Chapter 9 Exercise Solutions EX9.1

( )

2

1

1 2 1

15

20 K 15 15 20 300 K

v

i

RAR

R R R R

= − = −

= = ⇒ = = =

EX9.2

3 32

1 4 1

1 1

2 3

3

4

3 3

4 4 4

4

We can write 1

10 k

Want 50 Set 50 k

50 5 1 5

501 9 8

6.25 k

CL

CL

CL

R RRAR R R

R R

A R R

RAR

R RR R R

R

⎛ ⎞= − + −⎜ ⎟

⎝ ⎠= = Ω

= − = = Ω

⎛ ⎞= − = − + −⎜ ⎟

⎝ ⎠

+ = ⇒ = =

= Ω

EX9.3

We have 2

1 2

1

111 1

CL

d

RAR R

A R

= − ⋅⎡ ⎤⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

1 25 kiR R= = Ω Let 2

1

R xR

=

( )

( )

3

3

3

3

2

2

1211 1

5 10

1.00025 10

12 1.00025 10

12.0024 2.4 10

12.0024 12.03130.9976 25 k300.78 k

x

x

xx

x x

x x

Rx

R

−

− = −+ +

×−=+

×⎛ ⎞+ =⎜ ⎟×⎝ ⎠

= − ×

= = =Ω

= Ω

EX9.4

0 1 2 3 41 2 3 4

F F F FI I I I

R R R Rv v v v VR R R R

⎛ ⎞= − + + +⎜ ⎟

⎝ ⎠

We need 1 2 3 4

7, 14, 3.5, 10F F F FR R R RR R R R

= = = =

Set 280 kFR = Ω

Then 1

2

3

4

280 40 k7

280 20 k14280 80 k3.5280 28 k10

R

R

R

R

= = Ω

= = Ω

= = Ω

= = Ω

EX9.5

We may note that 3

2

3 21.5

RR

= = and 1

20 210

FRR

= = so that 3

2 1

FR RR R

=

( )

( )( )

( ) ( )

2

3

42

3 4

3 30 3 3 0

Then3

2 mA1.5 k

2 10 200 0.4 V

0.4 0.267 mA1.5 k

0.267 2 2.267 mA

2.267 10 3 10 0.4 7.2 V

IL L

L L L

L

L

L

vi iR

v i Z

viR

i i i

v i R v v

−

−

− −−= = ⇒ =Ω

= = × =

= = =Ω

= + = + =

= + = × × − ⇒ =

EX9.6 Refer to Fig. 9.24

1 12 5 kR R= = Ω Let 1 3 2.5 kR R= = Ω

Set 2 4R R=

0 2 22 4

1 1

Differential Gain 100 250 k2.5 k

v R R R Rv R

= = = = ⇒ = = ΩΩ

EX9.7

[ ][ ]

[ ][ ]

4 32 20 2 1

1 4 3 1

1 3 2 4

0 2 1

0 2

We have the general relation that

/1

1 /

10 k , 20 k , 21 k

21/1020 20110 1 21/10 10

2.0323 2.0

I I

I I

I I

R RR Rv v vR R R R

R R R R

v v v

v v v

⎛ ⎞⎛ ⎞= + −⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠= = Ω = Ω = Ω

⎛ ⎞⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+⎝ ⎠ ⎝ ⎠⎝ ⎠= −

a. 1 2

0 0

1, 12.0323 2.0 4.032 V

I Iv vv v

= = −= − − ⇒ = −

b. 1 2

0 0

1 V2.0323 2.0 0.0323 V

I Iv vv v

= == − ⇒ =

c. 1 2cm I Iv v v= = so common-mode gain

0 0.0323cmcm

vAv

= =

d.

( )

10

10

20 log

2.0323 12.0 2.0162 2

2.01620log 35.9 d B0.0323

ddB

cm

d

dB

AC M R RA

A

C M R R

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞= − − =⎜ ⎟⎝ ⎠⎛ ⎞= =⎜ ⎟⎝ ⎠

EX9.8

( )4 20 1 2

3 1

21 I IR Rv v vR R

⎛ ⎞= − + −⎜ ⎟

⎝ ⎠

4 2

3 1

2Differential gain (magnitude) 1R RR R

⎛ ⎞= +⎜ ⎟

⎝ ⎠

Minimum Gain ⇒ Maximum 1 1 50 51 kR = + = Ω

So ( )2 10020 1 4.92

20 51d dA A⎛ ⎞

= + ⇒ =⎜ ⎟⎝ ⎠

Maximum Gain ⇒ Minimum 1 1 kR = Ω

( )2 10020 1 20120 1d dA A

⎛ ⎞= + ⇒ =⎜ ⎟

⎝ ⎠

Range of Differential Gain 4.92 201= − EX9.9

( )( )4 61 2Time constant 10 0.1 10

1 m sec

r R C −= = = ×

=

01 2

10 1t v tR C

−≤ ≤ ⇒ = ×

0At 1 m sec 1 Vt v= ⇒ = −

( )01 2

10 2 1 1t v tR C

≤ ≤ ⇒ = − + × −

( )0

2 1At 2 m sec 1 0

1t v

−= ⇒ = − + =

EX9.10

0 1 2 3 410 25 80I I I Iv v v v v= + − − From Figure 9.40, v13 input to R1, vI4 input to R2, vI1 input to RA, and vI2 input to RB. From Equation (9.94)

1

25FRR

= and 2

80FRR

=

Set 500 k ,FR = Ω then 1 20 k ,R = Ω and 2 6.25 k .R = Ω

Also 1 1F P

N A

R RR R

⎛ ⎞⎛ ⎞+ =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ and 1 10F P

N B

R RR R

⎛ ⎞⎛ ⎞+ =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

where 1 2 20 6.25 4.76 kNR R R= = = Ω and P A B CR R R R=

We find that 10A

B

RR

=

Let 200 k , 20 kA BR R= Ω = Ω

Now ( )5001 1 1064.76 200

P P

A

R RR

⎛ ⎞ ⎛ ⎞⎛ ⎞+ = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Then 1.89 kPR = Ω 200 20 18.2 kA BR R = = Ω

So 18.21.89 2.11 k18.2

CP C

C

RR RR

= = ⇒ = Ω+

EX9.11 Computer Analysis TYU9.1

2

1

1 11

2 1

1

100 k 1010 k

0.25 V 2.5 V

0.25 0.025 mA 25 A10 k

25 A

10 k

CL CL

I o

I

i

RA AR

v v

vi iR

i i

R R

μ

μ

− Ω= − = ⇒ = −Ω

= ⇒ = −

= = = ⇒ =Ω

= =

= = Ω

TYU9.2 (a)

( )

( )

2

1

100min 4.92619 1.3

100max 5.07619 0.7

so 4.926 5.076

vS

v

v

v

RAR R

A

A

A

−=

+−= = −

+−= = −

+≤ ≤

(b)

( )

( )

1

1

1

0.1max 5.076 19 0.7

0.1min 4.926 19 1.3

so 4.926 5.076

i A

i A

i A

μ

μ

μ

= =+

= =+

≤ ≤

(c) Maximum current specification is violated. TYU9.3

( ) 30 2 1 10d dv A v v A= − =

a.

2 0

01 13

0, 55 5 mV

10d

v vvv vA

= =

= − = − ⇒ = −

b. 1 0

02 1

2 23

5, 10

10 5 4.99 V10

d

v vv v vA

v v

= = −

= −

− = − ⇒ =

c.

( )1 2

30

0

0.001, 0.00110 0.001 0.001

2 V

v vvv

= = −= − −= −

d.

( )2 0

0 2 1

02 1

1 13

3, 3

3 3 2.997 V10

d

d

v vv A v vv v vA

v v

= == −

= −

= − ⇒ =

TYU9.4

( ) ( ) ( )

[ ]

4 4 40 1 2 3

1 2 3

0

0

0

40 40 40250 200 7510 20 30

1000 400 1001500 V 1.5 mV

I I IR R Rv v v vR R R

v

vv μ

⎛ ⎞= − + +⎜ ⎟

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

= − + += − = −

TYU9.5

( )1 2 31 2 3

1 2 3

31 1 3

1Then 333 3

I I I FO I I I

F

F

v v v Rv v v vR

R R R R R MR

R M k

+ += = + +

= ⇒ = = ≡ = Ω

= Ω = Ω

TYU9.6

0 2

1

1 5vI

v RAv R

⎛ ⎞= = + =⎜ ⎟

⎝ ⎠ so that 2

1

4RR

=

For 0 10 V, 2 VIv v= =

Then 1 11

2 50 A 40 ki RR

μ= = ⇒ = Ω

Then 2 160 kR = Ω we find 02

2

10 2 50 A160

Iv viR

μ− −= = =

TYU9.7

( ) ( )0 2 1 1

0 01 1

0 111 2 2

1 2

or

and

od od I

I Iod od

v A v v A v vv vv v v vA A

v vvi i iR R

= − = −

− = − = −

−= = =

0 11

1 2

01

1 2 2

02 20 1

1 1

1Then

1 1

1 1 Iod

v vvR R

vvR R R

vR Rv v vR R A

⎛ ⎞ −=⎜ ⎟

⎝ ⎠⎛ ⎞

+ =⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

2

10

2

1

1So

11 1v

I

od

RRvA

v RA R

⎛ ⎞+⎜ ⎟

⎝ ⎠= =⎛ ⎞

+ +⎜ ⎟⎝ ⎠

TYU9.8

( )

( )

2 2 1

20 1 1

1

1

1

1 2 2

20 2 2

1

2

2

For 0, and

1

70 5015 50 25

10

For 0,

1

70 2515 25 50

5

bI I

b a

bI I

b a

I

I

aI I

b a

aI I

b a

I

I

Rv v vR R

RRv v vR R R

v

v

Rv v vR R

RRv v vR R R

v

v

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

=

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

=

( ) ( )0 0 1 0 2

0 1 2

Then

10 5I I

I I

v v v v vv v v

= += +

TYU9.9

S iR R so 1 2 100 ASi i i μ= = =

0 S Fv i R= −

We want ( )610 100 10 100 kF FR R−− = − × ⇒ = Ω TYU9.10 We want 1 mALi = when 5 VIv = −

( )

( )( )

2 232 2

3

4 42

3 4

55 k

10

10 500 0.5 V

0.5 0.1 mA5 kΩ

0.1 1 1.1 mA

I IL

L L L L

L

L

V vi R RR i

v i Z v

vi iR

i i i

−

−

− −− −= ⇒ = = ⇒ = Ω

= = ⇒ =

= = ⇒ =

= + = + =

If op-amp is biased at 10 V,± output must be limited to 8 V.≈

( )0 3 3

33 3

So

8 1.1 10 0.5 6.82 kLv i R v

R R−

= +

= × + ⇒ = Ω

Let 3 7.0 kR = Ω

Then we want 3

2 1

7 1.45

FR RR R

= = =

Can choose 1 10 kR = Ω and 14 kFR = Ω TYU9.11 a.

( )

[ ]

( ) ( )

( ) ( )

1 21

1

401 1 1 2 02 2 1 2 0 02 01

3

40 2 1 2 1 1 2

3

40 2 1 1 2 2

3

4 2 10 2 1 2 2

3 1

, and

I I

I I

I I

I I

I II I

v viR

Rv v i R v v i R v v vR

Rv v i R v i RRRv v v i R RR

R v vv v v R RR R

−=

′= + = − = −

′= − − −

′= − − +⎡ ⎤⎣ ⎦

⎡ ⎤⎛ ⎞− ′= − − +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

For common-mode input 2 1 0 0 Common Gain 0, I Iv v v C M R R= ⇒ = ⇒ = = ∞

b. ( ) 2 1min min, maxdA R R′⇒

( )

20 100 951 4.8220 5120 100 105max 1 20620 1

d

d

A

A

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

c.

0

d

cm

cm

AC M R RA

A C M R R

=

= ⇒ = ∞

TYU9.12

4 2

3 1

2Differential Gain 1R RR R

⎛ ⎞= +⎜ ⎟

⎝ ⎠

Let R3 = R4 so the difference amplifier gain is unity. Minimum Gain ⇒ Maximum R1

So ( )

2

1

21 2maxR

R⎛ ⎞

+ =⎜ ⎟⎜ ⎟⎝ ⎠

We want ( )2 12 maxR R= Maximum Gain ⇒ Minimum R1

So ( ) ( )2

2 11

21 1000 or 2 999 minminR R R

R⎛ ⎞

+ = =⎜ ⎟⎜ ⎟⎝ ⎠

If 2 50 k ,R = Ω let ( )1 min 100 R = Ω fixed resistor and let ( )1pot

max 100 k 100 100.1R = Ω + Ω =

Then actual differential gain is in the range of 1.999 1001− TYU9.13

End of 1st pulse: 6

10 sec0 0

1 10 10v tr r

μ−− − ×= × =

After 10 pulses: ( ) ( )6

0

10 10 105v

r

−− ×= − =

6

61 2

100 10So 20 sec520 sec 20 10

r r

r R C

μ

μ

−

−

×= = =

= = = ×

For example, 62 10.01 10 0.01 F 2 kC Rμ−= × = ⇒ = Ω

TYU9.14

( ) ( ) ( ) ( )01

01

2 2

2

R R R Rv VR R R R R R R R

R R R R VR R

R R R R VR

Rv VR

+

+

+

+

⎡ ⎤− Δ + Δ= −⎢ ⎥− Δ + + Δ + Δ + − Δ⎢ ⎥⎣ ⎦− Δ + Δ⎡ ⎤= −⎢ ⎥⎣ ⎦− Δ − − Δ⎛ ⎞= ⎜ ⎟

⎝ ⎠Δ⎛ ⎞= −⎜ ⎟

⎝ ⎠

For 33.5 V, 50, 10 10V R R+ = Δ = = ×

( ) 201 4

50 3.5 1.75 1010

v −⎛ ⎞= − = − ×⎜ ⎟⎝ ⎠

Need an amplifier with a gain of 02

5 285.71.75 10d d

i

vA Av −= = ⇒ = −

− ×

Use instrumentation amplifier, Fig. 9-25. Connect v01 to vI1 and ( )01v− to vI2.

4 2

3 1

21 285.7dR RAR R

⎛ ⎞⎛ ⎞= + =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Let 4 3150 k , 10 kR R= Ω = Ω Then 2

1

9.02RR

=

Let 2 1100 k , 11.1 kR R= Ω = Ω TYU9.15

( )( )

( )( )

( )

( )

01

01

01

12 1

1 22 1

2 2

45 For 0.01

0.01 5 0.01254

Rv VR R

R R RV

R R

R VR

v V

V

v

δ

δδ

δδ

δ

δ

+

+

+

+

+

⎡ ⎤= −⎢ ⎥+ +⎢ ⎥⎣ ⎦⎡ ⎤+ + −

= ⎢ ⎥+ +⎢ ⎥⎣ ⎦

= ×+

⎛ ⎞≈ ⎜ ⎟⎝ ⎠

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

Need a gain 0

01

5 4000.0125d

vAv

= = =

Use an instrumentation amplifier 4 2

3 1

2400 1dR RAR R

⎛ ⎞⎛ ⎞= = +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Let 4 3150 k , 10 kR R= Ω = Ω then 2

1

12.8RR

=

Let 2 1150 k , 11.7 kR R= Ω = Ω