Microelectronic Circuits Example Solutions

Exercise 1-1

Ex: 1.1 when output terminals ae open circuitedFor circuit a. oc : ,()Foi circuit b. uoc : i,(r) X R,when output terminals a.e shot-circuited

.,)ror crcut a. ,,. : For circui b. i,. : is(/)For ialencyRsis(r) : vs(r)

z- = 10mVX I :9-9mV" 100 + 1f R.: 19 1,^: lomx l0 -9l.v" 1+l _fR, = 1L11

.,-: lomx--!--:5." l+lf R. = 1

^:lomVx | -qt" !)+1K-8o9o ofsource oltae = lo mv x !Q _ mv' l)f& gives 8 mv when Rs = l k,then

8 = 10X_jr_=, : a +i,

Ex: 1.4 Using current divider

{"is : l0 RL

R.Rs + Rr-

Gien is : 10 A, R5 : l kf)For

Rr- lk.i- lOA X _-] :9.9' l(n+lFor R, _ o k. i- : 10 uA x l00' l+]0a 9.l AFor

R, _ tk.r- - x l,,:5A' l(n I loFo

R,: lM.i-_ l0uAX lK' l+M3 0.9 A80% of source cuent = 10 X g : 8 f a load R1 gives 80% of the sotce c.ent, then

8 r. : l0,', x ll+R,:+R. : 25 1

Rs

-I"b

Fire 1.1b

-l), l"

R,1'.(' : s()

" R' Gien ,s(r) : 0 mv and Rs = 1 kf n.: 1 un

-rflb

Figure 1.1a

,: G)

x| 1.2

Vr. : 10 mVi56 = 10

n: V :10mv - t kl l0

Ex: 1.3 Using volage diider

Rs

c)

+

o)

Xercise |_2

Ex:1.5 / = +: #: 10Hzl : 2zf : 2 101 radls

Ex: 1.6 (a) r

(b):l=t

161 7 = 1 =

Ex: 1.7 f 6 Mz is allocated for each channel'then 470 MHz to 806 Mz will accomodate806 470 : 56 channels

6

since i Stats with channel 14, i will go fomchannel 14 to channel 69

/ : 1v}D = 001U^:2v, = 0, : 15 vD : 1ll1(b) (i) + l v (ii) +2 v (iii) +4 v (i) +8 v(c) The closest discrete value represented by, is 5 v: hus D : ol1. The eor is -0.2 v or_0.a 5'2 1o0 : _4L

Ex: 1.10 Volage gain : 20log 1 : 40 dBcuent gain : 20 log l0oo : 60 dBower gain = 1log,: l0log(,,)

: l0 1og 05 = 50 dB

Ex:1.11 Pr. : 15x8: l20m

PL:6/J' = 18mw

Pl"*ipun",l : l20- l8 : l02mW

, = ar lrn - _!g l ' - lla' P,' l20Ex:1.12

,,, = l' _-l0 l'v - tovI0" + 0

P, : ,?o / R, - (lo x-10 6)'? : to Ir w

with he buffer ampifier:

n^= x R, xA x RL" R,+Rl ','' R+R":lx l xtx l0 = o.25l+l l0+1

2rp' : o :9J = ls-w'R.t0

volae ain _ b _ 0'25v - o.25 /,r lv= -12d8

P,Power gain (r) = 7iwhere P: 6.25 mw and P, = 'i' 'v,:0.5Vand

r. : -----l-v : 0.5 uA' iM+ lMThus.

Pi = .5 x 0.5 : 0.25 Wand.

1:6.25x10'=25xto1u 25xrool0 log ,', : 44 dB

:1fI----:;

l 'I

106

T

1[t'.,TJ R

: 1* : 16.7 .,.60

: l00O s

E:l.E P :

t v2. ^ v2=-x-x, : -TRRlteraively,P = Pt+ Pr + Pr +...

-/ 4v\,1,( 4v L, r 4v ),1*\il R - ]J2"J R' 52"., _ {ra y(*!- l -! r ..)R?r\92549t

t can be shown by direct caculation that theinnnie series in the parenheses has a sum tha

approaches 2 / 8; ths P becomes y2 / R asfound from direct caclation.F.action of energy in fundamental

= 8/'z : 0.81Fracin of energy in first five hamonics

:!(r*]+1)=.srort 9 25t

Fracion of energy in firs seen harmonics

:8(r*l+.1 +1): o.ss,rr\ g 25 49)

Fracion of energy in Rrst nine harmonics

s-8ll + .l *a*1*1): .s",r\ 9 25 49 8

/

Noe ha g07 of he energy of he square ae isin the first three harmonics: ha is' in he funda-mental and the third harmonic.

Er: 1.9 (a) D can represent I5 distinct aluesbetween O and +15 V Thus./ : ovD: o0

Exercise I -3

This fie belongs to Exercise 1.15

Ei: 1.13 opn_circuit (no load) ouut olage :

outp oltage with load connectedR.

= ] -------L"o'R+Ro

0.8: I aR-: o.25k = 25olRo+ I

Ex:1.14 ,o : 40 dB = 1 Vr'

P, : !eR, : (o,,"",ffi)", r,

-,| "[l'-!-|'/ lo = 2.5vj' \ l+ll22

nUi-|t- | l'oo n.7

A,=+ - '''l', = 2.s x to4 w\ry10 'i

l 0 log ,4" : ,4 dB

Ex: 1.15 without stage 3 (see figure aboe)

b_( r lo,/ |K ). \lK.lM/ '\lK,lK/

'('*(-t)

'o = 1.o11lo11o.99ol )(1)(.o9) : 8l.s v,J

Ex:1.16 Gien ," : l mv

1 : o.r, rozr, = 0.909 vs = 0.909X1 = 0.909 mV

oiz _ iz ""-jl: s.s X o.9o9

: 9v/v V' v".For,:1mVb:9xr:9x1:9mV

' . '' '- _ -x !!x!! _ go.g9.gyo'9o, oi, i' s:818 V/VFor ,, : 1mV.:8l8":8l8X1=818mV. '1. 1). '-- lt_' 12_- ll7]. ' ' '' ."l]7 1l

:.909 X 90.9 9.9 x o.9 =

7MForY":lmVv. :744 x l mV: 7,t4 mVL

Ex: 1.17 Using voltage amplifier model, it can represent aE

R"

R;:lMR" = l0- : A^ x Ai : 9.9 90.9 __ 900 v/vThe oveall oltage gain

V^ R, R,-- : -------Lx A l -------_v R, + R{ '"' RL+ R"For R : 0oerall oltage gain

: lM X 9(nX 10 :4N1M+lK 0+10For:16oerall voltage gain

= M XgX l0 : 8l0V/V1M+1K 1000+l0.'. Range of olage gain iS from 409 to 81 0 v/v

Exercise l-4

Ex: l.l8 Ex: l-21

6 : i6ro* ( + l)'rR":r[r"+(+)ft"]

But r, = ,. and i, : i.,thus

fl"=--=?=,-+(+)R"

1.22 Gain

10 z 60 dB10 kz dlkHz 20 dBl MHz dB

R",.: i.----:-' "R+ R,

i^= A i, Ro = tr.;" Rt R'" '"Ro_R' ""Rs+R,Rol RThs.

io,R"Ro'"-Rr+R,Rr+R.Er:1.19

&

, "R,+R,l' : G^'(Ro || Rr)

R, ,.=c.UsR(Ro|| R)

Thus.

,k : c.ufu'o11 r.l

Ex: l.20 Using transresistance cicuit model hecirci will be

3dBfreqency

v o : G.v iiRo ll R. ll c.l: G.V,a+a+,c,RoR''

'rr. 9 : G' Iy, lLll' JCRo R.' '_LJ

Ro Rhich is of he STc L yp.

Dc ain : G- - l)- l*lRo R,

!: R,i5 Ri+RJ

V: R.i1"#*vo ^ R

_ "'R,rR,

ro.,9: "! =,s i .lsR" R.:R -----i-"'R( + R, R, + Ra

+

R, R"R"X"-R, +Rn R,+Rs

( +l) ib

Ex:1.23

Exercise 1-5

ui.ul=*: #: o.rmA/v1=o.t -$ : o.o.lvn.=uft : 12.5 k

* : + * {)= z., l ,

C' =

(trt)\s , l' l2'5 y o3 )

Ex: 1.24 Refer to Fig. 8i.23v.: R, : R, ,s n.+a1R Rs+ R,*+ |" rC ' C{Rs + R,)which is a sTc fnction.

/,o, : ;;76|;1 = t l"

C>2?r(l+9)1o]x1

: 0.16 F

= 159.2 pF2 x |05

Execise 2_1

x 2'lThe miimum number of termia]s rcquired by a

single op amp is fie: to int erminals' one outputtermina|, one terminal fc,r positive po\er spply and

one terminal fo negatie power supply.The minimum number of terminas reqired by aquad op amp is 14: each op amp requies twoinp terminals and one output terminal (account-

ing for l2 terminals for he four op amps). naddiion. he four op amp can all share one termi-nal for positive power supply and one terminal fornegative power supply.

x 2'2Equation are q = (r_ l');

I

U,,j = ?] U|. Uicm :

'(l,

+ 2)

a)

' - ''! _ o_}': _0.02V -2mA Io'aia: t.- Ul =0 ( 0.2): + 0.02 v

= 2mI,n. = '(_2mv+):

lmV

b) lo : or(5 - v|) 9 r : 5.01 v7i11 : 7_ al:5_5.0l:0.0lV:lomV

tt4., _ 1,

_ vrl _ ;(5.0l r 5) = 5.rn5 v

:5V

c)

vj: A(r-UI) = lor(o.998 _ 1.2) : 4v7,i : 2 | : .998 _ 1.2 = _4 mV

rl',^ _ )(1 + ?,r) = ;(|.2

_0.q98) _ v

d)

_3.6: l03[ _ (_3.6)] : lor(v:+3.6)+ ,f2 = _3'6036

id = 2_ vl = _ 3.636 (-3.6): 0.36 v : _3.6 mv

o,"- : \{.o,

+ "r)

: Lt 3.6 + (-3.6)l: -3.6 V

Ex: 2.3From Fire E2.3 we have: yl : y,l andV,, = (G.V2 G.v'\R : G.R(V2- V1\

Therefore:

V' : G.R(V''V1)That is the openloop gain ofthe op ampis : cR.Fr G. : l0 mr' and : l) we bae:A = l l0 I0 = l04/ orequia-lentty 80 dB

Exz 2.4The gain and input resistance ofthe inertingamplifier circuit shown in Figure 2.5 are

_ & .ndR, resecivel. Therefore. e hae:R,

R : 1 k and_& = _1_4. = 16,Rr 'Thus:

Rl:10x1k:1M

Ei: 2.5

From Table 1. I we have:

R" _ + . i.e.. oupu is open circuit

The negaie inpu erminal of he op amp. i.e..

is a virtual gound, ths = 0

V6:V, Rii:0-Ri,: Ri,

R.=bl R1 - -R+R- - -Ri |o-n '

: _lok

R, = # and is a vinual gond ( = 0).

thusR,:!:oRi:Since ve are assuming tha the op amp in histransresistance amplifier is ideal' the o amp haszero oupu resisance and therefoe the outputresistance of his transresisance amplifier is also

zero. That is Ro : 0 .

R= l0k

R=l0k

Exercise 2-2

xz 2.7

connecing the signal source shown in FigureE2.5 to he input ofthis amplifier \rye hae: is a irtual goud that is = 0, thus he cr-rent floing through he 10kf,) resistor con_nect between ad ground is zero. Thereforevo= vi- Ro.5 m:0_ i0K 0.5 m

x 2'6

2 Ru=l0k

| Rr=lk

v"

Rr=lk

v is a virfual gound, thus yl = 0 v. lv y' _' R| lk

Assuming an ideal op amp, the current flov,/inginto the negatie inpu erminal of he op amp iszero. Therefore, i2 : i' = i2 = 1 mVo: V i2R2:o_ imX10k

;' =Vo: |OV - --l' Rr 1ki6: ir_ir= lomA lm= llmA

vo|ase sain _ Vo _ _l0V to/lv |or 20 dB

crrensain _ _ _lom _ _lo/' i' Imor 20 dB

Poer sain _ Pl - - lo( lo mA) = l w/v' Pl llmAor 20 dB

Noe hat ower gain in aa is lolog,n|!|."'"lP,l

vl

For the circuit shown aboe we have:

v^=(E!v.+&v"'l" \R, ' R, 'tsince it is ruired that V6 : -(V 1 + 5V).We want to have:

&: t "na&: sRr R2t is also desired that for a maximum output volt-age of lo v the cuent in the fedback resistordoes not exceed I mA.Therefore

JlJ 5 1 --.2 !9l R. okRf / lmt us choose R, to be 10 k(! , then

R' = R, = 1ok andf' = & = utl'5

Ex: 2.8

" : (ftff),,-(tfl"_(ff)n

\iy'e want o design the circuit such thaVo:2V1+V2 4V3Ths we need to hae

(L,&) - 2. (&]r&) = l ana& - +\R|,/Rr,/ \R2,'\Rr/ frFrom the aboe thee equations, we have to Findsix unknown resistors, herefore, we can araily choose three of these resistors. t uschoose:Then v,/e hae

v1

il

Exercise 2-3

R.: T(ft)(ff)

(ftfr)

e have (refer o the soluion ofexercise 2.9):

To fid he contribuion of y2 to yo we set

V : Vl : 0' then:. vo : 4v2To find he contribution of y] to yo e sey = y, = , thenv^: ek9v,= ev," I kl)combining the conributios of yl,y2 and y.] to

Vs we have: V6: 6V 1+ 4V2- 9V1

Ex: 2.11

Ext 2,9Using the super position principle' to find he con-ribution of to the outpu volage , we se y2= 0

The y* (the oltage at the positive input of he op

1amp is: V* : ;-v l = o.6v IThus

v,: (r +?tfi)r- = lox.ov, : v'To find he conribution of y2 o he o.lpu volt-

age y e se vt = 0.7

Then y* = iv , : 0.4 v,

ence

v" = (l 9'}r. _ loYo'42 _ 4v2lk/ 'combining the cont.ibutions of 1 and 2

To Yo we have Vo : 6V | + 4V7

Ex: 2.10

3kUsing he super posiion principle, to find the con-

ribution of y| to y \ir'e set y2 = V, = 0 Then

: !!: z.s4

: ,,* f x 1! : 2 = , : 5 1: =f,x]{ = l=r,: Io

-'*} = r-* = l-fl _ R:v, R, Rr ',f V, = 19 then it is desired thatl' = l0 .Thus.

i_ l0 _ lR, 'R', |0V

Rt+R, l0R,*Rr: l M andR, : Rr+R, : R, :0.5M

x 2.12a)

v^vo: Alv v l..1v-: v' --!

i'' i'=Vo_V _ _- =(! *)vL R, R, \R] R,)

v. = (, +fr), :(,.ft)(r-*)=I + R,/R.vo. ___?vo (+RzlR|)y/

vo_ lI RrlRt --_

l+ R./R|v , 'l'R2lR| , - l+R./R|AA

Exercise 2-4

(b) For Rl : l k and i2 : 9 k the idealo

aue for he closed-oo gain is l - . hal is

lo. The actual closed-loop gain is c = l0 ,l+

If = 101 then c = 9.901 and: G_10x:,|: _o'999 - |1

10

For V, = lV, Vo = Gx Vr : 9.901 V and

V^=AV vv' v yo _ 9'9o|' 0: 9.9 mVf : lOa then G : 9.99 artd. : _o.l%For y/ = l ' vo = GxV' : 9.99'therefoe,

v.-v = v o : 9.99 : 0.999 mv - I mv'l4

f A = 105 then G : 9.999 ande: _o.o|oFor y, : l'vo: G Vr:9.999thus'v. - v = vo = 9.9q9 : 0.09999 mv I0':. mv

Er:2.13i1 :OA, V1 : Vr: lV,

i,: v':JJ:1-4' lk lki, = i, = lm,

_ torne _ -i,0PL _ vox iL - 10X10 _ P vxi l10

x 2.14(a) load oltage

_ lk X| - lmv1k+ ! M(b) load oltage = lv

Ex: 2.15

(a) R| = R3 = 2 k, R: =R = 2 kSince Ry'Rj = R2lRl we have'

'' _!:- _ & _ 2 = tV/V" V,r-V,, Rr 2(b)Rid=2R|=2X2k=4kSince '.ve ae assuming the op amp is idealRo=0

Vtt

vD

2 9k

:[*],'ftfr

v.

lk

V6: V1* i29k = l + l X9 _ lov

i': Vo': l0 : torn.' tkt) tkl)io : iL+ i2 =

' nA

yo _ 0 = loY or2odBv, lv v

(c) ,. = h: #t4)( o& t)

Ro _L:Rt Rt

Ro*lRr

The \ols case common-rne gain 4m happens

when |",| has its maximum vaue.

f the resistors hae l tolerance, \,ehaeR4no'(1 - 0.0l) < &< R4*,(t +0.0l)

R1".n(1 + 0.01) Rr Rr"".(l - 0.01)where R3nom and ftano. are nominal alues fo R3

and Ra espctiely. ! e hae ;Rrno, = 2 k and Rno,n = 2 k, thus,

2 x 0.99 < R. < 2X l.ol2 l.0l R1 2xo.99

s'oz = 8!

= oz'mR]

Exercise 2-5

simlaly, we can sho ha

g8.o2=&=rc2.02R

Hence. _ 102.02 < _& = _q.oz

RTherefore,

a=&-&-a=lRo-R,l =aRr Ri lRr Rrln he \orst case

lR,, R:ll l = 4 =l 'l .1+& I+98'02

Rr

Note that the worst case cn happens when

& = 9s.o2and& = ro2.o2R1 RlThe differential gain Ad of he ampifie

isa,, = !. therefore. the corresponding value o["RlCMRR for the o.st case .- is :

cMRR _ 2 nn]jll _ 26 1nn!!1o2'" '"|eJ 'w t 0J4 j

CMRR = 20 tog(2550.5) :68 dB

Ext 2.16We choose R3 = i, and R4 = Rz Then for thecircuit to behave as a difference amplifier wiha gain of 1 and an input resistance of 2 k

,_(op mpr) : V : 5 +0.5sivl\J : 1\2_ v|t : 0.l sin/ -The o lage at the oupu ofop amp A.

- : V,' R' !g' 2R': 5 0_5 sin}t 50 k X '01 sintlk: (5 _ 5.5sint)

v2 - The oltage at the outpu ofop amp A2

V"' : l,'+'x]1' 2R'= (5 + 5.005 sin)v

.R,U,(p mp ) _ ,: ,. & * = ,,'?l -i

...R] _ R4 _ lk: !"-': !,s + 5.)5 sintl2'* 2'

: (2 5 + 2.5025 sin{rt)V? (op mp 1) : y-(op mp Al): (2.5 + 2.5025sint) v

",:ft(+&),,"|0 k(I * 0'5 M]

^ o.I.ln.,

l k\ .5 M.,,: l(t+0)X0.0lsiot

10.01 sint V

Ex:2.18

v, (+(+):],(+)d

The aeforms for one period ofhe input and theoutput signals are shown below:

10v

0

we requrre

,4.:&:16,n6"RRi,1 : 2R1 = 2kR! =Rr: AoR' : 1 X 1k :Therefore,R!:Rl:0kRu=Rl:l00k

E\t 2.17GienV,".: +5Vyd = losin mv2R, : 1 k, Rz : 0.5 MRr:Ra:lgk{)

t-U- : urc. )U,o:,-: 5 0_005 sint v

I2 = \\fr + _'D

= 5 + .oossin} v'(op AmpA,) : Vt : 5 o.Ssin

Exercise 2-6

liy'e hae

-20:

c: !!xmsXo.sms20

xz 2.19

v'(+

xz 2-2l

(+)J o (+)

C : 0.0t F s the input capaciance ofthisdifferentiator. ry'e want cR : 10 2 s (the timeconstant of the differentiator), thus,

,^ 2n=--.!lj-=I,t

0.01 FFrom uation (2.57), we know that the trasferfnction of the differentiator is of the form

j!) : _ i.cnv,( iw\

Ths, for = 10 rad / s he differentiato tans-fer function has magnitude

ll _ Io" lO 2 : 0.l v/v and haselv : _90'For w = 103 rad / s tbe differentiato. transferfunction has magnitde

ll : o'X lo-2 _ lov/v and phaseI Y,l

: _90"f '!e add a resistor in sries '\ith the capacitor tolimit the high frequency gain of the differentiatorto l' the cicuit would be:

(+) vo(+)

t high fr9qencies the capacilor c acts like ashort circuit. Therefore. he hi-freqency gain

ofhis circui is: To limi he maniude ofR

this high_frequency gain to 1, we should have:

- t-t, - n:l,=lnR] " 1 J

o

-tCR

I r'RI,

:lCR

-20

10 d

X 10X I ms

%(+)

The input esistance of this invering integaor is

R, therefore, R : l0 ksince the desired inegration ime constan

is lO-3 s, we hae: CR : 0 r s+

6 : !l--J = 6.1 ,ro k()

From equation (2.50) the transfer function ofthisintegrator is:

v o0') : _ v (j.) jCRFor 'r:10rad/s the integraor transferfunction has magnitude

l&l : __] . _ l v/ and phasel v,l l \ l0_'=90"For = 1rad/s theintegatortransferfunc-tion has magnitude

l&l : -L.-.

. _ l0) V/V and phaseI v,l lo / to-':90'Using equation (2.53) he fruency at which theintegator gai magnitude is unity is

-'': | - l : IOrad/s"" cR ro1

20

Exercise 2-7

xz 2-2|

9yl(mv)

V'o = 1 1Vo: V'_ Vo'_ V'wheny+=-=0then

V: o 5 mv: _5 mv. This inpt offsetvoltage causes an offset in the voage transfercharacteristic. Rather than passing through theoigin. it is now shifted to he left by yo

2'22From equation (2.41) we have:

SR : 15.915 kHz : 15.9 kHz

x]. 2'23

v1

v.

2V6^^^

Using equation (2.42), for an iput sinusoid \ith

freqency / : 5./' he maximumpossib]eampitude tha can be accommodated a the otpu\,r'ithout incuing sR distortion iS:

u" : u"*,(*): rox I : 2v(peak)

V: Vz v|Vro: V*-Vor-Vn order o hae zero differentia input for the off-set_free op am (i.e., y* - y' = ) we needV: V'_V _Vos:0 5mv=_5mvThus, he ransfer characeristic yo ersus d is:

v,,, (mv)

Exercise 2-8

x 2.AFom eqation(2'44) we have:Vo :

"1R1 1!"a

:lnXlM = 0.l vFrom equation (2.46) the alue of rsisor R3(p|aced in series with posilie inpu o minimizehe output offse voltage) is:

Rr: Rrll R,:: 9.9 k

R3 = 9.9k 1l0krly'ith this alue ofR3 the new alue ofhe outputdc voltage (using equation (2.47))is:

Vo : o"R2 : 10 nA 10 k:0.0l v

with the feedback resistor Rl to hae at leas:t10v ofoutput signal swing aailable, wehaveo make sure that the autput volage due to yos has

a magnitude of at most 2 V. From equation (2.43),vr'e kno that the otpu dc otage due to yo{ is

,/?lRz = lokf,rX lMR+i2 l0k+1M

t = Y: a = o.16Hz"22

x 2.26From equation (2.28) e hae:

| - Ao'b+ , ' Ao+ = |. una'Ao\e kow

2o |ogAo : 106 and /, = 3 Mz, theeforef"-l57

By definition the openloop gain (in dB) at 6 is:o(indB) _ 3 = 106 3: l03 dBTo find the open-oop gain at frequency/\e canuse uation (2.31 ) (especially when/>>/6whichis the case in this exercise) and wrie:x 2.25

Using uation (2.54) we hae:v^" 2 mvVo : Vos + 7ffl 9 |2

_ 2 mv :-.j],

=, _ 2Y_?mV l ms-6Dl _ 6)2mV

openloop gain atl: 20 b(+)

Therefore:OpenJoop gain at 3 Hz =

20 los3 Mz : 80 dB- 3)Openloop gain at 3 kHz =

20 on3 MH' : aB- -l kHzOpenloop gain at 12 kHz =

20 lo 3 MHz : a ag'12 kHz.Open-loop gain at 60 kHz =

20 lon3 Mz = ag"60 kHz

x 2.27

vi(+'v. *l:

._f, |- , *,a,

vo = vos(1 *\)-z : 2m(| l #)l+ R', = l0oR. - loMl k(

Tbe comer frequency of he resuling sTc

n"1*1;5, = J-CRt\e know Rc : l ms andR: 10k+C : 0.l Fhusw

--_L..._

_ lrad/s0.l F x l0 M()

The waefons fot o.te priod of the inut and theotput signals are shovn belo:

t0v

0

xercise 2_9

slope=SR

We have

-20:

cn: !9x20

Therefore:

"fn = -. and Il+--i

Rt

f':2Hz

_9 |-6 o-l X 1_6

cRJ,,

l dt

Ex2.frSince dc gain of the op amp is much larger thanthe de gain ofthe designed non-invertingamplifier, e can use equation(2.35).

1 V/S

=9 1. : l.28 s

Ex:2.30From equation (2.41) we have:

SR

-lCR

lms

2 z100

l0x lms

: 0.5 mst,

+& = landRt

2'rV o ^^"

: 15.915 kz : l5.9 kHz

Using equation (2.42), for an inpt sinusoid with

fruency / = 5 jrM, the maximum possibleamplitude that can be accommodated at the oupuwithout incurring SR distortion is:

,. : u.^-(Lu') = 10 l = 2 v (peak)Hence ./.,un : = 2o kz

Exz 2.29For the inpu olage step of magnitde he ou_put waveform will sill be given by he exponen-tial aveform of equation(2.40)

tf w,V < SR

162115 y 5 49 y:< JL2f'y < 0.16 v. thus' the agest possible input olage sep is 0.l6 v.From Appendix F we kno hat the lo% o90%rise time oithe otput aeform ofthe form of

eUaion (2.4o} is - - 2.2!' ' _ ,Thus, r, : 0.35 sf an inpu sep of amplifude l.6 v (l0 imes aslarge compared to the preious case) is applied,the the output is sle1v_rae limited and is linearlyrising with a slope equal to the slei-rate, assho\rn in he followig figure.

\y, (*)-20

xercise 3_1

Ex:3.1T:50K

^-3/2 [/(2K)n': l P '

: 7.3 10'l(5o)]" e- t'l2(2 x 8'62 X lo 5 x 5{))

1 9.6 x t0 3elcmlT=350Kn. = BT3l2 e-F'El(2'

= 7.3 X 1o5(35o)r/2e_l'l2(2x8'62xt 5x35)

: 4.15 x lOrr/cml

Er(: 3.2

ly'o : l0r7/cm3F.om Exercise 3.l i at

T:35oK:4.5xtO/cm]nl: ND : |ol1cm1

.2

- (4.15 x lorr)'?lor'7

= |"72 x |o6lcr3

x: 3.3

t 3 K' ,, : l-5 x l0l0/cm3

want eectron concentration

_ n_ - l'5 X o|o : l.5 r lO!/cm}" lo

.2

.''No: ": L' nr

_ ( 1.5 x toto)21.5 x to4

: 1.5 x 10r6/cm3

Ex: 3.4a. '-driff : _,E

ere negaive sign indicates that electronsmoe in a direction opposite to E'We use

.-diff = - ,E: |]5o J- ...tm - I0 acm

2 lo-': 6.75 X 106 cm/s : 6.75 X 1o4 /s

b. ]lme taken to cross 2 ,m

l"nnh : 2|0" - ]on.- 6.75 x n4

c. n n-si driffcuent density Jn in

J^ : qnp"

= 1.6x to-rex lo'ux 1350x 1V2x 10 4: 1.O8 x lOa '/cm2d. Driff curren , = Aqn "-diff

: Aqn": 0.25 x 10-8 x 1.08 x 104=21 A

Note 0.25 m2 : 0.25 x l0 8 cm2Ex3'5 J,: qD"From Figure E3.5

no : !0|1 lcml : lo5z(m)3D, : .15 cm'/s - 35 x (loa)2()2/s

: 35 X 1o8(m)2/sdn |5 _ -^5 -2a;= l : lum

.': qD,

: 1.6 x lo-re x 35 x to8 x 105: 56 X 10_6 /(m)'z= 56 /( m)'2

For/,:1m:J"X

*: lT: 10'l .=18 .,J" ss (m)'?

Ex: 3.6Usig uation 3.2lD" Db

D^ : "V = 1350 25.9 X 10 3-- ).

= -r) cm /s

Do : oV1= 480 X 25.9 10 3

= 12.4 cm2ls

Ex: 3.7Equation 3.26

\ : /2" *fr)v'q

Exercise 3-2

s one can see from aboe equation, to increaseminority carrier-concentration (pn) by a factor of2, one must lower ND (= n,) by a factor of 2.

Ex:3.10

Ex: 3.8n a p+ n diode 1v >> lVD

as Comlared to n

'' = oo'i(+- 9-,^)

. t010

* 1')v,

Equaion 3.3, J :

=of*(,\u"

lD1'l|i, + /L= ,l\ N^" "*z 2es1 ^ + o1u^q \ NoNo ) "

"': i(): (ffi;'' uatin l . = on'( Do D" \\LoNo'L'ol

"'n"" u" here approximately

simila aues, if Nn >> o, hen the erm

-2!_can be neglected as co marto ffi

". lr= qn 2_

Ex:3.11

Equaion3.26 " ff 1r";'")u"We can ne8lect the term 1since l{ >> Nr)

:"#=w ff,:w

tv.Equaion.1.28. X" = lv;;*since 1y',4 >> ND

=wy2: Wl(h)

uation3.29, a' : (i+"u )w

= . }y since l{ >> y'{D

I AqN 1W

since ly' >> N.,

: l J-z.gv oEx: 3.9

n example 3.5. ly', = lOlt/cm3 and

D: lol6/cm3n the n-region ofthis pnjuction die

n,:1Vr:1167"rl

" !; (l.5 X lo|n)2 = 2.25 l.a/cm,

, ll''

: 10 4X 1.6x 1o-''x1.5 x 1o'1'

s x lo-o x !! l0 x lo_'x lo'82

= l.45 X lo-l4

: r("u'u' _ 1)

=r"u : |.45 X lo l'e5/(25'9 x l

r)

: 0.2 mA

Ex:3.12

w: - Vt)

_ |zx .uy 9''(-l , _L1o.l+ o.sl,r/ .6 x to '' \t'' lo'o,,: t.66 x l0 5 cm = 0.166 m

Ex:3.13

E"1J4Xr l . l \- i' '-- " 'ir l-: + - - l(0.814 + 2), l.6 X l0 '' 'l0'_ l'' ,: 6.08 X 10-5 cm : 0.608 m

Using uaion 3.29

o,: n(ffi)w

2e,q

F"T#+)".

* fr)v,

* v,l

Execis 3_3

= l_4 } t.6 , ,o [ lo|8 ' lo|6l , o.o , t' 5 .,nI rot8 * rot6J

= 9.63 pC

Reerse Curre' =', : o o'|(}1* "

* fi): 10 la x .6 X 1o_|9 (1.5 X 1'0)'?

,( ro * t8 )\5xI0ox!'o lX l-4X lt,,/:7.31015

x:3-14Equation 3.48,

.,.= ^J(#h*)

Ex:3.16uation 3.5l

"_: !i." Dn(5 x lo al2

5

:25nsuation 3.57

c. = (h)r

n examle 3.6, N. : 10|8/cm] ,No: 10r6/cm3ssuming N^ >> ND

=p:25ns

:.Cd : ( 25 x lo-:) o.t x to '\25.9 X I "/: 96.5 pF

c': : !-,-l" lv Ev: !^[",'x l r{eu"t

: .-1. ! 1gu' _'" dv' /

: !! y 1"av'v

-l]l\v 1

I

: 3.2 pFuation 3.47

rrlt + --1( 0.814= 1.72 pF

Ex:3.15

- 1)l

1)

to'8 x tor6\/ I \

'nl * ,nlo /( 0.8 t4.

Exercise 4 l

Ex: 4.1Refer o Fig 4.3(a). for y/ 0' the diode con-ducs and presens a zero voltage dop. Thus

v o : v For y, < 0, the die is cu-off' zerocrrent ffows throgh R and V, : 0

' The

resuls is he transfer characteristic in Fig E4.l

x 4.2see Figure 4.3a ad 4.3bDuring the positive halfofthe sinusoid, he diodeis forard biased. so i conducs resu|ing in,, : 0 During he negaie half ofhe input sig-nal v,, the diode is reverse blased. The die doesnot conduct resulting in no cuaret flo\ring in he

circuit. so vo : 0 and , : o = This resuIs in he waveform shown in Figure E4.2

lO mA

dc component of b

1^ l0: _7 : -1

= 3.18 V

x 4.4(a)

ij:2m.

(c)

(d)

-5V

Ex:4.3

, ;, rov^ : : -

:" R lk,::2mA

1^

(e)

_]

2

(

(b) lk

: ]! 7

+v: tv

'f,,:

+3V

+2

+t v

Ex:4.5y = l

lo - o51 = -_lm.'.R:3.133k

5V

5V

Exercise 4-2

Ex: 4.6

uation 4.5

Vz_V : 2.3viol.j(1)t room temperatue V7 : 25 r

+

vD

V' V = 2.3x25y lo'X|oc(#): 115 mV

x 4.7

o-] :1se07l025

/. : 6.9, 'o

to

Nowi:.|m

v : v h(i) = 25 X 10 ,".(#iF): 0.64 V

For i = l m

V=25x!1rlXl0-r'l\6 g x ro-lo/

: 0.76 V

Ex: 4.8

: l25 25 : l0o"C1s = l0 lax l.l5

= l.17 1o'8

x 4.9

Ato'c = |V : t,-.almsince he reverse leakage cuent doubles for

eery 10"c increase, at 40"C

l:4l A:4=V : 4 X l M : 4.0V

@o'c r = 1*l4

+v:Lx1:0.25v4

Ex:4.10a. Using iterationDiode has 0.7 v drop at l m crrent.Assume V, : 0.7 V^:5_o'1 : o.4] m" 10 kUse eqaion 4.5 and note that

Yr:0.7V. /,:1mA

R: lok ,,

Vrr_ 5

vx- vt = 2.3 x vrloc(r!)

v.: v1+z.:xv,(,L)First iteration

V' = 0'' +2.3x25x 10 i"c() : .679vsecond iteraion

' _ 5_0'679 : 0.432 m' to kv2 _ o.7 +2..l x 25..1 \ lo-'l"c()

: 0.679 V:0.68 V\e ge almost the same olage

.'. The iteration yields

a 1 o.43 m, yD = 0.68 vb. Use consant volage drop model

V o : o.'1 constant oltage drop,^ 5_'7 = O_4] m" l0 k

Ex: 4-11

Diodes have 0.7 v drop at 1 mo71V-

.'. Im : /se: sen1V : 25 m1, : 6.91 ,

'o to

Exercise 4-3

For an output oltag of 2.4 V the voltage drop

across each diode : 11 : . v3

Now he current through each diode is. ,,*/l25 . . ,)

I = lse ' : 6.91 t 1o-'"e '= 54.6 m

^ to 2.454.6 x 10 r

: 139 {)

.,xz 4-12(a)

: 1.'2 A

+

_ 25l1 = rrnl X l '

_25Xlo-r-2.5loX I r

Ex: 4.14For small signal model, using equation 4.15

iD: D+2',1

io: b'6uo (lt

(d)

(e)

Ex:4.lJv-

/o = 0.l m

+

-0.7V:V

{' / = 0'l_r.5 5

= 1 .72 m

v=3-0.7: 2'1

V:1+O.7

25Xlo r 25() 0.lx10r

3V

2

(0

.)

2.5 k

5V

t/:0 + *

I

+3V

+2

+l v

(c)

/r:lm

1, : l0 m' r,

-5V

-5 V

Exercise ,1-{

For exponentia mode

io 2 |v ve :elo

vl -i: i22_ io. : i.e ' _ io,lv_: iDle ' _ l |2'

nthisproblemip, = 16 : 1 mUsing equations (l ) and (2) reslts and usingVr: 25mV

b. For V, : 3 V, voltage drop across eachdiode=i=0.7sv

_D '. = 4.7 Y l0 "'Ae

D

/s vve

a. ln his probtem {! _ zo.v _ ,o ' i, lm.'. Total small sial resistance ofthe fourdies

:20

V(mV) l'r(m) rD(m)small expo.signal model

-t0

-5

+5

+10

: ?Q:.4

25 mVD

2.4 k

c.fio:5 i-:5_1=4mcross each diode the voltage drop is

v" = v,h(?)

: 25 X 0 'x l"/-1-]!_' )\47 x l '6,i= 0.7443

Voltage dmp across 4 diodes: 4xo.7443:2.977

so change in Vo = 3 _ 2'977 : 23

Ex: 4.16For a zeer diode

V : V.o rr'10: V,"+0.01 x50v." : 9.5 For ,z : 5 mV, : 9.5 + 0.005 x 50 : 9.75 V

4.17

0to15m

The minimum zener current shoud be5 x lzk.: 5 Xl :5m.since the load current ca b as large as 15 m,we shoud select R so that with /. = 15 mA, azener crrent of 5 m is available. Thus the cu-ren shouId be 20 mA ,eading to

R _ 15 5.6 _ 4720 m

Maximm power dissipated in he diode occurswhen /.= 0 is

.", : 20 X lo_] X 5.6 = ll2 mv

a

b

c

d

Ex:4.15

.'. For each diode ,,,

v-But = --J+5:

'''1D:5mAx6a:15 3:

5m

_.4

-{J.2

+O.2

+O.4

-4.33

_.18

+O.22

+O.49

+l5v

15 V

xercise 4-5

Ex:4,18

t5vEx:4.19

FoR LNE REGULAoN

Line Regulation : 19 :i

For Load Regualation:

Vo _ ,r,' lm

- , ,nA

Vs=

a. The diode starts conduction a

'=Vo:o.7,s : yssint, hee vs : |2{2

5 : V5sin = Vo : 0.7 rtsi : 0.7 = .ln ,f 0'7 ) - z-+'\zrtl_conduction starts at and stops at l80 _ ..'. Totaconducion angle : 180 _ 2= 175.2"

( 0)t

b. ,r*n - :- | {y'\sin v D\d6'zJ

=;,'r,*_v,]t=; u

_ 11 v.cos yscos( _ ) yD{ _ 2)|

Bt cos :1, cos(zr ) :_1 and_2:

2V, VD'\E 2 2: vt vo

2For V, = 2aD and yD : 0.7 v

uo'*" = :s.sv

t no Ioad y7 : 5. v

200+1

Exercise 4-6

c. The peak diode current occers at the peak diodeoltage

" '' _ --T- = 163 m

P| = +Vs: rt: l7v

xz 4-2l

ys

a. shon in the diagam he output is zero

between (rr _ ) o ( + ):2

Here is he ange at which he input signal

reaches yo

.'. V5sin : V,

= .ln ,( V')\ yl

2 : 2s1-l!1\ y.,,

b. eage value of he output signal is given by

u"^"' _ !|r, '"

[' , r,,'n - ,,,ol_ ,nl ,, I

: 1[_yscos _ v o =',

='- '"

c. eak curren occers wben = !

Peak Current

_ v Ssin(l2\ _ vD : V'_ VoRR

f ,s is 2 v(rms)

then ys : .zx z : rt

eak current _ '2{2_o'7 - 163m1No zero output occurs for angIe : 2( _ 2)The fcaction of the cycle for which ,o > is

_ 2(1t _ 2) x rco2zr

2[ zsin '(ffi)]X1

2

= 91.41o

verage output oltage yo is

vo: 2\ vo:2x|2"2 -o.7 : 1o.l vneak diode currnt ?D is

_Vr_Vo nrt_o'7'o- n _ |

= 163 mAV: Vs_Vo+Vs: nrt _ o.' + 12{2= 33.2

xz 4.21

-ys

yo,", : +|(y.sin _ 2v d' 2J: t-vr"o' 2voli-=u"

1

=:-|2v. 2vn6 2)]Bt cos - 1cos(7, _ ): _t_2-

't t)

=V""":'-5 2vD

'l - 15: '::-::-:::: = 'l '4 : 9'4

(rr + ) 1

Exercise 4-7

(b) Peak diode curren _ Peak olageR

_ vs-2vp _ t2rt - 1.4Rl: 156 m

v: y5 vD: l2Jr- 0.7: |6.3V

Ex 4.22Fll wae peak Rectifier:

vvri^ _/')r = t -'J-: -:2. 2JRC 2JR=Yr

T/2

- v [ r * 4)- . RC|= l'- a)

2 fRCTo find he aerage crrent, note that the chargesupplied during conduction is equivalent to hecharge los during discharge.

Qsppto = Qosi",,,, = cv, sUB (a)

Dr

Ja, -t.V.t-

R

j'D7

Exercise 4-8

= '' *!!' !-1' v, ! v,J: ,|'-"ff]: t,l'*2"ffif a.r.o.

x', 4.23

The conduction angle , ca b obtained usinguation 4.3o

'o, = J-:ad = 20.7'

The outpu oltage, vo , can be express as

o = (V _ 2V og)e 'lrcAt he end ofthe discharge inervalo: (V'2vDo_ v')The discharge occrs almost over halfofthe timePeiod =12For time consta Rc >>

2

e 1l-

.'.Vp- 2VDo- V, : (Vo-2VDd(l -

+v, : (vP_ 2vo x L

HeeVo: nrt and v, : |Vr, : 0.8 V-II

f60t : (12rt - 2 ^ 0.8) x

The average and peak diode currents can be calcu-iated using equations (4.34) and (4.35)

,D,'": /.(l -"E)Herel. = !J,andVo: 2rt 2 Xo.8, y,:1v

i1,,." : 1.45 A

rr*o: (t +z: 2.74 A

v of he diodes

= Vs Voo : l2,D - 0.8 : 16.2 VTo keep the safety magin, select a diode capableof a pak current of 3.5 to 4 and having a Pvrating of 20 V

x4.A

A

r.. !2RC i*+

t1 1(

" |)2 RC)

The diode has 0.7 v drop a 1 m cent.

iD l,'D o'1v l

2X60X Xc=r^: y,ln(

i' )+.v" \l mA,'For z1 : l0 mV, vpt is idea op amp, so i*

:. iD : iR: 1'0# : o reD _ 25 x lo 'rnl--o '1) + o.7 = .58 v\m,/yr: D+ 10mv

= 0.58 + o-1: 0.59 V

1,o = =

C: *+*: 128l Fwithout cosidering he ripple oitage the dcouput voltage

: r2.la,2 x0.8 : 15.4vf rippie oltage is included the ouu voltage is

= nrt'zro._L: lasu2

Diode cuen without taking ripple voltage into

cosideraion : |2'.2.'.2 x o.8 = o.|5

^1

10 mv

2rt-zxo.s

Exercise 4-9

i^= o:-L:1.a" lk lk

v : o.7 + 1 k X 1 mA

For v, : _ 1 v, the die is ctoff

V: _12 because it is idea amplifier

Ex 4.tr5

irr>0 - diode is cutoffa: 0V

/, < - diode conducts and opamp sinks loadcurent.

x 4'i2'6

ForU=_5vDiode D1 conducts and

Io= _5 +;(+l,, + 5)

: [-zs-e)v\ 2.iForU/5vDiode D2 condcts and

Io: +5+1(l_5)

: lz.s * l!) v\ 2)x 4.27Reersing he diode results in he peak outputvoltage being clamped at 0V:

Here the dc component of 6 : V o : 5 v

!'5V

+

l k

Both diodes are cut-offfor _5 < rl s +5and ,, : 1

D2

+

5V

l0 k

] k

Exercise 5-1

l^ = !'Yv 2 = 0.25 m 2aL ovfor al ,Ds > Vr, : o.5 V.Ex: 5.6

v^= vL:50XO.8: V': l = 0.025 t

va

V25: l V>Vou: o.5+ saturation: o :

\''Yrv"ou1l + xv or1

l, : l' 2x l.s'?( + o.o25 X )0.8: 0.51 mA

, =V^ = = 7a'4 k -8ok'' D 0.5l

,' _ uo' + r^ - 3J _ o.ozs ,l" tn " 80KEx 5.1

+5v

,".-JJl{'

7'D

l= 6o A l 2

Y : rc= :6o0 uA/V2

(a) Conduction occurs for UGs < y : l v.ole =vP + yJ: +4v.

(b) Triode fegion occurs for ,cD S y,

or e - D< -|ot 1JD> e + 1(c) Convesely, for saturation

D< + |(d) cien 0

o = \o|vo"t = ls :.V ou| = o.5 = _cS+t

'

vo,

8os

: 450 cm2 / vS

,C,, : 388 A / V'?

= (ucs - Y,) : 0.5 V._ l _ iY^',=!: s.Is!k 'L "' L

: 0.18 m, so w : 0.93 mEx: 5.2

c,, : ',: _ 34'q pFlm _ 2.30 fFzm'7"' t", 4 nm

, : 550 cm2 z vsk"= "C.' : l21 A/2

l^ _ !Y' - o.2 ^.Y' 2o" 2"' ^ :.Vy:o'.Vo..'in : V9: o. v, for saturation

Ex: 5.3 /- : ! Yv 2 in saurai rn9 2J' oChange in /D is:

(a) doble L' 0.5

) doubl w' 2(c) doble V.,- 22 = 4(d) double V^' no change (ignoing lengthmodulation)(e) changes (a) - (d), 4case (c) would cause leaving saturation ifvr,,

Exercise 5-2

.. ,c : + 3'5 V.U9 U6 l l:4.5 V.

(e) Fo. : _o.o2 v ' ad |vov| = o.5 v.

^ _ 75 uA and r. l 667 k,' |2| o

()At v" = 1 .

"

:'k"|v ""r(l

+ ||lUDJ])

= 75 (.M) : 78 AAt VD : OV,/, : 75 (1.10) : 82.5 A

, : ^J-s:: 3v' : 667 k'' lo 4.5 Ex: 5-8

": \'c,,'fv''vo.3 =.. 120.,2 '^ -,r, -

V6y:0.5 vycs: ve + v, = 0.5+1= 1.5 V

v" vV" = _l5 3" = -J------

D

- r.5 - (-2.s)0.3

Rs = 3.33 k

o : voo vo :

f,x: 5.9

2.5 - O.4 = 7k

W o'12 m_:4' o.l8 m

sauration mode (cD : 0 < v ,'vo : 0.8 v. : 1.8 - loRo

t" : \"c.,!v " _ V-)1 :72

..-R_1.8_0.8:13.9k72 A

Ex:5.1

r-- *". s.slo. : o. v, J,, : o.s v,vov : 0.3 v.p : 72 A (saturation)t the triody'saturation boundary

Vo=Vou=o'3

.'. R, : l'8 V 0'3 v = 2.8 k12 A

Ex:5.11

Ro: 12.4x2 : 24.8 kAV,;" : 5 V, ssme ''ode egion:

lD k.l., V,lv", _ vJl-, =voo-vo, t-'o- n I

t/ / 1/2 \l \

(5 _ t)yDs )

+ y;s - 8.8yDs + 0.4 = o

= 7r, - 0.05 < ov lriode region

/^ 5 - 0.05 = o2mA" )4.8xz 5.12As indicded in Example 4.5,

vD2 vc 'y, for he ransisor o be insaturaion region.

1, 2 l)0

0.3

v,, : 0.5 v.

"C,,, = 0.4 nA / V2

+ l.8V

Execise 5_3

D: o.32 nA :

=Vr:0.8VyJ:0.8+1:V6 : V' Vo' :

^ Vo 3.4,

Assume

Vo : 3.4 V, then RD :

Ex:5.14

\uL" : !xxvbv

t.8 v1.6+1.8:3.4V

: 3.4 M,

v _D

5-1,4

Vo^,n: Vo Vt= 5-l

1o:0.5m9Rr.",:

: t%a = rzEx:5.13

Ex:5.15

u/ : 0: Since the circuit is perfectvsymmetfical yo = o and therefoe Vo, = o vhichimplies he transistos ae tumed offand ID] =op = o.

7r : 2.5 V: f we asum thi the NMos istumed on, then r,,o would be les than 2.5 v andthis implies hd MoS is off (V65 > 0)

'" = :k, o, ' v,)"

Io" : tx l(2.s - ys - 1)'/rr:0.5(1.5-V5)'

Also: vJ : R]o" : 1oo"

/r, = 0.5(l.5 _ loDN'2

2.5

+ l1}, _ 32 D + 2.25 : o+ D= o.lM m

Io" = 0, vo : 10x0.101 : 1.04v

V: _2.5 V: gain if we assume th QP\Stum on, then yo > -2.5 Vand V651 < 0 whichimpiies the NMoS o, is urned oflo=o

:5k

R-, 5_3'4: t.6M"' l

R": V' : s" 0.12Vos2 Vov-'+ Vo- Vou + ys yD 0.8+ l.6 = 24

I

I

IJ-

1.8 V

1

J l,,I

=

V,o: _o'4.

lo= ol m^Iv2

!Y _ |0m =+l _ _5.56 mA/VrL 0.l8 mVs: 0.6 + V,o: _1.0 v: _ 1.8 + rDR1rR:0.8-forvov = -0.6 V

!io: )koV'u = 0.t m

.'.R=80O

Exercise 5-4

-- _ !-v -- ly.l l' _ ! , DP 2"L's 2x (us + 2.5 - l)2Vs- loDP2DP _ ( lor" l l.5)'

= 1r, = 0.104 m^=a v = _10x 0.104

:_1.4Ex:5.16

vr,: 18vRD : l7.5 kY, = 0.4 v.

k' : 4 rn l v2=0(A) cutoff/saturation Boudary

6 : V,:o.4 v., ,, : UDs : l.8 v.(B) Saturai o/T.iode Boundary

1.leo = tcs = v=o.4.

= uo, _ |v oo _

!*" uo, _ v

'f n"] _ o'

Us tl.8 35(Uc'J o.8 ,cs + 0.16)] : o.4

351s_ 27 cs + 3.4 : 0es = 0'613 ., .1585

o = 9o.1 A

o : o'213 .(C) For u651. = Voo : 1.8 V., t iode,

rDS:(k"v",)':179

Vol, : Vor1r= roo -. o': 18 mV.

Vr, = 1.8 V.V65 : 0.6 V.k,= 0.4 m^ / v2

Y:oL

Ro : 17.5 kf)(a) V6 : o.2 .'

c. = k"v ou = 8 lV

for Au: -g.Ro = - 10, make

Ro : 12.5 kU65 = 0.6 V., /, : 0.08 m'Ve5 : 0.8 V.(b) kee R, : 17.5 k_8.Ro = l0+8- : 57l |!A'/

= k' v ^',

:.vv:0.|43'UGs : 0.54 V, /, = 6. ,4',Ds : l. v'Ex:5.1E

Ex: 5.17

oo'o---l

Exercise 5-5

0.7 v.

1mA/V2

Desig for ',=!b _ -25. n," - 5 1"i

:. E.Ro : 25 = k,VovRD

n,":9= o' R" Ui_ Uo= R6

: 26R'^ : 13 M

"n' : ()"v })n'1

= )^RoVo: l2'5 V6y

and

V6y: V6o_V, oRo:4.3 _ |2.5 vov

:. vv : o.3|9 .8.: 3l9 A'/vRo : 78.5 kvDs = vov + v

ico:0+26fi=V,

..-l'''j.L:z.v26

Ex:5.19

' , 1'.!,:!+i=-+g-

..R"q:1,,il *Ex: 5.20

Voo: 5.

:0l" : 2o Al v2Ro: l0 kOY=zoL(a)Vcs = 2v+vov = |.' : l'rvi": 2 AVos: Voo_DRD: +3v(b) g^ : k:"vLvv: 4 AlV

@) u = !!:_r.Ro : _4

(d) Us. = 0.2 sin , uUaJ : _0.8 sin t 1I.los _ Vos+a"+2-2 < UDJ

Exercise 5-{

= [200 + 80 sin, + (4 4 cosr)]/D shifs by 4 A

2HD _ - 4A = o.o5 (5.1)" 8o A

Ex:5.212^o\g.=.Iv

x 40 x (1.5 - l)'?/D:31 :

" _ 2r0.1.5, =Vo: ]: s" lo 0.3/, : 0.5 m=g. : ff''":rz""l"os"8- = 1.55 m/v

, :Vo =.! = l" ln 0.5x 5.22

1, : O.1 m, 8. : l rfiA'lv ' k'" : 5oAlv?

'^: ,'Vov = 2x'o'l : o.2

Ex: 5.24

":

'k'0,.-|v,|)":1xx-!9xt. l)'z2 0.8 : 216 A

'.: #,: ' , : 72o tLA

= o12 mt.l

- .oa+v. _ ! ^-l _ rs u/u." 0.04r v'^x - _ 25 o.8 _ 92.6 k}' o 0.2|6Ex: 5.25

n , : 2J2 "vo :2vo : AVou D Vou

v'Ax L = vl:0.8m+ _ 2X 2.5Y0.8

o.2

: l0O V/VEx: 5.26(5.'7o) A,o : .(Roll 'o')(s.'72) Ro : Roll ro

t = !9 : A R''' o, 'oRo + R L

= _ 8.Ro;+ R'-,

.'. A,, : _.(Roll R) : g.(R" llsame as (5.75)

\| 5.27

'oll n,t

: lxi"Yrv,o": 1x 60

0.3 m' Vy : o.5 Vl.2 m^ ,

" = \,!v'."- =2 o

k'"vu

2X0.l _ 1)50 .. ^^,10

Ex: 5.23

8^: "c ^yLv ou Same bias condiions, sosame Vy and also same and 8,,, for bth Mosand NMOS.

p"c "'w. = "C "'W " => t'

-Yt : 2.,w"

R'grci8 irv ' o gJ

100K -f>Rin

w= O.4 : --!

wo

Execise 5_7

o : o.25 mA' yov = 0.25 v,Vr:50V,^: \ : ,* ua

8.:2:2nS

A1)o = _8.Dll 16) 1_g.R : _4

i, : Rril 16 =R6=2Okd|Ar, : Gu : _c.(Ro ll ,oll R)

=

c.(Ro ll R) : 2lfor f'" _ (lv.\ 2Vou = 9.g5 ..ao: (vis"| : l v.Ex: 5.28

D

&'. G1*-1r---

lK +

ssuming Vo --) o

From Ex 5.27

g.:2ms". 1 5 mv# - :---: - _-::--:jj-j-=g.Rr:.3V o'" l + g-f,1 2 mv

.'. R5 : l.5 1

G', _ A', _ _s.(RD|l R) _ _20 - _sl+8-R{ 4

;, : lc,i^,*l -- t v.Ex: 5.29

R,":1:R,,.:1

98. = 10 m/

'": t= b=rD : 1 m

cu=b=}o"'.",

: (}) lo ,n l vlz l

= +i0Ex: 5.3OCD amplifier

R..,: l : l 8-: lO m/V8.

n _ 2o = 2lo =l^ _ l-25 mv,,u 0.25v

fn: f.'' 8'R' : o.9l v." l+g-R

a,: t,,*j;: ,' ,,.

Ex: 5.31CD (sorce follower)

R.,,, : 2 : !=g. : 5 m/v8.

g^ : k'"Y' ou : (o.4 mA / v'?)

(w).zs vl- w = so\L/ Ll' : |"Yrv'o": 0.625 mA

^ g^R"'- t+s-R.for ,(

Exe.cise 5-8

A : A-l1o .lR, --

a o _ _---_f n t -8.

(\t)

()-()2o , , V,,']-|=rv,+*_L'-rv,J

for Vo : 2o .

A,,o _ o'99 _ --_- j9 vl} - 40 v.t+ilg

4ov.Ex: 5.33

o : \'Yr{v o, - y,)2 + 0.5 mA12: I x t(Ycs - l)

Vo' = 2.F=I.5Vthen:

^=!|x(2 l.5l2 _ 0.125 mA

-'o : o.5 _ 0.125 : o.75 = 75%/,, 0.5

llx:5-34

^ voo vD^o' '

R. = vs v,s

- 2-(-5) - 6klr' ], 0.5'+ RJ : 6.2 k

f we choose RD = R, = 6-2 k then 4, wilsightly change:

l^: lxtr(y-"-l)'z.Also

Vs=_Vs:5-Rr/,2D: (4_ 6.2D)1+ 38.441D 5|'62D + 16 = o=+ D : o.49 mA, 0.86 m

/, = 0.86 results in % > 0 or % > y. which is notacceptable, therefofe 1" = 0.49 my.: 5 + 6.2x0.49=-t.96VvD:5_6.20.49=+1.96vR. should be seleced in he range of 1 M ol M to have low current.Ex: 5.35

1, : 0.5 m : "Yrv'o" = v'o"

_ .52 _ l-I

vo = |Ves: +1:2

= y^R^:5_2 = .5

.+ RD : 6.2 k standard vale. For this R,we hae to recalculate 1,,:

ID:lXl(ycs )'1t: )(v oo

_ RD D _ l'|-

(Ves: Vo: Voo _ Ro6)

l": \( 6.2 p)2 + o = o.! 11

vn : 5 6.2xO.49 : 1.96 VEx: 5.36Using q. 4.53:

8^rol + 8-r

J RD = 6.2 kC)

o = \'Yrv2ou .=> o.s

=+Vo, = Vu+Vr= 1' +1= 2V

9V' = '2

0..5

:1xt2

6k

x Lu

'=t"rn##,*/,.. - 0.5x1+ /irF : 0.1 mA

"uo : o.\ : )"(),v'"" - L"

' o'l x2 - o.25gv-,, - 0.5 v{} l

Ves: Vou + yl : .5 v+Vc: 5+1.5: -3.5 V

Exercise 5-9

p : Vot-Vo, 5 - (-3.5) - s5 k()

"uo 0.1

vDs7:'_vovVo'^'" : Vo": o'5 yD.i" = _4.5 v

Ex: 5.37

Y,:1,5vt!:lmlv!

Vl : 75 '/o = 0.5m :

\*''Y, v'o, ='nv : 1.0 v.

2.5

_2_5

- 10v

Vcs : V, + Vou : 2.5 VVc = 0Vs = '2.5 .V: Voo_ DRD = +2.5.

^: kYrnu: l mA/V

,^ = L = ls k" oVo _ 0a

-figa= 0aEx: 5.38

R'n+G

v, Vcn: 4.o.

s.: F': 1m/V.^: L = ls For ro -)

Rl" : Rc : 4.7 MAoo = E^Ro: -15Ro., = l?, : 15 kFor r, : 150 k, R. : 15 kQRi" = 4.7 MA"o = '^(Rol| r) = '13.6

R"- : R"ll .o: l3.6k

-Rr,R,c, _ ;'n*,oR;_ _7'o

7los = 1.lo : Vn5l l4,

= 2.5 + cv(o.4 P-P)o is a2.8 yP P sinusoid superimposed uoa2.5 dc o|ae.Ex: 5.39

8-g'

8-roc'

G8 ------o

(|| ^ : -- _roRD

- "'Ro *ro*RsR^+r^+R"(2\7t--: 1t- " " ""' "e'

_ "8 + ,o + Rr(1 };ro'!

'o _ vs;#, cr"''

we vant

I.5 inA

Exercise 5-lO

l l ol'* l, '' U* lR, = o

using (3)

8rr oR D(RD + r) + Rs(l +gnro)

2(R^ + r^\R. : --j-_-a_------gj : 2 l85' 1 + E-robasedon RD : 15 K, ro :

g.: rmS

Ex:5.40

4R.;g

I__1r_____l

+-*Ri"

EtroI + g^ro

8.oll RL)I]Jld-R,

Ex: 5-41

8 ^- 1- 8^roRo3 RD+ ro

k

150 K,

8.

R'n

A

: l50 k:R^

ri8

R"", = *il.,

8^: 1,A/For R.ie : 50 f,)k,"=a:ltou, : R, = 15 kA'r,o : + g.Ro :v = s.D| RL' =

R,.

+ 15

+ 7.5

G-: "' A" : 7.1' R +R. + Rt"^ v {b) c : R'" A,.: 0.768

'' R,. + R,," '

Ex 5.42see the nex pageFor oher R.ig

R*ie Gv

lk 1.7 51 k 0.68

l0o k 0.07

Exercise 5_l

using eq. (5.107) on page 324 ofth text:

V, : V,o +' 6' + v u_ ^!-'lV, = O.8+o.4J+3_J71V: | '23

Exercise 5-ll

using eq. (5.107) on page 324 ofthe text:

V, : V,o+ '6' +_ ^!-'lV, = 0.8+o.4J+3_J71V:1.23

xercise 5_12

Ex:5.43'.: +lv'v,: -2v%.-Y.=3VTo OERATE tN STURAON REGON:V : V^. V:aV

DS mir

': \''Yr{vo, v*\2

-! tz" t2 : 9 mA2

Exercise 6-l

Exi 6.1rD.v

:.c= ,e"' '

962 _ '51 : Vr|

"r':7+25|

= 642 m

"" : sc = !!a = 16-,, a""ls : lsc

[, - i]

: to_'6 x !!1: 1.0 l-16 A

V"u = V, [f] : "'[]#]: 25 x 29.9336= 748 mV

cz1t]

o.Illl

z-.. : 7 + zsln ]!'lLll: 758 mV

lxz 6.2B

+5 |5o

-

xercise 6_2

:.vc = vs_ v,c : vrn |f]For collector Area : l X Emier Areav6s : 25tn [f] : s .v

Ex: 6.9

..s/-Vc/Vc = se lsce

. . v' . !c'n: "

+ sce

,,,,"", : l .a l-,

Ex:6.12

FiE6.12

: l' Yr, : 0.8 v at 1. :V rz' V "',

: V'lnfrr/r']= 25 x 0.693 : 0.1733

:

:

- V r'V - |a''9Ise - lsce

;F'"p-,-"';- (vF v Bc)l vlse - 1sc

*n"'-"\ u*.-/

"u''"o"u' * p rr/ ,

,*a:lffi: l0O x 0.2219 :22.2

Ex:6.10

I: evv

2 mA : 11 "v/5

v", : 25tn [fr, ff, 'o'1: 650 mV-: p l-:Exz' +I" 5l= 1.96 m

": f : $=39.2 .

x:6.1l

. vuu :

" : !.5yr ln[1.5 X l ll]

25 x 25.734643 mV

lm

:' v B7 : 0.817

- : V- Vr :' -:5

o Vru_Vuu ^'_ /. (*

- l.5 _ 0.817 X ]! k2 101

:338

Ex:6.13

+tov

l.5 _ 0.5 k2

Fig 6.13B:50'vrr=6.7vE: vB-o.7

r 5v

l0v

Exercise 6-3

_0.7: 0.7v- _.7+l0

IO K: .93 m

Ex:6.15

V3g decreases approx 2 mV/ "C risefor 30 risev.,: _2x30

= 60 mVyE : _ mvsince /. is costant,. is constant.'. yc:0v

r: #r: O.91 mYc:10-0.91 x5

: 5.45 V' r 0.9l"5

: 0.082

Ex:6.14

5k

'v+t.7v

vc

' V_V'''---R-_ 10 - 1.7

5

: .66 m, V"-o"

_ I rrl

: 0.01 mA, = Iu u

: 1.65 mo:lr:1.65:o.gq

1.66g=]r:]'65'65' 0 0.0lvc : cc+ cRc: -10+1.65x5 = 1.75V

A + c, ic - 1""u"t'nt s !s!fo

^r^: \"

, ,v . vrr. r"u"''u,..rc_,s - _ ,."va' |1 1!S1" L V^JQD

i, = r"uo''u' + 1gg in rig (u)rO

i- = \ "u'u'

... i, : piu+ 9s asinFig(b)to

QED

-5 V

+t0v

-l0v

Ex:6.16

Exercise

Exr 6,17

,o: b

_l

Fig 6-18By similar triangles BD CE

l,,=+ .,= +

, vcc- vc'e - R,_ r0 - 5

l0: 0.5 mV"" : V"'* "R": 0.7 + l0 x 0.01:0.8v(b) edge ofsaturation 69 = 0.3 V

,^ : l0 _ 0._1 : 2J = 0.97 m' K 10: /: o.97/50: 0.0194 mV ""

: O.7 + 0-0194 x l0 : 0.894 V(c) sauraed Vc : 0'2Ic: (l 0.2)/0:0.98mAB: c/pF = .98/I0 : 0.8 mys = 0.7 + 0.098 l0 = 1.68 v

Er: 6.20

vBB + lv

Fig 6.20

For Yrs : 0I= 0Transisor is oFF

'"Ic = 0V, : V66 - R

: + 10 _ : +10V

(v)()

Ex: 6.18

1' _ cro V r'D ADC A : l X ll

t01

lm_ l+1 1+ Il: 1.099 m

.'. - 1.1 mAEx:6.19

Rc:10k,=50(a) active Y. - 5 V

/c(mA) . .0 l

l0oJ1

oJol

1000r

16 15 lo4

lM 00 k 1 k

Exercise 6-5

vc

Fol vRB= |.7 -: l.'| _0'7 = _l m'loc: 9s

:500.1 :5mvc:10_5X1k

: +5 v > yr, (so ctie)(a) ge of saturation Vc : 0.3 v

- = Vr, _ Vr, _ lo - 0.3 : 1.94 ktl' l, 5

(b) de satuation vc : 0.2 v1r = 0.1 mA (unchanged)

Ic : .n.""a : l00.l : l mAR-: |0_0'2 = 9.8' |m

*6.n+1v

6.21

vBB +l0v

.

lk

Fig 6.22t edge of saration ,c : 0.3 vV"r: rRro.3+ 'R'- I[Rc + Rr] + 0.3

. _ 10 _ o'2 = |'225 m' 4.7 + 3.3Vuu : 'R't o.7

: 1.225 x 3.3 + 0.1

Vr: Vo-O.1:4_0.7 :3.3

R-: 3'3V :6-k' 0.5 mvc = v0+2

=4+2: +6- = Vcc-Vc

\| 6.23

0_6

vc

v

0.5

Ex 6.A

ro.""a = 5Then1.:51,"=6".

= 8k

10 _ .2

: o.226 mv,= 6 "x 3.3

= 4.48 VV"": V" + O.1

= 5.8 V

5x4.7 +6x3.3

+10v

+10v

+t0v

l0v

Ex:6.25

Exercise

x 6.27

50=150n actie range

1" - 5 07 - 0.04:l mA' lK.lowest for larges , : ": 150 X 0.043 - .c 0'3p ----:--' t50 x 0.04.1

: l.5 kFor = 56. : l _ 5 X 0.043 = 6.78 vFor : 156

+.3 v] . : 0.3 vx6.

+0.7 v - 0.4 v_ +0.3

V"-ov': +o.' . to-0.7'L 2

: 4.65 m,': o.99 ,

- l0-0.4 +0.7c 99 x 465

: 2.2 kdt[.(max):0+o"1 _o.4:

x 6.26

Rclk

+10

-t0v

R. - l_.7 9.3k't

R_:l0 4=6k'1

:50y"":155 = sv'" 15Rr, : 5 1| 00

: 100/3 kV_ V

V""

i1

R+[Rsrl(+l)]4.3'^ l I3 5t

= 1.18 m

+t0v

Execise 6-7

,: ,#q change :

+ 9.8%

x 6.29

+15Vttt+0.i03 + |.252 + 2.78 m

Total cuent d'awn: 0.103 + 1.252 + 2.75 mA:4.135 mo\erconsum: y X 1

: l5 x 4.135:62m

Ex: 6J)

:10,2 unchang1c2 unchanged

v: va 0"l 1p, =

vC' - o7" o 470

: 101 1s3= tI z.ls _ lclL 2,7 JHence

"^^ - ll_"' L( + l)oj7 z7 kJ _

=Vcz:7'06v : vc2 _ 0.7 : 6.36 v, : !lx ! : 1.1.4n'ao'47 II

, 5 .7* 10+100l

: 0.039 m

Ans: V, : -3.94

x:.632

OFF

@l" ( + t)/,3.94 mA

ON:+

0v_'ur.:, ,r.*,

= (-)3.94 V

/6,: /9' X Saf

N_ ,a\L: ( + l)r, = 7.03 m y Sa'

-+

Ex: 6.31: l.l5 m

1.28 - 1.151.28

lk

OFF lkarr=o

2.75 + --JL-( + l )0.47

$vu : vu+o.l:5.5

, l0-5.5" : 0.45 m

@/.,.",, : lu l,: 4.8 - 0.45: 4.35 m

: 1 : 9.6 aq 30.45

: 7.03 v> 5 .'. Saturation

v -=V cc - O'2

1 + 4.8 m

v_-vcc-o.2I

= 4.8 V1, 4.8 m

+5V

+5V

Exercise 6-8

Ex:6.33

, Vcc _ Vc

=>v6: 10 -8 : 2.0vn.: o:Je = 0i2: YcE Swing : 2.0 - 0.3 : 1.7 V

Ex: 6-38

_ 10_ yc _.25

-.120 V/V

vb.

v"t

R

l0 k1m

= l00

. V-^ ^*

= lV,,| := _z : lL

V ",.ll : s..r .v

l2

Ex: 6.3

ciuen, g- : Jic I--'c=c

n-:r:lm:4om/vv 25 mVA : 1-:: = g^Rc

:-40 x 10: _4 V/V

V6 : Vr, rR6:15 lxl0:5V

Nc(0:vc+Ndo: (vcc cRd + Av1'be|)= (l5 _ 10) _ 4) 0.5 sinr:5 2sio(t)

ibQ) : R + beQ\= l + 2 sin , ()

Er: 639

But /c :

165 1llV

"t

Ex: 635^

Ex: 6J6/c : .5 m (conslan) = 5 :2

/. 0.5 m 25 mV

: 20 mA/V = 20 m/V' .5 0.5"5020

= 0 :2.5mB 5 2oo'" s- 20 20

: 2.5 k : !0 kf,

Ex:6.37:10 /c= 1mAn : -]-g = z111 m,l V

25 mV

':-: ]!9:z.s" 8- 40r : ---- - 25 " +]

lse_

-

0.5 m25 mV

: 2o ml

oe: g. : figien ic : i/,:'"'.()

and r. : -L

: (g-r.)i

+t5 v

Exercise 6-9

E: 6.40

+ l0v

vc

ic7.5 k

chage Rc to 7.5 kvc= -10 + o.92 x 7.5: - 3.1 vu :9 : g.R,

: 36.8 x 7.5:276N

:276xlbmV: 2.76

Ex: 6.41

/6:1mA

r- : llQ : o.qs rnr' lIl": | :0.0099 mA" |lVc = lo_ 8 0.99 : 2.18 vV : l0 x 0.0o99: _o.9 - -o.l VV: _0.1 _0.7: _0.8v

! = lq. ,nv0.25

l : 2.53 k39.6

f : tol - 0ss

, :v

:

v^ :c

V)

Rys

ltI

t : #tt (-)s.(.oltR.ll R.): _,

- r_l39.6 (3.s5) : _76.2 /

r, ll n. ll n. , 3.35il& _ 4.oo

hus effect of ro is - 3.9%

x6.41(2'lAro : - g.(roll Rr)R6 : Q6ll R4')

8'RoV - R^ _ 'Rr+R.R"R'

"^ Ro* R'.. (rol| R.)R'8. ^ a; ') + R

: - c^Ooll R. ll R.)x6.For/a:1,4

n : , : !- = 41 ,,o,7 v 0.025,-: = !: z.s"s-,^: V^ :]!9 : lrn" l, R11 : r. : 2.5 kAvo = - c^?ll Rc) = _,0(5 ll 1)

= _97.6N

Rs

),

l l,*'":

I 1""Rc

+lv

Exercise 6_l

6,42 (cont.)

1: R,* : 2.5 :L,s Rs+RN 5 +2.5 3o ' | .o = _1 x sz. : 32.5 v/vlrs us ul 3l^l _ A-^' - :z.s _la _ .q

lN)

For /c = 0.5 m and Rc : l0 k

s.:ffi=2omA/

,_ : ] : s. "20,^ : ]@ = urn " .5R,': r,:5kAvo : _20(2 k |l 0k)

: 190.5 vr'R9 = R:|| r.

: 10k|| 2k:9.5k

. _190.5 5 5+5: 65.6 vr'

1;'' : 'o : ! rs. : _l2.8v/v,,,2

'l : 5 = 1=i,.'": I.v/. 5+5 2 "''

li'ol : zz.s x lom : 0.33 vEx: 6.41

R,"==r.+(+l)R

Vs _ Rs+r"+(+l)R

' R., Rr" l"/(+ |)_,*Rrn&q

,,s:1:1*161*--{o l0 5/l0l

= " _ ]l . = o._,s k - .]5o r}' l 0l

R1 : 5 k + ( + 1)0.35 - 40.4k{'

- lrol1 f. l R,-.) t x loU R*-R" _

l4oj: - 19.8 V/V

x 6.44

n : I, = --]_: a."v 0.025,-:L:]!9:z.s'' 8- 4t,.. : .'" . -25 (t' +lR,*: r" = 25Ao : E^oo\l Rc) - .Rc

:X5:2V/V

A: Anx- = Ivr)+)c, _ R," .e,, = ---?L x t' Rs+R'* 50 + 25

: 0.5 V/V

Ex: 6.45Rs=50

R.., = r : !: zs mv -.nl, l,=+' = 2515n: 0.5 m

Ao = +9nRc:2X5 = 10 v/v

c.,:!xA'':4ol'2+,4u : 80 : g.Rj.'. R;: so/2o: 4kx 6.4

P nser 30

ib : !!_: and o - + g.v oR L

- , (g.vn + ihRL

b

-

r

R,:1: r.+(+1)R1

i"= b

*1-

Exercise 6-11

=0.5+l0l:101.5kG : |fR1 moed, ro: co]

R^ = "*R, _ .5 + 1o=9 M " +l loln _o_ RL ''v ; Rr'R':lo4+l-

x| 6.47

V.r: 12

: 0.995 m2 + ':.l5

: 0.995 _ 0.984 X lI

t.r %

Er: 6.,18

o

0.91 V/V

Design I

: 1R=3k

^ 80 x4"8' _ 8 +40

Av : s3 /6gien ... maximize Rc' Vt '"Vg = qRg}2+0.3+IR

v--:|2x="" 80+40-: !_91 : t.t m' 26.7.

^

l0

:501. : 3'3 = l.04 m' 26.7.

^

-t 1l5l

qo ch^'le : l'M _ 0'9'17 X )-l= |o.3

Design 2

: 00R, = 3.3 k

R-" = 8X :2'61 k" 8+4v"": ffi:

vc: v+o.4+2: +2.4

^ Vcc-Vc lO-2.4"._-T-": ' :-]-m," + 10l

R + Rr,/( + 1) RE + R"/( + )

: 26'' k

: 0.99 mA4-O.726.1 . ^-

Rr+Rrl(+ 1) = 4.3kFor independence from , set Rd : o (oK for cs)+Ro = 4.3 1Yc(min) = YE+0.3 V = -0.3VVCQ : V e(min) + 2V : +10 V

- = Vrc V.o - |0 : 8.48 k' /. .9Ex: 6.49

+10v

Rc

2A

: 10o

vt0t: l501- : 3'3 = 0.984 m' 2.6. ^^

- + -t.J

5t

:15: 7.6 =

1

0.3 v

7.6 k

xercise 6_12

1,_ l0_'7 0.5_.4_ 2: _2.9" l80 l 7.5( , |) y. - l()s _ q'92 '1:lolxs ,r:ffi: l

- 1.002 mA t" I myc : lo 7.5(l. 8- = i: ffi : 'lrv:2'5 .-:L=|:2.5k}'' R- m

6.50

vcc = 1 =zs

6.49 (conL)

R" _ v'_ v" _ lo{2.4 o.7) _ 17l.7 k" l"Usig 5% resistors:R.' : 7.5 k & = 180 ko.'1 + BRB Rc Vcc: o

_v v = 10=l R,= lkR,. : 7.5 kf'g : g/(p + l): / 1u'

: 0.01 mV: 0_ 13R9 = _lVVg: Vs O.1 V: -1.7Vv-_v-- _-L7171.5 ! 2'51

+|

R _ v-_o'7 + vL = t9.3 = l9._] k*,,. I

Ex; 6.51Refer to Fig 6.5lIc : (t m) r1r: 0.0l mVc : !0 _ 8 k(l m; : .27 = lk X (-o.ol ;) = '1 vV: 1 .7: 1.7v: l00: Upper limit.9 v cc _ Vc:8vower a|ue +v_ v'c: o.4v (wherev; = -1.4v)

Swing: -1.4 -2 = -3.4 V : 50: upper still 8 Y lower+ 1, : 6.9196.O so Y, : - 1.96 Vl.96v _ 0.4 _ 2: _4.4

: 2: uppe. still 8v'o'ver + 1, : 0.005 m so ys : _.5 v

Ex:6-52Example 6.50

vo

RL

R: r. n"= r"lln,

R,::r":2.ska

R," : R, ll r" : 1 |1 2'5 : 2'44 kAuo ! 8,Rc:_4oX8: 32oNRo: Rc: 8kAvo: -8^to ll Rc ll R.)- 40x 3.5

: l19V/VRo = rol Rc = l|| 8 = 7.4kAvo : - 8.Oo ll Rc)

: _4o '1.4 = '296N

^R,*.Rc+R;"'R"+R,7

= ---::-:-:( _ 2| 6------5 +LM '74+5

= 39.1 V/V1,.: Rs*Rri/: !5mV- R,"|;' : cn

:i19.1 x !5:586mV

Rs

),, *:

r rI,'$:|- v"|

Rc,

r 1 i

'l

Exercise 6_ 3

Ex: 6.5-l

Example 6.50

8.: 40 m/r' r- : 2.5 kro: lk : oov &: 1kR,;*:5 k &:8k &=5kR, = rtr + ( +1)&

R'" : Rrll R :4XR.ie:20k

"' R. _ 25 _ 2'5 _ o.22 k _ 223 ' 1l

' _8-Rc (8)luo: ii-,J:,- ffi _ _.12v/vRou, = 8 k

a, _ _Rcll R. _ _sKI| 5K - _D.4' r.R 25+223

G,, _ ( Rc i| R) = - 9.9 V/V' R"i* + ( + 1)(r. + R)oR c,, = R," y 1,, = 2!3 7 p,4' Rtie * R 25 oe: \hout &: Av : _g^(Rc| RL)

: -l23N

ih &:V.i :5+20smv 20

/o &:

: i- r",. = 6.25 mv

-r = s,:::ll ll-]+y. t5 m5mv (2.5|| l) l-'J|vr| : lv,l x Av : 512.4 : 62 fi

Ex: 6.9g.:40m/v = 1ro= lk : l0or,:2.5k :0.99r,=25 1.:lmAR.::2561Auo : + g.(Rrll ro)

:4010 3x111 t): _296N

n.", = R. ll r : '1.4 k, = +g.(R. || R' |l ,o)

:x3:'l2ott:tr=r"is Rsig + r.: 0.5 v/v^ (R. |l R)"'_ R"--: 0.6 v/v*_-_ : (Rc ll R)''" GR*;. : 54

5(m + 25

yr vo

R,

Rl

B c+

Rs E*"JLt.t: lRu

Exercise 6-14

Ex: 6.55

RJtnt

B

,.R,o = : r"+(P+ X/o|l Rr): 0.s + (10r)(2011 r): 96.7 kf)

R'* : fia ll R,, : 11 96.7 : 28.3 kc,, : vo : vt "vo' vs v\ vl:R,*,

Rs * Rl: 0.796 V/V

(R. |l Rs) + ( + l )(r. + oo||) RL)

^40uvo :

--

\

( + |)(roJl R)

.- : lQ : s ,, 5m,^ =Vo =]@: z" I, 5m

ib : b::-9 and yo = ( + l)i,(r, || R.)

. ' _ V l( + l)(ro|| R)

," l ,lo ( l K.|Llo )(5E . 5 20 K)l

Gvo = 0.8 V/V

R.,, = .oll (r.+tRr|| Rs])/( + l): 20 || t0.05 + .o79] k

-84,, V, x lroll r,1

For RI(k) 0.5 1.0G'.(vr') .68 0.735 0.765

5m

Vc

20K

o.l ./ 0.950.5

2.0

Exercise 7-1

Ex: 7..1 (a) The minimum value of 1, occurswhen

Vo,, = o.2 and U :0.1,thatis

l^ = \ ^c.'rviy 1O.8 AThe maximum value of ID occurs when

%,, : 0.4 V arrd : 1, hat is

^-: \.c.'!vLv:3.l mA(b) For a simila range of current in an npnrasisto we hae

1c.,, _ 3.l mA ' r"uu'^^''u'1cni 0.8 A

Se'mia/v

(v q-"' - v g 6^in) l v' _ a'' v,

_ 3.1 mA0.8 A

v,, : vln(ffi) and v.=(25) mv+ V"' : 2o1 m

Ex: 7.A.2 Fo an NMos Fabricat in the o.5

m pfocess. lih _ lo' ve wan o find he'rsconductance and the intrinsic gain obainfor the followig drain curents: ( : 0.5 m)ID:(ro) *o,r^:f*4!;",,,: (l9ol 4'-= J2y l90x l0 t -02 ja:-v,.:2- _'%iot'= l v

intrinsic gain

:8^r,' = o.z}4 xlM:2lFor ID : 1 A \e hae:

For/: 1 mA

: JzxrsXloX1-z!-v,.:q!. _ 20 x o'5 - l kD lmA

c.,. = 2 + X 10 k : 20 V/V

E: 7J For an NMos fabricated in the 0.5 mcMos aechnology spcified in Table 7.A.1 ith :0.5 m' W - 5 m. and V6y:o.3We hae

o- !p,c.')v'"" =l ,* 4x-xo.3']' z'' "'L "' 2 2 0.51 : 85.5

'^ = ,:

2 X_8. : o.57 mA / v

,^:ui'=20Xo.5-!?k" I' 85.5 A _A. = g^r. : 66.7 Ncr,: wc.' + cou

1:: 5 0.5 x 3.8 + 0.4 5cs, = 8.3 f F, C: Cox| : 0.4 X 5 :2 fF

o.57 4r- - 8'2 lCn ' C" 21 (8'3 a 2'/, : 8.8 Gz

Ex:7.1For his problem, use e4. (7.1):

{"'1vs 7'": l0,For ID

'._ F-\:"DX 19oX X !0on- : 6.62 ! - 6.6 " v- v,.:yo!- _ 20 x o'5 = l k,D 00 g.r. : o.62 1 166 1 : 62 /

".r..(r),,

,8,/5(1(1 , = O.28 mA/Vusing eq. (7.15):

" = ,'^ff-&

Ao : 5oSince g, vaies with o - o *itn ft

xercise 7_2

ForI

1, = 10O A ==r g. : 0'r8 rn(#)' : .89 m./

!

^:50(-!q.)':58/V" l/For 1, = l mA:

!,l\2

. = .28 mAlv(7.J = 2. mA/v

a_: sl.00'l, = svlv" \r,,x: 7.2

Since all transistors have the same

W _ 7.2 U-rol 0.36 m'we have = : la2: /or : l0o A

: /'; -^^{].,.),*

-^)

: L24 mAN

.' v"L, _ 5 v/m0.36 m) 18k!!''' o .l mA

, ^ _1voo|L, 6 v/m (o..1o m) = 2l.6k'n' n .l mvolage Gain is": g", (, ll .-)A'= _ (1 .24 mAN) (18 k l| 2l.6 k): - l2.2l

.,:/=tooAn ': r' : QJ-. = +.e"nvv 25 mvR _ r,= , _ l = 2s"' 8,r 4 mA./V, :Vo: 50V : sfflk''' 0_l m,":|v: 50 :50k()"' .l mAo : 8',' r,, : (4 m/v) (500 k) : 2 v/A': _ g^ (r. || r., ) = _(4 m/V)(5 k || 5 k) : _ l0 V/V

r: 7.4 f L is halved: 9-E!. 62

1v,1 : |v,,|.,, y.l _ s lom (0'55 rn ) : |..]75 '2

- lyol lvol 2 (t.37s v)," - v-, (o_1 \r,llo ): 126 k

Since , : \',c"'l(Yr)u*((' - ffi)

Ex: 7J

w_'

2 (l0 )

Y : z''

'.': ]z *"c.'(l), ';'

Exercise 7-3

x: 7-5

Vo = l.1 vYDD = +1'8 v

Q

vor = 0.8 v

f all transistors are identica and he gate votages

are fixed, |Vrr] :0.7_0.5: .2 vVor: Vr.- = V62 V,n-Von

= .o 0.5 _ 0.2 : 0.3 vthe oest yDs2 cangois |vru] : 0.2 V:. Vo-in : VDtt + VD.2 = 0.3 + 0.2 : 0.5 Vsimilaly, ysca = Yscl : 0.7 VVo = Vs: v+v||+|vo

: 0.8 + 0.5 + 0.2 : 1.5 VVsDj can go as low as lVoul, so

Vo.^' : Vo _ Vsr:'., = t.5 0.2 : l_1 v

: 2 |l,Ay'

,-:Vn: 5 :20k" ID 0.25 mA(a) From Fig. 7.13,

- I RL8- \8

^r ")

,q._ : | + R"' 2 m"/ (2 m/)(2 k):5)+3--+o

40

R:lM:R,_:50o+ l M: 25.5 k'" 4|

R" : l0o k:a,-:5+ l k _ 3 k'" 40R = 20 k:

R.- :5+ ?9_Q = l '" 40R: 0:

R.-:5+g = .5 k'" 40(b) Fram Fis. 7.13,Ro: ro+ Rs + (g.r.)RsRs: 0:Ro : 20k+0+ (40)(0) : 20kR" : 1k:Ro = 20k+ l k+ (40X1 k) : 61 kR, : 10 k:Ro : 20 k + 10 k + (40)(10 k) = 430 kRs : 20 lc:Ro = 2o ko + 20 k + (40X2 k) : 8 kis :' l0O k:Ro : 20 k + 10o k + (.0)(1 k) : 4.l2 M

Ex:7.8 g-, : E^1 : B^_ /o _ lA: i m/V

V nu o'2 l22

t'|:rr2:ro

Flx;7.7 s^: ta:2

0_25 m'25 l2

x.7.6 8^1

= lo :|v o

2

|:8.2=8.=8^=8.0.2mA _ 2 m/v02 2

fn=foz=r:to4:r

: lyol : 2v :lD 0.2 m

Ron : (g.2.2)r.1 : (2 m/v)(10 k)'?:2 kf}

R = (l r,,)o,^) = (2 m/v)(lo k)':2 k

R. : R., | R. :

= _t{',.)"A : _2 l

l0 k

: -}t{z ..){r )]'

xercise 7-4

-V;- 2V - ,nU,,ID 0.1 mAso, (8-r") = l m/v(2 k) : 20(a) For R,- : 20 k'

-' = R, *',, _ 20k + 20k _ |.9k"'' 1+e-tr,,7 1+20.'. vl : _8.l(r,I || R-2): _ l mr'(2 K || 1.9 K) : _ I.74 v/vorf we se he approximaion ofeq' (?.35)'

, =- R. _ | 2k- l"'n' 8.r..,., %' 20 ./v= 2k

hen.

Av : l m/v(20 ko 1l 2 k) : - 1.82 v/vEither method is coect.continuing, from eq. (7.31),

A : gmr|(gmrr.rr,,r)|| Rrf, : _ l m/V{[(20X20 k)] l| 2 k}: 19.04 /vA''' = Au _ 19'04 = 0.5 Vr'" Au' - 1.82

(b) No, for R, : 4 k,- R | ,lOok , I"'z = |^}l,-;' - 20 - l.A/v:2lkA : - | m/v(20 k |] 21 k) : - l0.2 v/v

A = 1 mA/v{[(20)(20 k)] ]| ,0 k}= 2 v/vl.^ : Au : :?Q9 : lq. vlv" Au, - to.2

Ex: 7.9 The citcit of Fig. 7.14 can bemodeled as

Where G-"' I + g-Rsand Rn: (l + g.Rs)r.,The open-circui (no Ioad) volag gain is

A'= -G-Ro = , 8. =.(|+8,Rs)r,,I + g-Rr

so, the gain remais the samef& is connected o he outpt',4, : #;t( + g.Rs)r.,] || R1.: 8. ( + grRr)r,,R

l + 8.RJ (l + 8Rs)r,, + RL

:-r." R,( I + g-Rr)/,

Ex: 7.10

(a) D|: and D1= since yovt = Vo, : 0.2 v ve have

,- :','''(),u""" , ,o' | "c '"(Yr),v""",

: b"(),4

o-+

vi

1

,, (w\ "(w\'" |l, "7,,,(b) The minimum olage allowed across current

source /' would be |vo = 0.2 if made witha single transisto fa 0.l yP signal swing is o beallo'vr'ed at the drain of o!, the highes dc biasolage ould be

v "' |vo0.1 vP = r.s - 0.2 1(0.1)

: 1.55 V(c)vsG2:|voyG"2 can b set at

+ lvtpl : O.2 + 0.5 = 0.7 V1.55 - 0.7 = 0.85 V

Vno = +1.8 V

Exercise 7-5

(d) since curret sorce 2is implemented ith a

cascoded current source similar to Fig. 7.10, he

minimum olage requi.ed across it for propr

operation is 2Vo : 2(o'2) : o.4 (e) From pans (c) and (d), the allowable ange ofsignal s\ing at the ouput is from 0.4 to l.55 v

Vov or 1.35 Vso. 0.4 V < Vo < 1.35 V

Ex:7.11 Referring to fig.7.9,Roo : (g.ro)o|| r.3) andR". : G^z'o)Q|l r"'')

f oI and oa can be selected and biased so that

ro, and ron are very high and have insigificant

effect (/o >> r-) hen,

R"^ = (g.rrot)rnt

Ro : (gfiro1)r.1

since 8.r. : ,Ron:9:ro:Roo = 9rro:Since i, : -g.r(R-ll R"e)'|v.",| : g.r(9zror ll ,.rr)

Ex: 7.12 For the npn trasistors,

v-,' occurs hen ol and o4 are selected ad bias

so that ,o1and ro4 are>> r

Then, Ro. : (g^7ra2\ro2 : 9zrozR.' = 1(25 ko) : 2.5 MRon : (8^ro')rn : FroRo : 5(20 k) : lMoFinally,,,", : _(8 m/v)(2.5 M || 1.0M)A,^'' = _5114N

Ex: 7.l3 - = 1/.l = J.4 -- 1 mA/v /) m' =9 = l : z.sR- mA/v' :Vo:10V = t/. lmRefeing to Fig. 7.20'

R.1 r6fr + s.(R.ll r.)l

lr -1-: .z = fi:'., = '.,

: Lgr=

lv..l :

0.2 m -

R" : i kl + 40 +(o.5 k l 2.5 kf' R": 176.7 kwithout R. (hat is, /?, = 0),

R, : rr: 191

Ex: ?.14 Fig. 7.21(a)v,rot:ro2:ro=i: 50 kf)

.l7;

"6(r7,),l -A)

25 mV

1 : tu.s 8 m/v

5V : zs0.2 mA

5V _0. m

From Fig.7.l9,R"^ = k,2r6)Oq1ll r "2)

= (s m/v)(25 k)(25 k || l2.5 k)R"" : 1.6? Mf)For the np tansistors,

c^':c.o=:vr =,ro:

*:

c., : 1#' G^: g. = 1m./v

" : lg1 _ o.lA:4mA/v 25 mV.-:-9_ = 1 :us

8^7 4 m/Vssuming an ideal cuent soulce,

Ro : (8'ro)(', o|| r oz)R" : (4 m/vxso k)(5o k || 25 k) : 3.33 M.6 : 'G^R6 : -(1 m/v)(3.33 M): _3.33 X 13 v/v

0.2 mA -25 mV

50-8 m/v

8 m/V

6.25 kfl

roj: ro4= H: #h: '*nR", : (.:rlXro ll r.:)

= (8 m/v)(2o k)(20 k | 6.25 k)

R., : 762 k, : _8.l(R"" 1| R"p): - (s mA/v) (1.6? M || .762 M),, : -4.186 Vr'

xercise 7_6

Fig. 7.21 (b)

From part (a),

g,'1 = g"2:4rA1ro\: r1 = rr,: rr:50knl : to2: r" = 25k

G.=gq:4mA/FromFig.7.19, Ro = (g.zro)Oorll ,

"z)R,, : (g-1ro1)R" so,R" : ( 1 m/)(50 k)(4 m/v)(50 k)(50 k || 25 k)R,, - 167 MA,n = _8-R- _ 4mA/v(l67M): _668 X 13 v/v

E: 7,15 n the cuent source of Example 7. 5we hae 1() : 10o and \e wan to redcethe change in output current, 1(), coresponding

to a 1 v change in output oltage' yo, o l7. of

o.

Tha is r, _ V o - 0.0l /, = !Jfoz roz= 0.0l l ,o.:JJ:161' |

,o^:V'o'L-tM:20Xa' Io l- l) vr- =-::rrm

20 V/mTo keep yov of he mached transistors he same

was ha of xamp|e 7.l5. 7 of he fansisorshuld remain the same. Therefore

W - l0m=W:50m5m lmso he dimensions of he mach ransisors 0!and 02 should be changed to:

W:50mand,=5m

Ex: 7.16 For he circuit Figure 6.7 ,/e hae:(w / L\,

z : n t'_;-_::-,. :

w / '\.nd . : ':-----------:' "w / L'4

. (w/L)\*'''@77'

Since all channel Iengths are equal

L:Lz:...:Ls:lmand

x1'1, : l0 A, 12 = 60 A' 11 = 20 ,: : 20 Aand /5 : 80 ,e have:

w' w' . |'' "'"--+-- _ -:-L _ - - 6' ^' ' 'y w | /*r. |o .,_W., _2-1,' _,R'' _ r-* |o -

l.=t'\-\=!=:+' 'Wo w I 10n order to allovr' the voltage at the drain of02 togo down o within .2 v ofthe negatie splyvoltage we need yo2 = 0.2

l' : \ ^c,^(l),' o", : \|'(l),v' -,60 : ;2$(}),orr-

[) _ Iz - _ l5=+w1_ 5\L2\ L 12 2 t (o.2)'y, - 15m. _ u=w' = _ 2.5mwt 6w-*' _ z=w\ ' 2x W,

_ 5m

n ordr o allo the oltage at the drain of5 ogo up to wihin 0.2 v ofposiive supply e need

vou, . = ,(l), =v"o"'

l : io$(}).o.zl'=l.y] _ zx - - 50--eWs 50s\ ,, 80 10.2;2Ws:50m

Y: = o.- v. _ lqJ*m _ l2.5 umw " 4

0._5 m - Inr (,+2_o.7)l+(2/l)\ 50 ,/

/a6p : 0.5 .Affi : 0.497 m1^,- =v- vuu-p = vcc-v"tR /"..R : 5 0.7 : 4.3 .s

0.497 m 0'497Vrr,n : V,.n.n. : 0.3 VFor % : 5 Y From equation (7.74) we have:

l = n' (, +V"_V"'\' l + {2/)\ vA )

I _ 0'497 (l + :-qJ) _ o.5.| mA" l + (2,i |)\ 50 /

Exercise 7-7

Ex: 7.19See next page:

Thus:

| : 2.5 q"m' Wz : 15 m, lv] : 5 mW a : l2.5 m and l{'' : 50 m

Ex: 7.17 From equation (7.72) we hae:

'.: '"..[;fuJ{'-'o)r\

. = l mA|- ,, l(,' #) - t.O2m\' '-i_/" : 'l.02 mA

Ro=,o,= = #* =98 k : l ko

Ex:7-18

Fom eqation (7.74) we have:

l^: Inr'r rtVo-V\=" 1+ t2/9)\ V^ t

Exercise 7-8

gnoring he effec offinite output resisances' wehae

: ]z : ... : / : /cOllr'

rQ"u, * : /r(*)

1 : /r",r,* u + ... + N

, /ar"., c, ' ."

/tl, = ,co*,,. + ,.r*,r,

- /" aa-l.oRF ' - ryl]

Thus:

For an error not exceeding l0% we need:

/^.,-a/r(| 0.l)t + il-j-.1

/t ?o.9r".' l >0.9!',ry+l l_+

- r + N-1-1 = I = I + {-L-.1 = I.r r0.9l-l

= o.It=9iv + l S 0.1 / + < ly'< 1The maximum number of outputs for an eor notexceeding to be lesstha 10% then we need N

Exercise 7-9

Ex: 7.20 Refering to Fig. 7.32,

Io : Io : ]ol = o: I*uo = l Since , : \'"c"'()v'."

lz a l v]l(J)

= 0.23 vThe minimum ouput voltage is

v*'+ 2vo" = 0.5 v + 2(0.23 v) = 96 vTo obain the output resistance, Ro, we need 8.3-

o ' o' _ 2(0. m) _ 0.87 mA/ vvov/2 o.23 V

: v(L) _ (5ylrn)(0.]6 m).o2_ o! h _ _-

o.l ma: 18 k. From eq. (7.77)Rn- g6rorrn7: (0.87 mA/V)(t8 k)'?: 282 k

Ex: 7.21 For the wilson mirror from he equation(7.80) e hae :

/,: l : .ggg/r ' + 2( + 2)

Thus !!:-sd x fiO : o.o2%

whefeas for the simple miro from equation(7.69) we hae :

1" = 1 = o.q1 l' + ?

"n"" ln _ *or| x |: 2%1*oa

For the wilson current mirror\e hae

R- =pr. _ lk_5Mandtor2

the simple mio R, : r, = 1 kEx: 7.22 For the two curent Sourcs designed inExample7.6. \re hae :

n = , : l0 _ o.om-ov 25 v

"^ _V^ _ l0v_ |o M..' = . 25ok" c l0A 8.

Fo the current source in Fig. 7.37a we have

Ro: roz = r,:10MFor he crrent sorce in Fig. 7.37b from equation(7.98) we have:

R.a tl +g.(Rrll r,)lr,n Example ?.6, R, : R1 : 11.5 1 ,therefoe.

..- l .a l t.s ll 25ok)'loM'-'-L - v ',J

=R,: 54 tr1

"',"'()

Exercise 8-l

Ex:8.1Refe.ring to Fi9.8.3,fR, is doubled to 5 K'

Vo : Voz: Voo_ Ro

= l.5 - 0'4 mA5 l j.s

v2

vc.u': v,+vD:o.5 +0.5: + l.0Vsince he currents /D|, and 2 ae still 0.2 mAeach-

Vo, : 0.82 VSo,V.r.,n = yJs+ yc + yoj

= - 1.5V + 0.4V + 0.82V = -0.28VSo, the common-mode range is

-0.28 V to 1.0 V

Ex:8,2(a) The ale of 1,ir' hat causes l to conduct he

enire current is ^/Vn,'l

"D x .:t : .5 v

th, yD : vDD xRD= 1.5 - 0.4 x 2.5 = 0.5VVo=Voo:+l.sv(b) For o2 to conduct the entire curret:

:-JiVov__o.45then.

Vo = Voo: + l.5VV.,2 = 1.5 -0.4x2.5:0.5V(c) Thus the differenial outp range is:

VD2- VDt.from 1.5 - 0.5 = + I Vto 0.5 - 1.5 = -l V

Ex: 8JRefer o anser tabe for Exercise 8.3 where val-ues were obtained in he follolving way:

v^'' = ^/I/ w/''')L = L v n2

":

l "'r' ' : 0.l --l ll , = 2 ^' "loj\ Vor )

EE: E.4

':'= -8 mA : 0.4 m

', 2 " 2(/ m)'- _ l"_ J;r(,.,v|) _ -''/^'7'

"-_ o _ o.4m(2) = 4rnA/vo/2 .2 v

,^:V^:20v : s" lD 0.4m4A1 : g.(Rrll re)o : (4 m/v)(5 K|l 50 K) : l8.2 V/V

Ex: 8.5with 1 = 2 , foral ransistors,

^= ! = 2P: I,,.a.'22L : 2(o.l8 m) : 0.36 m

- (10V.zm) (0.36 m) :,uo.l mA

+ Voo

Ro:5k Rl=5k

Since 1,, : lo, = L^ r^()r*' ,

(f),: (), = #tr}2( l00 l

(4 A / v'?)(.2 v)'

(y).: (9. :;#r2( l00 A) .n

( l / v'?)(.2) '

so,

= (t_4J(2) : 1m/V,

^,0o,l1 ,o) : 1( m / v)(36 K || 36 )18V/V

" : \'"(l)oul' so^t

Exercise 8'2

,o : |v _:

The drain crrent for all transistors is

' : : '*'*o : '* *o., = (10 V/ .)(!,36 m) _ ,u*nRefering to Fig 8.12(a),

Since D : \'^c"'()v""f foall NMoS '. = p"'lus''t---------------

= J2(0.2 m / v')( |0)(0.4 m)

Ex E.6L : 2(o.l8 m) : 0.36 m cMRR(dB'= 20 |ogl0 H -'oI"lo (ffi )

= 86 dB

Ex: 8.8From Exercise 8.7,

W/L: l,"C.'{o.2 mA /2)," = :=

08fl : o.a..

,5 0.7t,

nC.'(V o"

rw\ rw\_ |)._ \),_ 2( | )

4 A /2(o.2)2

8^ = 4 mA/sing eq. (8.64), and the fact that Rss = 25k

,"** _ (2 "'Rss\ _ 2(4 m/y)(25 K) : 20, oo('\ 0.0t\c-l

cMRR(dB) = 20log lo( 20,0) = 86 dB

Er:8.9

1k : 4.3 m

+0.7 v

v61 : -5V

+0.5V

Exr 8.10

62: o.7 ( 5+4.3xt)

lk 1k

- 5V+-43 mA(:l)

_25_ V

(g),:(g).=(y),:(:),

= 2o : 2(tA) :.npC.'(V o)2 1 p / v'?(o.2 vf

For al transistos.

,q:+P : 1m/VFrom Fig. 8.12(b),

Ron : (9616)16r: (l mA/V)(36k)'?= l.296 aRoo : {g-5 r65)rq: (l m/v)(36 k)'z= 1.296 MUsing eq.(8..18).

Ao : s-, (R..ll R.,): (l m/)1'296( M|| l.296 M)

= 648 N

Ex:8.7The ransconducance for each ransisor is

g^: "q'1w ll"l^ : ! = 0.8 m _ o.o.o"22from (8.35), the differential gain for matched

R^ alues is, _ Voz Vo _ n pV'"

fe ignore the 1% here'Aa : 8. Ro : (4 m / v)(5 K) : 20 v / vFrom eq. (8.49),

.'' = V.a = * R, _ (0.0l)(5 ) _ o.l v/vv"- 2 R'' 2(25 )

+5V

Exercise 8-3

c = c : l : z =: 0.2 m

From . (8.66)'vcMfr'-vc+o.4= Vr. gR+o.4

0.4 m

= 2.5 _ 0.2 m(5 K) + 0.4 v : + l.9VFrom eq. (8.67)

Vr' ^rn

= -v+vcs+vVct -rn : - 2.5V+0.3V+0.7V:-1,5VIpu rage is - l.5 v o + l.9 v

Ex:8.11subsifuting ig1 * i62 : / in Eqn. (8.70) yieds

!:2

Ex: t.13

v84

I

, * "(z t)/

/:20oASince>l,

I-'-^^-/ 2_22l

^:o : !:v

Voz

Vz

l0o

l0A:4.zv25 mV

lv^lc

50 k

25 mv : o.25 k.1 mA

0.99 / :' B1

D|\/

B1 _ B2: _v. l"(s_ t)= 2stn(r / 99)= 25 ln(99) = l15 mv

E:8.l2(a) The Dc curren in each ransister is .5 m.Thus yB for each wiIl be

V -. = o'7 + 0.025lnr)"' \L/= 0.683 V.+=5-0.683= + 4.3|1.nlo : 0.5 :2ovT 0.25 v

(c) jc : o'5 * g.13r,= .5 + 20 0.5sin(2 X r)

= 0.5 + 0.l Sin(2 X 1r)' mAicz = 0.5'0.lSin(2 X Ir), m(d)

61 : (V66 * 1cRc) _ .1 Rcsin( 2rr X l)0)= (l5 _ 0.5 l0) _ 0.l x l0sin(27rl0|)= l - l sin(2r' 10r)' vv.2 : l0+ lsin(2r'X 1000r), (e) 62 _ z., : 2'sin(2 Io0,), v

(0 voltage gain =c_ c

7A1 ' 7'A'

= 2LP"uk = zn vl.l vPeak

Rc:Rc2:Rc:ro

_ 10v :l)kl

. _ V, - IOV 7 _ 2 - _'

",="r:t':i:Since R.. >> r",

|A =R.l' ,o - |0 || l K _ 2o v/vr. 0'25 R',=2r-.-_- | _ z5l" 8^ 4 mA/YR'o : 2(25 K) = 50 Kfthe otal load resistance is assumed to bmismatched by l%,

l -l = R.

= (0.ol )(l0o ) _ o0|2R'' 2(50 )

cMRR(dB) .: 20 loc,,, lj:l

::86d8

Note: fonly the oad ransiso$ are mismatched.andsince-1.

+Vcc

Il=zrl"

V

Exercise 8--4

u _ _ Rrll Vo u _ _( K || l K)yl..'ol 2Rfr. r,' ''- ,(5o ) , o25 : _o-499 vi._,, _[(Rc + R.)|| r"]v'-"o= 2l;;T;7-_ _(l.0l)t K]|| l.v = l5(t|y.

2(50 ) _ 0.25 K

CMRR: l| _ 21,.l 0.50l 0'499

:10o.00--)10dBUsing eq. (8.103),

r*&R;.. : Rr. ---"- _ { l)(5o K)

ro

lK. l(l ). IK+2(5K)

lR;.. : l.68 M

Ex:8.14From Exercise 8.4:v*: o.2Using u. (8.l08) e obtain y due to & / RDas:

,.,: () (*)

_ Q x o.oz _ o.)2 v i.e 2 mv2

To obain y^. due Ayl- wlLUse n. (8.l l3)

"*:()(#)- v^" : (92) x o-o2 = 0.2"" \2 t

--r 2 mVThe offset oltage arising from vt is obtainedfrom n. (8.l 16)vs= vt = 2mFinally, from n. (8.117) th total input offset is:

t(? *' (y!! ^+#)' * (',)'l'':

^/12 x ro t1'+ q2x 1o-r;2 + 12 x lo-312

: i'" 10: 3.46 mV

Ex: E.15From Eqn. (8.127)

,,, = ,.ie:*(J:256.f * ,1r: 2.5 mV

l_ : 1 : l =s,,a'" 2(+l) 2X1l

u,: u():0.5 X0.1 =50n

Ex: t.16

(w L),x ^C'": o.2m x 1:20 m I

(w l L) x : 0. m x 2 : 20 m $since all transistors have the same drain cuent(1/ 2) and the name product vl then alltransconductances 8. ae identical.

" ou _ ^!fiir -thus,

: o'2

" { .8 m-/2) 8.=;:-l-=*.vFom Eqn. (8.138)

G-: g- : 4 m/Ro = r orll ron

,^' : yo: 20 : so"' o' (0.8 /2),^' = y-' = 20 = 50k"' lo' (.8 m/2)thus,

Ro:50|| 50: 25kFrom Eqn. (8.141)

: G'Rq: o"25k: lFrom n. (8.l48a)

A-- | = O.OOsv/v-"- 2g-.Rr, 2x4x25

1"rr = lAo| : l = 2.0{)l."l 0.5'i 86 dB

Ex:8.17From Eqn. (8.156): :3"

I/2 (-8 m/2)""- v, 25 mvFrom Eqn. (8.159):

Ro = r nrll roo

'ul

Exercise 8-5

Vou Vo 1 V7l ='-2

: lV = 125 k0.8 m

A,t : G^x Ro : 16 x 125From n. (8.l62)R,": 2x ro

-2x vt g =_ |/2)' "= 20 k

2x25mx160

For a simple cuent mior the otput resistance(thus RJ is r"'_,R"' _\ = |V - l25 k

0.8 mFrom Eqn. (8.67):

"R*

' -2 / |25 "'- _ l6i, la5 A : .OI25y

- _ l20 lL MRR lJ tr5l

Cr*^ : 160, 02 og0(160,) = l04 dB

Et: E.18

^ l/2 lm/zU-:8m:':'-:-:=- 2) mA l v/2 0.5mA

-) Roa = aro4 : 50 x 2 K

R^. _ . ls : l, ?9L! _ 16 112

From qn. (8.174)

n, : [.

,^l| a']= (l0|1 10) M : 5 MAl: 8. x R': 20 50 = 15 v/v

i.e. l dB

Ex: E.19Refer o Fig (8.41)(a) Using Eqn. (8.178)

.=(W'L\o(/2,'' (w / L')=1111 =

(W/L)n r anl)

thus. (lVl L) 6 : 2

Using Eqn. (8.179)

. = !! |' (w/Lt\ '= 11p :

(w/L), ; 16112

thus. (}v/ )' = 20o

similarly for 0,, %a : 0.29 vFor q,trn: x920 y^''2

yov6:0.105v

a,2-=!

a,

a,

O^

50 0.129 V 0.775 mA.V50 A 0.129 v 0.775 m./vl0A 0.105 v 1.90 m/v

(d) r", = 10/0.05 :2ko& = lo / 0.05 : 2 kn :10/0.1 : lkr" :1o/0.l = lk

(e) n. (8.l76):A : _8. Qo'|l ,o): -775 12 ll 2} : _?7.5 yvEqn. (8.177):

Az : g. (roll rol)

oerall volage gain is:A' x l,: 11.5 X 95 : 7363 Vr'

Ex: E.20Refe.ring to Fig. 8.42, al| /D aues are the same.so' '.|. : yct + /AR,Usig the equation deeloped in the ext'

:2oy(b) Fr o''

l : \,c,,'(l),v*",(0.8 m/2) : 0.129 V

:20m

2 k

: l0 M

r"_ffi"tr]n-- 2 ( l.l"

^E ,lzlll, l ,a, :0 l

1rlrzn2

= 5'21 k

Exercise 8-6

8^:

Ex E.21/,:p'C-: 1 pA12p"C-: AN2Fr Q."nd Q,:w L: 40 l0.8(as given in xample 8.5)

lv oul :

)lv oul".n : = 0.3 V

then.

2I" 2X9ou8.s'9:|v=_'-

since 8. of o, o11 and o,, ae identical to & of and 0, hen:Y.,, = 0.3 vThus, for O,.

o.1)2: 29' l 'lw/L)n)(W/L)0: 12.5

i.e. (10 / .8)Since Q,, is 4 times as wide as Q,,, then

rw\ 4xl0 40i,, _ ol _ ^s

( -1)

v-,,: v,"," - v",.', : 0.l5 + 0.7 = 0.85 Vthus, y",1,, = vcs 2 + , _ yss

= 0.8s + 0.15 - 2.5: - t.5vV on ''

: |Vou"| = 0.3 tVcsl: o.3+0.7: lvVe : _ 1.5+l: _0.5VFinaly'V.": V," - v*.: +2.5 + (-0.3 - 0.8): +1.4v

Ex E.22R'=2o.2kd- : 8513 v/vR.:152with&: 10kand&: 1kA,': 20'2 x sl:l x ----!-' 20.2 + r0 (r + 0.rs2)

:4943N

Ex E.23

il : r+t: Iotba

irr: R, - 15.7 , o.u9zi,l lQ5 + Ri4 l5.7 + 33.5

=,:l

): 0.38 m/v

-+Rr: 1.67 111The oltage drop on R, iS:t.67 k x A : 150 mY

: 0.26i.s Rr + Ri3 3 +234.8

:'=lb5

i5 _ il+R2 _i"z Rr + R2 + Ri2

?:,:r|

Thus the overall current gain is:

_ Il x-M92X 10ox.0126X lo..|

X 0.8879 X 1: 55993 ./Aand the overall oltage 8ain isvo , Ru -i,"V,, R,, i,

: a256 N

il R3

: 0.8879 + 5.05

: 0.15 V : -- 591't't

C,''(w / L)

40o;

2x'|xlgxqo

160uX:Y' .8

Exercise 9-1

Ex:9.1

'4M :

R-&- x s-(R.ll RD)c,, : \wc,,' * cou

: ?x t, 1 X 3.45 + |-72 : 24:2 fFCg: C6: | .72 fF

l0 .-^.-l0t0+0.1 2

9.9 V/VI

,cc(Rc + RC'" = =:P_ :

/'* ll,! voC ooo 10 : 4.1 fF

Ex 9.4Peak current occurs AtVl= Vn : 5

ea : |'(l)' v, rv,^l'_ 1\2y205_2)'z- 18

2

Ex: 9.5

n-=r:lmA_40mv 25 mV vCo":.r'8.:20x 10 12 l0 ]

: 0.8 pFC\.:2Ci",,:2x20: ffc,: c," + Cl" = 0.84 ppC: Co

( * :'e.l-'u'\ vc

lo :6l m

: 0.016 Hz2n| x (l0 + 0.I M)I - I _.: ;e '' '-

=

--:_- : 3l8 z

_1

-'z_lJl 0.6

f =-fn: 318 Hz

x 9.2Cc:C=C62:lF

3. : ao$-+l. : 4mX25rn : l m

r, : 2.5 k = =: ]8-

. :V':25mv ,'" l, ImA,:l

2 zC., IR,ll r, + R,,r]

2cc2(RL+ RD) 2rl X (l0 + 0):8Hz

2z1[1 K|| 2.5 K+5 K]: 21 .4 z

2 fF t2 fF,o ' r-, * Rll R 'rl"L' + J (, * u1)""

8." _ ,;G;'

_ ,0 X l0 l = 7'47 cH2(0.84 + 0.|2) Y |0 ,

Ex: 9.6

l'l - f- _ 'o : fJ

.fr : 50o MHz

] _ i-;;\4 l r( ,'7 + ('U" = ---j:----_j-:- - 12.7 pF' 2x5fi)\ 106

C: l2.7 _ C: 12.'7 _2 : |o.'1 pF

Ex:9.7 Diffusion componen of cat /, of l m:10.7-2:8.7pFSince C,i, is proortional to /., then:

2 ' Cc2' (Rc + R'\

I

2"l(8K+5)f.: |2.2 zEx: 9J

- n' 3.45 X 0

'"'_ ; \ l0': 3.45 X 10 rF,i m2 : 3.45 fF / m

C6v: WL6vC",: l0 x 0.05 x 3.45:172 tF

l * vo

Exercise 9-2

cd,(: .'| m) = 0.87 pFC(r: 0.| m) : 2.87 pFf-- : o.| m) = -----_g!-' 2'(c + c!,)

: 4x10 r2(2.87+2)X10-12

: l3O-7 MHz

Ex: 9.E

^ :- Ro - o'M

Ro + R,,.u'"'R': 7.l4 k, g. : 1 .47

Rg, : 10 t_4'7 l1 Xl7.l4

(4.7 + 0.0 ) Mf}_7'12 /

cia:'' + |(1 + X 10_3 + l.5 + l)=68pF

C., -- 4,26 oF2C'"(R';.||R61

2 y 4'26(lo K | 4.7 M)

= 1 .42 Hz2 68p . 1.65

Ex: 9.11 Usig equations 9.6l and 9.63, we canwrite th genera form of the transfer function of adirect-coupled amplifi e. as:

A): Aoc here r is the Dc gainl+ s

2nfo"of the amplifier and/3J, is the upper 3dBfruency of the amplifier.n this case we hae : 1000 and/Md : 00 Kz = 1 HzTherefore A(s) - 10

t+ J2 lo5

Ex: 9.12For this amplifie we have:

t():('.*)('.*)

By dennition at = ,we have

|a11,,"1f : +A1

(,-(ff)')('-(#)

Ex: 9.9

cs":1pFC.q: (l + g.R)C"a = (1 + 1 x 7.14)C ra : 8.l4Cra

/r > 1 MHz.e

C^: C*+ C,

1MHz , A'M ,_'2rCii(RsisllRc)= I pF + 8.14 CslD

1

2(l + a.l4cga'pF(l K || 4.7 M)1.63> + 8.|4C"

C < 0.o17 PF or C< 77 ff

Ex:9.10A": _3912: _19.5 V/V

=1MHz ['.(ff)']l' .(#)'l='f 72 : p1and fl = 0.9 l, then

['-()']['"(**)'] :'

(l +o.g'?)(l -()') = ,

' -()' : l.1=(l)' : 0.l +K : 2.78

f , - 0.99 t, then :

['-eP]['-(#] :,1r *o.oo,1(r -()') = ,-

, -()' : r.ol =+(s)' : o.or =K : 9.88

' R^" _ nr+*,0

- r"' 8-' Rr+r+(Rr+Rsig)

2.540.o-]XR;1 + 5 2.5 + 0.05 + (10 || 5)

: -0.013 x R.=n. = t.s k : ro|i Rc|l R.

l.5 k = (10ll 8 1| R.) k: '7 .4 kll RL

-J R, = l.9 k

( f '' : I R_.^ __ l.65 k2C,. ' R"1" ''"

Cin : C + c(l + s. R;)

Exercise 9-4

Ex: 9.16 Refering to the solutio ofExample9. l 0, the ale of| determined by the exactanatysis is:

f : f"1 : 143.4l1zlso'Au: _c'.RL: _l'25x|o: _12.5NTherefoe the gain-bandwidth pmduct (f,) is:

f, : 143.4 x l2.5 : l.79 Gzsince, is less than, :2.44Glzuld:40 GHz, theefore it is a good aproximation ofthe unity gain frequency.

Ex:9.17 Referring to the soion ofExample9.10, ifa load resistor is connected at the outputhaling the value of R; , then e have

n;: rolL roz and herefore

|A'| : s. 'o' 'o, : l.25 x : 6.25 N

Using equation 9.92 and assumingJ,s/",,

I

0 k( in example

2.{{C'" + C "o(|

+ .RL)I. R:ie + (c. + csl)R;.]

"" _ 2LL20 [ + 5f|l + l'25 5)l. l0 K

+(2sl+s/)xsKl'' =

223 |Hz

1, = |M|.f l : 6.25 x 223= l.4 Gltz

Ex: 9.1t Refering to the soluion ofExample 9.0

s,, = +, is1,,is increased by4 atdV*by 2

then :

'^=ffi:2,.,: 2.5 4

To calcuat Rilf Rr, = ronrll ron, =9.0

'Since ro J + increasing /, by 4 reducesboh roQ and roq2by 4Thus :

,o,|1 r o = lo-sil(rrll'o)

nj : lxroK = 2.sko-4|ul = g.. R, : 2.5 x 2'5

: 6.25 AUsing uation 9.93 and assuming/,1/",we hae:

2|[Cr' + Cr7(l + g.,Ri)iR"I* + (cL+ csd\R;}

21[[20 f + 5 f(l + 6.25)]l0 K + (25 + 5) f X2.5 Kf H : 25o l1jz -+ f,' !f" : 250 Hzf,=lAtl' 7" : 6.25x25O: 1.56GHz

Ex: 9.19

,.""^=+: ffi: r:o",.=l?l

:#:son; : ,",r"ll roono: 130k| 50kR;:36k

,a lmv 25 mv

'-:E-: M :5k" ' 4n v

(a) From uation (9.97) we have:

lu: - ____.'--G-n;),s|3rf/: _ --_j__{ 40 X 36 k) - - 7516+0.2+5 ',-

A,, = -fisy(b) Using MiIIer's theorem e have:

Cin: C'+C'(|+.RL\= 16 pF + 0.3 pF(l + 40 X 36) : ,48 pFC;" : 448 pFr- |'"_ 2nc,.R'," 2C,^tr"|| (R.ie + r,)]

_ l :8z.6kz27 448 pFt5ll (36 + 0.2)]

4.3 k(c) USig the method ofopen_circuit time con-stats, from equation (9.l00) we hae:

r: C,R.'. + crt(l + g.Ri)R.l. + RL+ 1LRLWe hae i.,. = r- ll (R.ie + r,) 3 4.3 ki = "-ll ro,n, : 36kO

:46m4

v

= zx t.zs !

t!92 : L44

xercise 9-5

f :.

I

Thus:

to : |6x 4.3 + 0.3[(1 + , x 36)4.3 + 36]+5x36o = 2.12 nsec

I

f, : :L :75.1 kHz(d) Using equations (9.l02)' (9.103), and (9. 04)e have:

40 4, 18. I v - -,.- .;Hz2Cu 20.3 pF

2c "

+ CP( + g.R.)]R.', + (c + c)R;

= f pt = 75.1 k{z

'fz =

l [c, + c,(l + 8.R;)]f;i + GL+ c.)RL2z IC"(CL+ C) + c.c]R.iRf : 25.2 HzSincef

Exercise 9-6

+1H: css' Rsig + .25 x cs,X 21 XRsi=Rsigcs,X6.25

cscoDE - Amplifier:Using 9.l 8 and neglecing the terms that donot inclde Rsis

a : R.i[C, + Cr/l + 8^R\lR,,: r^ll (t'+Rtl : ,^11 2'o"' "'' \ g^ro l "' -ro

/'#1---l , : R.;r|Cr' + 0.25 X cs,l +

= Rslg ' cs, .75

"fa cscon : '""-/ cs cAscoDE

_ R.ia.Cs, x 6.25 _ 3.6R*ip.Cz, )/ L75

c) f = |Au|' f .fr.n..,,,n. /lvlc,scoo\,lt'"fac.scoo\,- 1- ., ' /,,,- J= 2x3.6

Ex: 9.24 Referring to he solution ofExe.cise 9.19 we have:

s.:40 and r. = 5kNoe that for the cascode amplifie considered inthis exercise:

t|:to2=r.:5k,^ mA

8n : 8-z : 8- : ,'J -Rn : r.t + r : 5 kr, + 0.2 k = 5.2 kf}A. : 8-.'. : x 130 = 52 /Ro : ro: ro : ]30k^ r^1+ R 5,(. ^r 5.- R' 20+ l

"' F+1._ 130+50

13 + lq20

R1"2 : 35 Ro

=p2r92 : 2 X 130 k = 26 M

o' : ;* r'* ,_L"'c'('" ll R.,)

A.. - azy

To calculate, we use the method of open-circuittime constants. Fron Figre 9.30 e have:

R',,. = r.1||(r,r + R.1.)= 5 K|(0.2 + 36)R'.,. : 4.4 kR., : R'",. : 4.4 kO

Ro, _ ,o,l R,", - ,,ll [,"J ', - {. - I -l R,. ll

L (",*ffiJlRol : 130 I| 35 :35 Rnl : R',r*(l +g-,Rot\*RotRr1 : 10.6 kn: C..Rot * C*l .Rn1 +(C""r+C.:)Ro| + (CL + C,", + C"). (n ll Ro)r, = i6px4.4K+0.3 px 10.6K+(0+ 16p)

X 35 + (5 p +0+0.3 p)(50 | 26 M)t, : 339 nsr,': | _ l - 469Hz2' 2 x 339 nS _

f ' =|u1' 1o : 242 x 469 1-lz

: l l3.5 Mzcompared o the cE amplifier in Exercise 9.l9,

l.4rl has increased from 175 V/V to242 / 'J"has icreased from 75.1 KHz to469 z and, has ineased from 13.1 Mz ol l3.5 MzTo incease, to Mz we need:

.,, : -_L : 159 ns_ hus" 2of "

16 p x 4.4 K + 0.3 p x 10.6 K + 16 p x 35+(C.+0.3p).(50Kll 26M): 159 ns+CL: l.4pf

Ex: 9.25 R'. : R ll , = 2 K |1 20 K= l0K

From Eq. 9.121 we have:

: ,"ll'# ='fi

, 8^R'^"

_ i^R,

"8-J ;q"'':8Gz

1.25 x 10 : 6_93 Y1+1.25l0 vl _- 1-25 n2 (2of +5f)

r' = | .8. _ 1 . !-9 - 196112-' 2 C"" 2 2ot _R3a = R,i3 : 10 - R.i"+R',. 10+l0K

t-s-.R! t+t25xto: 1.48 kO

Exercise 9-?

R.. : R.ll R. : n"1 (."U ;):20 K || 20 K 1| 8

R.. : 0.74 kTo find r, :

79 = R'7C3a : 10kX5fF : 5ps., : R., X Cr. : 1.48 k 20 fF:30 ps7cL: RcLC = o'74 kX 15 fF:ll ps'rH: '* r,+ca: 9l psThe percentage contibution of ime consants to

, associaed ih c"i, c.", and c are:

E x t = 55ga: 19 x l = ]_37:9t 9t!!'ul:lzco9l

|- - | - l _ l.75cz2' 2 9|

Ex: 9.26 e have

,'. _ 9V' _ l 25 mv - 2.5 k-lm

n :L: J-: n,'ll 25 mvnoting that in this case R, -+. , \e hae:

G"= ( + l )(r" || R1)R'.ig + ( + l)tr. + (r,1| R.)1

" + lR'*i : R.lc * r, ssume r, : +c - 0.965 Yv

zoc -!:-' + l8-

2nlc "

+ C!\

mc" :4" -c,, - ^---=:- -2pF'' 2 l. n 2,4Hzc" = 14 FThusF : +l :459MHz' 2C.r.'e hae:

R, = R'.', ll tr" + ( + l )R'.]R'.,* : R.,.*r,: l k+.l k

: l.1 kR'' : R,||r,,= l.l k]| l0o k : 0.99 k

R : 1.l kl| [2.5 + (l + l)0.99]R = 1. k

we hae:

= R''i*R'+R''ip+R'r r.

R.:5 flwe hae:

_ l.l + 0.99- 1.1 . o-99

2.5 2.5l

2n(cp + C.R,)/ : 55 MHz

x':9.27 1a1 Vo =

e have:

r: 1" 2C,r,+ l

2zrC -r-

n2Yt n - -L o8m - o aVo o'2 'v(b) 7 : g-(Rr|| r") where,-: Vo : 20 = s" (/2\ 0.4m- A,t: 4 m(5 K l! 50 K) : 18.2 Y

(c) For a CS amplifier, when R.i8 is low:

r. : I l E-9- l 5l2(C, + C

"nlR'1 '

here

R'. = Roll r, : 5 K|l 50 : 4.5 kand c. = l fF + CDB : |lo fFhs.

f. 295 MHz2?r( |1 f + l0 f.) 4.5 K

(d) Using the open-circui ime-consans method

for R. : l0 k

/, : _L(Eq e.8s)where, from Eq. 9. 84:

, : Cr,' i, + csa[R,(l + 8.f'.) + R']* C'R'

thus' 11 :50f l0K+...10fuo K(l +4 m x 4.5 K) +4.5 Kl +...(lf+lf)X4.5Kf, : 0.5 ns +1.96 ns +

0.5 ns : 2.96 nsfH: 1/(2 2.96ns) :53.- z

l x 0.2 m/V2

Exercise 9-8

Ex: 9.28

,: IJ z 2' C.5' R55

- I :15.9MH22z ' (0.4 p)25

Ex: 9,29 For a loaded bipolar differential

ampli6er:1

Ad : tgn rowhere,

/2 _8^:_

/2

Using (9.l53) to (9.157) ve have:n'.: R,.ll Ro:2oKll K = i5 kRr2 = R.ig(l +G.R'L\+ R'. = 235k

R +R. R"i" + Rs..: -----_---------_ 20"' l u 8.R, ";u,

|] 2

':_o, R.ie * Rs

').

2+ lR._ ! = kRs. _'"j'-t'sk

T, = Cl?g, + C' Rr1 CR':2o f(1o.5K + 5 f 235 K + 5 f 15)

" : 1.46 ns+ f 11 = }a: '* '""GBP = 10 X lo9 Mlz:1. Gz

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