛ ⎞= − = − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠== ⋅ = ⇒ =

TYU16.6 a.

2

2 2

50 /

2.5 V

(max) ( ) 50(2.5 0.8) (max) 145 A

n p

It

D n It TN D

K K A V

V

i K V V i

μ

μ

= =

=

= − = − ⇒ =

b. 2

2

200 /

2.5 V

(max) (200)(2.5 0.8) (max) 578 A

n p

It

D D

K K A V

V

i i

μ

μ

= =

=

= − ⇒ =

TYU16.7 a.

0

0

5 2 (1)(0.8)1 1

1.9 V3.9 V1.1 V

It

It

Pt

Nt

V

VVV

− +=+

===

b.

{ }

{ }

0

0

0

0

30.8 [5 2 0.8] 1.63 V8

1 2(1.63) 5 0.8 224.73 V

50.8 (5 2 0.8) 2.18 V8

1 2(2.18) 5 0.8 220.275 V

IL IL

HU

HU

IH IH

LU

LU

V V

V

V

V V

V

V

= + ⋅ − − ⇒ =

= + − +

=

= + − − ⇒ =

= − − +

=

c. 0

0

1.63 0.275 1.35 V

4.73 2.18 2.55 VL IL LU L

H HU IH H

NM V V NM

NM V V NM

= − = − ⇒ =

= − = − ⇒ =

TYU16.8

TYU16.9

( )

( )

NMOS 2 transistors in series2 2W

PMOS 2 transistors in series2 2 4W

n

p

W W

W W

−= =

−= =

TYU16.10 The NMOS part of the circuit is:

TYU16.11 The NMOS part of the circuit is:

TYU16.12 Exclusive-OR A B f 0 0 0 1 0 1 0 1 1 1 1 0

TYU16.13

NMOS conducting for 0 4.2 VIv≤ ≤ ⇒ NMOS Conducting: 0 8.4 st≤ ≤ NMOS Cutoff: 8.4 10 st≤ ≤ PMOS cutoff for 0 1.2 VIv≤ ≤ ⇒ PMOS Cutoff: 0 2.4 st≤ ≤ PMOS Conducting: 2.4 10 st≤ ≤

TYU16.14 (a) 1 32 32K ⇒ × array Each row and column requires a 5-bit word ⇒ 6 transistors per row and column, 32 6 32 6 384⇒ × + × = transistors plus buffer transistors. (b) 4 64 64K ⇒ × array Each row and column requires a 6-bit word ⇒ 7 transistors per row and column 64 7 64 7 896⇒ × + × = transistors plus buffer transistors. (c) 16 128 128K ⇒ × array Each row and column requires a 7-bit word ⇒ 8 transistors per row and column

128 8 128 8 2048⇒ × + × = transistors plus buffer transistors. TYU16.15 From Equation (16.84) ( )( )

( )( ) ( )

2 2

2 21

/ 2 3 2(2.5)(0.4) 3(0.4) 0.526/ 2 2.5 2(0.4)

DD TN TNnA

n DD TN

W L V V VW L V V

− −= = =− −

From Equation (16.86)

( )( )

( )( )

2 2

2 2

/ 2 3 2(2.5)(0.4) 3(0.4)(2.5) 0.862/ (2.5 0.4)

p DD TN TNn

pnB DD TP

W L V V VkW L k V V

−′ ⎡ ⎤−= ⋅ = =⎢ ⎥′ −+ ⎣ ⎦

So WL

⎛ ⎞⎜ ⎟⎝ ⎠

of transmission gate device must be 0.526< times the WL

⎛ ⎞⎜ ⎟⎝ ⎠

of the NMOS transistors in the

inverter cell. The WL

⎛ ⎞⎜ ⎟⎝ ⎠

of the PMOS transistors must be 0.862< times the WL

⎛ ⎞⎜ ⎟⎝ ⎠

of the transmission gate

devices. Then the WL

⎛ ⎞⎜ ⎟⎝ ⎠

of the PMOS devices must be 0.453< times WL

⎛ ⎞⎜ ⎟⎝ ⎠

of NMOS devices in cell.

TYU16.16 Initial voltage across the storage capacitor 3 0.5 2.5 .DD TNV V V= − = − = Now

dVI Cdt

− = or IV t KC

= − ⋅ +

where 2.5 ,K V= 1.5 ,t ms= 2.5 1.25 ,2

V V= = and 0.05 .C pF= Then

3

12

11

(1.5 10 )1.25 2.5(0.05 10 )

4.17 10 41.7

I

I A I pA

−

−

−

×= − ⇒×

= × ⇒ =