Chapter 14 Exercise Solutions EX14.1

a. ( )4

50 49.94911 51

5 10

CL CLA A−= ⇒ = −⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥×⎝ ⎠⎣ ⎦

b. 4

4

5110 0.0102%5 10

50 49.943511

4.5 10

CL CL

CL CL

CL CL

dA dAA A

A A

= × ⇒ =×

−= ⇒ = −⎡ ⎤+⎢ ⎥×⎣ ⎦

EX14.2 a. For 0 0R =

( )4 3

3

1 1 1 1 10 0.1 1010 10

10 k 1i f

i f

R

R −

= + + = +

⇒ = Ω = Ω

b. For 0 10 kR = Ω

( )4 4

3

1 1 1 1 10 1 100.110 10 1 1 1 3 10

3 10 k 3i f

if i f

R

R R−

⎡ ⎤+ += + × ≅ +⎢ ⎥+ +⎣ ⎦= × Ω ⇒ = Ω

EX14.3

( )4

5 3

3

4040 1 10 99 11

9911

4 10 4.059 10100

4.04 10 k 4.04 M

i f

i f i f

R

R R

⎛ ⎞+ + +⎜ ⎟⎝ ⎠=

+

× + ×≅

= × Ω ⇒ = Ω

EX14.4

2

1

1 100RR

+ =

a. 5

00

1 1 10 10 0.1 100 100 f

f

RR

⎡ ⎤= = ⇒ = Ω⎢ ⎥

⎣ ⎦

b. 5

2

0

1 1 10 1010 100fR

⎡ ⎤= =⎢ ⎥

⎣ ⎦

20 010 k 10 f fR R−= Ω ⇒ = Ω

EX14.5 From Equation (14.43)

( )( )

( )( )

0

0 0

44

1

25 25

1 12 1050 10 25

CLCL

PD CL

AA ffj

f A A

f fj j

=+ ⋅

/

= =+ ⋅ + ⋅

×/

a. 2 kHzf =

( )0025 peak 1.25 mV

I

v vv

= ⇒ =

b. 20 kHzf =

( )00

1 25 peak 0.884 mV2I

v vv

= ⋅ ⇒ =

c. 100 kHzf =

( )0

2

0

25 25 4.905.0991 100 / 20

0.245 mV

I

vv

v

= = =+

⇒ =

EX14.6 Full-scale response 1 5 5 V= × =

5 2.5 s2

t t μ= ⇒ =

EX14.7

a. ( ) ( )

6

05

0.63 102 max 2 11.0 10 100 kHz

S RFPBWV

FPBW FPBWπ π

×= =

= × ⇒ =

b. ( )

640.63 10 1.0 10

2 1010 kHz

FPBW

FPBWπ

×= = ×

⇒ =

EX14.8

142

0 141

0

1.85 10ln (0.026) ln2 10

2.03 mV

SS T

S

S

IV VI

V

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠⇒ =

EX14.9 We need

1 2 3 4 1 2, 0.6 V, and 10 VC C EC EC CE CEi i v v v v= = = = = By Equation (14.60(a))

11 1

33

10exp 150

0.6exp 150

BEC S

T

EBS

T

vi IV

vI

V

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

By Equation (14.60(b))

22 2

44

1 2,

10exp 150

0.6exp 150

take the ratio:

BEC S

T

EBS

T

S S

vi IV

vIV

I I

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

=

31 2

4

exp SBE BE

T S

Iv vV I

⎛ ⎞−=⎜ ⎟

⎝ ⎠

( )

31 2 0

4

0

ln

0.026 ln 1.051.27 m V

SBE BE S T

S

S

Iv v V V

I

V

⎛ ⎞− = = ⎜ ⎟

⎝ ⎠= ⋅⇒ =

EX14.10

( )2

12 2

1 1500.0202 2 50 50

1.63 /

1.63 3.26%50

Q nOS

n n

n

n

n

n

I KVK K

K

K A V

KK

μ

⎛ ⎞Δ= ⋅ ⋅⎜ ⎟

⎝ ⎠

Δ⎛ ⎞= ⋅ ⋅⎜ ⎟⎝ ⎠

⇒ Δ =

Δ⇒ = ⇒

EX14.11

Want 5

5 4

5 m VR

VR R

+⎛ ⎞=⎜ ⎟+⎝ ⎠

55 4

4

so 0.005RR R VR

+× =

( )( )5

5

0.005 1000.05 k

1050

R

R

= = Ω

⇒ = Ω

EX14.12

1

2

25 1 0.9615 k75 1 0.9868 k

RR

′ = = Ω′ = = Ω

For 1 2100 A 50 AQ C CI i iμ μ= ⇒ = = From Equation (14.75)

( ) ( )( )

( ) ( )( )

( )

6

14

2

4

2

4

50 100.026 ln 0.050 0.961510

0.026 ln 0.050 0.9868

0.58065 0.048075

0.026 ln 0.04934

C

S

C

S

iI

iI

−

−

⎛ ⎞× +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟

⎝ ⎠+

⎛ ⎞= +⎜ ⎟

⎝ ⎠

2

4

69

414

4

ln 22.284

50 10 4.7625 10

1.05 10 A

C

S

S

S

iI

II

−

−

⎛ ⎞=⎜ ⎟

⎝ ⎠× = ×

≅ ×

EX14.13 From Equation (14.79)

20 1 2 2 3

1

1B BRv I R I RR

⎛ ⎞= − +⎜ ⎟

⎝ ⎠

For 0 0v =

( )( ) ( )( ) ( )( )

6 63

3 3

1000 1.1 10 100 k 1.0 10 110

11 1.1 100 k 10 k

R

R R

− − ⎛ ⎞= × Ω − × +⎜ ⎟⎝ ⎠

= Ω ⇒ = Ω

TYU14.1

( ) ( )( ) ( ) ( )

1 1 1

1 4 1 1

maxmin

CM SD SG

CM DS SD SG

v V V sat Vv V V sat V sat V

+

−

= − −= + + −

We have: 2 2100 , 80 / , 40 / ,

25

REF n pI A k A V k A VWL

μ μ μ′ ′= = =

⎛ ⎞ =⎜ ⎟⎝ ⎠

For 1 :M

( )( )21

4050 252D SG TPI V V⎛ ⎞= = +⎜ ⎟

⎝ ⎠

So ( )21 150 500 0.5 0.816 SG SGV V V= − ⇒ =

( )1 0.816 0.5 0.316 SDV sat V= − = Then

( )max 0.316 0.816 1.13CMv V V V+ += − − = − For 4 :M

( ) ( )24

80100 252D GS TNI V V⎛ ⎞= = −⎜ ⎟

⎝ ⎠

So ( )24 4100 1000 0.5 0.816 GS GSV V V= − ⇒ =

( )( )

4 0.816 0.5 0.316min 0.316 0.316 0.816 0.184

DS

CM

V sat VV V V− −

= − == + + − = −

So 0.184 1.13 CMV v V V− +− ≤ ≤ −

TYU14.2

( ) ( )( ) ( ) ( )

8 10

4 6

maxmin

o SD SG

o DS DS

v V V sat Vv V V sat V sat

+

−

= − −= + +

Now

( )( )( ) ( )

8 10

8 10

50 0.5 0.816 40 2 25

0.316

SG SG

SD SD

V V V

V sat V sat V

= = + =/

= =

So ( )max 0.316 0.816 1.13ov V V+ += − − = − Also

( )( )

( )( )

6

4

50 0.5 0.72480 2 25

100 0.5 0.81680 2 25

GS

GS

V V

V V

= + =/

= + =/

( )( )

6

4

0.724 0.5 0.224 0.816 0.5 0.316

DS

DS

V sat VV sat V

= − == − =

So ( )min 0.316 0.224 0.54ov V V− −= + + = +

Then 0.54 1.13oV v V V− ++ ≤ ≤ −

TYU14.3

( ) 500ideal 2520CLA = − = −

Within ( )( )0.1% 25 0.001 25 24.975CLA⇒ − + ⇒ = −

0

2524.975261

LA

−− =⎡ ⎤

+⎢ ⎥⎣ ⎦

0

0

26 25 1 0.001024.975

25.974L

L

AA

−= − =−

=

TYU14.4

a. ( )

( )0

1

CLCL

CL

L

AA

AA

∞=

∞⎡ ⎤+ ⎢ ⎥⎣ ⎦

( ) 2

1

4951 1 1005CL

RAR

∞ = + = + =

( )5

100 99.90100110

100

CL CL

CL

A A

A

= ⇒ =+

∞ =

b. 5

10010 0.01%10

CL

CL

dAA

= × =

( )( )99.90 0.0001 99.9099.89

CL

CL

AA

= −⇒ =

TYU14.5

( )( ) ( ) ( )

( )

( )

( )

( )

( )

( ) ( )

0

0 0

0 0

0

40

11 11

1 1So 0.001

1 1

0.001 0.999

0.001 0.001 (10 ) 10.0100.999 0.999

CL CL CL

CLCL CL

L

CL CL

L L

CL CL

L L

CL

L

CL L CL

A A AAA A

AA A

A AA A

A AA

A

A A A

∞ −= − = −

∞∞ ∞+

∞ ∞+ −

= =∞ ∞

+ +

∞= ⋅

∞ = ⋅ = ⋅ ⇒ ∞ =

or ( )( )1 0.001 10.010 10.0CL CLA A= − ⇒ =

TYU14.6

1

ifi

i

RiI R

⎛ ⎞= ⎜ ⎟⎝ ⎠

a. 54

1

0.1 1 1010

Iii

−= = ×

b. 34

1

10 1 1010

Iii

−= = ×

TYU14.7 Voltage follower 2 10, R R= = ∞

( ) ( )50

6

1 10 1 5 10

5 10 k 5000 Mif i L

if

R R A

R

= + = + ×

≅ × Ω ⇒ = Ω

TYU14.8

( )( )

( )

5

30

max 30

10 1020 kHz

50

2 max

TdB

CL

dB

ffA

S Rf fVπ

= = ⇒

= =

( ) ( )( )

6

0 33

0

0.8 10max2 2 20 10

max 6.37 VdB

S RVf

V

π π×= =

×

⇒ =

TYU14.9 a. ( )( )6 3

0 1 3

0

10 200 10

0.20 VBv I R

v

−= = ×

⇒ =

b. 4 1 2 3

4

100 50 200

28.6 k

R R R R

R

= =

⇒ = Ω