Chapter l-l

l.1 aε)1 = !: 10V: tο.a' R IkΩ(blR: y: 10V : lοιΩ' Ι 1m,4

(c' v _ ΙR: 10mΑXlokΩ: lωv(d)1:Y=loY:o.raR l(π) ll,ryor..'volts, miΙliaπps, and kilo-ohms constitute aconsistent set of uΠits.

1.2(a\ P : /'?R : (3oΧ 10 312x 1 x 10r

:0.9wThus, l? should haνe a l-w

'ating.(b)Ρ : /?R : (4οX ιo_])2X 1 Χ 1ο3

= I.6 ΨThus. ιhe resisιor should haνe a 2-W raιing.

(c) P = Ι2R: (3Xlo-3)2XιoXlor= 0.09 W

Thus. ιhe resisιor should have a | -W rating.8

(ΦP : /'?R : (4X 10'1'x lox 10':0.16w

Thus' the resistor shoυld have a ] -W raιing.4

(e') P = v2/R = 2ox/(|Χ lο3): o.4w

ThuS, lhe resistor shoυ|d haνe a | _W raling.2

σ' P = v2/R = 1ι2l(ι X 1011 :6.12r,Thus, a raιing ot I w should ιheoreaically Suftice-8

ιhoush ] W ψould be Drudenι ιo aιιo\,ν t'or con_"4sisιent toterances and measurement erτors.

r3(a) y = ΙR = l0mΑx ι kΩ = 10V

P: Ι2R = (lo mΑ)'?Χ l kΩ = 10OmW

Φ)R: v/Ι: l0 V/1 mA: lokΩP=VΙ= 10VX1mΑ:10mΨ(c)Ι: P/V = l W/Ι0v=0.l ΑR: V/Ι = l0 V/o.lA: lωο(d)v : P/Ι : 0.1 W10mA

= 100 mW10 mA = 10 VR: v/Ι: l0V/lOmΑ= l kΩ

(e)P: Ι7R=>Ι : JP/R1 : Jloω mwl kΩ : 31.6 mΑ

V : ΙR : 31.6mΑΧ 1 kο = 31.6vΝol?.' ν, mΑ. kΩ, and mW consιifuιe a cοnsis-tent set οf units.

1.4Thus, theΙe are ι7 possibιe resistance νalues.

1.5 shunting ιhe l0 kΩ by a resistor ofvalue ofi result in the combination having a resistance

R,q'

R : lOR'e R+t0

Thus,fo.a l reduction,

R = o.99+R:990kΩR+ l0For a 57a reduction,

R : o.95=R: lΦkf}R+ 10

For a l 0ρlo reduction,

--&- : o.qο= n : qoιΩR+ l0For a 5Φlo reduοtion,

R : o-50=R: tokΩR+ to

shunting the !0 kΩ by

(a) tMΩ τesυlt in

R-- _ !9,Ι lΨ = -]9 _ s.sιο.'l lαn + l0 l.0la ιolo r€duction;

(b) lω kΩ resυlts in

R-' : l0 Χ ιω = -!9 : q.ω ιο." Iα) + I0 l_l

a 9.19lo reduction;(c) 10 kο Iesults in

R-' : l0 : _5 kΩ.'{ 1ο+lοa 507o reduction.

t.6R2vo _ v DDl|ττ,

To find Rρ, we short circuit VDp and look back

inιo nωe X,

R^ : R. II R, : RIRZ

Rr+R2

1.7 Use νolωge divider to find y,

V:9 .3'3 = 2g4Υ" l.:l + 6.8

Equivalent ouΦut resistance Ro is

R, = (3.3 kΩ l| 6.8 kΩ) : 2.22 kΩ

Τhe extreme values of y. for 15 o/o toleτance

τesistor are

9 3.3( I - 0.05)3.3(1 - 0.05) + 6.8(r + 0.05)

2.'l3 ν

chapιer 1_2

This figure belongs to 1.4

lοq/v\_ο20

"--tts40

*_{ηΓ-Ξl0

ιo 20o-lWΗΛΛ-=o

r0 40o-ΔWΗΛΛr ο

o-_1λΗΛΛr_{

10 20 40*_1Λ^r1ilΗΛΛr-

l0

20

4ο 30

5ο

60

'Ιo

4020

lο

20Γ____1Λλ--------r

*_{ lο 4ο μηΛ^r-rΛΛr]

40

Γ--a\ι---lηlo20 l'--'-oq^^Γ__η^^r-l

23.3

46.7

-__ΔM-r"_+ 2o |- o ε.τ

-1rγV_r"__+ 40 }-----ο ε .o

Ψ$rllo20

-1ΛΛ.--l -_t\ι--------r"-+ 40 l-'- l].:ι *{ zo 40 μ 8.6

Ψ$r-_l qΛΛr-lΛΛ;

5.7 14.1

28

l7.l

-1.3( I + 0.05)3.3(l + 0.05) + 6-8(r - 0.05)

1.7 continued here

+9V

t68kΩ *

L,----------.,

I

"*t IΞlRo

: 3.l4 νTh€ extτ€me valu€s of Ro for x5o/o

toleranc€ resistors are

. .]..1ι l _ o.o5) Χ 6.s(l _ o.o5)rιo mlπ _ .1J(| 005) re.s(l _ 5)

: 2t1kΩ.r..1( l - 0.ο5) X 6.8(I + 0.05)

'ιo mJι _ 3

'(| , 0J5) l 63(|

: 2.33 kΩ

+9V

(a)

1.8

10 kΩ

+9Vt

I

tokΩtI

l-:o'u'ooοf ι

+ Ro='tott to

=5kΩΦ)

Voltage generded+3v (tι{o vays: (a) and (c) wiιh (c) haνing lower

ouιpυt resisbnce)+4.5 V (b)

+6V (two $ays: (a) and (d) with (d) having a lorveroutput resistance)

t.9

l0 kΩ

Y^:15 l0 :10.2V" lo + 4.'Ι

To reduce y, to l0.00 v we shunt the ι0-kΩresistor by a r€sistor R whose value is such that

l0 ι| R: 2X4.7.

Chapter I --l

+6VRo=20lΙ l0 kΩ

= 6.7 kΩ

+3Vfo= 10 kΩ // 20 kΩ

= 6.67 kΩ 10

10 kΩ

kΩ

+3VRo= lO ll lO ll lO

= 3.33 kO

10 kΩ

(c)

+9V

10 kο 10 kο

1ο kΩ

Ro= lo ll |o ΙΙ |0

= 3.33 kο

4.7 kΩ

lo.ωv

10 kΩ

Thus

tttl0 R 9.4

=eR: 156.7-1.57kf)Now,

Ro: lokΩ I| R 11 4.7kΩ

: g.4 ll 4.7 : 94 = :.t:: ιο"lTo make ζ = 3.33 ιve add a s€ries resisbnce ofappΙoxim*b 2ω Ω , as shown.

(d)

15 vt

j:., -'

r_= ''

t=

+9V

+15 v

+t5v

chapιer l-4

l57 kΩ

To obιain y,, : 10.ω y and Ro : 3 kΩ

we have to shunι boιh the 4.7_kΩ and the lο_

kΩ resistoπ as shoιγn. To yield an ouιpuι

voltage y., : lο.ωv we musι haνe

v = /(Rr ll Rr)

. Rι f:- 'R, +R,

1, : V : 1 R'' f, Rr +R:

l. : JL: ι-l'' R, Rr +R2

l'Ι 1 connecι a rΘsistor R in paralIel wiιh R..To make /. = ο.21 (and ιhus ιhe current

ιhroυgh R' 0.8/ ), R shoυld be sυch

0.2/xIkΩ:0.8/R9R : 25o kΩ

(R ll t0) :

R: 2R, (l)

R2

2(Rt ll 4.'1)

R;

l.l2

4.7 kΩ

,a-1

For Ro : 3 ιο ," rnlrtιuu"

R; ll R; :3Solving (l) and (2) yields

l?l : 4.5 kΩ

R; : e.o kcr

Ro

which caη be ιιsed to find R| and R2

respectively,

Rl : l57 kΩ

R::90kΩ

Ι -l0

To make the current through R equal to 1/3 we

shuntRbyaresistance RI ofνalue such that the

current through it will be 2rl3; thus

ln = 4n' -ρ' = E3 .t' ' 2

Τhe inpul resislance ofthe divider. Ri" . is

R,=Rlln =r11 f :lnNow if R, is Ι0% too high, i.e.,

R, = llE'2the problem can be solved in tιι'o ψays:

(a) Connect a resistor R, across R, ofvalue

suchtbat R2 ll R, = R z2,thus

R2(l.l R / 2) _ RR2+ (t.tR /2) 2

t.tn.: π.+!82

(2)

ιι=o.zΙΙ

+ ι5 v

Chapter 1-5

+R, : ll4 : 5.5 p

R,:Rll ΨιΨ:*lf:ξ

\+-iz

(b) conne.t a resistor in series Ψith the load resis_tor R so as ιo raise the resistance of ιhe loadbraηch by l0%' thereby Ιestoring the current diνi-sion ratio to its desired value. The added seriesresistance must te |(9ο of Rile.'o.|R.

Rr, = 1.1R 1l

l.1R3

i/e., 107o higher than iη cas€ (a).

1.t3For R. : l0 kΩ' ιvhen signal source generates

ο_ l mΑ, a voltage of 0 _l0 V may appeaτ acrossthe source

Rι= 1o kΩ

To limit vs s ι v, thο Ιet resistance has to b€

= 1 kΩ. To aοhieve this \γ€ have 1o shunt R.with a resistor R so that (R || R.) < 1 kΩ.

R ll n.- 1 111.

RR. =lkΩR+R.

ForR.:1614R3l.ll kΩΤhe resuΙting ciιοuit neωs only on€ additional

r€sistanc€ of 1.1l kΩ in parall€l lvith R. so that

yr< I v1-11

1.5 V

tR,,

l.tR2

O-1mA

Rι= l0 kΩ0 lmΑ

2ΙΙ3

5Rτιl

0.5 kο

Chapter 1-6

v.,::(--!L) = rsv

RΙ, = l k]| 1k : 0.5ksame procedυιe is used for b) & c)

, o.176. t5 + 1.5

: 0.l mΑ

1.16 (a) Νode equation at the common mod€yields

Using ιhe facι that ιhe sum ofthe voltage dιops

across Rl and R1 equaIs 15 ν, we write

15 : / 1i1 +/1R1

:loΙ|+U1+Ι)x2: |2Ι l + 2Ι1

+ t5 v

+lονRllο kΩ

f,

That is,

12Ι|+2Ι2: |5 (ι)

similarly, the νoltage drops across R2 and R1 add

υp to 10 Y thus

l0: /2R2+11R1

= 5Ι2+ υ| + ι)x2which yieιds

2Ι1+',ΙΙ1 :10 (2)

Equations ( ι) and (2) can be solνω together bymlltiplying (2) by 6, '

12Ι|+42Ι2:6J (3)

Νow, subtracting (1) from (3) yields

40Ι.. : 45

Ξ/, : l.l25 mA

subsιitutini in (2) giνes

2I| = |o_ 7 X 1.125 mΑ

Ξ 1I = l.0625 mΑ

Ι1 : ι1+ ι2: 1.0625 + t. t250: 1.l875 mΑ

: 1.1875X2:2.3'150ν

u

Thdvenin equival€nt: (10//16) = 6.l5 kΩ 4

2x !Lι0+16

=o'77 ν

o.71 ν |.5 kΩ

Now' \ir'hen a resisιance of l.5 kΩ is connecιed

between 4 and ground,

Γ-_{v\r-'-----ΙΙ

2kΩ

l.l5

(b)

10 kΩ

Ι*'

(c)

10 kf)l0 kΩ

1ο kΩ

Ι

16 γ _JLl0tlο

0.5 kΩ

i ξ9 z '9.k9

6-|5kΩ

chapter ι_7

To sυmmarize:1,:1.06mΑ /,3 1.13mΑ/,:1.19m4 v

= 2.38ν

(b) A node equaιion at the common nωe can bewritten in terms of Y as

15-v+ l0-y _Rr

Thus,

R2

15 y'Ι0 v _105Ξ0.8v:3.5+v :2.3'15ν

Now /1, 12, and 13 can be easiΙy found as

, 15-V 15-2,375r':--Τδ-: |o

= 1.0625 mΑ a l.06 mΑ

, 10-v t0 -2-375'55: 1.125 mΑ 3 l.13τnΑ

Ι' _ V = 1ΞΞ _ t.1375mΑ - l.Ι9mA'Rr2-Method (b) is mυch p.efeΙred; fasteι moΙeinsightful and less prone ιo errors. Ιn geneEΙ, oneattempts to identify the least possibιe numbe. ofvariables and \λ,riιe ιhe corresponding minimumnumber of equations.1,17Find Theνenin ηuiνaΙent circuit to the left ofnode 1

Bet'ween node ι and ground

R,,, = (lk|| 1.2k ):0.545kΩv.":g\ t2 : 4.909 V'" I + 1.2

Find Theνenin φuiνaιent circuit to the right ofnode 2

R:=9.1kΩ

Rι= l lkΩ

Between node 2 and ground

Rri : 9.1k || llk = 4.98 kΩ

v'':9x || : ι.gzsν'' 1l + 9.1

The resulting simplifiω circuit is

0.545 kΩΙ5

2 4.98 kΩ

4.9ο9 ν

vR3

vt

ΙΞ4.925 - 4.9(9

-vs +

(4.98+3+0.545)k: 1.88 μA

V5:1.88μΑX3k: 5.64 mV

1.1E From the symmetΙy ofthe circuit, there ψillbe no current in &. (otherwise the symmeιrywoυld be νiolated.) Thus each branch will carry a

cuπent yΙl2 kο and 1, will be the sum ofthe two

cuπent,

, 2V, V,' 2kΩ lkΩ

Thus.

vR.,=7j = lkΩ

No\r, if Ra is Γaised to 1.2 kο the symmetry wiιl

be broken. To fiηd ι ιγe use T't€νenin's theorem

as folloΨs:

1. : o'*5V'-o'SV, : 6.622γ' ο.5+ l +ο.545

vvt: i+0.022v,x0.5

R3

chapιer l_8

0.5 kΩ Vl

/R, \v'ι--R.. R,lv,t2

,-(

: 0.5V, Χ 1.022 : 0.5ΙlV.

V2: V,* 1.R' : 0.533V.

ν 'ν^/, = _ikΩ- = ο.489 yt

v _ ν^t,:' , : o.461V' l kΙιI' = I|+ Ι7 = o.956v,

v= R.= Τ: I.05 kΩ

1.19 (a) T : 10 ams = l0 7s

1: !:1y7 γ1.'T

ιι : 2τf : 6.28 x lo1 lΙz(b)/ : l GΗz: 1o9 Hz

r = 1 = lο 'q"

ι,l = 2τf : 6.28 X l09rad/s

(c)ω:6.28XlO2raιl/s

f : f;: ιo' rι.

r:l:ro?"f(d) T = los

1. : l : 16-ι11,"Tιι : 2ιtf : 6.28 Χ lο ιrad,/s

(e\ f : ωΗzT : !:1.67x ro 2s

ω : 2τf : 3.'17 x lo1 radιS

(Γ.1 ω: l krad,/s: l03 radls

f : ω : l.59Χ l02Ηz"2τ

1:1=6.2gy1gt"

(g) .f = 19ω MΗz : t.9 Χ l09 Ηz

7=l:6.526y16e3fω : 2τf : 1.194 x l09 radls

1.20 (a) Z : l kΩ at al| frequencies

(b)z:1/jωc = ;^ -:-|-,21τfΧ10xl0At/: ωΗz, z: _ j265 kdΣ

At/: lωkHz, Z: _ j|59 {'

^tf: lcΗz' z= j0.016Ω

ιcJz _ l/jωc _ i-= ! -

π21τfx2xl0

^tf = ωHa Z: jl.33GΩ

Atf : \oa.kΗz' Z: j0.8 MΩΑt/:1cHz, z: _ i79.6 dl

(d) Z = jωL: j2τfL : j2τf x |0x l0 \

Aι f : 60 Ηz' Z : j3.77 {ιΑt/:1o0kΗz, z = j6.28k{ιAt/:1GHz' z= j62.8 Ω

(e)Z: jιoL = 'i21τfL: j2τf(l x10 9)

f : ΦHz' 2 = j3.77 X lo'7 = jo.377 μΩ/: 1ω kΗz,

Z : j6'28 x10 a = ;0.628 mΩ

f = 16 Hz, Z : j62.8 t7

Ιx

uΙ2

υ

R1lkΩ

R4l-2kΩ

y, o.s+i iο

0.545 η

1.21 (a) Z

: lο3 +j2τx|ox

: (l j l.59) kο

ρ6γ : L+ i-cR"

z:

J--+ izr-x lo Χ lo] Y ο.0| X 10_6!o'

1o_3( 1 + jo.628) Ω

!= lω0Υ l + j0.628

lοοο( 1 _ j0.628)

I + 0.6282

: (7t7.2 j4s.O4) {t

ιcly=!+iωCR'= |

, + j2rr Χ lo X lο] Χ lω Xl0,ο X lο'

: lo 5(ι + jo.628)

- lοr/.=-I + j0.628

: (71.72 _ j450.4) k{ι(d'lZ: R+ j<ιL

: l0o+ j2τx 10Χ lo]x lox lo 3

: 1ω + j6.28 X 1ω: (lω + j628) Ω

1-22

NoatonEquiνaΙent

z5 : i5R5

Thus,

R,: Ψ(a) v, : vρc. : 10 V

iΙ: i!. = lω μΑ

R. - y9! - lOv =0.lMΩ_ lωklr' i.r. l00 μΑ(b)v5: τt. = 6.1 γi5: i5g: l0 μΑ

R._oo'- 0'lν - o.oιMΩ _ lokΙr" ir. l0 μA

1.?3

υo: Rιv. R1 *R5

"^:""2(r+&)" '\ Rr)

Thus,

vs :301+jΣ

1ωand

vs =10

l+&10

Diνiding (l) by (2) gives

! + (Rs/!0)t + (Rr/too)

: r

=R, = 23.6 1ρsubstifuting in (2) giνes

υ" : 38.6 mV

The Norton cuπent i, can be found as

;. = 's : 386 mV t-ra ,',' R. 28.6 ιΩ

1.24 The obseΓνed output voltage is 1 mv/"c\νhich iS one half the voltage specified by the sen-

soι presumably under open-circuit conditions ιhatis without a load connected. ιt follows that thatsensorintemal resistance must be equal to Rι, i.e.,

ι0 kΩ.1.25

&

i"+&υo

= R+ 1

jωC

1o3Χ10Xlo 9

Chapter 1-9

!o '2

-{Ιλ_.rloThdνenin

ηυiνalent

Chapter I 10

f=ι i"

tshort-circυit (1r, : 0) cuπent

t.26(l) 1.26(2)

Rl represenιs the inpυt resisιance of the processor

For υ, : 0.9 υ,

09 = Rt =R, = gR"

R/ +RsFor i., = 0.9 i.

0.9 = Rs +R, : R"/9. R{+ RΙ

1.27

ι9 : Q_ i)R,

Open-circuil(io : 0) --> 1,,

voltage

1.24

(a)Vo".ι='l0xrt(b) v.., = 33.9 x Jt(c) V o"'ι : 22ox rtG)v.,* = zzo x J1

t -29

(a) , : 10 sin (2r-X loar)' ν(b) v : l2O"lsin (2ιτ x Φ)' Υ(c) v: 0.l sin(Ιωor), v(d) v : 0.l sin (2τ x 10*3r), V

1.30 Comparing the given Ψaνeform to thatdescribed by Εq. I .2 \νe observe ιhat ιhe giνenιγaνeform has an amplitude of 0.5 v ( 1 v peak-

to-peak) and iιS leνel is shiftω up by 0.5 v (the

first term in the equation). Thus ιhe wavefoΙmlook as follo\γs

Average νalυe : 0.5 vPeak{c.peak valυe : i vLowestvalue:οvΗighestvalue:lv

PeriodΤ:1= 1Ξ: tο',fo ωo

lJl The two harmonics have the ratio126198 : 9Ι7 . Thιs' ιhese are ιhe 7th and 9th har-monics. From η. l.2 Ψe note that the amplifudesofthese tιγo harmonics will have the raιio 7 ιo 9,which is confiΓmω by ιhe measurement Γepo.ted.Thυs ιhe fuπdamentai will haνe a frequency of 98/7 or 14 kΗz and peak ampΙitude of 63 X 7 : 441

mV The rms value ofthe fundamental \vill be

441 / J2 : 3l2 mv To find ιhe peakτo-peak

amplitude of ιhe square \'ave we note that

4V/τ : 441 mv. Thus.Ρeak_ιo_peak amplitude

:2V:441xΞ:693mv1

Period r _ l _ , _ 1l'4μsl 14' ιο'

1J2 To be barely audible by a relaιively younglisteneι the 5th harmonic must be limited to 2οkΗz; thus the fundamental wilΙ be 4 kΗz. Αt theΙow end. hearing exιends down ιo abouι 20 Ηz.For ιhe fifth and higher to be audible the fifιhmust be no lower than 20 Ηz. Coοespondingly.the fυndamental \rill be aι 4 Hz.

: 165 V

:24Υ

= 3llV: 311 kV

Case ω (rad/s) f (llz't r (s)

a 6.28 x 1σ 1x10" lxl0,b !xto, l.59 X lο3 6.28 X 1ο'

c 6.28 Χ |on lX10Φ l X 10ω

d 3.?7 X lα 6ο l -67 x 10I

6.28 x 10r IXΙ0r I Y 10'

f 6.28 Χ lσ lXlσ I X 1ο6

chapter 1_ι l

t" : 0 t.{zt'RR

l 33 Ιf ιhe amplitude of the squaτe waνe is y.then the power deliverω by ιhe square waνe to a

resistance R wiIl be v]r/R. Ιfthis poιιer is to

equal thaι delivered by a sine waνe of p€ak amli_

ιude , then

vφ

0

_yη

Thus, V,, : 0 / ^E. This resulι is independant

of frequency.

1J4Decimεl Binary

0 0

5 lο1

8 lωo25 ll(nl57 l l lα)l

Note that there are ιwo possible representation ofzero: ο0ω and 1000. For a 0.5-v step size, analog

signals in the Ιange :ξ3.5 v can be represented

Ιnput st€ps Code

+2-5 ν +5 0101

:].ο v 6 1Π0

+2.'l +5 0ι0ι

-2.8 -6 1ltο

136 (a) For Ι bits there '\μill be 2λ possible lev-els, frorn 0 ιo yFs. Thus ther€ will b€ (2Ι _ 1) dis-

crete steps from o to η, wiιh ιhe sιep size givenby

Step size :

This is ιhe analog chan8e corιesponding to achaπge in the LSB. lt is the νaΙue of the .esolutionofιheΑDc.(b) The maximum error in conversion occuΙswh€n the aηalog signal value is at the middιe of astep. Thus the maximum eιror is

ΙStep

T!Χ.t"n.ir.: ! Ιr,2 22N-lThis is kno\Ιn as the quantizatioπ er.oL

ι"l !9y =s.v'2ι_|2Ν _ |>2m2r = 2OOl =N : 11,Forl{: 11

Resolutiοn 10 : 4.9 mV21\-l

Ouantization .r,o' = Ψ _ 2.4.γt2

1J7 when b, = l, the ith s\λ,itch is in position 1

and a current (y../2 R) flo\γs to the output. Thus i,will b€'the sum ofalt the currents correspondingto "1" bits, i.e.,

'": ?(?.u;. .u#)

Φ) rΙ is ιhe LsBb, is the MSB

l<-r--->i

l-35

b, b" b, b" valu€Repr€sented

0 0 0 0 +0

ο 0 o I +l

0 ο I 0 +2

0 0 ι Ι +3

ο 0 0 +4

o 0 I +5

0 1 0 +6

0 ι I +7

1 0 0 0 0

I ο 0 I -tI 0 I 0 -2I 0 1 -3I o 0 -4I o I 5

I ι 0 6

I ι

Chapter l-12

(c)

. l0ν/l l t I ! I\5 kΩι2| 22 2' 24'21 2n)

: 1.96875 mACorresponding to the LSB changing from 0 to I

the output changes by l0/5 x l/2" : 0.03125mA.

1J8 TheΙe will b€ Ζι4'1ω sampιes per secondψiιh each sample represented by 16 bits. Thυs the

ιhrough-puι or speed will be,l4'lω Χ 16:7-056 X lσ bits per second.

139 (a) A,, : !9 :υi

l0vlω mv

: lΦ v/V

oη 20log lω:4ο dB

' iο ιolRι l0ν/l0oΩ 0.ι Α' i, i, |00 μΑ lΦ μΑ

= l0ο0 Α'/Αoι 20 log 1ω0 : 60 dB

A, : 7'oi'o : "o x':: lω Χ lοωυli !ι |l

or 10log 10r = 50 dB

ιblA =!9= 2v - 2'/lo\ν/ν" u, l0μVor'2olo92 Χ 1σ : l06 dB

' io υ6/R1 2v/1okΩ^'_τ_ ι

_ 1ω"Α

_0.2mA _ ο.2Xl0: - 2αn A/A100 nΑ Ι(n X l0 9

or 20logΑ, = 66 dB

, νoio υo '' io^P----------^-:υιιι 0i li

:2xl0'X20Φ:4x 1σ vr'Λv

or l0logΑP = 86 dB

ιclA'': b =]!l: lo vzv" o, tvoη 20 log l0 : 20 dB

. i^ ν^/R' lον/|0ΩA:=:' i, i, lmA

= ιΑ : κxn A/ A1mA

oη 20 ιog lοοo : 6ο dB

, υoio υo -' io^P----------.--ιlι !ι |ι

: 10 x l0Φ: lσWWor l0 log,ο Α, = Φ dB

A = !ρ:4" u, O'2

or 2ο log 1l : 20.8 dB

^ io 2.2v / l00Ω^'_τ_ lmΑ

:22!Δ : zzlll1mΑ

oη 20log Αl : 26.8 dB

^ _ Po - |2'2 / {Δ'/ lω^"_ i _η^φr

It L

:242WlNoη lο logΑ. : 2].8 dBSupplypower:2 X 3v X 20mΑ: l20mwOuιput povr'eΙ :,ξ,.' : (z.z t "'D\' = 24'2 τnWRι lωΩ

lnpuι power Ψ = o'' m\ry (negligihlel242

Amplifier dissipation 1supply power _ ouιpuιpoψeι

= Ι2o _ 24.2: 95.8 mw

Αmotifier efficiencν _ ouιpuι po\νer y tωSupply power

= 42 x κ[' : 2o'2qo120

1.41 For V,,, = 5 γ;The largest undistorted sine-wave output is of

3.8-v peak amp|itude or 3'8 / .Γ2 = 2.7 V..,.Ιnput needω is 5.4 mv_..

Supplies are Y,D and - Y""For V,,,, : l0 V ιhe largest undistored sine-waveouιpuι is of 8.8-v peak amplitude or 6.2 v..,.Ιnput ne€ded is 12.4 mv,.,.

+3V

__] ν

chapιeι l_l3

Foι V,, = 15 V the largest undistorted sine-Ψave ouιput is of 13.8_v peak amplitude or 9.8v_. The input needω is 9.8 V/5Φ : 19.6 mv,-.

1.42 (a) For an output whose extremes are just aιιhe edge ofclipping, i.e.' an output of 9_vo*, ιheinput must have 9 v/ι0Φ = 9 mV,*.

(b) For a.n output that is cliρping 90% of the time,

θ = 0.1 ΧΦ' : 9'and%sin9" : 9 V aVp : 57.5 Y whichofcourse

do€s not occuΙ as the ouιput saturates at :ξ 9 v.To produce this result, ιhe input peak must be57.5/lο00 : 57.5 mv

ι^ lOR" lOR^(a)__:r: " xA x

---:'" η IoR. 'R.

" _'" loR, + R,

:]!rlor,9:z'zεvtv11 1ι

oι 20 Ιog 8.26 : l8.3 dB

rυl 19: Rs xl ^ --4ηus Rs+Rs "n Ro+Ro

: 0.5 x 10 X 0.5 : 2.5 VΔr'oη 20 log 2.5 : 8 dB(c)

R"/ to R^/ Ιo" _ --------Ξ- v. v -----------Ξ--Ul (Rl/ ι0) . Rs "" (Ro/ lo\ ι Ro

:1rtox!:o.oεινlvιl ll

or 20log 0.083 = -2| '6 dΒ

1.44

20 lo9 A,b : Φ dB 9 Α,o = lω v./v

Λ', : oo

__ l0ο X lωIω+l0

: 9ο.9 vΛr'oη 20 log 90.9 : 39.1 dB

A^ - υbl lω Ω : Α] X lga = 83x |0ιW /W' υ!/1l'ιΩor i0log (8.3 X 1α) : 79.1 dB.For a peak output sine_waνe current o{ 1 ω !ι, thepeak output νoltage \ryill be ιω mΑ X lω !l =l0Y Corτespondiπgly !i ιvill be a sine wave \ρith

a peak value of t0ν/λ : 10/90.9 or aη rms vaΙue

of lo/(Φ.9 Χ "D) = o.oε v.corιesponding oυΦut po\rer :(o / Ja''x / |ω Ω:O.5W1.45

ιoτrs

X tαnx |ωΩ .l(nl}+tkΙι

= -!9 , tαn, |Φ = 8.26 V lvll0 1lω

vDD - 1.2

"l,"1L

Ξ

+

7ri

(c) For an output ιhat is clipping 997o of the time'

0:0.01 x90'= 0.9'

V, sin 0.9" : 9 V

+vP: 571ν and the iηput must be 573 v/1α)ο or 0.573 v**.

1.43

9V-

+

υi

10 kΩl0kο+ 1ω kΩ

10Ω

ΙωkΩ 1kΩ

νo = R' XΛ X Rιv, R,*R, ''' Rι+Ro

chapιer l-l4

This figure beΙongs ιo 1.47

l.45 coηtinυed hereThe signal loses about 90% of its strength whenconnected to the amplifier inpυt (because & : R./10)- Αlso, the output signaΙ of ιhe amplifier losesapproximaιely 907o of iιS strength ιγhen ιhe loadis connectω (because R, : &/l0). Not a gooddesign! Neνertheless, ifthe source were con_necιed diΓectly to the load,

υo: RιuJ R.+ Rs

l0o Ω

Polιer ρain _ υb/ lΦΩ - -]578W/w_ ι,i / l.l MΩ

or l0 log 7578 : 38.8 dB(This ιakes into acct. ιhe power dissipated in theintemal resistance of the source.)

l.47 Ιn example ι.3 \γhen ιhe firsι and the sec_ond stages are inιetchanged, the circυiι looks likeιhe figυre above

ι1l _ lΦ kΩ - o.5 v./ νι1 lΦ k{} + lω kΩ

A':υiι = 16 1 tMΩ

= 99.9 V/V

A':ι,ι = 16 1 l0kΩ'' u,, l0 k() + l kΩ

: 9.ο9 v/v

A ''' = ιL _ l X -_..φq Q- = ο.9υ9 v/v' Ui]l l00Ω+l0Ω

Τotalgain : A'' = υL : A"'x A"1>< A,,1aιι

: 99.9 x 9.Φ Χ 0.909 : 825.5 V/VThe νoltage gain from source to load isιιL υL '' νiι

' 0tι

0s \ι ?s Ul: 825.5 x 0.5: 412.'7 νNThe overalΙ νoltage has ιeduced appreciabιy. Ιt isdue to the reason because ιhe input imμdance ofιhe firsι sιage. R.. is compaΙabΙe ιo ιhe source resis-ιance R.. Ιn example l .3 the input impedance of thefirsι sιage is much larger than the souΓce resisιance

1.48 a. Case S-A-B-LVo : Vo Χyi xvi" -ys v,o v,o yJ

(r , --l!!-), ( rrn, |ω 'l

" r lο )\ lω ] lα},/ \ l00 + l0./ \|0ο _ |o,,

Υρ = ι'ιlv/ν and gain in dB 20 log 4.l _v.l2'i}2 dB (See fiμre below)

lΦΩ+1ωkΩ- 0.ω1 v/v

Rs = l0o kf)

which is cιearly a much worse siιuation. Ιndeedinserting the amplifier increases the gain by a fac-tor 8.3/0.οο1 : 83ω.

1.46

1ωf}

l,^ _ lVx lMcl λ|X ΙωΩ" I MΩ.lωkΩ lωΩ+l0Ω=lr!Φ:ο.εlv1.1 110

voΙtagegain: k: o.ε: V/V or_l.6dBιrs

Current gain =

νollΦQ:o.8]Xl.l Xlο4v./ l.l Ml): 909l Α-lΑ or 'l9.2 dΒ

This figure belongs ιo 1.48a

&:1ωkf)

l0Ω

This figure belong to 1.48b

b. case s_B_Α-LVo _ Vo .V'o -V'υyJ vio v ib vs

- l'ιαl , |ω ) x l'r " ---.!$-] "\ lω + l0κ./ \ loK+lω/

/ Ι(n K \ιlω K + lω/p : o-oν V / S and gain in dB is 20 log 0.49 :vs

-6.19 dB case a is preferred as it provides higherνolιage gain.

1-49ReqυiΙed oveΓall voltage gain : 2 v/l0 mv : 2ωv/v. Each sιage is cφable of proνiding a rnari,lι..rfl volιage gain of l0 (the open-circuit gainvaΙue). For,, stages in cascade the maximum(unattainable) νoltage gain in ια. we thυs see thatιγe need at Ιeasι 3 stages. For 3 stages, the overallvoιtage gain obtaiηed isυo: |o x lο x -.1!_ x lο x -JLv. 10+10 1+10 l+10

xtox-Ll+l:206.6νΝThus. ιhfee sιages suffice and ρmνide a gainslightly laΙge. ιhan requirei. The ouιput νohageacιualιy obtained is 10 mV X 206.6 = 2.07 V.

This Rgu.e b€longs to !.50

1.50 Deliver 0.5W to a tΦΩ loadsource is 30mν RMs \λ,ith 0.5MΩ source resis-tanc€. choose from 3 ampΙifiers types

R;= lMΩ &= ,0kΩ R;= lOkΩ

Ao= lo νlν Α,= lΦ v/v A; | νlν

ζ= lokο ζ= lkΩ R"= 2a{ι

Choose order to elimirιate loading on input andouιpυιΑ _ l st{o minimize Ιoading oπ 0.5 MΩ sourceB - 2nd-to boost gainc _ 3Ιd - to minimize ιoading at 100f, output.(see figure b€low)

ιo_ 2V - urs.l<( lμ )rtοrq 3omV \0.5μ + l μ,/

(π#τ}'*,(#*τ}' (π-t*)235.1 < 253.6

vp : (253.6)(30mV) : 7.61v RMS2 .^,-.2

ρ : 39- = ll9lΙ = ο.sε ιlvRL 1ω

l.sΙ (a) Reouired νoltaρe sain = k'vι

= -Jl- = -rrn vzv0.01 v

(b) The smallest R, alloψed is obtained from

o.ι oΑ : l0lV = R. + R. _ ιω k!}' R!+Rr "

Thus R_ : ΦkΩ.For ξ : Φ kΩ. .] : 0. 1 μA peak, and

Overall current gain : Φιi

.li *tι;,\;+

Chapter 1*15

R"= lο kΩ Rl=1kΩ

Stage I

0.5 MΩ 1O KC) lKΩ 20Ω

1ο KΩ

_ 3mA _ ]Χ 1ο4Α,/A0.1 μΑ

oνerall power gain =#

(ft,5'''o'/loX 10 '\ l.ο_l X lο ^\l-lxl-lι '/2,,\ 'l2 l

:9 X 1σW/WCΓhis takes into acct. the poιγer dissipat€d in ιheinιemal r€sistance of the source.)

(c) Ιf (l,, τ/) has its peak value limited to 5 v. ιhe

largest value of R, is found from

_ , -!ι_ - _'].+ R- - ?n' - ooz οRι+Ro " 3'(Ιf Ro Ψere greater ιhan this νalue, the output vo!t-

age across Rι would be ιess ιhan 3 Y)

(d) For R' : Φ kΩ and & : 667 Ω, the requiΙedvalue l- can be found fιom

}ωv/v _ 90 "a

x I

90.t0 i I . 0.667

303 mv - 30.3 v/vιο mv

lrnο, lΦ : 333_3 V/V1α) + 20o

connect a resistance & in paτallel ψith the inputand select its value from

(Rp 1l Rr) : I R,(RP ll Rr) + R. 2Ri + Rs

chapιer l-16

Θ) ls1,J

(") 9υi

(d)

=t+ Ii :22-+R^ llR" ll R,

1ι2lRo R, ιω

R-= l = 9.l kΩ' ο.2I _ ο-1

1.53

R.'21

1οo2l

+

&

: 555.7 V /V1ω kΩ (l Χ lo'Ω)1ωΩ(1 X tσ Ω)

lω Iω0Iα) + l0 ''" l0o0 + 1ω

\ r- lωιΩ ,']-/rΙdηmv | ιd

+

(a) cuπent gain =

R^- ^"Rn R.

: 1οoψt!

: so.gΔ : 39.2 dBΑ

(b) Voltage gain : 19τ/s

=b R'iiRs + R,

: qο.q , -L1ο1

: 0.9 vΔr' : _0.9 dB

(c) Ρower gain : Aρ: "o'1olsl,

: 0.9 X Φ.9: 8l.8 wΛ = 19.1 dB

+

ιo

(e) R, :

300 :

1.52

η= lωkΩ+

νi

ΞΞΞ(a)

z^: lomV* l0 ," lο + l00: 303 mV

t(ntir(irx..-t(n + 2rn

peal(

&= 2ωΩ

Chapter l-17

+

axι

Use suμφosition (see figure above)

,,, : c.,,,(, *ξft7)ιs κl+

t,, = ""(Ξτξ+τ) ιsκ{,,(ο7;sξl0K+R,r )

c.-:ω.πR":20kΩ&=lkΩ

R' "Rs + R;

2υ,= τ,c- =

j

'2+2 2

ιo : G^ιι(RL || Ro)

:4020xlι)20+l': q12! o,

)t ')

oνerall voltage gain =!ρ = ιg.osν tνιS

1.55 Need yo : loι' t Δ1'Rxl and lto aΙe used to get ahe appropriaιe gains

on v, and v,

This figure belongs to 1.55c

loK R,a

ι: ix R

lr.t : tο : |z0 mΑ)|lo κ). Kr lο κ )u, \ v ,,\15 K/ \20 K + R,r,/

0.ιs : lo K20K+R,l

"' R,r : 46'67 kΩ

Use same procedure for v2 but you will find 1

stage is not enough again, thus: (see Rgυre beloΨ)

"-ιι = zo : |20 m.η)( l0 Κ )15 K tr20 mΑ)v' \ v .r\5κ+ l0κ./' \ ν /

( ιoK )ι!o Κ)( loK )\l0K+ l0K/ \!0K+ l0K+R,)/

2o : 6.6j, tnr( loK ')

\20 κ + R.r,'

.'. R', : 3.3μρ '"n",

or caη us€ of ιr2

then

R,r: 30Ω

+

Vtι+

vrι

&=2kΩ

Chapter l-18

FinaΙly' For & νarying in ιhe range ι to 10 kΩ , the

change in i" can be kept to l0% ifRo is seΙectedsufficiently large;

R, > R..,,Thus R, : 166 1ρFor v5 : 10 mV,

,RR^i"mir _ Ι0 _RΙ+ft;σ---_------g-

t03:l02 lω c lοοl0o + 1ο _I0o + l0

G. : l'2l X lo ι Α/v: 121 mΑ/V

Ιn o*--rl+Iωιο $ v

Ι

i' : υ'/ R' Ι g.υ'

. .(t \ll=τ/t+ι_f8,J

ot: I

i, l/Ri+8.

= R. :n-I + g-R,

1.5E Tran-sresistance iτptifi erTo limit ΔU. to l(% coπesponding ιo&varying

in ιhe range , to 10 kf,ι ' we seiect η suflicieπtlylow:

R Ξ :.Ι"Ej-ι' to

Thus, R, : 1φ ρTo limit Δv,, ιo l09o while &νaries oνer the

range l to l0 kf,}, we selecι R.,sufficienιly iow;

R^ < :-ι!l!" 10

Thus, Ro : lω ΩNow, for i. : l0μΑ.

1.57 Transconductance amplifi er

R"

l to lοkο .- _ ;' -,': ρ R,',"R,-," .' R, "',R,.i" I Ro

t=1ρi Ι0Φ R loοol00ο + lοo 'l0ω | lω

=t Rn = 121 x 1Os

: l2l kΩ

{' 'r

+

υi

Rι!tol0 kΩ

For & νarying in the range l ιo l0 kΩ, and Δl,limited ιo lο% we haνe to selecι η sufRcienιlyIarge;

R, > loR..,rRi: lω kΩ

Out

Where zo= loι + 201η

---+ι2

0" R,lο mv

1οο Ω

Cbapter I-19

1.59 voltage Αmplifier

&1ιo lOkΩ

ψιRLΙ ktol0 kΩ

For R, varying in tιe rilge 1 K to 10 kΩ and

Δi, νariation limiι€d ιo 10%, select R, ιo be suffi-

cienιly large:

R, Ξ 1ο RJra,

Ri : 10x lOkΩ = l0okΩ : l X lo5 Ω

For R. νarying in ιhe Γange 1 to ι0 kΩ , ιhe load

current variation limited ιo l09r, seΙect Ro suffi-ciently low:

R^ = Rι"'

" t0

R^ = lkΩ = l0oΩ : 1X tο2 Ω" |ο

Νow find,4lo

R.ιi".," = lomvΥπlj;xΑ-rr+ R.*lX103=10X1ο3Χ lωkf}

l00 kΩ + l0 kΩ

XΑ x l'" l0ΟΩ+ l0kΩ

lxl0 ι - lo-Ιo-ιX!Φre', I

ιl0 _' lIω

voltage amplifier equivalent ciΓcuit is

Ri : 1X lοJΩ, Α,. : l2l vΛr' and

Ro:1Xιo2(}

l.60 currenι Αmp|ifier

RLlk tol0kΩ

Forlζ varying in the 1ange 1 kΩ to 10 kο Ιange

aηd load νoltage νariation limited to |o%' *le.tR, to be sufficiently loιγ:

R - Ιj!.,,' tο

f.: tkΩ:1α)Ω: tXlο)Ω' 1ο

For Rι νarying in the range 1kΩ to 10kf} and

load νoltage νariπion |imited ιo l0%. & isselected suffi ciendy large:

R. Ξ l0 Rιr,,R": l0Χ lokΩ

= l00kΩ: lX1o5ΩNow we find Α,,

Vr,', = l0 μΑ )( τ5;

XΑi, X Ro|ι R.rin

= lο X lo 6-&Ξμ- , o' RoRl''nR..,n , f, ^Ro + Rι.,n

l0Χ]οo lkΩ XΑ lωKXlΚlkΩ+l0οΩ ^Ι0οK|1Κ

+ Α,, : 1l1.l Α'/Α

currenι amplilier equiνalent circuit is

lo2 Ω, Α,, : l1l.1 Α/s,

lo5 Ω

+

vi

lΧ1Χ

η_ R=lkΩ i.

ffii ffi,, 1 x to 'x (tωll

= 888 VΛr' or 58.9 dB

:b: oo/Rt =i; 10 3, lΦ

Iω+ l0: 220o Α./Α or 66.8 dB

101 x rO'

8 z(ι x lo')1n'3 1!Φlt0

Ai

ι^oνeraιI currenι ρain = :--:--:" tμA

ιto/ Rι _ εr(η x to')|μλ 10'

: 20oo Α-lΑ or 66 dB

1.63 Using ιhe voltage divider ruΙe

8ιz Ιz

Chapte. 1-20

1.61

Apply kirchhoff's current law in the mesh on the left

u?o I n, 8'z 14 x ιo'1

iiR' (lο , x |Φ 1'lο z to'\ |0ο + l0./

: 19.36 Χ 1o5 \γΛv or 62.9 dB

_τb+ r,ib+ RE(βi, + ib) : 0

ιo= i6fro + (β + 1)Rε]

Now r.: _βΧirXR.!, : βR.!6 roΙ (β + ι)RE

υ" = (i6 + gi)Rε:ir(β + l)R,

η _ (β+ 1)Rευh rτ+ (β+l)Rε

:i._!ι + n"β+ !

t.62

^ Ooen-circuit outDuι νolιase,(^:..-" shorι-circuit outpυι cυrrent

=1kΩ

ll^: loλ 4 :εν" I +4

a. ιο : 8nι1_ 8mι2

υ6 : ioR.. = g^Rι(ιι _ υz) : υo

b. ι1 : ι2 :.υn : 0V

υ1 : l.ol J...?h= |0yιι: o'Φ J

\.64

-- -- ---o+

υ' s^= lO0nA/Rι = 5kΩ

υo

: t0 vι0 mΑ

ι? R,

ΙΓ--->

Figuτe l. ι6a

chapter 1_2ι

Ι' = g''V't g''Ι,Vr: g.'V'Ι grrlu

thus

L| :8ιz= Roι2|\= n

L| :g7l=AνI2|ιa=o

Ι'l l

π|,r=o:''':&Ι^l:l :8ιz=Φ,||UΙ=0

=R,Rs*sC,R,R,+R,

V, :vs

Ri

_ (Rs + R,)

1.65

-ry\ r----rlR l-ηΘ

'#uo u'

ttJ-=(a)

dυe to unilateraΙ nafure of Fiμre 1.l6a

( Rs + R,)+ rC,R,Rs , . "l/

C,R,R, \\RΙ + R,/

tγι"r" κ : --&-(Rs + Ri)

. = ξ+^& from ιable |.2 low ρass for giνenciRiR.

values ωo : ι2.5 MHz

1.67 Using the γoltage-diνider rule.

Γ-Ψ--+l-] -

ηo '^,*r,

-LΞTιsl:Υ9: R'

v, R.*t,*asC

//\,,,,: (ot*)- -.-l, . ιs+-'\ c(Rr + Rr)'\νhich is fmm Table 1 .2 is of the high-pass tyρe with

Κ: R'RΙ+R, " c(Rl +Rr)

As a further νerification that this is a high-passnetwork and f(s) is a high-pass transfer function,

vr'e assume as 'r90, T(J) + 0; and as s--lοο,Γ(s) : R, / (R| + R2).Αlso, frorτι the circuit

obserνeas J Jcο, (l /sc) 90 and

Vo/V, : R, / (R, + R2). Now,for

R, : l0 kΩ, R2 : Φ kΩ, and C = 0.1 μF.

2τ 2nxojx ιo 6(10+Φ)Χ 1o3

: 31.8 Hz

lrιl,o"ll -! 40 l = 0.57VΛrt 10+ 4Δ {2

+

vo

a

\

for(a) yo : ,,('#fτ)vo: 1

vι 1+scRwhere k =l

I"RC

for (b) yο

VO : SRCvι l + scRvo: svι .s+ _L

RCwhere k = I

6; = Ι166 1a616 1.2 it is hiρh τ,ass."RC1-66

from table 1.2 it is lo\μ pass.

:,{+]

+

vi

Usiπg the voltage divider rule,r. l

'sC,

" *ΙV, ' sC,ζ_ r-_ιi

π"nl ''c' I' l*,*41\ Jζi'l

&9Ri

_ I + sciRi

"-(Γ'-j.π)

(b)

Chapter 1 22

vo = R,vs ρ.*ρ.*a

(Ι

_Rι

2τ x |0(21+ 5) X lo3

Thus. ιhe smallest value of c that 'v/ilι do thejobis c : 0.64 μF.

1,69 The giνen measured data indicate that thisamplifier has a low-pass sTc freqυency responsewith a Ιow-frequency gain of,l0 dB, and a 3-dBfreqυency of l04 Hz. From oυr knoψ|edge oftheBode plots fbΓ loΨ-pass sTc neιworks (Figure1.23a) we can complete the Table entries andsketch the ampΙifier frequency response

1.70 From our knoιvledge of the Bode ploιs ofsTc ιow-pass and high-pass net\γo.ks \ιe see thatthis amplifier has a mid-band gain of4ο dB. alow-frφuency response ofthe high-pass STctype \γiιhλ. : 102 Hz, and a high-freqυencyresponse ofthe low-pass STc ιype ιγithλ! : IΦΗz. we thus can sketch the amplifier frφυencyresponse and complete the tabιe entries as folιows

lΤl' dB

lο lo? lo3 ld lο5 tο6 to?

|+-:-aε εanοwlaιι _+{

f, f.,

1.71 Since the overall transfer function is that ofιhree identical sTc LΡ circuits in cascade (but

with no loading effects siηce the bufferampΙifiershaνe input and zero oυtpυι resisιances) ιhe overallgain ψill drop by 3 dB beloΨ the νalue at dc aι thefrequency for which the gain of each STC circuitis l dB dowη. This frequency is found as folloν,,s:The transfer funcιion of each sTc circυit is

If(r): ---_:-1+Σ

ωο

\,,/here

ιo1 : l/CRThus.

lr(j.Ι :

R,+R. I

c(R. + Rs)

ψhich is of the high-pass sTc type (see Τhble 1.2)wilh

Κ : R'Rι+Rs

For /,, Ξ l0 Ηz

--l- Ξ l0

2ιτC(R1+ R5)

ΞcΞ

1

c.(R, + R.)

40

30

20

lο

t0" /(Hz)

20 loρ l = _1',Πl9-τ{ \ rυ,, /

=l*[9ψ]'= loq'

ιo1u, : 0.5l ω6

οl1;n : 0.51 /CR

t-.τ{t ! 10 10, 1ιlβ 10r 1σ lσ 10, 1α1rl(dB) o 20 37 40 40 40 37 20 o

-,tΗz) lrl(dB) LT(."\

ο 40 0

10ο 40 0

ιo00 40 0

t01 37 - 45"

to. 20 - 90.

lο6 o

Chapter l-23

l.72 RS = 1Φ kΩ, since ιhe 3-dB frequency is

reduced by a very high factor (from 6 MHz to 120kΗz) c, must b€ much larger than c,. Thus,neglecting C, we find C, from

120 kHz - 1_ 2τC2R3

Th€veninequiνalenι aι

node Α Rl node A

21tc2Χ l05

e C2 : 13.3 pFΙfthe original 3-dB frequency (6 MHz) is attribut-able to C, then

6MHz: I

2ιτC'R"

2r'X6Xlο6X105= 0.26 ρF

1.73since Ψhen c is connecιω ιhe ]_dB frequency isreduced by a laΙge facιoη the νalυe of c must b€much larger ιhan whaιever parasitic capacitanceo.iginally existed aι node Α (i.e., between A aηdground). Furthermore, it must b€ ιhat c is now thedominant deteΙminant of ιhe amplif,er 3-dB fre-quency (i.e., it iS dominating over whateνer maybe happening aι node B or anywhere else iΠ theampliier). Thus, we can write

l50kΗz: 1 ..

2τCι'R,,' Ι| R'l\

g (R", 11 n") :10]X1Χ1o92τx|5ox

: l .06 kΩ

Now R,, : 1φ 161.

Thus R.' = 1.q7 1ρsimilarly, foΓ node B,

15 kHz :2πc(R") || R,])

Ξ R,r 1| Ri3 =2τx15x lo3 X l X lo-'10.6 kΩ

R", : 11.9 kΩshe should conηect a capacitor of value q tonode B where ζ can be found from,

lOkHz: 1 ,,

21τc ρ( R"r|| RiJl

shυntcapacitor

Ιnitialcapacitor

*co2τx10x 1or X 10.6X 103

ιη

= 1.5 nFNote that if she chooses to use node A she wouldneed to connect a capaciιor l 0 times larger!

1.74For the input circuit' the comer frequ€ncyjξl isfound from

,tJ Δ' ,.cJR' + R')

i"t0 kΩ Cl

R.ιkο c2

ll-r,.T-tΞ *

Ι_ΞFor /,l < 1ω Hz ,

I Ξ Ι(n2zτC ]lo + lω) X 1o3

cbapteτ l _24

+cιΞ =1.4Χ10'12'rXlΙoXl03Xlo2

Thus'\ir'e select C, : 1 x 10 7 F : o.1 μr.The actual comer frequency resulting from C,\γi11 be

Ι',, _ _-_-l ' = l4'5Hz2zτx lo ' X lι0λ I0'

For ιhe output circuit,

. : l --l : β'9HzJο7 znc,ιn"+ n'l 2nC|R| 2τrlο lι106

BetteΓ approximation(3-dB frequencies)

Band\νidth : 1σ _ 1σ : 99Φ Ηz

1,.16

Yi(s) l/sC, I

' l'J' η(r) yrc|+ R, _ jcf , + I

LΡ3 dB frequency

For To(s) , the following quiνalenι circuit can be

υsed:

Foτ /", < 1Φ Ηz.

I

2τC2(l

Ξ Cl2

+ l) x l0Ξ Ι0ο

: O.8x 10 6

2τx2xl03x1o2selectc2:lΧlo6:|μFThis will place ιhe corner frequency aι

2nX1ο6Χ2Χ10]

rι.t : l0(] s

1.;)(,.zft)1.75 The LPfacιor 1(1 +/10β) resυlts in a Bodeploι like thaι in Fig. ι.23(a) \γiιh the 3dB fre-

quency /, : lo4 Hz. Tbe high-pass factor

1/(l + 1o4/rf) resυlιs in a Bωe ploι like that inFig.l.24(a) \γiιh ιhe 3dB frequency

fo : Ι0a 11z'

The Bode plot for the overall transfer funcιioncan be obtained by sυmming the dB νalues oflheι\,,,o iηdividual plots and then raising ιhe resu|tingploι νeπically by Φ dB (corresponding to the fac-ιor l0o in the numerator). The resulι is as folloνls:

ι4,,l, dB

:80H2

+20 dB/decade -20 dB/decade

T,,rs) : _G^R,R1# νsc,

= _ G,,ιR, l n,l____l_I" clπ, + Rn

3 dB frequency :2τCr(R2 + R.)

= 1 :53H22π1Φ X l0 'x 30 x lo'

.'. Τ(s) : ri(s)To(s)

2τxl5.9Χl0r

+ 20 dB/decade

x -666,7 x ss+(2πx5])

-20 dB/decade

1+

30

2A

10

ο

10

-211

f31

f =10lAJ,.ZO

+37

lο ld lo1 lo1 Ιo5 lo1 f (lΙz)

Bandψidιh = l6 kΗz_ 53 ΗzΞ 16 Ηz

R1.77 v, : vsiia,a) To satisfy const.aint (l ), nameιy

v-(l +-Ι)v"' \ r(n., "

we substifuιe in η.(l ) to obtain

R, -l_ jRs + R, lο0lΦ lο3 lΦ

40 40 40105 l06 lο7 (Hz)20 o _2ο (dB)

al0ο

l -Ξ-1ω

Chapter l-25

ThusRr+Ri< I

R, l,1ω

Rs= I -1=R; ,_ Ιlω

which can be expressed as

R, -' lφRs -Ξ-100

resulting in

Ro<

Τhe 6rst factor on the LHs is (ffom constaiηt (l))greaιer o, equal to (l r/lω). Thus

AoG.>(l _

ft)ιno l| π.l

substituting Rs : 10 kΩ andι =20% in(|)results in

R.- 'or.!!q

_ I'| = aοkΩ" \20 )

subsdωting/3dB = 3 MHz, C1= l0 pF and

(3)

RL __

Ro<

l0 kΩ in η. (2) result in

I

2πX3X106X10X10

: I 1.3 kf)substitυtingΑ" = 80,, = 2V7.' R': 10 ko, and

Ro = 1l.3 ιΩ , eq. (3) results in

G^-- : 18.85 mΑ./V

-t2 1

;i

R,ΞRs(lψ ι)

b) The 3-dB freqυency is determinω by ιhe par-

allel RC circuit at the output

,1ttJo _ 2τωo

_ 2'c7RΛ| %-)

Thus.

- 1 rrl l\l" : 2'c,\πt π)

To obtain a vaΙue for, greaιer than a specified

valuejrkB we seΙect i, so ιhaι

;q(+: *.)=","

{r+ 1_*-_zτcιt'ιo

{-ι''c'f .o" !"

Ιf ιhe moΙe practicaι νalue ofio:1ο kο is us€d

then

G^> : 2o mAΛl

(r _ ffi)ι to lι l1.3 ) X 1o3

(ι - ffi)ιrοll rol, ιo'

vo :vi

1sCι

+ s(Cr +

s+_c!

(2)

1.78 Using the voltage_diνider rule we obtain

vo = z,vi zt+22

'ιryhere

z,: R,ll +,^nO r, - *, ll iΙt is obviously more conνenient to work in termsof admittances. Therefore lve express Vulζ iπ thealtemate formVo : Υ'vi Y1+Υ2

and substitute η : (1/R|) + Jc' and η = (1/&) +

.rc. tο οbtain

2ιτ f ,o" + cr_Lc) To satisfy constraint (3), we firsι determine thedc gain as

dc sain = nj;;o.,r,11 *. ,

For the dc gain to b€ gΙeater than a specified νalueAo,

*f7;o't"'ll *'.'='o

c, , czs +,.j,(.a _ i)This transfer function wiιl be independent of fre-quency (s) if the second factor reduces ιo υnitΙ

I

RtIIRr R2

Ct)

I

C,R,

chapιer ι_26

This in tum \irill happen if! Ι /l t\

C,R, Cr + C2\Rr Rr)

which can be simplified as follows

C, +Cl:n,Γ|*1] {t)c2 '\R' Rrt

r+Q: r+&Ct R2

orC 1R1 : C2R1

when this conditioη applies. the attenuaιor is said1o be compensaιed. and its transfef fυncιion isgiven by

vo = ctvi cl+ c2

ψhich, using Eq. (l) can be expΙessed iπ the aΙteι-naιe form

vo: | : Rtu' t*\ Rr+R2

R1

Thus when the attenυator is coπpensatω (c,R' =c,R,) its traηsmission can be determined eiιher byits tψo resistors R,, & or by iιS ιwo capacitors. c'.c,, and ιhe transmission is fo, a funcιion of fre-quency.

1.79 The ΗP sTc circuit whose rcsponse deter-mines the frequency response of ιhe amplifier iηιhe Iow-freqυency range has a phase angle of11.4" arf: lω Ηz. Using the equation for

Ζr(jω) from Table ι.2 we obtain

ιan-|j-: - |l'4"Ξ t-' 20'16 Hιt(n

The LP STc circuiι ινhοse response deιerminesιhe amplifier response at ιhe high-frηuency end

has a phase angle of -11,4" aLf : 1 kHz. Using

the ιelationship for ΖTUω) given in TabΙe l.2we obtain for ιhe LΡ sTc circυiι.

on-, d= -1.4"=+ t = 4959.4s2Ι-

Aι/: 1α) Hz thedrop in gain is due to the HPsTc neιwork. and thus its value is

20 loρ | : -0.l7 dΒ

/r + r 20. t6).

η/ \]0o/SimiΙarly, atf = I kΗz the drop in gain is causedby ιhe LP sTc neιwork. The drop in gain is

2ο ιog : -0.17 dB

The gain drops by 3 dB at the comer frequenciesof the two sTc net\νorks, ιhat is, atl: 20. I6 Hzandf: 4959-4Ηz'

-ι#Ι