Exercise 1-1
Ex: 1.1 when output terminals
: b. uoc :
aΙe open
For circuit a. υoc
υ,(ι)
Foi circuit
i,(r) X R,
circuited
100
ν.ι,)
:
,,.
For circuiι b.
i,. :
:
ΙfR, =
is(/)
.,-: " lomνx--!--:5.γ l+l
-rfl
Ιf
vs(r)
Ιf&
Figure 1.1a
IΙ"
Εx| 1.2
i56
n:
mV
Ιf
n.:
RL
R.
:
is
Giνen
Rs + Rr10
μA,
R5
: lω
:9.9μΑ Rr- lkΩ.iο- lOμA ' X _-]ω l(n+l l00 For R, _ ιo kΩ. i- : 10 uA x ' lω+]0
-
a 9.l μA
t kΩ
For
R,
lΦ,,:5μA _ tωkΩ.r- - ιοιΑx ' l(n I lοo
Foι +
lωK R,: lMΩ.i-_ l0uAX ' lωΚ+ιM
νo
3 0.9
μA
:
R'
Τπ
Ι0 mv and Rs
1φ un
X
g : ιω
8
μΑ
Ιf a load R1 gives 80% of the soυtce cυ.Γent, then
R,
"
kf)
For
80% of source cuπent = 10
υs(ι)
Giνen ,s(r)
l0μΑ
b
Φ )
:
:
{ι"
1.1b
Rs
1'.(ι'
Rs = l kΩ,then
: aιο 10X_jr_=ρ, Ι +i,
is
Ex: 1.3 Using volιage diνider
ηc)
_ ε mv
Ex: 1.4 Using current divider
Γ-Ψl), l"
= 10 μΑ V :10mv l l0 μΑ
gives 8 mv when
8=
δ-Γ
Fiμre
10
R. = 1φρ
lα)
b
,: G)
1L11
|Φ υ^:lomVx " !α)+1K--οqtν 8o9o ofsource νoltaρe = lo mv x !Q '
Rs
Vr. :
R.: 19 1ρ l0 ,^: " lomνx 1ο+l _-9l.v
τ
For φυiνalency
Rsis(r)
1
Ιf
when output terminals a.e shoΙt-circuited ror cιrcuιt a.
z-" = 10mVX Iω :9-9mV +
8 ιr.Α
= 1 kΩ
:
:+R.
l0,',η
:
25
x
1ρ
lΦ lω+R,
ΕXercise |_2
/
Ex:1.5
ιl :
=
2zτf :
Ex: 1.6 (a)
r
(b)Τ:l= t
7=1
161
=
+:
#:
2π Χ
101
:1
f
radls
: 1* :
16.7
60
:
I
----:;
lο '
: 1vΞ}D = 00ο1 U^:2vΞ, = ωΙ0 ,Α : 15 vΞD : 1ll1 ι/Α
10ωHz
(b) (i) +
.,.
,
470
6
:
l00O s
56 channels
t v2. ^ : v2 =-x-x, TRR P = Pt+ Pr + Pr +... 4v r 4v ),1* -/ 4v\,1,( \il R - ι]J2"JΥL, R' ι5Γ2"., π
y(ι*!- l {ra R?τr\92549t
-μ! r
..)
direct caΙculation that the
innniιe series in the parenιheses has a sum thaι approaches π2 / 8; thυs P becomes y2 / R as found from direct caΙcυlation. F.action of energy in fundamental 0.81
Fracιiοn of energy in first five haΙmonics
:!(r*]+1)=ο.sr
Fracιion of energy in firsι seνen harmonics
,rr\
o.ss
Fracιion of energy in Rrst nine harmonics
",r\
+ .l
*a*1*1): 8Ι /
9 25 49
ο.sο
Noιe ιhaι g07 of ιhe energy of ιhe square ινaιe is in the first three harmonics: ιhaι is' in ιhe fundamental and the third harmonic.
Er: 1.9
(a) D can represent I5 distinct νalues
between
O
ι/Α
and +15
: ovΞD:
V Thus. oο0ο
1og Ι05 = 50
: 15x8:
dB
l20mψ
PL:6/Jη' = 18mw Pιl"*ipun",l : l20- l8 : l02mW
,,, =
l' _-l0
l
_!g
l20
P, :
,?o /
I0" + Ι0
ΚΧ'
-
llΨa
- toμv
Ξlο'v
R, - (lo x-10
6)'?
: to Ir w
with ιhe buffer ampΙifier:
R, RL " ιx R,+RlxA ',''x RΙ+R" :lx l xtx l0 = o.25ν l+l l0+1ο n^=
p' :
2r
υo
'R.t0
:9JΣ
volιaρe ρain
=
εls-w
_ τb _ 0'25v -
lv
,r
o.25
= -12d8 Power gain (Αr) =
25t
:8(r*l+.1 g 25+1): 49)
Ex:1.11 Pr.
l0
Ex:1.12
Αlteraιively,
s-8ll
:
:
,' = ar lrn Pλ,'
Eι:l.E P : 1[t'., TJ R
ort 9
:
20log 1Φ : 40 dB 20 log l0oo : 60 dB Ρower gain = 1οlogΑ,: l0log(Α,Α,)
T
:
olο1. The eπor is -0.2 v or
cuπent gain
channel 14 to channel 69
= 8/π'z
v
5'2 Χ 1o0 : _4L
106
Ιt can be shown by
:
Ex: 1.10 Volιage gain
I
since iι StaΙts with channel 14, iι will go fιom
_
is 5 v: ιhus D
_0.a
Ex: 1.7 Ιf 6 MΗz is allocated for each channel' then 470 MHz to 806 MΗz will accomodate
806
l v (ii) +2 v (iii) +4 v (iν) +8
(c) The closest discrete value represented by
P,
7i
Pι: v,:0.5Vand : r. : -----l-v ' iMΩ+ lMΩ
6.25 mw and P,
where
Thus.
Pi =
ο.5
x 0.5
:
= ι'i'
0.5
uA
0.25 μW
and.
1:6.25x10'=25xto1 u ο25xroo l0 log ,'ι,
:
44 dB
'
ν/ν
Exercise I -3
This
Ei:
fiμιe
belongs to Exercise 1.15
(no load) ouφut νolιage
1.13 op€n_circuit
:
υ' _ υ. ι'' υ' _ !!x!! -Ι -Ιx o, oi, νi' νs
outpυι νoltage with load connected R.
=λ
ι] -------L
I aR-: Ro+
For
P, :
!eR,
:
40
dB = 1Φ
' "[lω'-!-|'/ \ l+ll
22 nUi-υ| ιt-
lο'οοo
o
n.7
0 log ,4"
- 10'''l', 'υi
:
2.5vj
= 2.s x
to4
w\ry
(see figure aboνe)
|ωK lo,/ '\lωK,lK/
1ο.oω11lo11o.99ol )(1ω)(ο.Φo9)
Ex:1.16 Giνen ι,"
1:
o.r,
zr, =
0.909 vs
oiz
υiz
R;:lMΩ
)
R" = l0Ω
Α- :
'('*(-tξτ)
.
ω
R"
Ι
υ.
υ
mV
,ι4 dB
ιrι b_(\lΦK.lM/
_
7,t4
7M
Ex: 1.17 Using voltage amplifier model, it can
Ex: 1.15 without stage 3
= 'o ,J
ForY":lmV v.ιL :744 x l mV:
=
representω aE
π|
A,=+
VΔr'
loω =
1mV
'ι1. υ.ιι ιι'-- lt_' 12_- 1).ll 7]. υ' υ' ι''1l υ. "l]ι7 :ο.909 X 90.9 Χ 9.9 x o.9Φ
: (o,,"",ffi)", r,
-,|
,, :
υ.:8l8υ":8l8X1=818mV
o.25kΩ = 25olι
I
Ex:1.14 Α,o
l
:818 V/V
"o'Rι+Ro
0.8:
go.gΥ9.gyo'Φ9
:
:
= 0.909X1 =
Forυ,:1mV
s.s X o.9o9
υb:9xνr:9x1:9mV
Ai :
9.9 Χ 90.9 __ 900
v/v
v
V^: vΙ
--Ξ
R, A l R{ '"'
-------Lx
R, +
Rι : Ι0Ω
R,
-------Ξ_ RL+ R"
oνerall νoltage gain
lM X 9(nX 10 :4ΦΥN 1M+lωK ι0+10 Forξ:1φ6ρ :
ro
V' ""-jl: v"
8l.s
For
l mv
A^ x
The oveΓall νoltage gain
0.909 mV
:
9v/v
oνerall voltage gain
ιM XgωX lω0 : =1M+1ωK 1000+l0
8l0V/V
.'. Range of νolιage gain iS from 409 to
v/v
81 0
Exercise
Ex:
l.l8
l-4 Ex: l-21
(β +l) ib R" ,.: ' i.----:"RΙ+ R, Ro i^= A i, " '"Ro_R'
=
Rt R' tr.;" ""Rs+R,Rol Rι
υ6
:
Thυs.
io,R"Ro Ι'"-Rr+R,Rr+R.
But
Ειι
& +
ιl'
:
r, = ,.
and
i, :
i.,thus
fl"=--=?=,-+(β+ι)R"
Er:1.19
,
i6ro* (β + l)'rR"
:rδ[r"+(β+ι)ft"]
1.22
ι Ηz 10 kΗz lωkHz l MHz
Gain
10
60
dB
20
dB
ΦdΒ οdB
"R,+R, G^υ'(Ro || Rr)
R,
,.
=c.UsRΞτ(Ro|| Rι) Thus.
,k
:
c.ufu'ιπo11 r.l
3dB freqυency
Ex: l.20 Using transresistance ciιcuit model ιhe circυiι will be
Ex:1.23
v
!:
i5
:
R,
Ri+RJ
τ ιro.,9: ,s Ψ"! ιi R" :R -----i"'R( + R,
.ls
=
R.
R, + Ra
:
G.v
R, R"X" -R, +Rn
iiRo ll R.
G.V,
ll c.l
a+a+,c,
RoR'' : τι'rr. Ι9 y,
Vρ: R.i1"#* vo _ ^ Rι "'R,rR,
o
G'
I
lLll'
JCι
Ro R.' '_LJ
Ro Rι
R"
R,+Rs
ιγhich is of ιhe STc LΡ ιyp€.
Dc -ρain
:
Gl*l Ro R,
lα)
Exercise 1-5
ui.ul=*: #: 1=o.t
-$ :
n.=uft
ιο :
* : +θ C' =
*
(trt) \sο , lο'
Ex: 1.24 Refer to Fig. 8i.23
o.rmA/v
v.: R, : νs n.+a1R " rC '
o.oε.ιlv 12.5
{)=
2τ x
l2'5 |05
which is a ΗΡ
kΩ
z., y
lω
ιo3 )
=
/,o,
ιπ, 159.2
C> pF
:
sTc
R, , R,*+
Rs+
fυnction.
;;76|;1
|
C{Rs + R,)
= tοο
2?r(l+9)1o]x1ω
:
ιl" 0.16
μF
Exeιcise 2_1
Εxι 2'l
Therefore:
The miπimum number of termiΠa]s rcquired by a single op amp is fiνe: tιγo inμt ιerminals' one output termina|, one terminal fc,r positive po\νer sυpply and one terminal foΙ negatiνe power supply. The minimum number of terminaιs reqυired by a quad op amp is 14: each op amp requiΓes two inpυι terminals and one output terminal (account-
V' : μG.R(V''V1)
That is the openloop gain ofthe op amp
isΑ
μ:
:
Exz 2.4
Εxι 2'2
_
q
=
U,,j
= ?]
U|.
Uicm
λ(ιr_ :
ιl');
I
'(l,ι
a)
_0.02V -2mν
- υ''!A _ o_}': Io'
υ'
aia: t.ι-
=0 ( 0.ω2):
Ul
+ 0.0ο2
v
= 2mΥ
,n. =
orequiνa-
dB
& .ndR, R,
resοecιivelν. Therefore. ψe haνe:
1Φ kΩ and
_& = _1ρ_4. = ' Rr
16ρ,
Thus:
Rl:10x1ωkΩ:1MΩ
Ei:
2.5
R= l0kΩ
lmV
I
and
The gain and input resistance ofthe inνerting amplifier circuit shown in Figure 2.5 are
Rι :
+ ν2)
l0 mΑΛr'
A = lΦΥ l0λ I0 = l04ν/ν lentty 80
ing for l2 terminals for ιhe four op amps). Ιn addiιion. ιhe four op amp can all share one terminal for positive power supply and one terminal for negative power supply.
Equation are
:
μcπR.Fοr G. lσ) we baνe:
'(_2mv+ο): b) lo : Ιor(5 - v|) 9 rη : 5.01 v 7ιi11 : ι7_ al:5_5.0l:0.0lV:lomV tt _ r 5) = 5.rn5 v vrl _ 4., _ 1ιν,
:5V
;(5.0l
From Table c)
vj: A(ιr-UI) = lor(o.998 _ 1.ω2) : 4v 7,iι : υ2 υ| : ο.998 _ 1.Φ2 = _4 mV rl _0.q98) _ ιv + ?,r) = υ',^ _ ;(|.ω2
)(υ1
d)
_3.6:
l03[η _ (_3.6)]
:
:
o,"-
:
0.ω36 \{.o,
v:
+
"r)
lor(v:+3.6)
_3.6 mv
:
Lt
V,,
3.6 +
(-3.6)l
yl : μy,l and = (G.V2 G.v'\R : G.R(V2- V1\
From
Fiμre
E2.3 we have:
R" _ +ι
. i.e..
ouιpuι is open circuit
The negaιiνe inpuι ιerminal of ιhe op amp. i.e.. η is a virtual gτound, thυs η = 0
V6:V,
Rii:0-Ri,:
R1 ι'
|o-n
Ri,
- -R+R- -
-R
: _lokο
R, = #
and η is a vinual gτoυnd (η = 0).
thusR,:!:oΞRi:οΩ
-3.6 V
Ex: 2.3
I we have:
R.=bl ιi
+ ,f2 = _3'6036 ν ιid = υ2_ vl = _ 3.6ο36 (-3.6)
:
1.
Since ιve are assuming thaι the op amp in ιhis transresistance amplifier is ideal' the oρ amp has zero ouιpuι resisιance and therefoιe the output resistance of ιhis transresisιance amplifier is also zero. That is
Ro
:
0 Ω
.
Exercise 2-2
Εxz 2.7
R=l0kΩ
vl v1
connecιing the signal source shown in Figure E2.5 to ιhe input ofthis amplifier \rye haνe: η is a νirtual gιouπd that is ξ = 0, thus ιhe cυrrent floψing through ιhe
For the circuit shown aboνe we have:
v^=(E!v.+&v"'l " \R, ' R, 't
10kf,) resistor con_
nectω between η aπd ground is zero. Therefore
vo= vi-
mΑ:0_ i0KΧ
RΧo.5
0.5 mΑ
since it is rηuired that V6
:
-(V
1
+
5V).
We want to have:
&: Rr
Εxι 2'6 ι2
ι|
age of lo v the cuπent in the fe€dback does not exceed I mA.
Therefore
Rr=lkΩ
JlJ
. '
lv
R|
y'
ι_ο
5
Rf
Rr=lkΩ is a virfual gΙound, thus yl = 0
s
"na&: R2
Ιt is also desired that for a maximum output
Ru=l0kΩ
v"
vι
t
v
1
ΞR.Ξ / !9l -η-ρ.2 lmΑ
Ιεt
us choose
R'
=
R,
to
volt-
resistor
ιokΩ
be 10 k(! , then
R, = 1okΩ andf''5= & = uιtl
Ex: 2.8
lkΩ
η
il
Assuming an ideal op amp, the current flov,/ing into the negatiνe inpuι ιerminal of ιhe op amp is
: i' i2 = 1 mΑ = Vo: Vι i2R2:o_ imΑX10kΩ
zero. Therefore,
i2
;'' =Vo: |OV - -ιο-l Rr 1kΩ i6: ir_ir= lomA lmΑ= llmA Vo _ _l0V vo|ιase sain _ toν/ν lv |ν or 20 dB
_ ι _ _lomΑ _ _loΑ/Α cυrrenιsain ' i' ImΑ or 20 dB
Poψer sain
'
_ Pl - - lo( lo mA) = lω Pl lνΧlmA
or 20 dB
Noιe ιhat ρower gain in aa is lolog,n|!ι|. "'"lP,l
"
: (ftχff),,-(tχfl"_(ff)n
\iy'e want ιo
design the circuit such thaι
Vo:2V1+V2
4V3
Thυs we need to haνe
(L,γ&) - 2. (&]r&) = l ana& \R|,/ιRr,/ \R2,'\Rr/ fr w/νv
+
From the aboνe thΙee equations, we have to Find six unknown resistors, ιherefore, we can aδι
ιraΙily choose three of these resistors. choose: Then
v,/e
haνe
ι€t
us
Exercise 2-3
R.: T : !!: z.sιο 4 (ft)(ff)
: ,,*
(ftχfr)
: ι=f,x]{
f
x 1!
Ψe have (refer ιo the soluιion ofexercise 2.9):
: =
2
ρ, =
:
l=r,:
5
1ρ
Ioιο
Ext 2,9 Using the super position principle' to find ιhe conιribution of η to the outpuι volιage η, we seι y2= 0
To fiπd ιhe contribuιion of y2 to yo we set
Vι : Vl : 0'
then:.
vo
:
4v2
To find ιhe contribution of y] to yo ινe seι
yι
= y, = ο,
then
ek9v,= v^: " I kl)
ev,
combining the conιributioοs of yl,y2 and
V6:
Vs we have:
6V 1+
y.] to
4V2- 9V1
Ex: 2.11
The y* (the νoltage at the positive input of ιhe op amp is:
V*
Thus
v,:
(r
:
1
τ;-v l
= o.6v
+?tfi)r- = loxο.ov, :
7
= ΣiΞv ,
:
ο v'
0.4 v,
=
(l ι 9Ψ'}r.' _ lkΩ/
γ
then it is desired that
Thus.
l0ν _ lοιιΑΞR, Rt+R, 'R', R,*Rr: l MΩ and R, : Rr+R, : R, :0.5MΩ
|0V
i_
l0μΑ
Εxι 2.12
Ηence
v"
19
R:
',
l' = l0 μΑ.
To find ιhe conιribution of y2 ιo ιhe oι.lιpuι voltage yο Ψe seι vt = 0. Then y*
V, =
Ιf
I
_
Ψ v, -'*} R, = r-* Rr = l-fl
loYo'4ν2 _ 4v2
a)
combining the cont.ibutions of ν1 and ν2 To Yo we have Vo : 6V | + 4V7 Ex: 2.10
vo: Alvι v l..1v-: vι'
v^
--!
i'' i'=Vo_V _ Υ_-Υρ \R]*Ι)v R, R, =(! R,) L v. 3kο Using ιhe super posiιion principle, to find the conιribution of y| to yο \ir'e set y2 = V, = 0 Then
= (,
+fr), :(,.ft)(r-*)=
I + R,/R. vo. ___?vo
(Ι+RzlR|)y/
vo_ lI RrlRt vι , 'l'R2lR| --_ AA
l+ ,
-
R.Σ/R|
l+R.Σ/R|
Exercise 2-4
Rl : l kο
(b) For
i2 :
and
νaιue for ιhe closed-ιooρ gain
is
the ideal
o
lτ.
,
€:
101 then
10
= lV, Vo = Gx Vr
For V,
V^=AιV
:
vΙΞv' ' v
ι _ torne _ i,0 PL _ vox iL 10X10 _ Pι vιxΙi - lΧ10 (a) load νoltage
_
|1ο
V and yo _ 9'9o|
9.901
Α
ιο0ο
lkΩ
1kΩ+
X|ν - lmv = lv
! MΩ
(b) load νoltage
Ex: 2.15
9.9 mV
ΙfΑ :
G : 9.99 artd. : _o.l% For y/ = l Υ' vo = GxV' : 9.99Υ' lOa then
Vtt
therefoΙe, vo : v.-v 'Αlο4 =
Ιf
A=
e:
105 then
9.99
G
:
:
vD
0.999 mv - I mv
9.999 and
_o.o|Ψo
2Φ kΩ
: lν'vo: GΧ Vr:9.999thus' vo 9.9q9 : v. - v = 0.09999 mv Α=
(a) R| = R3 = 2
:ο.ι mv
(b)Rid=2R|=2X2kΩ=4kΩ
For
y,
I0'
:OA,
kο, R: =Rι
=
Since Ry'Rj = R2lRl we have'
_ & _ 2ω = tΦV/V ι''" _!:V,r-V,, Rr 2
Since '.ve aτe assuming the op amp is ideal
Er:2.13 i1
ω
Εxι 2.14
c = 9.901 and G_10x:,σ|: _o'999ο -
IfΑ =
ιhal is
c = l0 l+ψ
lo. The actual closed-loop gain is
:
kΩ
9
V1
Ro=0Ω
: Vr: lV,
v':JJ:1-4 i,: ' lkΩ lkΩ
(c)
Α,. =
h:
#t4)( t) o&
:[*],'ftfr
i, = i, = lmΑ,
9kΩ
ι2
Ro
:Rt
Rt
Ro*l
v.
lkΩ
_L
Rr The \ιolsι case common-rnωe gain 4m happens when
|Α",|
has its maximum vaΙue.
Ιf the resistors haνe
l
tolerance,
\λ,e
&< R4*,(t +0.0l) haνeR4no'(1 R1".n(1 + 0.01) Rr Rr"".(l - 0.01) 0.0l) <
V6: i': '
V1*
i2Χ9kΩ = l
+
l X9
Vo': l0Υ : tkt) tkl) torn.
io : iL+ i2 = τnA 'Ι yo _ ι0ν = loY or2odB
v, lv
v
_ lov
where R3nom and ftano. are nominal νalues foι R3 and Ra Ιesp€ctiνely.
!
e haνe
;
Rrno, = 2 kΩ and Rηno,n = 2Φ kΩ, thus,
2Φ x
0.99 < R.ι <
2Χ l.0l
sε'oz
8!
= R] =
R1
2ωX l.ol
ιoz'm
2xo.99
Exercise 2-5
simlaιly, we can shoιγ ιhaι
ι,_(op ΑmpΑr)
g8.o2=&=rc2.02
l\ιJ
RΙ
_&
Hence. _ 102.02 <
Rι =
τ/οι
_qι.oz
lRr
4 =lΑ 'l Ξο.οι =I+98'02
πl
1+&
9s.o2and& = R1 = Rl
CMRR
!.
0.οl sinωι
-The ιo νοlιage at the ouιpuι
is
2R'
: τl,'+π'x]1ι ' 2R' + 5.005
U,(οp Αmp
sinωι)v
.R,ιο _ η,: Αι)
ro2.o2
Α.-
:
: !"-': !,s 2'* 2'
:
:
CMRR =
:68 dB
",:ft(ι+&),,"
0J4
|0
We choose R3 =
i,
lο
and R4 = Rz Then for the
circuit to behave as a difference amplifier wiιh a gain
of
1ο and an
input resistance of 2ο
kΩ
we requrre
ο-iτ
y-(op Αmp Al)
(2.5 + 2.5025sinωt)
v
k(I * 0'5 MΩ]
k\ ο.5 MΩ.,, ^ o.οI.ln., : l(t+ιω0)X0.0lsioωt
10.01 sinωt
V
Ex:2.18
,4.:&:16,n6 "Rι Ri,1 : 2R1 = 2οkΩΞR! = l0 kο and Rr: AoR' : 1ο X 1οkΩ : lω kΩ
Therefore,R!:Rl:Ι0kΩ
_ lοkΩ
,,'?l
(2 5 + 2.5025 sin{rt)V
? (op Αmp Α1)
:
26 tνδ 1nn!!1o2 jΞ
Ext 2.16
= & *π
+ 5.α)5 sinωtl
_ cMRR _ 2ο '" ιnn]jll '"ε|eJ 20 tog(2550.5)
,.
...R] _ R4
'w
ofop amp A.
V"'
= (5
therefore. the corresponding value o[
for the ψo.st case
+0.ω5siηωιv
The νoltage at the outpuι ofop amp A2
The differential gain Ad of ιhe ampΙifieΙ
isa,, = "Rl
5
vο2 -
Rr
Note that the worst case Αcn happens when
&
:
: 5 0_Φ5 sinο}t 50ο k X ο'01 sinΦt lkΩ : (5 _ 5.ω5sinωt) ν
Ιn ιhe \νorst case
lπ
:
1\2_ v|t
'
a=&-&-a=lRo-R,l Rr Ri Rrl =a R:l
Vιι
υ- : V,' R' Χ Ι!g
Therefore,
lR,,
:
:
and
v, (+
Ru=Rl:l00kΩ
ξ(+):τπ],η(+)dι
E\t 2.17
GiνenV,".: +5V
The ψaνeforms for one period ofιhe input and the output signals are shown below:
yΙd = losinΦι mv
: 1 kΩ, Rz : 0.5 MΩ Rr:Ra:lgk{) tU- : urc. )U,o:,: 5 0_005 sinωt v 2R,
10v 1 x 0.0l sinωt 0
I ιι2 = ι\\fr + _ι'ιD
=
5 + ο.oossinο}ι
ι'(op AmpA,)
:
v
Vιt
:5
o.ΦSsin
V
Exercise 2-6
Εxz 2-2Ιl
η(+)αJ
20
o
(+)
liy'e haνe
I r' -20: :l CR RI,
10
o
-t
-20
CR
C:
dι
0.0t μF
differentiator.
X 10X I
Ιs the input capaciιance
Ιιry'e
want
cR :
10
2
s
ofthis (the time
constant of the differentiator), thus,
ms
,^
2
n=--.!lj-=Iι,tο
cπ: !!xιmsXo.sms
0.01 μF
20
From φuation (2.57), we know that the traηsfer fυnction of the differentiator is of the form
Εxz 2.19
ΥΔj!)
: _ i.cn
Thυs, for
ψ=
v,( iw\
10 rad /
s ιhe differentiatoΙ
tτans-
fer function has magnitude
v'(+
%(+)
lΙρl _ Io" lO
lvλ Φ For
The input Ιesistance of this inverιing integτaιor is
Rι, therefore, R : l0 kΩ since the desired inιegration ιime constanι
is lO-3 s,
we haνe:
6 : !Ιl--J Ιo k()
CR
:
Ι0 r s+
= 6.1 ,rρ
From equation (2.50) the transfer function ofthis
:
:
2
0.l
v/v
and ρhase
_90'
w=
103 rad /
s tbe differentiato. transfer
function has magnitυde
lΙρl I
Y,l
Φ:
: ιo'X
lo-2
_ lov/v
and phase
_90"
ιf '!Ιe add a resistor in s€ries '\ρith the capacitor to limit the high frequency gain of the differentiator to lω' the ciΙcuit would be:
integrator is:
o0'ι)
v v
ι(j.)
For
ι : _jννCR
'r:10rad/s
the integraιor transfer
function has magnitude
l&l : __] . _ lω v/ν lο \ l0_' l v,l
Φ=90"
tion has magnitude
I
v,l
Φ:90'
. _ l0σ) V/V / to-' lo -L.-.
Αt high fr9qυencies the capacilor c acts like a short circuit. Therefore. ιhe hiΦ-freqυency gain
ofιhis circuiι is: and phase
Using equation (2.53) ιhe frηuency at which the integτator gaiη magnitude is unity is
| l : -'': "" cR - ro1
vo(+)
and phase
For ψ = 1rad/s theintegιatortransferfunc-
l&l :
η(+)€
IOοοrad/s
Ξ
Rι
To limiι ιhe maρniιude of
this high_frequency gain to 1ω, we should have:
n:lΥ,Ω=lοιn ξ- tοο-t," - 1ω ιΙJ
R]
Ι
Exercise 2-7
Εxz 2-2|
9ylρ(mv)
Εx]. 2'23
V'o = Υ1
Υ1
Vιo: V'_ Vo'_
V'
wheny+=ν-=0then
Vη: o
5
mv:
_5 mv. This inpυt offset
v1
v.
voltage causes an offset in the voιιage transfer characteristic. Rather than passing through the oΓigin. it is now shifted to ιhe left by yoδ
Ειι
2'22
From equation (2.41) we have:
SR 2τV6^^^
:
15.915
kHz
:
15.9
kHz
Using equation (2.42), for an iηput sinusoid \νith freqυency
/ : 5./γ'
ιhe maximumpossib]e
ampΙitude thaι can be accommodated aι the oυtpuι \,r'ithout
u"
incuπing sR distortion
: u"*,(*):
iS:
rox I
:
Vιι: Vz v| Vro: V*-Vor-V Ιn order ιo haνe zero differentiaι input for the offset_free op amρ (i.e., y* - y' = ο) we need
Vιο: V'_V _Vos:0
5mv=_5mv
Thus, ιhe ιransfer characιeristic yo νersus ηd is:
2v(peak)
v,,,
(mv)
Exercise 2-8
Εxι 2.A
t = Y: a "2π2τ
Fτom eqυation(2'44) we have:
Vo : 1!"βa "1R1 :lωnΑXlMΩ
= o.16Hz
Ι
v
= 0.l
Εxι 2.26
From equation (2.46) the νalue of r€sisιor R3 (p|aced in series with posiliνe inpuι ιo minimize ιhe output offseι voltage) is:
,/?lRz Rr: Rrll R,: Rι+i2 = lokf,rX lMο l0kΩ+1MΩ : 9.9 kΩ
R3 = 9.9kΩ
1l0kΩ
rly'ith this νalue
ofR3 the new νalue ofιhe output
dc voltage (using equation (2.47))is:
Vo
:
Ιo"R2
:
10
nA Χ
10
kΩ:0.0l v
-
|νι \νe
Ι, '
AoΙι+
= |. Ιι'Ao
106 and
/, =
MΗz,
Aoι'b+
kηow
2o |ogAo
:
f"-l5Η7
3
theτefore
By definition the openloop gain (in dB) at
Αo(indB) _ 3 =
106 3:
una
6
is:
l03 dB
To find the open-ιoop gain at frequency/\ιe can use ηuation (2.31 ) (especially when/>>/6which is the case in this exercise) and wriιe:
Εxι 2.25 Using φuation (2.54) we haνe:
Vo :
From equation (2.28) ιγe haνe:
Vos +
v^"9
7ffl
|2
_ 2 mv ι
2 mv :-.j]Ι,
_ Ι2Y_?mVΧ l ms-6DΞl _ =, 2mV
6ι)
openloop gain
atl:
20 bΕ(+)
Therefore:
OpenJoop gain at 3Φ Hz 20
:
80 dB
MH' :
εο aB
MΗz los3 -
=
3α) Openloop gain at 3 kHz = 20 ιon3
-
-l kHz
Openloop gain at 12 kHz 20 loρ
3
'12
MHz
:
=
aε ag
kHz. Open-loop gain at 60 kHz =
MΗz
"60 kHz =
20 lon3
μ ag
Εxι 2.27 with
the feedback resistor
Rl
to haνe at
leasι
:t10v ofoutput signal swing aνailable, wehave ιo make sure that the autput volιage due to yos has V. From equation (2.43), knoΨ that the oυtpuι dc νoιtage due to yo{ is
a magnitude of at most 2 vr'e
vo = vos(1
*\)-zν :
2mν(| l
l+ lοR', = l0οoΞR. - loMΩ k(Ι Tbe comer frequency of ιhe resulιing
n"1*ρ1;5, =
J-
CRt \γe know Rc : l ms and R: 10kΩ+C : 0.l μF
Τhusw
0.l μF
x
l0 M()
--_L..._
vi(+
'v. ι
._f,
|-ι , *,a,
#) The waνefoπns fot o.te p€riod of the inρut and the oυtput signals are shoιvn beloΨ:
sTc
t0v 0
_ lrad/s
*l:
Εxercise 2_9
\y,
(*) slope=SR
-20 We have
Π
-20: cRJ,,
lο dt
-l
l0x lms
CR
cn:
!9x lms :
0.5 ms
20
=9
Since dc gain of the op amp is much larger than the de gain ofthe designed non-inverting amplifier, Ψe can use equation(2.35). Therefore:
=
-ξ.
and
l+--i
I
./.,un :
2
ΜΗz
o-l X V/μS
1_6
l.28 μs
100
=
2o
kΗz
tf w,V < SR
y5
Σ49
y < 0.16 v. age sιep is
y:<
JL
2τf'
thus' the Ιaτgest possible input
νolι
0.l6 v.
From Appendix F we knoψ ιhat the lo% ιo90% rise time oithe oυtput ψaνeform ofthe form of eοUaιion (2.4o} is
'
:
ι-' _ - 2.2! νν,
0.35 μs
From equation (2.41) we have:
:
SR 2'τrV o
15.915
kΗz
:
l5.9 kHz
^^"
Using equation (2.42), for an inpυt sinusoid with frφuency = 5 jrM, the maximum possible
,. :
Exz 2.29 For the inpuι νolιage step of magnitυde ν ιhe ouι_ put waveform will sιill be given by ιhe exponential Ψaveform of equation(2.40)
Thus, r,
1
amplitude that can be accommodated at the ouιpuι without incurring SR distortion is:
f':2ΜHz
162115
:
|-6
/
+& = lΦand Rt
Rt
Hence
1.
Χ
Ex:2.30
Exι2.fr
"fιαn
ο_9
t,
Ιf an inpuι sιep of amplifude l.6 v (l0 ιimes as large compared to the preνious case) is applied, the the output is sle1v_raιe limited and is linearly rising with a slope equal to the sleνi-rate, as sho\rn in ιhe followiηg figure.
u.^-(Lu') =
10
Χ
l
=
2
v
(peak)
Εxercise 3_1
Ex:3.1
b.
T:50K ' ^-3/2P [Ι/(2Kτ) βl
n': :
n. =
=
t'l2(2 x
8'62 X
lo
5
x
5{))
X 1oΙ5(35o)r/2e_l'l2(2x8'62xtο
x
4.15
:
J^ :
5x35ο)
:
x
1.O8
: :
lOrr/cml
0.25
x
10-8
T:35oK:4.ι5xtOΙΙ/cm]
Ex3'5
-
(4.15
=
|"72
x
lorr)'?
x
x
:
1.08
0.25
x
104
x l0
8
cm2
J,: qD"Ψ
,, :
l-5
_ n_ -
x
:
l0l0/cm3
lοo
L
ρ": '
:
:
l.5
r
.2
nr
6.75
X
J-
106 cm/s
x
105
10_6 Α/(μm)'z
μ Α/( μm)'2
D"
D^
:
Do
υ.-dιiff = - μ,E Χ lo-'
to8
Usiηg φuation 3.2l Db
Ηere negaιive sign indicates that electrons moνe in a direction opposite to E 'We use
2
X
x
Ex: 3.6
: _μ,E
|]5o Χ
56
35
For/,:1mΑ:J"XΑ 10'βl .=18 μ., *Α: lTΑ: J" ss μΑ(μm)'?
x toto)2 1.5 x to4 : 1.5 x 10r6/cm3 υ'-driff
: -^5 luμm -2
x lo-re x
1.6
= 56
lO!/cm}
1.5
Ex: 3.4
ο
a;= l .Ι': qD,Ψ
l'5 X ιo|o
"
|ο5 _
dn
|o6lcrι3
want eιectron concentration
:
ν"-diff
: !0|1 lcml : lo5z(μm)3 D, : .15 cm'/s - 35 x (loa)2(μΠ)2/s : 35 X 1o8(μm)2/s
lor'7
Αt 3ω K'
:
4
no
Εx: 3.3
a.
10
From Figure E3.5
.2
(
2x
Aqnμ"Ε
Note 0.25 μm2
_
in
lOa Α'/cm2
=21 μA
ND : |ol1Ιcm1
]on.
lo'ux 1350x 1V
d. Driff currenι Ι, = Aqn
l0r7/cm3
.''No:
n4
qnp"Ε
F.om Exercise 3.l πi at
nl:
x
c. Ιn n-si driffcuπent density Jn
Er(: 3.2
ly'o
6.75
= 1.6x to-rex
BT3l2 e-F'El(2Κτ'
7.3
:
e-
3elcml
x t0 T=350K 1 9.6
: 2Υ|0" -
l"nnιh -
Χ 10'l(5o)]"
7.3
]lme taken to cross 2 μ,m
...tμm - I0
:
6.75
X
1o4
π/s
acm
Χ 25.9 X 10
ιι"Vτ =
1350
:
μoV1=
480 X 25.9
=
12.4 cm2ls
-- ). = -r) cm /s
Ex: 3.7 Equation 3.26
\ν
: /2" q
*fr)v'
Χ 10
3
3
Exercise 3-2
lD1'l|i, + /ζL = ,lτ\ N^Ν" γ"
2, one must lower ND (= n,) by a factor of 2.
+ Νo1u^
*z
2es1Ν
"':
i(ξ):
q
Αs one can see from aboνe equation, to increase minority carrier-concentration (pn) by a factor of
^ \ NoNo ) "
Ex:3.10
(ffi;''
Εοuatiοn
Ex: 3.8
"'n""
Ιn a p+ n diode
Equaιion3.26
1vΑ >> lVD
" ff
We can ne8lect the term
1r";'")u" as
1ξ
Comlared to
-2!_
nξ
''
=w ff,
:w
1y',4
>>
ηuation3.29, w
=
ΜΨ
I
tv.
a' : η(i+"u }y since l{Α
10
x lo-o x
:
t0
!!Ι
l0
>> y'{D
:
|.45
X
lo
F"T#+)". N.,
>>
_ :
2e, q
*
=
Vt)
, _L1ιo.εl+ |zx ι.uy ι9''(-l ,r/ ι.6 x to '' \tο'' lo'o,, t.66
x l0 5 cm =
o.οοsl
0.166 μm
ofthis pnjuction diωe
(l.5 X lo|n)2 lοl''
=
2.25
ι
*
v,l
fr)ιv, E"1J4Xστr l +. l \ i' '-- " 'ir l-: - l(0.814 - ,ν l.6 X l0 '' 'l0'_ lο''- , : 6.08 X 10-5 cm : 0.608 μm
lOlt/cm3 and
n,:1Vr:1ρ167"rl η,
x lο r)
Ex:3.13
lol6/cm3
!;
l'ιeαω5/(25'9
1')ιv, -
*
Ιn the n-region
lo'8
Ex:3.12
Ex: 3.9 Ιn example 3.5. ly',η
x lo_'x
2
0.2 mA
w: οJ :
4X 1.6x 1o-''x1ι.5 x 1o'1'
=Ιr"uΝτ
η)
: l J-z.gηv o
"
10
9-ω,^)
= l.45 X lo-l4 Α Ι : Ι r("u'u' _ 1)
=of*(Ψ,\u" since ly'Ι
ΝD:
:
s
AqN 1W
Equaιion 3.3ο,
oo'i(+η-
lv;;*
Wl(h)
.
=
.ι
ND
=wy2η:
2ι_
lr= ιqn
ffi
Ex:3.11
:"#π
since
here approximately
ξ
can be neglected as co mνarωto
".
X" =
u"ι
ξ
similaτ νaιues, if Nn >> Νo, ιhen the ιerm
since l{Α >> Nr)
Equaιion.1.28.
l ιι ι. = ιon'( Do ι D" \ \LoNo'L'Νol
l.a/cm,
Using φuaιion 3.29
o,:
n(ffi)w
+ 2)
Exeιcis€
= lο_4 }
=
t.6
, ,o
9.63
la
10
I
lo|6l
'
rot8 * rot6J
x
=', :
o
"
pF
'nlε
x tor6\/ I \ * ,nlo /( 0.8 t4.Ι
3.47
rr
--1 (lt + 0.814 pF
1.72
Ex:3.15
: !-ι,-lι c': " Φ ιlv ιEv :
!^[",'x l r{eu"t
:
.-1.
! '" dv 1gu'ντ ' Ι ν/ντ
: !! -Ιl]l \v
5
.,n
Ex:3.16 ηuation 3.5l
!i.
"_: "
ξfi)
Dn
x lo
(5
y 1"av'vτ
τ1
_
-
1)l
1)
al2
5
:25ns )
lοt,,/
ηuation
c.
=
No:
^J(Ψχ#hχ*) to'8
=
tο'
3.57
(h)r
Ιn examρle
Equation 3.48,
ηuation
t8
lοX lο-4X
Εx:3-14
3.2
,
o'|(}1* *
:7.3Χ1015Α
:
o.oε
Ι.6 X 1o_|9 Χ (1.5 X 1ο'0)'?
ro * ,(\5xI0ox!ο'o
.,.=
,
pC
Reνerse Curreπι'
:
ιο[ lo|8
3_3
3.6, N.ι
:
10|8/cm]
,
10r6/cm3
Αssuming N^ >> ND
ττ=τp:25ns :.Cd : ( 25 x lo-:Ξ)
:
\25.9
96.5
pF
X Iο "/
o.t
x to '
Exercise
Ex: 4.1
4l (c)
Refer ιo Fig 4.3(a). for y/ Ξ
0'
the diode con-
ducιs and presenιs a zero voltage dΙop. Thus vo v ι For y, < 0, the diωe is cuι-off' zero
:
cυrrent ffows throυgh R and
V, :
-5V
0 The '
resulιs is ιhe transfer characteristic in Fig E4.l
Εxι 4.2 see Figure 4.3a aηd 4.3b
During the positive halfofthe sinusoid, ιhe diode is forινard biased. so iι conducιs resu|ιing in : 0 During ιhe negaιiνe half ofιhe input signal v,, the diode is reverse blased. The diωe does not conduct resulting in no cuareπt flo\ring in ιhe
5V
,,
circuit. so vo
:
0 and
(d)
υ, : υι υo = ιι
This resuIιs in ιhe waveform shown in Figure E4.2
Ex:4.3
, : Ξ;, : R
ι^ "
dc component
rov lkΩ
-
of
:
τb
,:Ψι
lO mA
:2mA
1^
1^: l0 : _7η 1ττ
5V
= 3.18 V
(e)
_]ν
Εxι 4.4 (a)
2ν
ij:2m.ι (η
lkΩ
(b)
'f,,:
+3V +
v: tv
+2ν +t
v
Ex:4.5
y
=
lο
lo
-o 5ρ1ρ = Ξ-_ lmΑ .'.R:3.133kΩ
:
]! 7τ
ιο
Exercise 4-2
Ex: 4.6
ηuation
:
Vz_Vι
2.3viol.j(Ι1)
:
Αt room temperatuΙe V7
Vι' :
R: lokΩ
4.5
+
115
mV
vx- vt
v.:
Υ
vD
Vι'
tεo
Nowi:ο.|mΑ 'o
:
vΙ 0.64
h(i)
=
25
X 10
,".(#iF)
v1+z.:xv,ωε(,L)
i = lο mΑ V=25x!ο1rlοXl0-r'l \6 g
=
0''Ι
+2.3x25x 10
: Ι'' _ 5_0'679 to k _
:
For
0.76
x vrloc(r!)
2.3
i"c(Ψ) :
second iteraιion
v2
V
x
=
First iteration
:1se07lο025
Ξ/. : 6.9,
:
5
Vι = 2.3x25y lo'X|oc(#)
ι Χ ιo-]
:
Vrr_
25 rιιΥ
Εxι 4.7
v
,,
x
o.7 +2..l
0.679
0.432 mΑ 25..1
\ lo-'l"c(Ψ)
V:0.68 V
\γe geι almost the same νolιage
ro-lo/
V
.'. The iteration yields Ι
a1
o.43
mΑ, yD =
0.68
v
b. Use consιant volιage drop model
Ex: 4.8
ΔΤ : l25 1s
25 :
lax l.l5ΔΙ = l0
V
l0o"C
Atλo'c Ι
:
o.'1
ν
5_ο'7 ,^ " l0 k
= l.17 Χ 1o'8Α Εxι 4.9
o
constant νoltage drop
=
O_4]
mΑ
Ex: 4-11
|V : = lmΩ
t,-.a
since ιhe reverse leakage cuπent doubles for eνery 10"c increase, at 40"C
l:4Χl μA:4μΑ : : =V 4 μΑ X l MΩ
@o'c r = 1*l 4
4.0V
+v:Lx1:0.25v 4 Ex:4.10 a. Using iteration
Diode has 0.7
Assume
v
V, :
drop at l mΑ cυrrent.
0.7
: Ι^:5_o'1 " 10 k
V
o.4] mΑ
Use eqυaιion 4.5 and note that
Yr:0.7V. /,:1mA
v
Diodes have 0.7 .'.
ImΑ : /se : Ιsen1Vτ :
Ξ1, :
drop at 1 mΑ
o71V-
6.91
,
25 mΥ
tοo
'o
ο.679v
Exercise 4-3
For an output νoltagΘ of 2.4 V the voltage drop across each diode
Now
ι
^ :
3
ο.ι v
ιhe current through each diode
I = lse ' : =
: 11 :
6.91
(d)
.
,,*/l25 '
t 1o-'"e
+
is .
Ι.
-0.7V:V
,)
54.6 mΑ
to
54.6
2.4
x
10
{' /
r
=
0'Ζl_r.5 5
= 1 .72 mλ
139 {)
-5V
Ι.,xz 4-12
(e)
(a)
3V
v=3-0.7 : 2'1Υ
2ν
:
1.'Ι2
πA
(0 Φ.)
5V
/:0Α
t +
* 2.5 kΩ
V:1+O.7
+3V +2Υ
λ I
ΙΞ
+
+l v
Ex:4.lJ v-
25Xlo r 0.lx10r
(c)
/o = 0.l mΑ
/r:lmΑ 1,
:
l0
η
mΑ'
r,
_
25ιlξ1
25()
=
l X lο '
rrn
_25Xlo-r-2.5Ω loX Iο
r
Ex: 4.14 For small signal model, using equation 4.15
-5 V
iD:
ιD+2'υ,1
Δio: b'6uo
(lt
Ω
Exercise ,1-{
For exponentiaΙ modeι
b.
loι
V, :
V,
3
ιD Δvlν
i22_ io. : iρ.e
Δiρ:
: iDleΔνlv_' _ ιl Ιnthisproblemip, = 16
:
-_ ' io,
/s _ιD e
|2' 1
25mV
ΔV(mV) Δl'r(mΑ)
small signal
-t0
a
ζ'.
vιvτ
_ι.4
-4.33
b
-5
-{J.2
_ο.18
c
+5
+O.2
+O.22
d
+10
+O.4
+O.49
4.7
Y l0 "'A
iΙ-:5_1=4mΑ
Αcross each diode the voltage drop is
v" = v,h(?)
:
ΔrD(mΑ) expo. model
=
e
c.Ιfio:5
mΑ
Using equations (l ) and (2) resυlts and using
Vr:
voltage drop across each
diode=i=0.7sv
ιν2 ν |ιvτ Δνιvτ :e e
ioι
For
=
25
ι0
X
0.7443
ν
'x l"/-1Ξ-]!_' \47 x lο
)
'6,i
Voltage dmp across 4 diodes
: 4xo.7443:2.977ν
in Vo = 3 _ 2'977
so change
:
23
πΥ
Ex: 4.16 For a zeηer diode
: V.oΙ Ιrr' 10: V,"+0.01 x50 v." : 9.5 ν For ,z : 5 mΑ V, : 9.5 + 0.005 x 50 : Vρ
Ex:4.15
+l5v
Ειι
9.75
V
4.17 15
V
0to15mΑ a.
.'.
'
ln ιhis probtem Total small
:20Ω
{! _ zo.v _ ,o ρ Δi,
siμal
lmΑ
resistance ofthe
.'.
For each diode ,,,
But
vη = --J+5:
: ?Q:.ο 4 25 mV ΙD
'''1D:5mA
3: xη6a:15 5mΑ
2.4 kΩ
fourdiωes
The minimum zener current shouιd be :5mΑ. 5 x lzk.: 5
Xl
since the load current caη b€ as large as 15 mΑ, we shouιd select R so that with /. = 15 mA, a zener cυrrent of 5 mΑ is available. Thus the cuιrenι shouId be 20 mA Ι,eading to
R _ 15 5.6 _ 47οΩ
20 mΑ Maximυm power dissipated in ιhe diode occurs when /.= 0
Ρ.", :
is
20
X lo_] X 5.6 = ll2 mv
Εxercise 4-5
Ex:4.19
Ex:4,18
t5v Vs=
Αt no Ioad y7
:
5.ι
v
FoR LΙNE REGULAτΙoN a.
The diode starts conduction aι
τ'=Vo:o.7Υ ,s : yssinωt, heΓe vs : :
ν5
V5sinθ = Vo
:
ΩrtsiιΘ Line Regulation
:
19 ιi
For Load Regualation:
:
0.7
Υ
0.7
ο = .ln ,f 0'7 ) -
\ιzrtl_
200+1
:
|2{2
z-+'
conduction starts at θ and stops at l80 _ θ.
.'. Totaιconducιion angle
=
:
_
180
175.2"
(τ
0)
ιt b. ,r*n - :- | {y'\sinΦ 'zτJ
v
D\d6
=;,'r,*λ_v,Φ]t=; _ 11 v.cosθ
yscos(τ _
:1, cos(zr τ_2Θ:ιτ 2V, VD νυ'Δ\E 2π 2 : vt vo ιτ2
Bυt cosΘ
ΔVo
ΔΙ' -
_ ,
ΔΙ,r,
lmΑ
For
,nA
2Θ
V, =
uo'*"
=
Ι2aD and yD
Φ
u
θ)
θ)
:_1
:
0.7
Ψ:s.οsv
v
yD{π _ 2Θ)| and
Exercise 4-6
c. The peak diode current occers at the peak diode
νoltage
"
'' _ --T-
= 163 mΑ P|Υ = +Vs: : l7v
_
Ρeak current
'2{2_o'7 1ω
The fcaction of the cycle for
_
Ωrt
2(1t
_ 2θ) x rco
2zr
2[ττ zsin
Εxz 4-2Ιl
'(ffi)]
2τ =
:
2(τ _ 2Θ) which τ,o > ο is
NoΠ zero output occurs for angIe
Ιω
163mΑ
91.41o
Αverage output νoltage
X1ω
yo is
vo: 2\nτ vo:2x|2"Γ2 -o.7 : (rr +
Θ)
ι _Vr_Vo_ nrt_o'7
'o-
=
Αδ shoιγn in the diagΙam ιhe output is zero
_
between (rr
θ) ιo
:2Θ
(π
n
163
|ω
mA
ΡΙV: Vs_Vo+Vs : nrt _ o.'Ι + 12{2 = 33.2 ν
ys
a.
?D is
Ρeak diode curr€nt
1ι
+ θ) Εxz 4.21
Here Θ is ιhe angΙe at which ιhe input signal reaches .'.
yo
V5sinθ
ρ = .ln 2θ b.
:
: V, ,(
\
V') yΙl
2s1Λ-lΥ!1 y.,, \
-ys
ΑνeΙage value of ιhe output signal is given by
'"
[' r,,'nφ - ,,,oφl __ !|r, ,nl ,,
u"^"'
:
1[_yscosΦ _ v oΦΙΙ =',
='Ψ-
Peak Current
v Ssin(τl2\ _
,s is ι2 v(rms) then ys : .Γzx
:
vD
Ιf
ιz
: V'_
: ρrt
_ +|(y.sinΦ 2τJ
ξt-vr"o'ψ 2vn6 =:-|2v. τ Bυt cosΘ
occers wben Φ = !
RR
I
:
1
'"
c. Ρeak currenι
_
yo,", '
,
Vo
cos(7,
-
2v
dΦ
2voΦli-=u" 2θ)]
1
_ Θ): _t
τ_2θ-τ
't t)
=Vο""":'-5 ν ι-ι 15 : 'l'::-::-::::
2vD
= 'l '4 :
9'4
Υ
1o.l
v
Exercise 4-7
(b) Peak diode currenι
_
vv ri^ _/')Δr = t. -'J-: 2JRC -:2 2JR
Peak νolιage R
_ vs-2vp _ t2rt -
Rlω
=Yr
1.4
: 156 mΑ ΡΙv: y5 vD: l2Jr- 0.7:
<οR
|6.3V
+1,
angle.
Note ιhe conduction angle is the same exprossion as for ιhe ha|f waνe recιiner and is giνen iπ EQ(4.30)
Fυll waνe peak Rectifier:
.o, -- E!-t
Dr vo
-t
.V.
"
ιυΔlR where ωΔ, is the conducιion
Exι 4.22
Ja,
Vττ
i^ :
R
j'ι t-
(b)
substituting for ωΔ, we geι:
ξ!-+Ι,
-io^":
[2u,.r 4ν"
since the ouιpuι is approximately held at VΡ v,-
D7
;:-/ι.Thus:
Ξ
iDΔ' =
π_ πιLlff,+ ιι
: ι,|t+'fff]a.z.υ ιf,
=ο
is aι ιhe peak, the maximum diode currenι
occurs at the onset ofconduction or aι
,:
During conduction' ιhe diode cuπenι is giν€n
The ripple voltage is ιhe amounι of discharge that occurs'q,iheη the diodes arΘ noι conducting. The outpuι νolιage is giνen by:
v,,
v,.
:
voe
ft <-
discharge is only
halfιhe peΙiod.
v,
: vr(t
τι -e D.-t
e
*)
RC|
v
= ,ι
-;:
+ 1ι
ΙL
fRC
Qsυppιtεo = Qιosτ
sUB
Γj
C l!ωv"+ ι'
: 2τl vn = 2t
SUB ω
ιa)
To find ιhe aνerage cυrrent, note that the charge supplied during conduction is equivalent to ιhe charge losι during discharge.
cv,
iι
: C4ι VP cοsιοιl + /, dι : _csinωt X ωyp + /. : -csin( ωΔr) X ωy,
in-*:
Ι4)
l'2
i",,,Δ, =
is consι.
Sub (b) to get:
forCR>>T/2
=
iι
sin( ωΔr): ιoΔl. Thus: Ξ io.,,, : cι'Δtxιnvρ+
RC
v [ι -- .ι
assυming
for a smalΙ condυction angle
Tl2
r*
ι1τ" cii+iι
,D.mιι:
(a)
sυb (a)
for/
2V,RC
'
:
c
Σv.
2nγ rz + ιL
lτπie : .!",, E. * t, v.-ηv''
=)Ιo.mα
ωΔr. by:
Exercise 4-8
Γ!-1 = ι'Γ' 'γ *!!' v, ι! v,J
:
ι,|'-"ff]
ΓJ-:
'o, =
: t,l'*2"ffif
a.r.o.
The average and peak diode currents can be calcuiated using equations (4.34) and (4.35)
-"E)Herel. = !Ξ€J, Ι2rt 2 Xo.8, y,:1v
,D,'": /.(l andVo: i1,,."
:
1.45
rr*o: :
A
ι(t +zτ
2.74
A
ΡΙv of ιhe diodes
Vs
=
Voo
:
:
l2,D - 0.8
16.2
V
To keep the safety maΙgin, select a diode capable of a p€ak current of 3.5 to 4Α and having a PΙv rating of 20
The outpuι γoltage, vo , can be expressω as (V
caη b€ obtained using
2rt-zxo.s
ιad = 20.7'
Βx', 4.23
νo =
ωΔ,
The conduction angle
φuation 4.3o
V
_ 2V og)e 'lrc ρ
Εxι4.A
At ιhe end ofthe discharge inιerval
ιo: (Vρ'2vDo_ v')
The discharge occυrs almost over
Peiod =Τ12 For time constaηι Rc >>
e
1l-
r..
halfofthe time
Ι 2
!
2RC
.'.Vp- 2VDo- V,
i*+
:
(Vo-2VDd(l
+v, : (vP_ 2voδ x L HeιeVo: nrt Vr, : 0.8 V
-II t
:
v,
and
2
The diode has 0.7
: |ν
0.8)
C: *+*:
x
2X60X ιωXc
ouιput voltage
r2.la,2 x0.8
:
15.4v
nrt'zro.ε_L: 2
For z1
:
lasu
|2'.Γ2.'.2 x o.8 1οο
Ω
drop aι 1 mΑ cυπent.
o'1νv
:
τ
l0 mV, vp
= o.|5
^
: iR:
1'0#
ιD _
25
yr:
νD+ 10mv
=
:
Diode cuπenι without taking ripple voltage into
coηsideraιion
1(Ω
i' =r^: " y,ln(\l mA,')+ο.τv
:. iD
Ιf rippie νoltage is included the ouφuι voltage is
=
v
ιt is ideaΙ op amp, so i*
128l μF
without coηsidering ιhe ripple νoitage the dc
:
l,'D
iD
lηΑ
^
1
|) - Ι" 2 RC)
f60 (12rt -
tι
υA
1,o
x lo
0.59
V
= ιΙ = Ιν
:
mv
ο
ιo
'1) \ιmΑ,/
'rnΓl--o
0.58 + o-ο1
10
re +
o.7 = ο.58
v
Exercise 4-9
υo:-L:1.a i^= " lkΩ lkΩ
ForUΙ=_5v Diode D1 conducts and
υo= _5 +;(+l,, I
vλ :
o.7
ν + 1 kΩ X 1 mA
For v,
:
_1
Vι: _12Υ
v,
the
+ 5)
: [-zs-e)v \ 2.i
diωe is cυtoff
ForU/Ξ5v
because it is ideaΙ amplifier
Exι 4.tr5
Diode D2 condυcts and
υo: +5+1(νl_5) I
:
lz.s * l!) v \ 2)
Εxι 4.27 Reνersing ιhe diode results in ιhe peak output voltage being clamped at 0V:
irr>0 -
diode is cutoff
aρ: 0V τ/,
< ο - diode conducts and opamp sinks load
curτent. Here the dc component
Εxι
4'i2'6
]ο
kΩ
!'
5V
lο kΩ
l0 kΩ
Both diodes are cut-off and υ,, :
+
5V +
for _5 <
D2
τrl
ι1
s
+5
of υ6
:
V
o
:
5
v
Exercise 5-1
!ι'Yv ov2 = 0.25 mΑ
l^ μ= for
μ' :
vS
450 cm2 /
μ,C,,
:
_ 34'q pFlm _ : c,, "' ',: t", 4 nm μ, : 550 cm2 z vs
s.Is
2.30 fFzμm'7
-
Yv 2 /-9 : !ι 2ηJ' oν
Change in
/D
(a) doυble
L' 0.5
Φ) doubl€
w' 2 =
4
case (c) would cause leaving saturation
vr,,<2vov
ιo: \l'fv
1u
, w., : ι"7vov:
k-" =
k'γ .Ι' : 2
if
Uos2Vov, So 2Vov
ι 'so
Ex:5.5 yov = ο.5
xv
l, : l' 2ωx 0.8 βlο.s'?(Ι + o.o25 X : 0.51 mA ,'' =V^ = Φ = 7a'4 kΩ -8okο ΙD 0.5l _
or1
Ι)
o.ozs
,l
:
l v.
Exι 5.1 Δ
,".-JJ
+5v
l{'
lρ= 6o ιιA l Υ2
modulation)
Ex: 5.4 Ιn saturation
+
7'D
4
(e) changes (a) - (d),
t
V25: l V>Vou: o.5Υ + saturation: ιo :
in saιuraιi rn
(d) double V^' no change (ignoτing length
'''
ν
is:
(c) doνble V.,- 22
,o,
= 0.025
,'" _ Δuo' + Δr^ - 3J " 80K Δtn
" 2"ι' ^ν o.2 ^ι.Y' ι 2o :.Vρy:o'ΦΥ. Vo..'in : V9γ: o.Φ v, for saturation
Ex: 5.3
va
ΦV'
\ι''Yrv"ou1l
l21 ιιA/Υ2
!ιYν'
l
ι:
Ex: 5.2
l^ _
o.5 V.
v^= vλL:50XO.8:
388 μA / V'?
- Y,) : 0.5 V. _ l _ iYν^',=!: 8os !kΩ 'L "' L ι : 0.18 μm, so w : 0.93 μm
:
Vr, :
alι ,Ds >
Ex: 5.6
vo, = (ucs
k"= ν"C.'
2aL
lo.
v
Y : rc=ι :6o0 uA/V2 (a) Conduction occurs for UGs
olιe =vιP
+
yJ:
+4v.
(b) Triode fegion occurs
or υe
ot
< yφ
for ,cD
S
y,ρ
- ιD< -| ιe
1JD>
+
1
(c) Conveτsely, for saturation !
,ffr τnλ/Υ2
For ,Ds = 0'5 Υ'= Vov
υD< (d)
ιο
+
ciνen λ
|
Ξ0
ιo = \ιoξ|vo"t :.ΙV ou| = o.5
=
=
ls
νι
Υ = _υcS+νtρ
_υo + υs + vιρ = 4 _ υc
Exercise 5-2
:
.. ,c
W
U9Ξ U6
l l:4.5
(e)
Fo.λ
:
Ι^
_
uA and
75
ι
V.
v '
_o.o2
r.,'
=
o.5
667
kΩ
aηd |vov|
l
|2|Ι o
v.
3v' : ,'' : ^J-s:: Δlo 4.5 μΑ
667
Ex: 5-8
V
-
r.5
Rs =
:
v. :
1.8
\ν"c.,!ιv
"
-
loRo
_ V-)1 :72
μι
kΩ
οο
ιD
- (-2.s)
r--
ε*".
vov
:
Ιp
3.33 kΩ
:
s.slo.
0.3 72
:
μA
:
l'8
:
o.s
v,
(saturation)
Vo=Vou=o'3Υ R,
v, J,,
o.ε
v.
Αt the triod€y'saturation
.'.
0.3
o : voo
0.8
ι,'
Ex:5.1Ο
v" v = -J------Ξ
V" = _l5 γ3ρ"
vo :
0 < v
72 μA
\υ'c,,'fv''vΞo.3 = 2 lα)0 .. 120.,2 ' ^ -τ,r, V6y:0.5 vΞycs: veν + v, = 0.5+1 1.5
:
..-R_1.8_0.8:13.9kΩ
1,
ι":
o.l8 μm
saιuration mode (υcD
t"
v" = 1 γ. :'k"Υ|v + |λ|lUDJ]) ι " ""r(l = 75 μΑ(Ι.M) : 78 μA At VD : OV, /, : 75 μΑ(1.10) : 82.5 μA (Γ)At
=
ιm
_:ΞΞ4ι'o'12
+ 3'5 V.
V 12
0'3
μA
boundary
v
= 2ο.8 kΩ
Ex:5.11
vo
:
2.5
f,x: 5.9 +
l.8V
-
0.3
O.4
= 7kΩ
Ro: 12.4x2 : 24.8 kA V,;" : 5 V, Αssυme ιΙ''ode
Ιegion:
lD k.Υlψ., V,lv", _ ιvλJl ι, =voo-vo, t'on I
ξ
t/
+ y;s -
/
l\
ι(5
_ t)yDs
8.ο8yDs + 0.4
=
7r, - 0.05 ν < νovΞ 5 - 0.05 /^ = o2mA " )4.8 =
v,,
:
0.5
ξ)
o
lriode region
Εxz 5.12 As indicded in Example 4.5,
v.
μ"C,,, = 0.4 ιnA /
1/2 \
V2
vD2 vc 'y,
for ιhe ιransisιor ιo be in
saturaιion region.
Exeιcise 5_3
Vo^,n: Vo Vt= 5-l
t%Ξa
=
ΙD
rzιο
Ιop = o.
7r :
Ex:5.13
ΙD:
o.32
u/ : 0: Since the circuit is perfectv symmetfical yo = o and therefoτe Vo, = o ιvhich implies ιhe transistoτs aΙe tumed offand ID]γ =
v _ν
1o:0.5mΑ9Rr.",:
:
Ex:5.15
ιnA :
\ηΥuL"
: !xιxvbv
=Vρr:0.8V yοJ:0.8+1: t.8 v V6 : V'Ι Vo' : 1.6+1.8:3.4V Vo 3.4 : ^ 3.4 MΩ, , Ιμ 5_3'4: t.6MΩ R-, "' l
2.5 V: Ιf we asδum€
thi
the
NMos is
r,,o would be lesδ than 2.5 v and this implies ιhd ΡMoS is off (V65ρ > 0)
tumed on, then
'"
= :k,Υσ
Io"
:
Ι
o,
' v,)"
μ
V' : R": " 0.12 sιΩ
Vos2 Vov-'+ Vo- Vou + ys + l.6 = 24ν Assume
Vo
:
3.4
V,
Ξ
yD
Ξ
0.8
: 5-1,4 :5kΩ
then RD
Ex:5.14 1.8
V
1
οJ l,,
tx
l(2.s
-
ys
/rr:0.5(1.5-V5)' Also: vJ : R]o" :
/r, = 0.5(l.5 _
-
1)'
1oΙo"
loΙDN'2
I
I I
JΞ I
ξ
Ι
V,o: _o'4Υ. lo=
ol
=
_ |0μm _ L 0.l8 μm =+l _5.56 mA/Vr Vοs: 0.6 + V,o: _1.0 v: _ 1.8 + rDR 1rR:0.8ν-forvov = -0.6 V !Y
Ιo:
2.5ν
m^Iv2
!i
0.t mΑ )koV'u =
.'.R=80OΩ
+ lΦ1}, _ = o.lM mΑ Io" = 0, vo
Vι: tumω
32Ι DΝ +
:
2.25
10x0.101
_2.5 V: Αgain if on, then yo >
:
ΙDΝ
1.04v
we assume thΛ QP\S
-2.5 Vand
impiies the NMoS oλ, is ιurned
Ιoπ=o
: o+
V651 < 0
ofl
which
Exercise 5-4
_! -- _ !ι-Ψιv DP 2"L'sο-- ly.l l' 2 , ι x (us + 2.5 - l)2 Vs- loιDPΞ2ιDP _ ( loΙr" l
Ex: 5.17
ι
_10x = 1r, = 0.104 m^=a vο =
l.5)' 0.104
:_1.ο4ν
oo'o---l
Ex:5.16
Vr, = 1.8 V. V65 : 0.6 V. k,= 0.4 m^ / v2
vr,: RD : Y, =
Y:ιo L
18v 0.4
k' : 4 λ=0
for
=
v., ,, :
UDs :
l.8 v.
U65
_
|v
oo
_
!*"ι
Uοs tl.8 35(Uc'J 351λs_
27
υcs + 3.4
υes = 0'613 ν.,
Ιo = υo
:
Ve5
= tcs υο= vι=o.4Υ.
uo,
9o.1
:
_
uo,
v
'f
n"] _ o'ι
o.8 ,cs + 0.16)]
:
0.6
V., /, :
0.8
V.
=
:
= k'
: ,Ds :
Ex:5.1E
:
0.08 mΑ'
R, : 17.5 kΩ _8.Ro = l0+8- :
UGs
1.8
V.,
t iode,
rDS:(k"v",)':179Ω Vol, : Vor1r= roo
kΩ
make
γv
57l |!A'/ν
^',
:.vοv:0.|43ν'
0
ο.1585
= Voo
μΑlV
ou = 8Φ
12.5
o.4
Υ.
(C) For u651.
Υ.'
(b) keeρ
μA
o'213
o.2
Au: -g.Ro = - 10,
Ro :
(B) Saturaιi oι/T.iode Boundary 1.leo
:
c. = k"Υv
rnλ l v2
V,:o.4
kf)
17.5
(a) V6γ
v.
(A) cutoff/saturation BouΠdary
υ6 :
:
Ro
l7.5 kΩ
πΞ-. o':
18 mV.
0.54
V,
l.Ι v'
/,
= 6.μ
,4'
Exercise 5-5
0.7
v.
υ, 1'. ' !,:!+i=-+g-υι
1mA/V2
..R"q:1,,il
Ro.*ii
*
Ex: 5.20
Desigπ for :.
ι',=!b "υi
_
-25. n,"
-
5φ 1ρ
E.Ro : 25 = k,VovRD
o' n,":9= ιι Ui_
Uo
= R6
:
26R'^
:
13
R"
MΩ
ι"n' : ()ι"v })n'
Voo: 5Υ.
1
= )ε^RoVoν: l2'5 V6y and
V6y: V6o_V,
:
:. vοv
8.: Ro :
o.3|9
ΙoRo:4.3 _ |2.5 vov
ν.
3l9 μA'/v 78.5
kΩ
vDs = vov + vι
ico:0+26fi=V,
..-l'''j.L:zτ.v 26 Ex:5.19
λ:0 l" : 2o μAl Ro: l0 kO Y=zo
v2
L (a)Vcs = 2v+vov =
ι' : lι'Ψrvi":
|ν.
2Φ μA
Vos: Voo_ΙDRD: +3v (b) g^ : k:"vLvοv: 4Φ μAlV ιu
@)
(d) Us.
= !!ι:_r.Ro υμ
= 0.2
sin ω,
:
_4
u
: _0.8 sin ωt Ι1 I.los _ Vos+ιa"+2-2 UaJ
(e)
< UDJ<3.8 v.
Using (5.43)
ι,
io = )k,(vor+
k"(vcs
io = 2Φ μA
V,)'
v,1"r, *
l*,tfi"
+ (80 μA)sinω,
+ (8 μA) sin2ιoι
Exercise 5-{
(4
= [200 + 80 sinω, + /D shifιs by 4 μA
2HD _
4μA Ψ - 8o μA η"
Ex:5.21
x 40 x
-
(1.5
=
ι":
:1xωx-!9xιt.ο l)'z 2 'k'Υ0,.-|v,|)" 0.8 Ιο : 216 ιιA
o.o5 (5.1)
: lxi"Yrv,o":
2Ι^ o\g.=.ΞIρ ιοv
Ex: 5.24
4 cosωr)]μΑ
1x
60
0.3 mΑ' Vρy
l.2 m^ΙΥ
ο.5
:
o.5
V
, :Vo =.!Ι = lοιο " ln 0.5
'^: ξ,'Vov
ι"
=
2X0.l _
50
1ω0
..
o
ι- _
25 Χ
o.8
0.2|6
_
_ rs u/u. 92.6 kΙ}
Ex: 5.26
8. : l rfiA'lv ' k'" : 5oμAlv? = 2x'o'l : o.2ν
\ι,!v'."-ξ
Ι
^-l 0.04
n
mΑ/v
O.1 mΑ,
72o tLAΙν
2J2 :2vo : A , : Vou ιD Vou "vo v'Ax L = vλ _ 2X ι2.5Y0.8 l:0.8ιm+Α o.2 : l0O V/V
: 0.5 mΑ=g. : ff'ξ'" :rΓz"ω"lο"os"ω
1, :
v'^x
Ex: 5.25
/,
Εxι 5.22
λ - ο.oa+v." _ !λ r'
,
]Ξ: ," =Vo: lo 0.3 sοιο
1.55
Φ' , :
= o12 mt.lΥ
l)'?
/D:3ωμ1 : 2r0.1 " _
8- =
'.: #,:
Aι,o : ε.(Roll 'o') (s.'72) Ro : Roll ro (5.'7o)
2Ι o
=
k'"vλu
1α)
^^,
Ex: 5.23
8^:
μ"c
^yLv
ou Same bias condiιions, so
same Vρy and also same and
p"c
ι
and
8,,,
NMOS.
"'w.
-Ytw" :
=
μ"C
"'W "
=>
tι
μ'
for bοth
ΡMos
w = O.4 : --!
wo
2.,
t
''
=
=
_
.'.
A,,
!9 : Aι'oRo R'+ υo,
8.Roτ;+
same as (5.75)
R'grc ' iλrv 100K -f> Rin
o
τgJ
R'-,
: _ε.(Roll Rι) :
Ε\| 5.27
τιi8
R L
g.(R"
ll
'oll
n,t
Exeιcise 5_7
: o.25 mA' yov Vr:50V ,^: \ : ,* ua Ιo
=
0.25
R,":1:R,,.:1ΦΩ
v,
98. =
16)
1_g.Rρ : _4
i, : Rril 16 =R6=2Okd| Ar, : Gu : _c.(Ro ll ,oll
Rι)
2βνlν c.(Ro ll Rι) : for fι'" _ (lσv.\ 2Vou = 9.g5 γ.
1 mΑ
:
(})ι lo
,n l vlιz κοl
= +i0
=
.ao: (Αvis"| : l v.
Ex: 5.28 D
Ex: 5.3O CD amplifier
R..,: l : lω ΩΞ8-: 8. 2Ιo
G ν1*-1ΛΛr---ο
lO
mΑ/V
n _ v,,u= 2lo =l^ _ l-25 0.25v fn: fι.'' 8'R' : o.9l v. " l+g-Rι
aε,:
&'.
lωK
:
cu=b=π}o"'.",
_8.αDll
=
mΑ/ν
b=rD t= '":
8.:Ξ2:2ΙnS A1)o
10
t,η,*j;:
mΑ
,' ,,.
+ '"ig 1V
Αssuming Vo --)
οo
Ex: 5.31 CD (soυrce follower)
From Ex 5.27
g.:2ms υ". 1 # - :---: V
o'"
.'. R5
fl
G',
_
;, :
5ο mv - _-::--:jj-j-=g.Rr:.3 + g-f,1 l 2ω mv
:
l.5 1ρ
A',
_ _s.(RD|l Rι) _ _20 - _s 4
l+8-R{
lc,i^,*l -- t v.
Ex: 5.29
R.,,, g^
:
:
2ω Ω k'"Y'ν
(w)ιο.zs
\L/ l' : ^ "'-
: !=g. : :
ou
vl-
8.
(o.4 mA / v'?)
w= L
ξι|"Yrv'o": g^Rι
t+s-R. for ι ,(
5
so
0.625 mA
mΑ/v
Exe.cise 5-8
vs v,s ι], '+ RJ : 6.2 kΩ
2-(-5) - 6klr 0.5
R.' =
Ιf we choose RD = R, = 6-2
kΩ then
4,
sιightly change:
wilΙ
l^: lxtr(y-"-l)'z.Also Vοs=_Vs:5-Rr/, 2ΙD: (4_ 6.2ΙD)1 + 38.44Ι1D 5|'6Ι2D + 16 = o : o.49 mA, 0.86 mΑ =+ Ι D
A1ιo :
A-l.lR,
-a νo _ _---_Τ f n t 8.
/, = 0.86 results in % > 0 or % > y. which is not acceptable, therefofe 1" = 0.49 mΑ
8^ro
l
+
8-rο
y.:
5 + 6.2x0.49=-t.96V vD:5_6.2Χ0.49=+1.96v
R. should be selecιed in ιhe range of
1, :
(ξ)-(ξ) :
A,,o
_
2o
_ --_t+ilg 4
j9 vl}ν
-
ο40 v.
-
\ι'Yr{v o,
12
: Ix
t(Ycs
^ ^o'
J RD = ι
)(v oo
_
l.5l2 0.5
:
vD
.=>
o.s
0.125 mA
o.75
=
9V' = '2γ
RDι D
_ l'|-
Voo 6.2Ι
_ RoΙ6) +
Ι
o =
o.ξ! 11μ
1.96
V
Using Εq. 4.53: 75%
'=t"rn##,*/,.. + /irF :
x νLu
"uo
'
:
o'l {}
1'
Ι)'
Ex: 5.36
Ι
:1xt 2
= Vρu+Vr=
1,,:
p)2 \(ι vn : 5 6.2xO.49 :
6kΩ
0..5
6.2 kC)
o = \ι'Yrv2ou =+Vo,
_
(Ves: Vo:
o.5 _ 0.125
voo ι'
1t
: l":
ι^=!ι|x(2
llx:5-34
0.5 mA
- l)
Vo' = 2Υ. ΙFζ=I.5Vthen:
: -Δ'o /,,
|νΞVes: Ι+1:2Υ = ειΩ = y^ΞR^:5_2 ο.5 .+ RD : 6.2 kΩ standard valυe. For this R,
ID:lXlΧ(ycs
+
y,)2
v'o" ξι"Yrv'o" =
we haνe to recalculate
Ex: 5.33
:
:
voν =
ov.
ιo
mΑ
I
Υ.
o'99
0.5
_ ο.5Χ2 _ l-
2Υo Γ, , V,,ν']-| =rv,+ν*_L'-rv,J Vo
MΩ ιo
Ex: 5.35
(\t)
for
1
ο MΩ to have low current.
l
+1= 2V
o.\
-
0.5x1
0.1 mA
:
)κ"(ξ),v'""
-
x2 - o.25gv-,, ιl
Ves: Vou + yl : Ι.5 v +Vc: 5+1.5: -3.5
νL"
0.5
V
v
Exercise 5-9
p : Vot-Vo, "uo
- (-3.5) 0.1
5
Ι
s.: Fξ':
s5 k()
.^: L
: Vo": o'5Υ
vDs7:'_vovΞVo'^'" Ξ yD.i" = _4.5 v
= lsο
ιο
Φ
: Rc : 4.7 MΩ Aoo = E^Ro: -15 Ro., = l?, : 15 kΩ For r, : 150 kΩ, R. :
Rl"
Y,:1,5v t!:lmΑlv! Vl :
-)
For ro
Ex: 5.37
1mΑ/V
75Υ '
/o = 0.5mΑ :
\*''Y,
v'o,
='νnv :
1.0
v.
15
kQ
Ri" = 4.7 MΩ
A"o = 'ε^(Rol| rο) = '13.6
R"- : R"ll .o: l3.6kΩ
_ -Rr,R, c, π;Ξ'n*Α,oRΞτ;ι_ 2.5ν
_2_5ν
ο.5
-
Vcs
:
V, + Vou
:
Vc = 0 Vs = '2.5 ν. Vο: Voo_ ΙDRD
ε^:
I
inA
7los = 1.lo : Vn5l ιl4,
= 2.5 ν + cv(o.4 νP-P) τo is a2.8 yP P sinusoid superimposed uμo a2.5ν dc νo|ιaΕe. Ex: 5.39 8-νg'
10v
2.5 V
=
+2.5ν.
kλYrνnu: l mA/V
,^ = L = lsο kο " Ιo _ Vοo 0ιa -figa=
0a
_7'o
8-roυc'
ι8
G
------o
v, Vcn: 4.oν.
Ex: 5.38 R'n
ι+G (|| υ^ -
: υ--
"'Ro *ro*Rs
(2\7t--: _ 1t"' "e' "8η
θ'υo _ we ιvant
_ζηroRD
R^+r^+R"
"
"
+ ,o + Rr(1
vsπτ;Ρ#,
"
};ro'!
cr"''
Exercise 5-lO
υοl l υol '' lR, '* lπ,
Ex: 5-41
U*
= o
using (3)
(RD +
r)
8rr oR D
+
Rs(l +gnro)
-
8
1- 8^roRo
^ιμ
3 RD+ ro
2(R^ + r^\ R. : --j-_-Ξa_------gj : 2 l85 kΩ ' 1 + E-ro basedon RD : 15 K, ro : 150 K,
g.:
rmS
Ex:5.40
8.
: l50 kΩ R'n :R^
4Η R.;g
Etro
A
I + g^ro
8.σoll
RL)
I]Jld-ΓR,
rιi8 __1Μr_____l I
+-*
R"", =
Ri"
*il.,
8^: 1ιΙ,A/ν For R.ie : 50 f,)
k,"=a:lιΩ : R, = 15 kΩ + g.Ro : A'r,o : λv = s.βD|ι RL' =
+
R,. "' A"v : G-: ' RΦ ++R. Rt"^
7.1
tou,
15
+ 7.5 {b)
c : ''
R'" R,. +
Ex 5.42 see the nexι page
For oιher R.ig R*ie
Gv
lkΩ
1.7 5
1ο kΩ
0.68
l0o kΩ
0.07
A,.: 0.768 '
R,,"
Exercise 5_l
using eq. (5.107) on page 324 ofth€
V,
:
V,o +'
β6'
+
text:
u_ ^!-ω'l V, = O.8+o.4ΙJΦ+3_J71 Vι:
|
'23Υ
v
ι
Exercise
using eq. (5.107) on page 324 ofthe text:
V,
:
V,o+
'β6' +ζ_
^!-ω'l
V, = 0.8+o.4ΙJΦ+3_J71 Vι:1.23ν
5-ll
Εxercise 5_12
Ex:5.43
η'.: +lv'v,: -2v %.-Y.=3V
To OΡERATE tN SΑTURAΤΙON REGΙON:
VDS mir : V^. V:aV
ι':
\ι''Yr{vo,
-!2 tz"
t2
:
v*\2 9 mA
Exercise
Exi
: Ιsc ""βlω = !!a = 16-,,
6.1 rD.ιvτ
:.Ιc=
ι,e"'
ι962 _
Ι""
'
tπ]
lτl
lxz
7Φ
:
758 mV
+
lsc
[,
-
i]
Γo.Il
V"u = V,Ιπ
= 642 mΥ
:
:
Ιsε
ι'51 : Vr|π ΓΙcz1
ι"r':7Φ+25|ι z-..
6-l
:
zsln Γ]!'l
Lll
25
x
:
:
:
[f]
to_'6
x !!1
1.0ι Χ lσ-16 A
"'[]#]
29.9336
= 748 mV Ex: 6.6
6.2 B
VcC= +5
β+ι
5ο 50+1
|5o
150+l 0.98<α<0.993 -<_α
Ιg :
Ιρ- Ιg
: l.4ω
:
mΑ _0.01446 mA
1.M6 mλ
o:Ι":l.Φ:o.99 Ιε |.4ω g : /. : l ''146 ' Iλ ο.014,ι6 υ'-1ν_
Ιc: lse"' Ι. ' |sFl ντ =
tα)
: 690 mV 19:lmA
urE
'
For acιiνe range
l.446
"
-1Φι
!ΞΦ1 = 2*
25
R-lmaxl =
16-l5a
vc Ξ
Ycc
-
y,
0'690
Ι-
_5-0.69 I
= 4.3ι kΩ
Ex: 6.4
β = .Ξ- and/c= lοmA Forα:0.99
B: '
:k ι""β99
!9
:
Forα:0.98 B '
0'99
:
I_ο-99
0l
'Εx: 6.7
:
Α Αreac = 1ωXΑreaE Ιs
=99
mA
15g
= l 0.98 =49 ο.98
10-15
:
loοX 1s = t0-|3A
Er: 6.E
'.Vιε'Vτ.vιc'Vτ _Isce ιc: ιse
I, 1" : "β49 : !! : o.z.l
foΙic=0
. ιε'V τ : Ise ΙscP VιrlVτ , |
Ex; 6.5 Giνen:
Is
a
_ l0 16Α, β: 1ω, Ic:
Isε:/sc+1r,:l,.[r*|]
1mΑ
Ι'-u"cu'
ιvBE- νBc)/vτ
VεclVτ
ν
Εxercise 6_2
:
:.vcΕ = vsΕ_ v,c For collector
:
v6s
Area : lω
:
[f]
25tn
vrιn
|f]
Ex:6.12
X Emiιιer Area
υs
.v
Ex: 6.9
..Υsε/Υτ-Vιc/Vτ ιc lsce = ιse vιε'ντ Ι. . . !ιcιντ + ιsce 'n: β"
: Ξl .ρ
ρ,,,,"",
Ιa
l-,
- ιr'Vτ Ise V
:β
- (vΒF lse
v Bc)l
vτ
*n"'-"ι\
- 1sc
u*
Ι.-/ Ι * pΙ
Ι,
βι,*a:lΦffi "u''"o"u'
:
rr/
l0O x 0.2219
:22.2
Ex:6.10 ΙΞevΒειvτ
IΕ:
: Σ11
mA
v",
5v
;F'"ηp-,-"';η
:β
2
r
- |aι''9τ lsce
FiE6.12
β: lΦ' Yr, :
0.8
v
at
l.5 _ 0.817 X
2
]!ρ kΩ 101
:338Ω
5ο
:
25tn
:
650 mV
=
1.96 mΑ
[fr, ff,
Ex:6.13
'o'1
+tov
p l-:Exz Ι-: ' 5l β+I"
Ι":
Ιf
:
$=39.2 ι.Α
Εx:6.1l . vuu
:
"
:
yr
ln[1.5
25
x
!.5Α X lο ll]
25.734
: lmΑ
ιrz' V "', : V'lnfΙrr/Ιr'] = 25 x 0.693 : 0.ο1733 :' v BΕ7 : 0.817 ν _ ρ- : V- Vr : l.5 0.5 kΩ ' Ι2 :5ΦΩ o Vru_Vuu β ^'_ /. (β*η V
ιι"vυε/ντ
1.
l0v
643 mV
B:50'vrr=6.7γ vE: vB-o.7ν
Fig 6.13
Exercise 6-3
ο_0.7: 0.7v _ο.7+l0 - IO K :
Ex:6.15
ο.93 mΑ
ιr: #Ιr:
O.91
mΑ
Yc:10-0.91 x5
:
5.45
V
' Ιr 0.9l "β5ο : 0.0Ι82 μΑ -5 V
Ex:6.14
+t0v
rise
V3g decreases approx 2 mV/ "C
for
30τ
rise
Δv.,Ε: _2x30
5kΩ 'vΕ
+t.7v vc
ΔyE
= 60 mV : _ω mv
since /. is coπstant ,. is constant
.'.
Δyc:0v
Ex:6.16
-l0v Vεε_V' ' ''---R_ 10 - 1.7
A+
5
: .66 mΑ _ V"-o
, " : Ι,
Ι
I
rrl
"Ι ^ιr^:
ιΩ
Iu
Ιu
1.65
mΑ
o:lr:1.65:o.gq Ε 1.66 g=]r:]'65'Ι65 ' Ι0 0.0l vc : νcc+ ΙcRc : -10+1.65x5 =
1""u"t'nt
s !s! fo
\
..rc_,sε,
0.01 mA
=
:
c, ic -
νιε,vτ . vrr. Ιr"u"''u,
-
η
_ ,."vaε'ντ |1 1!SΞ1
"
L
V^J
QΕD
Ι
i, 1.75V
= ι r"uo''u' + 1gg rO
i-'β= \ ...
i, :
QED
in
rig
(u)
"uεε'uτ piu+ 9sΞ asinFig(b)
to
Exercise
, vcc- vc 'e R, r0 5 _ -
Exr 6,17
,o:
(v)
bη
(Α)
ιω
l0ο
1ο5
Ω
lo4 Ω
lMΩ
ι00
kο
1ο
: V"'*
ν.,=
ι69 =
0._1 : 2J : = 0.97 mΑ ' ιοK 10 Ιρ: Ιρ/β: o.97/50: 0.0194 mΑ : O.7 + 0-0194 x l0 : 0.894 V V "" (c) saιuraιed Vcε : 0'2ν Ic: (lο 0.2)/Ι0:0.98mA ΙB: Ιc/pF = ο.98/I0 : 0.Φ8 mΑ
,^
+ι
0.7 + 0.098
Χ l0 =
Er: 6.20 vBB
+
Fig 6-18
lοv
By similar triangles ΑBD ν ΑCE
1' _ ro ΒD CΕ
ΔΙc ΔV
r'
AD
AΕ ΞΙ : l X lιl t01
lmΑ_ lω+1 Ι 1ω+ Il : 1.099 mΑ
.'. Ex:6.19
-
1.1
mA
For Yrs
:
Fig 6.20
0
Iυ= 0 Transisιor is
oFF
'"Ic V, :
-
: :
Rc:10kΩ,β=50 (a) active
Y.
-
5
V
0.3 V
l0 _
ys =
l,,=+ιν
ofsaturation
(b) edge
kΩ
Ex: 6.18
_lω
Ι"R"
: 0.7 + l0 x 0.01 :0.8v
0Ιr ο
Ω
1ο6
0.5 mΑ
V""
100
oJol
oJω1
l0
:
lο
ι.0
ο.ι
/c(mA)
=0 V66
ΙρRρ
+ 10 _ ο +10V
1.68
v
Exercise 6-5
Ειι
Ε\| 6.23
6.21
+10v
+l0v
vBB
Ι.
lkΩ
vc
vc
Fol vRB= |.7
Ι-: 'lo
l.'|
vΕ
ν
_0'7
Vr:
= ο_l mΑ
:50Χ0.1 :5mΑ
:4_0.7 :3.3ν R-: 3'3V :6-εkΩ
:
vc = v0+2ν
Ιc:
9Ιs
'
vc:10_5X1kο (a)
Vo-O.1
v
yr,
(so Αctiνe) ωge of saturation Vcε : 0.3 +5
>
1r =
=4+2:
v
_ ρ- = Vr, Vr, _ lo - 0.3 : l, ' 5 (b) de€ρ satuΓation vcε : 0.2 v 0.1 mA (unchanged)
0.5 mΑ
1.94
Ic : β.n.""οΙa : l0Χ0.l : l mA |0_0'2 9.8ιΩ R-: = ' |mΑ
ktl
+6Υ
ρ- = Vcc-Vc
Ι0_6 = 8kΩ 0.5 Exι 6.A
+10v
Ε*6.n +1οv
βro.""a = 5
Fig 6.22 Αt edge of saιυration ,cε : 0.3
V"r: ΙrRrΙo.3+ Ι'R' - Iε[Rc + Rr] + 0.3 10 _ o'2 Ι.' _ 4.7 = |'225 mλ + 3.3 Vuu : Ι'R't o.7 : 1.225 x 3.3 + 0.1
Then1.:51, Ι"=6Ι". 10
v :
_ ο.2
5x4.7 +6x3.3 o.226 mλ
6 Ι"x 3.3 = 4.48 V V"": V" + O.1
v,=
= 5.Ι8
V
Exercise
Μ Εxι 6.27
Ex:6.25
+t0v
v
+0.7
-
+t0v
_v +0.3
0.4
Rc
lkο 50=βΞ150
l0v
. 'L
:
Ι,':
Λc
:
5 07 ' - lωK -
1"
V"-o
v':
Ιn actiνe range
+o.'Ι
Υ
η.lowest for largesι β Ι, : β Ι": 150 X 0.043 Α p- Ξ ν.c 0'3
to-0.7 2
4.65 o.99
'
mΑ Ι,
l0-0.4
+0.7
kdt
[η.(max):0+o"1
x kο
t50
: :
η.
_o.4:
----:--
l.5
0.04.1
For β = 56
α99 x 465
2.2
0.04:l mA
+ο.3
v]
lο _ 5ο X 0.043 =
For β
η.
:
:
6.78
156
0.3
v
Εx6.Σ
Εxι 6.26
+10ν V""
i1 -t0v R.'t - lο_ο.7
R_:l0 '1
β:50 y"":15Χ5ο
'"
9.3kΩ
4=6kΩ
Rr,
=
15ο
:
:
5ο 1|
ι00
100/3
kΩ
Vιι_
sv
Vιε
Rε+[Rsrl(β+l)]
'^
4.3
lω 3
= 1.18 mΑ
I
5t
v
Exeτcise 6-7
ι,: qο
Ι,# :
change
+
:
Ex: 6.31
l.l5 mΑ
-
+5V
1.28 1.15 1.28
9.8%
Θ
OFF
Εxι 6.29
@l"Ξ
π+15V ttt
+0.i03 + |.252
+
2.78
Total cuπent d'awn 0.103 + 1.252 + 2.75
:
:4.135 mΑ
mΑ
mA
3.94
ΙΕ
ON: lkΩ
Θ+
Θ 5
mA
0v _'uΘr.:,
ο.7 ,* 10+100Χl
Ρo\ιerconsumω: y X : l5 x 4.135
:
1
:62mΨ
Ex:
(β + t)/,
0.039
Ans: V,
6J)
:
-3.94
,r.*,
= (-)3.94
mΑ
V
γ
Εx:.632
+5V
/6,: β/9' X Saf
Θ
οN _ ,a\L: (β + l)r, Ψ = 7.03 mΑ y Saι'
-+
lkΩ
OFF aΣ
Ιrr=o
β:1ο0
:
,ε2 unchangω 1c2
unchanged
vΒ: va
ν
0"l
$vu
vC' - o7 1p, = "
:
L
lcl 2,7
, " J
Hence
ι ll_ "^^Γ "' L(β + l)oj7- z7 kJ
_
=Vcz:7'06Υ vΒ : vc2 _ 0.7 : 6.36 v μ, : !Ξlx !Φ : 1.1.4n'a
o'47
IοI
:
vu+o.l
v
.'.
v
Saturation
-=V cc -
O'2
:5.5ν
o 470
101 1s3
= tοI Γz.ls _
7.03
>5
2.75
+
--JL-
(β + l )0.47
l0-5.5
:
ιο
1ε
+ 4.8 mΑ
0.45 mΑ
@/.,.",,
: lu l,
: :
4.8
:
1ΞΞ
-
I
=
0.45
4.35 mΑ
0.45
v_-vcc-o.2
:
1,
9.6 aq 3ρ
Ξ
4.8
V
4.8 mΑ
Exercise 6-8
Ex:6.33
Vcc _ Vcε
,
10_ ycε _ -.120 V/V ο.ο25
_
η
Ex: 6-38 +t5
=>v6: 10 -8 : 2.0v n.: Ιo:Jeε = ι0i2: ειο YcE Swing : 2.0 - 0.3 : 1.7 V . ΔV- = _ιzο : lL
^ν
ΔV
^ν* : .ll lΔV,,| = l2ο
Ex:
6.3
ciuen, g-
:
:
s..r
.v
16μ5
/c
β= l00
vb.
ΙΕ
v"t
n-:Ιr:lmΑ:4omΑ/v vτ 25 mV 1-::
=
g^Rc
10
: _4Φ V/V V6 : Vr, ΙrR6 :15 lxl0:5V
Nc(0:vc+Ndo : (vcc ΙcRd
Ι-
=
(l5 _
:5
τ
ibQ)
Ex: 635
:
=
0.5 mΑ 25 mV
Ι^
:
kο
1mΑ
:-40 x
Ξ1l _ηlse "t
l0
Aν :
Jic --'c=ιc
:
lV
R
",
I
But
v
2o
mλlΥ
10)
+ Av1'be|)
_ 4σ) Χ 0.Φ5 sinιυr
2sioωι(t) ΙR + ιbeQ\
lο
+ 2 sin ω,
(μΑ)
Er: 639
Ex: 6J6 /c : ο.5 mΑ (conslanι)
5ο β:2ω /. 0.5 mΑ ντ 25 mV : 20 mA/V = 20 mΑ/V 0.5 ' ιΓ ο.5 "β502ο0 :2.5mΑ = Ι0μΑ B 5ο 2oo '" s- 20 20 : !0 kf, : 2.5 kΩ
β=
Ex:6.37
β:10Ο /c= 1mA n : -]-gΔ = mΑ,l z111
25 mV
" 8':β-:
V
]!9:z.sιο
r" : ---- β+]
40
25 Ω
πoιe:
g.
:
: :'"'.(ξ)
giνen
ic
fi and r. βi/,
:
€-
L
: (g-r.)iι
Exercise 6-9
Ι, : !€Σ
EΙ: 6.40 +
vτ
l0v
= lq.ε ,nυv
0.Ο25
β: lΦ :
2.53
39.6
v^ Ιc
ic
ys
chaΙΙge Rc to 7.5 kΩ
vc= -10
:
-
3.1
v
+ o.92
:9 :
λu
x
7.5
g.R,
: 36.8 x 7.5 :276νN :
- rι_l39.6 ll n. ll n. , _
ilΓ&
ιhus effect of ro is
:276xlbmV 2.76Υ
Ex: 6.41
+lοv
ιο Vι)
1""
-
(3.s5)
:
_76.2
3.35
4.oo 3.9%
l
Aro : - g.(roll Rr) R6 : Q6ll R4')
I
^ν
:
_
8'RoV ι
"^ '8. .. ^
η
Rι -'Rr+R.
R"R' Ro* R' (rol| R.)Rι a;Γ ν') + Rι
- c^Ooll R.
ll
R.)
Εxι6.Φ
/6:1mA
r-'
l": "
:
llQ lοI
|
|οl
:
o.qs rnr
:0.0099 mA
Vc = lo_ 8 Χ 0.99 : 2.18 v Vε : l0 x 0.0o99: _o.Φ9 - -o.l V _0.1 _0.7: _0.8v Vι:
For/a:1,4 n : Ι, : !-ΞΔ vτ 0.025
β = !Φ: ,-: "s-Φ
= 41 ,,o,7
γ
z.sιο
V^ :]!9 : ,^: lrnιο " l, ι
: r. :
2.5 kΩ Avo = - c^?οll Rc) = _,ι0(5 ll 1ω)
R11
= _97.6ΥN
R.)
ν/Υ
Εx6.41(2'l
t
Rι
Rc
(-)s.(.oltR.ll
: _ι, r,
ιοο
l
I
#tt
:
t
-
,*'"Φ:
)κ,
kΩ
tol
ηl
Rs
vc 7.5
: 0ss fΦ :
kΩ
Exercise 6_lο
6,42 (cont.)
= π" '
R,* : 2.5 :L 1: τ,s Rs+RΙN 5 +2.5 3 υo ' υ| .ιo = _1 x sz.ο : lrs
lι^l
us ul
_
A-^ι'
R1μ 32.5
v/v
ο.ηq
ν
3
-
:z.s
For /c = 0.5 mΑ and
Rc
ι _la lαN) :
_
l0 kΩ
s.:δffi=2omAΛ/
: ]Φ ,_"20 ,^
:
:
]@
s.ο
R9 =
Rι:||
zz.s
= o._,s
0.33
.]5o r}
Uι
Εxι 6.44
n : vI, = --]_: aο." τ 0.025
.'" . -25 (t ' : β+l R,*: r" = 25Ω Aνo : E^oo\l Rc) - ε.Rc :ΦX5:2ΦV/V
r. 2ωk
:
-
- 40.4k{' βlrol1 f. l R,-.) _ tω x lo R*-R" lοι4oj : - 19.8 V/V
,..
x lom
kΩ
5 k + (β + 1)0.35
ιο
5 = 1=i,.'": Iο.v 'lι/. : 5+5 2 "''
:
.Ξ 0l
,-:L:]!9:z.sιο '' 8- 4t
: 10k|| :9.5k _190.5Χ 5 . Λν 5+ι5 : 65.6 vΛr' : ι'o : ! rοs.ο : _l2.8v/v 1;'' ,,,2
li'ol
Ζ]l Ιοl
ιο
" ο.5 = R,': r,:5kΩ Avo : _20(2ω k |l Ι0k) : 190.5 vΔr' urn
:
_
v
Aν: Aνnx-ξ )+) = Iωvrν
R," .e,, c,' _ Rs+R'* :
0.5
x tου ---?L 5ω0 + 25
=
V/V
Ex: 6.45
Rs=50Ω R.., =
r : !: l,
zs mv
l,
-.n
=+Ι' = 2515n: 0.5 mΑ Aνo = +9nRc:2οX5 =
10ο
v/v
c.,:!xA'':4oΥlΥ '2 +,4u : 80 : g.Rj .'. R;: so/2o: 4kΩ
Ex: 6.41
Εxι 6.4
Ε i"= βιb
*1R,"=Ψ=r.+(β+l)Rε Vs _ Rs+r"+(β+l)Rε Rε ' R., r" l"/(β+
_,*Rrn&qευ
|)
,,s:1ω:1*161*-Ιε5/l0l {o l0
P Ιnserι 30
ib
: !!_:ρ
ιb
-
and
υ, (g.vn
ιo
-
+
g.v
oR L
+ ihΙRL
rτ
ΞR,:1: -
r.+(β+1)R1
Exercise 6-11
=0.5+l0ιΧl:101.5kΩ Gγρ : |fR1 moνed, ro: co]
"*R, _ β+l
R^ = "
ΙM Ω
lol
RL
n _υo_
''v
ο.5 + 1o=9 ι
Rr'R':οlo4+l-
;
2β l5ι
:
Ψo
0.91
V/V
t.r Er:
:
+ ι':.
0.995 mΑ
0.995 _ 0.984 I
X
lω
%
6.,18
Βx| 6.47
V.r: Ι12Υ
0.3
Design
I
β:
1ω
^ "8'
_
Av'
Rε=3kΩ 80 x4ο 8ο +40
:
26''Ι kΩ
v--:|2xΦ=ιν "" 80+40 !_91 : ι-: ' 26.7.
β:50 3'3 1. : ' 26.7. =
t.ρt mΑ
l5l
qo ch^Ι'lρe
-l
: l'M
= |o.3
ιs3Ξ
'" Vt /6giνen
l.04 mΑ
_ 0'9'17 X ια)
ΞRr+Rrl(β+
ι00
3.3
For independence from β, set Rd
1-
'
3'3
2.6.+ -t.J ^^ -5t
β:15ο
o
(oK for cs)
/.
ο.9
+10v Rc
kΩ
2Aν
ιν :
=
β:
0.99 mA
10o
οv
t0t
:
:
1ρ Yc(min) = YE+0.3 V = -0.3V VCQ : V e(min) + 2V : +10 V ρ-' = Vrc V.o - |0 : 8.48 kΩ
kΩ
4-O.7 26.1 . ^-
β:
R"/(β + ι)
4.3
Ψο
ffi:
l50
RE +
Ex: 6.49
R-" = 8Xω :2'61 " 8+4
v"":
1)
1) = 4.3kο
Design 2
β:
... maximize Rc
Vg = ΙqRg}2+0.3+IΕRΕ
+Ro =
^1
-t
:
Rε + Rr,/(β +
^
l0Ι
R, =
v
0.984 mΑ
vc: :
vΕ+o.4+2 +2.4Υ
Vcc-Vc
^ "._-TΙ": "
Ι'
β+ι
lO-2.4
:
7.6 1
:-]-m,η 10l
= 7.6
kΩ
Εxercise 6_12
6.49 (conL)
v'_ v" _ loι{2.4 R" _ " l"
o.7)
_
17l.7 kΩ
Usiπg 5% resistors: R.'
:
7.5 kΩ
& =
180
kΩ
ΙεRc Vcc: l0_ο'7 1,_ " l80 l 7.5(β , |) ιs _ q'92 μ o.'1
+ ιBRBΙ
o
-
1.002
:
mA
:2'5Υ
V Y lower
6.9196.O so Y, : - 1.96 V l.96v _ 0.4 _ 2: _4.4Υ β : 2ω: uppe. still 8v' Ιo'ιver + 1, : 0.005 mΑ so ys : _ο.5 + 1,
2:
_2.9Υ
y. '
l(χ)ν
8- =
t" I m : i: ffi ιο 'lrv
,r:ffi:
: lo 7.5(l.ωη
6.50
:
0.5_ο.4_
1ε:lolxΙs yc
Swing: -1.4 -2 = -3.4 β 50: upper still 8
v
lωιο
.-:L=|Φ:2.5kΙ} '' R- ωm
ι
vcc
=
φξ1 =zsο
Ex:6-52 Example 6.50
Rs
_vεΕ
vΕΕ
=
β=lω R,= lωkΩ : 7.5 kf' Ιg : Ιg/(p + l): ΙΕ/ 1u' : 0.01 mΑ Vι: 0_ 13R9 = _lV Vg: Vs O.1 V: -1.7V v-_v-- _-L7171.5 β+|
vΓι-_o'7 R_
+
Ι*,,.
vεL
= t9.3 =
!
2'51
l9._]
: α(t m) Ξ ι Ιrιη 1r: 0.0l mΑ Vc : !0 _ 8 k(l m; : .ι2γ 7ε = lΦk X (-o.ol ;) = '1 v Vε: 1 ο.7: 1.7v β: l00: Upper limit.9 v cc _ Vc:8v Ιower νa|ue +vΒ_ v'c: o.4v (where Ic
v; = -1.4v)
ν
kΩ
I
Ex; 6.51 Refer to Fig Ε6.5l
r r
Rc,
rιι
|ε- v"
Ε|
Rι: r.
10ν
R,.
,'$: I
),, *:
vo
1
i
'l
RL
n"= r"lln,
R,:Ψ:r":2.ska R," : R, ll r" : 1ω |1 2'5 : 2'44 kΩ Auo ! 8,Rc:_4oX8: 32oνN Ro: Rc: 8kΩ Avo: -8^to ll Rc ll R.)- 40x 3.5 : l19V/V Ro = roΙl Rc = lω|| 8 = 7.4kΩ Avo : - 8.Oo ll Rc)
:
_4o Χ'1.4 =
'296ΥN
^R,*.Rc Γ+R;"'R"+R, 7ΔΔ
= ---::-:-:( +LM 5
_ 2|
= 39.1 V/V Rs*Rrιi/: 1,.: |;'λ
:
R,"
cλnλ
:i19.1
ξ
6Ι-----'74+5 !5mV
x !5:586mV
Exercise 6_ ι 3
Ex: 6.5-l
c
B
yr
+ Rs
vo
t:
V.iι :5+20
smv
R, lRu
η
:
6.9
g.:40mΑ/v η
mλ/r' r- : 2.5 kο ro: lωkΩ η: ιoov &: 1ωkο R,;*:5 kΩ &:8kΩ &=5kΩ : Rrll Rι :4XR.ie:20kΩ _ R.' _ 25 2'5 _ o.22 kΩ _ 223 Ω 1οl "' _8-Rc ω(8) _ _.12v/v ' luo: τii-,J:,- ΓΞffiΞ Rou,
= 8 kΩ
_Rcll R. _ _sKI| 5K
a,' _ r.ΙRε
G,, ' oR c,,'
_ R"i* +
25+223
β( Rc
i|
- _D.4νΝ
Rι)
9.9 V/V (β + 1)(r. + RΕ) = -
y 1,, = 2!3 7 =RtieR," * Rιι 25 Κ
Νoιe: \νιhout
&: Av
:
_g^(Rc|ι RL)
: -l23ΥN
p,4
6.25 mv
= 1ω
R.:η:2561 Auo
R, = rtr + (β +1)&
R'"
r",. =
lωkΩ β: l0o r,:2.5kΩ α:0.99 r,=25Ω 1.:lmA ro=
Example 6.50 8.: 40
i-
s,ι:::ll lωl-]+y. -Ιrι 5mv= (2.5|| lω) l-'J |vr| : lv,l x Av : 512.4 : 62 fiν
Ex:
Rl
20
&:
ψ/o
E*" JL
t.
&:
ψiιh
:
g.(Rrll ro) :40Χ10 3x1ιι11 : _296νN +
tωι)
n.", = R. ll rο : '1.4 kΩ Α, = +g.(R. || R' |l ,o)
:Φx3:'l2oΥΝ
tt:tr=
r"is
:
Rsig +
0.ω5
0.6
R"--ι
v/v
''" : R*;. : *_-_
v/v
α(R. |l Rι)
^ "'_
:
r.
α(Rc ll Rι) Gν
54
Ω
5(m
+ 25
t5 mν
Exercise 6-14
Ex: 6.55 RJ
,.R,o =
tnt
:
B
:
Ψ:
r"+(P+ ΙX/o|l
0.s + (10r)(2011 r) 96.7
kf)
: fia ll R,, : ω c,,' : vo : vt vs v\"vo vl
R'*
:R,*, : βVτ Ιc
: .-,, : lΦQΣΦ 5m
0.796
sοο
ο
zοιο " =Vo I, =]@: 5m ib : b::-9 and yo = (β + l)i,(r, lι(β + l)(ro|| Rι)
Rlι
11
96.7
:
28.3
kΩ
(β + |)(roJl Rι) (R. |l Rs) + (β + l )(r. + oo||) RL)
V/V
Ιο l
,lo\
20K
(
lο K.|Llo κ )(5E . 5 lοΙ
-- V/V Gvo = 0.8
R.,, = .oll (r.+tRr|| Rs])/(β + l) : 20 || t0.05 + ο.o79] kΩ
,^
. ' _ Vι
Rs *
^40 uvo ," :
5mΑ
Rr)
-84Ω
||
R.)
,,
V, x lroll
For
RI(kΩ) 0.5
G'.(vΛr')
o.οl ./
r,1
0.95
0.ω5 1.0
ο.68 0.735
2.0 0.765
ι
20 K)
Exercise 7-1
Ex: 7.Α.1
(a)
when
o.2Υ
Vo,, =
l^
=
The minimum value of 1, occurs
U
and
:0.1,thatis
\ μ^c.'Υrviy 0.4
ι^-:
:
Ψ
V arrd
ID
1Φ, ιhat is
mA
(v ιq-"' -
vg
v'
ιv,, : vτln(ffi) + ΔV"'
:
3.1 mA 0.8 μA
'^
and v.=(25) mv
2o1 mΥ
Ex: 7.A.2 FoΓ an NMos Fabricatω in the o.5
_ lo' Ψ ι'
μm pfocess. lρiιh
ιve wanι ιo find ιhe
ιrιιηsconductance and the intrinsic gain obιainω for the followiΠg drain cuΓrents: (ι : 0.5 μm)
ID:(ro)
μ,ζ,:
*o,r^:f*4!;",
(l9ol
,.:2-
intrinsic gain
:8^r,' For ID
:
=
6.62
,.:yo!,D g.r. :
o.62
19oX
lΦ
kΩ
ΞΔ
1ρ
:
166
V/V
=
2
ξ,:
X_8Ξ.ΣμΑ
:
o.57 mA /
cr,: ξwιc.'
::1
Χ
cs, =
5
Χ 0.5 x 3.8
8.3 f F,
Cμ:
8' 2τ lCn ' 8.8
+ 0.4
Χ
Coνx|
:
5
0.4 X 5
o.57
C"ιΙ
21τ
:2 fF
η4
(8'3 a 2'
GΗz
{"ι'1vs For ID
: l0μΑ,
7'"
,€8η,Α/τ5(1η(1ο
πX
62 ν
v
+ cou
μ, =
O.28
using eq. (7.15):
x o'5 = ι00 μΑ
1
20
For ιhis problem, use e4. (7.1ι):
\γe haνe:
20
:
lο kΩ
,^:ui'=20Xo.5-Ι!?kΩ " IΙ' 85.5 μA _ A. = g^r. : 66.7 ΥN
!0o
ι"
=
,'^ff-&Ξ
Ao
:
5oΥΝ
!Δ - 6.6 ΙΔ vv _
kΩ
xlMΩ:2ωl
'._ F-\ψ:"DX
"n- :
10
= 2
-
Ex:7.1
ξ_'%iot'= l vο
1ω μA
X
/, :
4ν'
o.z}4
+
r-
J2y l90x l0Χ tο -v -02 ja:
ρ-=
x o'5
lmA
ιo' !p,c.')v'"" z'' "'L "' =l2 ,* 4x-Ξxo.3'] ν2 0.5 1ρ : 85.5 μΑ
_ aιν'' ι v,
_
20
We haνe
ΙSeν'Εmia/vτ
6^in) l
_
EΙ: 7ΑJ For an NMos fabricated in the 0.5 μm cMos aechnology sp€cified in Table 7.A.1 Ψith ι :0.5 μm' W - 5 μm. and V6y:o.3Υ
Ιr"uu'^^''u'
'
,.:q!.ιD c.,.
(b) For a similaΓ range of current in an npn ιraηsistoι we haνe
1c.,, _ 3.l mA 1cniο 0.8 μA
mA
".r..(r),,
occurs when
\ν.c.'!vLv:3.l
1
: JzxrsοXloX1-z!Δ -v
1O.8 μA
The maximum value of %,, :
For/ρ:
/ν
νηο μΑ
Since g, vaτies with
Γιo -ι ιo
*itn
ft
mA/V
Εxercise 7_2
Ex:
For
1, =
10O
μA
==r
7J
I
g. :
0'r8
rn(#)' :
.89 mΑ./ν
!
Α^:50(-!q.)':Ι58ν/V " ι lω/ For
1, = l mA: ,l\2
!
mAlv(7.J
ε. =
.28
a_: "
sοl.0ι0'l, =
ι
\r,,
=
2.ε mA/v
ι.,:/=tooμA Ιr' : QJ-.ιιΔ n
svlv
': vτ R"' _ r,ι=
Εx: 7.2
25
mv
= +.e"nv
_ 4 lΦ
β,
= 2sιΩ
8,r mA./V ,''' :Vo: 50V : sfflkΩ Ι 0_l mΑ 50ν :5ο0k() ,":|vΛ: "' Ι ο.l mΑ :
Ao
A':
:
(4 mΑ/v) (500 kΩ) 8',' r,,ι _(4 mΑ/V) _ g^ (r.ι || r., ) =
(5ω kΩ
Since all transistors have the same
W
l
_
Ιρρρ =
Ιη:
'.':
]z
: :
la2:
/or :
1v,1
l0o μA
,
_ 5 v/μmι0.36 ο.l mA
volιage Gain is g",
(η,
ll
μm)
18k!!
Since
(18 kΩ
l|
2l.6 kΩ)
Ι,
:
w_
ι
9ΞΞ-E!. η6 2
(0'55 μrn )
Ι'
:
zο'ι
: 2
(o_1 126
|..]75
(l0ο μΑ)
ν
(t.37s v),
\r,lloο μΑ) kΩ
\ι',c"'l(Yr)ιu*((' 2
Ι'
2οω v/ν
lΦ0 V/V
:
Y
.-)
A'= _ (1 .24 mAN) : - l2.2ΥlΥ
_
|v,,|.ι,
-^)
, ^ _1voo|L, 6 v/μm (o..1o μm) = 2l.6kΩ 'n'ι Ιnι ο.l mΑ
λ":
is halved:
y.l _ s νlom '2
L24 mAN
.'''' vι"L, Ιoι
:
:
lyol lvol "- - ιv-ν, τ
*"c.'(l), 'π;'
-^^{]Ξ.,.),*
/';
5ω kΩ)
Εr: 7.4 ιf L
7.2 U-ro
0.36 μm' we have
||
:
-
ffi)
Exercise 7-3
Εx: 7-5 Voι = l.1
vor = 0.8
YDD = +1'8
v
Flx;7.7
v
ta:
s^:
0_25 mΑ
ο'25 νl2
2
Qι
: 5ν
,-:Vn: " ID
v
0.25 mA
2
Ι|l,AΙy'
:20kΩ
(a) From Fig. 7.13,
I 8-
:
,q._
"'
RL \8
|
2
^r
")
Rι
+(2
mΑ"/ν
mΑ/ν)(2ο kΩ)
:5σ)Ω+3--+οo
Rι:lMΩ:
l MΩ: R,_:50oΩ+ '" 4Ι|
Ιf all transistors are identicaΙ and ιhe gate voιtages are fixed,
Vor:
=
|Vrr]
:0.7_0.5:
v
R"
V62 V,n-Von 0.5 _ 0.2 : 0.3 v
Vr.- =
ι.o
the ιoΨest yDs2 cangois |vru]
:. Vo-in
ο.2
:
:
= Yscl : 0.7 V Voι = Vsι: vΦ+ιv||+|voλ : 0.8 + 0.5 + 0.2 : 1.5 V
:
0.5
V
similaτly, ysca
VsDj can go
Vo.^'
:
as low as
t.5
0.2
:
l_1
v
8^1|:8.2=8.ι=8^ι=8. lo : 0.2mA _ 2 mΑ/v = 02ν Ι 2
oλ
2
fnι=foz=rδ:to4:rο
: lyol : 2v :lοιο
ΙD 0.2 mΑ Ron : (g.2ι.2)r.1 : (2 mΑ/v)(10 kΩ)'? :2ω kf} Rη = (l r,,)o,^) = (2 mΑ/v)(lo kο)' :2ω kο R. : R., ι| R.Ι : lο0 kο
μ
= _t{ε',.)"
Aν
: _2Φ
ΥlΥ
:
l0o kΩ:
R.-
-}t{z ..ιπ){rο ιο)]'
kΩ
20 kΩ:
:5ΦΩ+
?9_ξQ = 40
l ιο
0:
R.-:5ΦΩ+g '" 40
Εx.7.6 |v
Rι =
'" Rι:
lVoul, so
Voι _ Vsr:'., =
:
25.5 kΩ
lω kΩ _ 3 a,-:5ΦΩ+ '" 40
0.2 V
VDtt + VD.2 = 0.3 + 0.2
40
= ο.5 kΩ
(b) Fram Fis. 7.13,
Ro: ro+ Rs + (g.r.)Rs Rs: 0: Ro : 20kΩ+0+ (40)(0) : R" : 1kΩ:
20kΩ
Ro = 20kΩ+ l kο+ (40X1 kΩ)
: Ro : Rs : R,
Ro =
is :' Ro :
10
20
:
kΩ: 10
kο
+ (40)(10
kο) =
+ 20
kο
+ (40X2ο
kΩ)
kΩ +
20 lcΩ: 2o
ko
l0O kΩ: 20
kο
61
kΩ +
10o
kΩ + (.ι0)(1ω
: E^1 : B^ _ /o _ lΦμA: i mΑ/V V nu o'2 νl2
Ex:7.8 g-,
2
tι'|:rr2:ro
:
430
kο
8Φ kΩ
kο) :
4.l2 MΩ
Εxercise 7-4
2V - ,nU,,
-V;ID
0.1 mA
so, (8-r") = l mΑ/v(2ο kΩ) (a) For R,- : 20 kΩ'
:
so, the gain remaiηs the same
ιf&
20
,4,
R, *',, _ 20kΩ + 20kΩ _ ρ-' "'' = 1+e-tr,,7 1+20 .'.
:
Αvl : _8.l(r,I _
l
||
R-2)
K)
mΑΔr'(2ο K || 1.9
:
is connected ιo ιhe outpυt'
_ I.74
:
|.9kΩ
l
:
+
#;t(ι
|| R1.
(ι + grRr)r,,Rι (l 8.RJ + 8πRs)r,, + RL
8.
:-r."
v/v
g.Rs)r.,]
+
R, (
I+
g-Rr)/,
or Ιf we υse ιhe approximaιion
R. _ | , "'n'=-8.r..,., %'
ofeq' (?.35)'
Ex: 7.10
2οkΩ- l 20 ι .Α/v
V
Vno = +1.8
= 2kΩ ιhen.
:
Avι
l
mΑ/v(20 ko
1l 2
kΩ)
: - 1.82 v/v
Either method is coπect. continuing, from eq. (7.31),
Aν
Α, :
:
:
_
gmr|(gmrr.rr,,r)|| Rrf
l
19.04
mΑ/V{[(20X20 kΩ)] l| 2ο kΩ}
ν/v
A''' = Au "
Au'
(b) Noιγ, for
_
-
19'04 1.82
=
Ι0.5 VΛr'
R, : 4ω kΩ,
Rι | ,lOokΩ , "ι'z = |^}l,-;' - 20 - l.A/v :2lkΩ Aνι : - | mΑ/v(20 kΩ |] 21 kΩ) : - l0.2 Aν = 1 mA/v{[(20)(20 kΩ)] ]| ,ω0 kΩ}
(a) ΙD|: Ι and ΙD1= Ι
since yovt = Voν,
I
=
v/v
:
0.2
v
ιve have
,- :ξ','''(Υ),u""" , Ι
Ιo'
,
| ν"c '"(Yr),v""",
2ω v/v
l.^ : Au : :?Q9 : " Au, - to.2
lq.ε
vlv
:
b"(Υ), 4
Ex: 7.9 The citcυit of Fig. 7.14 can be modeled as
o-
,, '"
+
(w\ |τl,
"(w\ "ι7,,,
vi
(b) The minimum νolιage allowed across current
1
a
source /' would be
v
"'
Rn: (l
made with
νolιage ψould be
Where Gand
|voιλ = 0.2ν if
single transistoι Ιfa 0.l yP signal swing is ιo be allo'vr'ed at the drain of o!, the highesι dc bias
:
I + g-Rs +
g.Rs)r.,
The open-circuiι (no Ioad) volιag€
"'
|voΔ 1.55
0.1 vρP
V
(c)vsG2:|voΔ gain is
Aν'= -G-Ro = , +8. =.(|+8,Rs)r,, I g-Rr
+ lvtpl
yG"2 can b€ set at 1.55
-
= r.s - 0.2 1(0.1)
:
O.2 +
0.7 = 0.85
0.5 = 0.7 V
V
Exercise 7-5
(d) since curreπt soυrce Ι2is implemented Ψith a cascoded current source similar to Fig. 7.10, ιhe
minimum νolιage requi.ed across it for prop€r o.4 ν 2(o'2Υ) operation is 2Voυ
:
:
(e) From pans (c) and (d), the allowable Γange of signal s\γing at the ouιput is from 0.4 ν to l.55
Vov or 1.35
v
V
so. 0.4 V < Vo < 1.35
Ex:7.11 Referring
to
V
and
and biased so that
oI and oa can be selected
and ron are very high and have insigηificant
effect (/o >> r-) ιhen, R"^ Roρ
= (g.rrot)rnt
8.r. :
Ron:9:ro: Roo
:
R" R"e)'
ε.z =
=
'.,
'.,
:
fi:
0.2
lr -1
Lgr=
lv..l
25
mΑ
mV
-
8 mΑ/v
"2)
= (s mΑ/v)(25 kΩ)(25 kο || l2.5
tu.s
ιΩ
R""
1.6?
c^':c.o=Ψ: vτ r
=,ro:
kο)
Mf)
For the ρnp tΓansistors,
0.2 mA 25 mV
50-
8
*:
roj: ro4= H: R", : (ε.:rρlXroι
mΑ/v
-
8
mΑ/V
6.25
r.:)
kΩ) = (8 mΑ/v)(2o kΩ)(20 kο |ι 6.25
R., Α,
:-
: :
kΩ _8.l(R"" 762
1|
R"p)
(s mA/v) (1.6? MΩ Α,, -4.186 VΔr'
:
kfl
#h: '*n ll
||
ο.762
+
Ζ1Ο
mA/ν
s.(R.ll r.)l
kοιl
iο
176.7
+ 40
+(o.5
kΩ ιl 2.5 kf'
η
kΩ
: rr:
191ρ
Ex: ?.14 Fig. 7.21(a)
: 5V : zsιΩ 0.2 mA r
r6fr
:
R,
1Φ :
= k,2r6)Oq1ll
:
1.0MΩ)
without R. (ιhat is, /?, = 0),
From Fig.7.l9, R"^
7.l3 ρ- = 1/.l =
R":
g.r(9zror ll β,.rr)
Ex: 7.12 For the npn trasistors,
ε-ι:
||
-vτ J.η4 /) mγ = lΦ : z.sιΩ ' =9 R- ω mA/v
Ex:
R.1
iτ, : -g.r(R-ll
|Αv.",|
5ο(20
Refeσing to Fig. 7.20'
β,
= 9rro:
Since
: Fιroι : kΩ) lMo
(8^ιroι')rnι
tοιο ' :Vo:10V /. lmΑ =
: (gfiro1)r.1
since
: Roρ : Ron
Α,,", : _(8 mΑ/v)(2.5 Mο A,^'' = _5114νN
fig.7.ι9,
: (g.ιro)σoι|| r.3) R". : G^z'o)Qρι|l r"'')
ro,
Then, Ro. : (g^7ra2\ro2 : 9zroz R.' = 1Φ(25 ko) : 2.5 MΩ
Finally,
Roo
Ιf
Αv-,' occurs Ψhen ol and o4 are selected aΙd bias so that ,o1and ro4 are>> rτ
MΩ)
rot:ro2:ro=i:
v,
5V
0.ι mΑ
_
50
kf)
kΩ)
:
.lττΞ7τττ;π -A)
"6(rω7πτ,δ),lω c., : 1#' G^: g.ι = 1mΑ./v _ o.lπA:4mA/ν " : ιlg1 vτ 25 mV
1Φ :usιΩ .-:-9_ =4 mΑ/V 8^7
Αssuming an ideal cuπent soulce, Ro : (8'ιro)(', oι|| r oz)
R"
:
λ.6
:
(4 mΑ/vxso kο)(5o kΩ || 25
:
'G^R6
_3.33 X
1ο3
:
v/v
-(1 mΑ/v)(3.33 M)
3.33
MΩ
Εxercise 7_6
W l0μm=W:50ιm 5μm - lμm
Fig. 7.21 (b)
so ιhe dimensions of ιhe maιchω ιransisιors 0! and 02 should be changed to:
W:50μmandΙ,=5μm
Ex: 7.16 For ιhe circuit Figure 6.7
Ιz
:
Ι
nτ
. Ιι : ι*'''@77Ι'
/ L\, t'_;-_::-,.
ιw
λnd ι.' :
ν,/e haνe:
(w/L)\
(w
/ ι'\.
ι ':-----------: "ιw
L'4
/
Since all channel Iengths are equal
Lι:Lz:...:Ls:lμm
and
: l0 Ιι: Ιι :
Ιx1'1,
ψe have:
From part (a), g,'1 =
ι'''
g"2:4rπA1Υ
ro\: rο1 = rr,: rr:50kΩ Γnl : to2: r" = 25kΩ
: (g-1ro1)R" so, R" : ( 1 mΑ/ν)(50 kο)(4
,
"z)
R,,
-
MΩ
167
_
A,n = _8-ιR-
:
_668 X
EΙ: 7,15
1ο3
mΑ/v)(50 kΩ)
:
μΑ
7.
ι5
and \νe wanι to redυce
the change in output current, Δ1(), corτesponding to a
1
v
change in output νoltage'
Δyo, ιo l7. of
Ιo. Thaι
is
Δr, _
= 0.0l Χ
lω
0.0l /, = !J roz foz -
ΔV
o
μΑ
,o.:JJ:1μ61 ' |μΑ
To keep yov of ιhe maιched transistors ιhe same
7.l5.
w
7
of ιhe ιfansisιor
shοuld remain the same. Therefore
= 20 μΑ,
80 μΑ,
-
6
ι Ψ.,_W., _ Ιι _2ρ-1 ,'ι _,RΙ'π'π r-* |o
-
ι
w
| /*r.
|o
Ιn order to allovr' the voltage at the drain
of02
to
o", \ ν^c,^(l),ν'
:
\ι|'(l),v'
-,
:
;2ω$(}),ιorr_ Izο - _ l5=+w1_ ι5\L2 [Ψ) \ L 12 2ω t (o.2)' ιy,
-
w-
_ z=w\
*'
Ιn ord€r
15μm.Ψ _ u=w' = wt
'
vou,
ι.
ιl
_ Ψ 6
=
:
2.5μm
2x W, _ 5μm
ιo alloψ the νoltage at the drain
go up to wiιhin 0.2
ω
,o^:V'o'L-tMΩ:20Xa ' Io lΦμΑ lα) v Ξr-- =-::rrΙm 20 V/μm
as ιhaι of Εxamp|e
11
w' w'_ -:-L ι. _ α|ι'y
60μΑ
v/v 10o
μA'
20 μAand /5 :
Ι"'"--+-^' '
l' :
4mA/v(l67MΩ)
Ιn the cuπent source of Example
we haνe 1()
= 60
go down ιo within ο.2 v ofthe negatiνe sυpρly voltage we need yoι2 = 0.2 ν
(50 kΩ || 25 kΩ)
R,,
12
l.=t'\-\=!=Ψ:+ ' 'Wo wΔ IΔ 10
G.=gq:4mA/Υ
FromFig.7.19, Ro = (g.zro)Oorll
μA,
v ofposiιive
ξι,(l),
ofρ5 ιo
supply ψe need
=v"o"'
iεo$(}).ιo.zl'=
_ zxεο - - 50--eWs 50ιs l.ιy] \ ι,,ξ 80 ι 10.2;2 Ws:50μm Y: = o.- ιv. _ lqJ*m _ l2.5 um wι
"
4
Exercise 7-7
Thus:
: Wa :
|Υι
2.5 q"m' Wz
:
15
l2.5 μm and l{''
μm, lv]
:
Ex: 7.19
:
5 μm
50 μm
Ex: 7.17 From equation (7.72) we haνe:
'.: '"..[;fuJ{'-'oΡ) ι.
r\
l mA|- ,, \' '-iω_/l(,'
=
ι" :
#)
-
kΩ
: lΦ ko
t.O2mΑ
'l.02 mA
Ro=,o,=ξ
=
#*
=98
Ex:7-18
Fτom eqυation (7.74) we have:
Inr'r γrtVo-Vιε\= V^ t t2/9)\ Inεr (,+2_o.7)Ξ 0._5 mΑ l+(2/lω)\ 50
l^: "
1+
,/
/a6p
:
0.5
.Aπffi :
0.497 mΑ
=v- Rvuu-p = vcc-v"t /".. R : 5 0.7 : 4.3 ε.εs ιΩ 0.497 mΑ 0'497 1^,-
Vrr,n
:
V,.n.n.
:
0.3
V
For % : 5 Y From equation (7.74) l = Ι nε'ε (, +V"_V"'\
'
I" _
{2/β)\
l
+
l
+ (2,i
0'497 (l
vA
we have:
)
Σ:-qJ) _
|ω)\ + 50
/
o.5.| mA
See next page:
Exercise 7-8
/ιtlι, = ,co*,,.
-l.oRιF Thus:
ιgnoring ιhe effecι offinite output resisιances' we haνe
Ιι : ]z : ... : /ι : Ι rQ"u, * Ι : /ιυr(*) 1
,
: /rρ",r,* Ιu /ar".,
+ ... + ΙΒN
Ιc,
βββ
/cOιllr'
'
Ι."
,:,"n**(λ-i- -J) ι : ι-^ ..Rl l ryjJ β
From (*) 1coR|.|
\ir'e have:
+/:
/RFΙ,Ξ
+
,.r*,r,Ψ /" aa
'
- ryl] β
For an error not exceeding l0% we need: /^.,aΞ/πrτ(|
0.l)
t + il-j-.1
β
/ιιt ?o.9r".' Ξ l !',ry+l l_Ν+Ι ββ
Ιl-l
I = I + {-L-.1 = I.r = β0.9β
r + N-1-1
β = /ν + ι <
>0.9
o.It=9iv + l S 0.Ι1
ιι Ξly'<
Β
r
Ξ
1ο
The maximum number of outputs for an eπor not exceeding to be lessthaο 10% then we need N
Exercise 7-9
Ex: 7.20 Refering to Fig. 7.32,
: Ioι : ]ol = Ιoι: I*uo = lΦ μΑ Since Ι, : \'"c"'(ξ)v'." Ioι
= 0.23
ιlεz ιa l v]l(JΙ)
"',"'(Υ)
v
0.5
v
+ 2(0.23
v) =
ο96
v
To obιain the output resistance, Ro, we need 8.3-
_ 0.87 mA/ v o' _ vov/2 o.23 V vι(L) _ (5ylμrn)(0.]6 μm) _.o2_ ιo! : o.l ma h : 18 kΩ. From eq. (7.77)
o
'
2(0.ι mΑ)
Ι
Rn- g6rorrn7: (0.87 mA/V)(t8 kΩ)'?
:
1"
1
1ιιι = l' + ? μ"n""
_
lΙn
= o.qε
β
Ι*or|
1*oa
x |ω:
2%
For the wilson current mirror\ρe haνe
=pr. _ ιωΧ 2lωkΩ_5MΩandtor the simple miπoΙ R, : r, = 1ω kΩ R-
The minimum ouιput voltage is
v*'+ 2vo" =
whefeas for the simple miΓroΙ from equation (7.69) we haνe :
282 kΩ
Ex: 7.22 For the two curτent Sourc€s designed in Example 7.6. \re haνe
:
n = Ι, : l0μΑ _ o.omΑ-o v vτ 25 πν _V^ _ l0οv_ |o MΩ..' = β.
"
"^
l0μA
ιc
8.
25okΩ
Foι the current source in Fig. 7.37a we have
Ro:
roz =
r,:10MΩ
For ιhe cυrrent soυrce in Fig. 7.37b from equation
Ex: 7.21 For the wilson mirror from ιhe equation (7.80) Ψe haνe
/,: /πεr
Thus
'ι
+
:
l 2
:
ο.gggε
:
11.5
1ρ
,
therefoΙe.
β(β + 2)
!!:-ΙsΞd x fiO :
(7.98) we have:
R.a tl +g.(Rrll r,)lr, Ιn Example ?.6, R, : R1
o.o2%
κ..'-'-LΓl
ι ο.a - ΞΔι ',J v l t.s ιο ll 25okΩ)'lιoMΩ
=R,:
54 tr1ρ
Exercise
8-l
Ex:8.1 Refe.ring to Fi9.8.3,
ΙfR,
is doubled to 5
Voι
: Voz: -
= l.5
+ Voo
K'
ι Voo_ ΞRo
0'4 mAι5
jο.s
κl
2
Ro:5kΩ
Rιl=5kΩ
v
vcν.u': v,+vD:o.5 +0.5:
+
l.0V
since ιhe currents /D|, and Ιρ2 aΙe still 0.2 mA each-
Vo, : 0.82 V So,V.r.,n = yJs+ ycΙ + yoj =
-
1.5V
+
0.4V
0.82V = -0.28V
+
So, the common-mode range is
-0.28 V to
V
1.0
1,ir'
ιhat causes
ρl
to conduct ιhe
enιire current is ^/ΣVn,
'l
th€Ι,
=
:
yDι
ο.η5 v : vDD ΙxRD
-
x 2.5 = 0.5V
x ο.:tε
"D
1.5
0.4
ι"
Γ2
Γ
l"ιγι_
2(ο/ mΑ)
: g.(Rrll re) Αo : (4 mΑ/v)(5
K|l
50
K)
:
l8.2
Voι=Voo:+l.sv
Ex: 8.5
(b) For o2 to conduct the entire curreπt:
with 1 = 2ω μΑ, foralΙ ιransistors,
then.
Voι = V.,2 =
Voo: 1.5
V/V
! 2ΦPΑ: Iω,,.a. ι^= '22 = L:
l.5V
+
-''
Ιo _ o.4mΑ(2) "-_ voν/2 ο.2 v = 4rnA/Υ : ,^:V^:20v " lD 0.4m4 sοιΩ A1
υι:-JiVov__o.45Υ
_
J;r(,.Α,vπ|ω)
η/^'ι7'
Ex:8,2 (a) The νalυe of
','- _
2(o.l8 μm)
:
0.36 μm
-0.4x2.5:0.5V
(c) Thus the differenιial outpυι range is:
VD2- VDt.from
-
to 0.5
Ex:
1.5 =
-l
-
1.5
0.5 = + I V
V
ο.l mA
Since
8J
Refer ιo ansιγer tabΙe for Exercise 8.3 where values were obtained in ιhe follolving way:
v^'' =
ι ^/I/ Κw/ι''')L = L Κv nν2
1,,
(f),: 2(
(4ω μA
":Ι l "'r'
(10V.zμm) (0.36 μm)
-
'Ι :
\ Vor
)
lo, =
(Ψ), = l00
μΑl
0.l --l ll , = 2 ν^'
"loj
EE: E.4 ο-8
mA
:
ι" : \ι'"(l)υoul'
lω
2(
l00
μA)
μΑ / v'?)(ο.2)
=
0.4 mΑ
Lμ^
v)'
.n
'
(tω_μ4J(2)
ε^,0o,l1 ,oι)
18V/V
,
#tr}
so,
soιι^t
r^(Υ)r*'
:;#r
/ v'?)(ο.2
(y).: (9. (
Ι':'=
:
:,uoρ
:
1( mΑ /
:
1mΑ/V,
v)(36
K
|| 36
κ)
Exercise
Ex E.6
L
:
:
2(o.l8 μm)
ι ιιι ,o : |vλ_:
cMRR(dB'=
0.36 μm
ι' : ξ: '*'*o : '* .,
=
\'^c"'(ξ)ιv""f
W/L:
Ι" =
foιall NMoS
\τ),
t---------------
2(tΦμA) :.n / v'?(o.2 vf
,qδ:+P :
_
(2 ε"'Rss\
(Δε'\
cMRR(dB) = 20log Er:8.9
: (9616)16r: (l
K)
2(4 mΑ/y)(25
lo(
20,0ω) = 86 dB
0.0t
:
+5V
1mΑ/V
,5 t,
1kΩ
0.7
:
4.3 mΑ
+0.7
mA/V)(36k)'?
Μa Roo : {g-5 r65)rq: (l mΑ/v)(36 = 1.296 Mο
= 25kΩ
_
From Fig. 8.12(b), Ron
o.a..ι
\c-l
1ω pΑ
For aιl transistoΙs.
/ν2),
p"α'lus''
,"**
(g),:(g).=(y),:(:), 2Ιo : μpC.'(V o)2
08fl :
:=
mA
υsing eq. (8.64), and the fact that Rss
|ω μΑ) _ 4ω μA /ν2(o.2ν)2 μnC.'(V oλ" 2(
=
=
'.
lΦ,μ"C.'{o.2
= J2(0.2 mΑ / v')( |ο0)(0.4 mΑ) 8^ = 4 mA/ν
rw\ rw\ |τ)._
_
v
= l.296
k)'z
+0.5V
Using eq.(8..18).
Ao
:
=
:
s-, (R..ll R.,) (l mλ/Υ)1'296( Mο|| l.296 MΩ)
648
v61
: -5V
υ62:
1kΩ
- 5V
Ex:8.7 The ιransconducιance for each ιransisιor is
Exr 8.10
g^: β μ"q'1w lηl" l^ : ! = 0.8 mΑ _ o.o.o "22
from φ(8.35), the differential gain for matched
isΑ, _
Voz Voι _ n V'"
Ιfιγe ignore the 1% here' Aa : 8. Ro : (4 mΑ / v)(5 K) From eq. (8.49),
V.a Λ.'' = v"=
*
ΔR, _ 2
R''
:
p
20
v
/
(0.0l)(5 Κ) _ 2(25
κ)
o.7
Υ
( 5+4.3xt)
lkΩ
ΥN
R^ νalues
)
From Exercise 8.7,
*o
V/ μ.Ψ)(!,36 μm) _ ,u*n
:
(ffi
-'oI"εlo
H
Ex: 8.8
Refering to Fig 8.12(a),
Since ΙD
20 |ogl0
= 86 dB
The drain cυrrent for all transistors is
(10
8'2
v o.ωl v/v
_25ν _
Vεε
mA +-43 (α:l)
20, ooο
Exercise 8-3
Ιcι = Ιcι : Ιεl :
Ιεz
=
:
!: 2
0.4 mΑ
Ex: t.13
+Vcc
0.2 mΑ From φ. (8.66)'
vcMfrι'-vc+o.4ν
=
Vr.
ΙgRρ+o.4Υ
_ = 2.5 0.2 mΑ(5 From eq. (8.67)
Vr'
Vct
^rn
K)
is - l.5
subsιifuting ig1
v
* i62 I
,* /
:
"(υιz ιυB1
' υB1 _ υB2:
2stn(r 25
=
+ l.9V
Voz
= -vεΕ+vcs+vΒΕ
Ex:8.11
=
v:
Vιz
: - 2.5V+0.3V+0.7V:-1,5V -rn
Iηpuι raηge
0.99
+ 0.4
v84
/
_v.
ιo + l.9
:
/
v
Il=zrlο"ι
in Eqn. (8.70) yieΙds
ιιt)/ντ
VΕΕ
/:20oμA
υD|\/ντ
l"(s_
Sinceβ>l, t)
I-'-ι^^-/ 2ΦμΑ_ 22
99)
:o
vτ
ln(99) = l15 mv
Rcι:Rc2:Rc:ro
in each ιransister is ο.5 mΑ. Thus yBε for each wiIl be
_ 10v :lσ)kΩ
V
Dc currenι
-. = "'
o'7 +
0.025lnrψ)
\L/
lω μΑ . _ V,ι -_ ΙιΕε
7
= 0.683 V
.+ιΕ=5-0.683= + 4.3|1ν .nlo : Ε 0.5 :2oηΔ vT 0.ο25 v (c) jcι : o'5 * g.1Δι3r, = ο.5 + 20 Χ 0.ω5sin(2π X ιοωr) = 0.5 + 0.l Sin(2τ X 1οωr)' mA
icz = 0.5'0.lSin(2π X Iοωr), mΑ (d)
ι61
:
(V66 * 1cRc) _ ο.1 Χ Rcsin( 2rr X lα)0ι)
_ _ = (l5 0.5 Χ l0) 0.l x l0sin(27rlω0|) = lο - l sin(2r' Χ 10ωr)' v
v.2 : l0+ lsin(2r'X 1000r), ν (e) ι62 _ z., : 2'sin(2σ Χ Ioο0,), v (0 voltage
gain =υcι_' 7ιA1
2LP"uk
= ο.l vPeak = zιn
vlν
ιcι 7'A'
IOV 2οο μΑ-
25 mV
lv^l
Ιc
_
ν'
",="r:t':i:
l0ΟμA:4.Αzv
lΙ ^Ι ι !ι:
:
EΙ:8.l2 (a) The
l0o μΑ
50
kΩ
:
25 mv ο.1 mA
o.25 kΩ
Since R.. >> r",
|Aλ
=
R.l' ,o
|0ο
κ|| lω K _
r. 0'25 Κ R',=2r-.-_β" 8^ R'o : 2(25 K) = 50 K
2οo
v/v
|Φ _ z5ιlι
4 mA/Y
ιfthe ιotal load resistance is assumed to b€ mismatched by l%,
lΑ
ΔR. -l = 2R'' =
cMRR(dB)
.:
(0.ol )(l0o κ) 2(50 κ)
20 loc,,,
:86d8 Note:
Ιfonly
the Ιoad
andsinceα-1.
lj:l
_
o0|
:
ιransisιo$ are mismatched.
Exercise 8--4
K)yl.. u'ol _ _ Rrll Vo u _ _(Ιω K || lω 2Rfr. r,' '',(5o κ) , o25 κ : _o-499 vi._ ,, _[(Rc + ΔR.)|| r"]v'-
"oι=
2l;;T;7-
_ _Ι(l.0l)tω K]|| lωκ.v
= ιl5(t|y.
2(50 Κ) _ 0.25 K
2ω CMRR: ιΑl| _ 1Α,.l 0.50l 0'499
.
r*&
lωK
lω(lω
_
{
lω)(5o K)
ro
:
l.68 MΩ
Δ&
/ RD
as:
_
QΞ x o.oz 2
Use
ηn.
y^.
-
due
ιο
o.α)2
v
i.e 2
mv
wlL
: :
:
^/12
3.46
Δvt
(y!! ^+#)'
mV
10τ
4
Ro = r orll
ron
o'2
Α
mλ/Υ
: yΔo: 20 : (0.8 Ιo'
πι/2)
20
soιΩ
= 50kΩ
*
25kΩ
G'Rq: oΨ"25kΩ:
η: is obtained
(8.117) th€ total input offset is:
x ro t1'+ q2x
δi'"
$
From Eqn. (8.141)
(8.l 16)
*Ι'
G-: g- :
Ro:50|| 50:
mV
The offset νoltage arising from
t(?
20 m
thus,
--r 2
ηn.
I
m
:
{
lo' = (ο.8 m/2)
- "" : (92) \2 t x o-o2 = 0.Φ2
Finally, from
:
-
^!fiiιrν
,^' = y-Δ' "'
v^"
vιs= Δvt = 2mν
0.ι m x 2ω
_ Γη Ι
"'
(8.l l3)
ηn.
xμρζ :
x 1Φ:20
ο.8 mΑ-/2) " 8.=τ;:-lν-=*.v
,^'
AylΞ
"*:(Ψ)(Ψ#)
from
o.2m
Fιom Eqn. (8.138)
(Ψ) (*)
To obιain
ΙL),x μ^C'":
thus, Ψe obtain yσ due to
ηu. (8.l08)
_
(w
ι" ouι
o.2ν
,.,:
Ex: t.16
ιvlι
From Exercise 8.4:
v*:
:25ι6.Φf * φ,1r : 2.5 mV 1Φ : lΦ l_ '" : 2(β+l) 2X1οl =οs,,a
since all transistors have the same drain cuπent (1/ 2) and the name product Χ μζthen all transconductances 8. aΙe identical.
Ex:8.14
Using
,.ie:Ι*(ΨJ
=
(w l L)ρ
Κ)
. IωK+2(5οK) lωκ
R;..
,,,
:0.5 X0.1 μΑ=50nΑ
: βRεr. ---Ψξ"-
R;..
From Eqn. (8.127)
u,: u(Ψ)
:10o.00Ο--)10οdB
Using eq. (8.103),
Ex: E.15
(Δ',)'l''
1o-r;2 + 12
x
lo-312
From
ηn. (8.l48a)
lωγ
| O.OOsv/v 2g-.Rr, 2x4x25 = 1"rr = lAo| : lΦ = 2ο.0{χ)
A--"'i
lΑ."l
86
0.ω5
dB
Ex:8.17
ξ:3"
From Eqn. (8.156): (ο-8 mΑ/2)
I/2 ""- v,
25 mv
From Eqn. (8.159):
Ro = r nrll
roo
'ul
Υ
Exercise 8-5
Vo 1 Vι Γ7lι π='-Π2 : lΦV = 125 kΩ
Using Eqn. (8.179)
Vou
Ψ!!
Ι.' = (w/Lt\ιΙ|'
0.8 mΑ
A,t : From
Ro :
G^x
ηn.
16
x
125
:2ωoy
(8.l62)
2x25mx160
=
|ΦV = 0.8 mΑ
l25 kΩ
:
2
l20οο
θ. lsΞ
= l04
O^
dB
lοο, ?9L! _ 2
16
a'ξ]
MΩ : 5 MΩ x Al: 8. Rι': 20 Χ 5ω0 = 1ο5 v/v = (l0|1 10)
i.e.
l
Φ dB
Ex: E.19 Refer ιo Fig (8.41) (a) Using Eqn. (8.178)
Ι.=(W'L\o(Ι/2, '' (w /
=1111 = thus.
L')Δ
(W/L)n
(lVl L) 6
0.775 mA.V 0.775 mΑ./v 1.90
mΑ/v
&
= 10/0.05 = lo / 0.05
r" :1o/0.l = lωkΩ ηn. (8.l76): Aι : _8.ι Qo'ι|l ,oι)
(e)
:
ο-775 12ω
:
ll 2ω}
_?7.5
y v
Eqn. (8.177):
:
,^l|
0.129
:2Φko : 2ω kο :10/0.1 : lΦkΩ n
(d) r",
From Εqn. (8.174)
[β.
v
V 0.129 v l0ΟμA 0.105 v μΑ 50 μA 50
a,
0Φ
l/2 lm/z :20mΔ ^ U-:8mιι:':'-:-:=ντ 2) m νA lω v 2ω kΩ Ι/2 0.5mA -) Roa = βaro4 : 50 x 2Φ K : l0 MΩ
:
y^''2
a,
Et: E.18
n,
Ι x9οΧ2ο0
V
a,2-=!ι
l
2ο ιogι0(160,οω)
R^. _
0.Ι29
q,
lJ tr5l 160,
:
similarly for 0,, %a
0.129
Ξyov6:0.105v
-2 / |25 Κ ' _ l6i,Χ "'la5 κ A : .OI25y
Cr*^
:
1rlοrzιn trn:
β"R*
- MRR_
thus. (}v/ ι)' = 20o (b) Fοr o''
For
From Eqn. (8.Ι67):
L
1611
2ω
(0.8 m/2)
= 20 kΩ For a simple cuπent miπor the oυtput resistance (thus RJ is r"
'_,R"' _\
: (w/L), ;
l : \υ,c,,'(l),v*",
R,": 2x ro
vt g _-2x |ι/2)' "
11p
=
lα) : 2Φ
r
an
μ11
Az
:
g.ο (roοll rol)
oνerall volιage gain
A' x
l,:
is:
11.5 X 95
:
7363 VΔr'
Ex: E.20 Refe.ring to Fig. 8.42, al| /D νaιues are the same. so' η'.|. : yct + /AR, UsiΠg the equation deνeloped in the ιext'
r"_ffi"tΓr] n-" =
2
^E ιφ 5'21
kΩ
,lzllιεοl,
(Γφ
lο
,a, ιι:0
l.l
l
Exercise 8-6
v-,,: η,.', :
- v" 0.l5 + 0.7 = 0.85 V v,","
thus, y",1,, = vcs
ι2
+
Η, _ yss - 2.5
= 0.8s + 0.15
:-
8^ιι:
0.38 mΑ/v
):
Ex E.21
FinaΙly' V.": V,"
/,:ΦμΑ p'C-:
1Φ pA1Υ2
p"C-: Φ μAN2 Fσr
Q."nd Q,:w
Ι
L:
oul
)lv
:
μρC,''(w
oul".n
:
/
- v*.:
:
+2.5 + (-0.3
+1.4v
-
0.8)
Exι E.22 R'=2o.2kdΙ
40 l0.8
(as given in Εxample 8.5)
lv
t.5v
: |Vou"| = 0.3 Υ on '' tVcsιl: o.3+0.7: lv Ve ιι: _ 1.5+l: _0.5V V
Α- :
v/v
8513
R.:152Ω with&: 10kΩand&: 1kο
L)
=
40μΧo;
0.3
V
20'2 x εsl:l x ----!(r + 0.rs2) r0
A,': '
20.2 +
:4943νN
then.
2I" 2X9ouΑ 8.s'9:|vλ=_τ'τ-
since 8. of oΦ, α and 0, ιhen:
Y.,, =
0.3
o11 and
o,,
aτe
Ex E.23
identical to & of
v
Thus, for O,.
i,l
lQ5 +
Ri4
l5.7 + 3ο3.5
Ri3
3 +234.8
o.u9z
Ξ=β,:lω
2Χ9ομ ιo.1)2: ' lω ι'lw/L)n )(W/L)0: 12.5
iυl i.s
i.e. (10 / ο.8)
:
0.οι26
ιb5
40
iδ5
i"z
ο^s
-1)
(
R3
Rr +
Ξ:β'=lω
Since Q,, is 4 times as wide as Q,,, then
rw\ _ 4xl0 _ ιΖi,, ol
il : ρr+t: Iot ιba irr: R, 15.7 ,
_
il+R2 Rr + R2 + Ri2
_
ω
+ 5.05
:
0.8879
?:β,:rω ι|
Thus the overall current gain is:
Ξ _ ι|
2x'|Φμxlgxqoμ -+Rr: 1.67 111
X
:
:
160uX:Y '
ο.8
0.8879
X 1ω
55993 Α./A and the overall νoltage 8ain
The νoltage drop on R, iS: t.67 kΩ x Φ μA 150 mY
:
Iοl xο-M92X 10oxο.0126X loο..
0.15
V
vo ,
Ru
: -Ξ'tο't
Χ ξξ591
is
-i,"
V,, R,,
i,
:
a256
ΥN
Exercise 9-1
Ex:9.1
: '4M
c,,
x s-(R.ll
RD) R-&l0 .-^.-l0Κ t0+0.1 2
,τcc(Rc
τ;e
_1
I
''
'-
=
+
0.I M)
I
,!
:8Hz
0.016 Hz
_. : 3l8
r, :
2.5
kΩ =
β=β:
. :V':25mv '" l, ImA ,:l
2
:
,ξο
zC., IR,ll r,
+ R,,r]
2zτ1μ[1Φ K|| 2.5 21
]οο
8-
K+5 K]
:
,o ' r-γ, *
10 : Γ -'z_lJ
ooo
Δl
24:2 fF
m
4.1
fF
0.6
AtVl= Vn : 5Υ
|κ'(l)'
ιv, rv,^l'
_ 1\2οy20ι5_2)'z- 18Φ 2
mΑ
Ex: 9.5
n-=Ιr:lmA_40mΑ v vτ 25 mV 12ΧωΧ Co":.r'8.:20x 10 l0 : 0.8 pF C\.:2Ci",,:2x20: Φff c,: c,ι" + Cl" = 0.84 pp Cμo Cψ: (ι
\
'ιrlJ
"L' β+ι
]
* Υ:'eι.l-'u' vοcΙ 2Ο
R
:
lo :6l
:
Peak current occurs
.4 Ηz
Rεll
fF
.72
Exι 9.4
Εxι 9.2
Ccι:Cε=C62:lμF 3. : ao$-+l. : 4οmX25rn : l
3.45 + |-72
vo
ι νeaι
318 Hz
X
l * Υιι
Ηz
2rτl μ X (l0 + Ι0)
1
vo
C
--:_-
RD)
2τcc2(RL+
:
|
* cou
=:ΞP_ /'* Ιll
Rω
+
2nΧ| μx (l0
fι =-fn:
Cgι: C6ν: C'" =
I
:
\wιc,,'
: ?x tο,
V/V
9.9
:
fF
t2 fF
(, * u1)""
_ " ,;G;Ξττ' 8.
_
2τ ' Cc2' (Rc + Rι'\ I
2π"lμ(8K+5Κ)
f.: Ex:
|2.2
9J
'"'_
Ηz
ξn' ;
3.45
X Ι0
ΙΙ
ιο\ l0'
: 3.45 X 10 rF,i m2 : 3.45 fF / μm C6v: WL6vC",: l0 x 0.05 x 3.45 :172 tF
,ι0
X l0
l
2π(0.84 + 0.ο|2)
= Y |0 ι,
7'47
cHι
Ex: 9.6
lι'l
- f-ι _ι 'ιo
Ξ.fr : ]τ _
: fJ
50o MHz
Σi-;ττ;\
4ο Χ lο r = ---j:----_j-:-
2πx5fi)\ 106 Cιτ: l2.7 _ Cμ: 12.'7 _2 : |o.'1 pF (
,'7 +
('U"
'
Ex:9.7 Diffusion componenι
:10.7-2:8.7pF
of
12.7
cτat
Since C,i, is proρortional to /., then:
pF
/, of
l mΑ
Exercise 9-2
cd,α(:
:
f-ιΙ-
o.| mΑ)' =
4x10
:
r
ι'M ^ :-
7.l4
68p . 1.65
Aoc
Aβ):
l+
ιγhere
s
1
ΜHz
.42
Αr
is the
Dc gain
2nfιo"
frφuency of the amplifier. Ιn this case we haνe
/Md :
XlΧ7.l4
:
Απ
1000 and
ι00 KΗz = 1Φ Hz
Therefore
C., --
Κ =
of the amplifier and/3J, is the upper 3dB
tΩ
_4'7 lν1Ω (4.7 + 0.0ι ) Mf} _7'12 ν / ν
lσ)
+ l.5 +
10_3
write th€ generaΙ form of the transfer function of a direct-coupled amplifi e. as:
o
2πC'"(R';.||R61
ΦX
Ex: 9.11 Usiπg equations 9.6l and 9.63, we can
R,,.u'"' kΩ, g. : 1 .47γ
Ro
10
Ro +
+
2τ
l3O-7
Ex: 9.E
R'ι: :
-----_g!2'ιτ(cτ + cι!,)
2τ(2.87+2)X10-12 MHz
:
Rg,
cia:''ι + |(1 =68pF
ο.'| mΑ) = 0.87 pF 0.| mΑ) : 2.87 pF
Cιτ(Ιr:
A(s)
-
t+
4,26 oF
10ω J
2τ Χ lo5
Ex: 9.12
2τ y 4'26(lo K
|Ι
For this amplifieΙ we have:
4.7 M)
tΙ(ο):
Ex: 9.9
cs":1pF C.q: (l + g.R)C"a C ra : 8.l4Cra
= (1 +
1
x
7.14)
1
+
a.l4cga'pF(lω K
Ξ1.63>ι Cμ
+ 8.|4C"ι
< 0.o17 PF or
Cμ<
77
|| 4.7
M)
=1MHz
:
5
2.5Χ40.ιo-]XR; 2.5 + 0.05 + (10ο || 5)
-0.013 x R. = t.s kΩ
=n.
:
l.5 kΩ = (1ο0ll 8
:
R, =
-J (Σ
Cin
'7
f
'' :
:
Cτ
.4
kll
1|
ro|i Rc|l R.
A'M , ,_τ'
(,-(ff)')('-(#Ι)
['.(ff)']l' .(#)'l=' :
Κωp1and ωfl = 0.9 ωΡl, then
RL
I
R_.^ __ l.65
+
'
''"
s. R;)
+o.g'?)(l
:'
kΩ
-(Ψ)')
=
,
-(Ψ)' : l.1=(Ψl)' :
Ιf ωΙ,
-
0.99 ωΡt, then
0.l
+K :
['-ePΙ]['-(Ψ#Ι] :, ,
2πC,. ' R"1"
cμ(l
(l
1r *o.oo,1(r
R.) kΩ
l.9 kΩ
+
: ξ+
['-(Ψ)']['"(**)']
Ex:9.10
1ω +
ω = ω,we have
at
A1
Ιf ω72
ff
ΘA": _3912: _19.5 V/V Rι r"' 8-' Rι ' _ nr+π* ^" rτ+rΙ+(Rr+Rsig) ,ο0
By dennition
|a11,,"1f
/r > 1 MHz.e 2rCii(RsisllRc) Ξ1MHz C^: C*+ C, = I pF + 8.14 CslD 2π(l
('.*)('.*)
-(Ψ)')
-(Ψ)' :
K:
9.88
:
r.ol
=
,-
=+(s)' :
o.or
=
2.78
Exercise 9-4
nj-4:
Ex: 9.16 Refering to the solutioη ofExample
of|
νalυe anatysis is: 9. l 0, the
fιι : f"1 Αlso'
:
determined by the exact
143.4Μl1z
Au: _c'.RL: _l'25x|o: _12.5νN
Therefoιe the gain-bandwidth pmduct (f,) is:
f, :
x l2.5 : l.79 GΗz since, is less than, :2.44GΙlzuldΙ:
of
Ex:9.17 Referring to the soιυιion ofExample 9.10, ifa load resistor is connected at the output halνing the value of R;
roιlL roz
n;:
,
then
2τ.{{C'"
x 2'5
we haνe:
l.25
x
:
ξ
6.25
(c.
+
ε.RL)I.
R:ie +
5f|l
+ l'25 Σ
5)l. l0 K
"o(|
νN
+ csl)R;.]
+(2sl+s/)xsKl Ι'' = 223 |ΛHz
1, =
|λM|.f
ιl :
6.25
x
223= l.4
Gltz
is1,,is increased by4 atdV*by 2
+,
:
'^=ffi:2,., 2.5ο
t.zs
!Δ
,oρ,|1 r oρι
=
(rρρrll'oρι)
K
n;
k(Ι
in example
+ (25 + 5)
250
44
:
L
ΜHz
1.56GHz
r:oιο
,",r"ll roono: 130kΩ|ι 50kο
" Β' '-:E-: (a) From
M :5kΩ ΞΔ
4n
v
φuation (9.97) we have:
- ____.'ι--G-n;) ,τs|3τrτf/τ
: _ --_j__{ 40 X 36 kΩ)',-16+0.2+5 A,, =
-
Ι75
v
-fisy
(b) Using MiIIer's theorem Ψe have:
16
r-
:
pF + 0.3 pF(l + 40 X 36) 448 pF
:
,ι48 pF
|
'"_ 2nc,.R',"
_
27 Χ 448
2τC,^tr"|| (R.ie + r,)]
l
pFt5ll
(36 + 0.2)]
Ξ 4.3
t!92
f X2.5 K
R;:36kΩ ,a lmΑ :46m4 vτ 25 mv
increasing /, by 4 reduces
loκ-sil
(cL+ csd\R;}
:#:soιο
",.=l?l :
boιh roQι and roq2by 4 :
f,' !f" :
ffi:
,.""^=+:
C;"
ro Jν' +
+ 6.25)]l0
-+
Cin: C'+C'(|+ε.RL\
Ri = Ι0
f(l
Ex: 9.19
=
Rr, = ronrll ron,
Since
zx
η4
To calcuΙat€
9.ι0
=
+ Cr7(l + g.,Ri)iR"I* +
25o lν1Ιjz
lu:
Ex: 9.1t Refeιring to the soluιion of Example 9.Ι0
s,, =
2τ|[Cr'
f,=lAtl' 7" : 6.25x25O:
I
+ C
"" _ 2ιτLL20 [ +
Thus
2.5
Using φuation 9.93 and assuming/,1/",
fH :
Using equation 9.92 and assumingJ,s/",,
lf
|ιul = g.. R, : : 6.25 ΝA
21τ[[20 f + 5
ιιe have
and ιherefore
: s. 'o'ι 'o, :
|A'|
:
= 2.sko
143.4
40 GHz, theτefore it is a good aρproximation the unity gain frequency.
then
lxroK
:8z.6kΗz
kΩ
(c) USiηg the method ofopen_circuit time constaηts, from equation (9.l00) we haνe:
τr:
C,R.'. =
crt(l
+
g.Ri)R.l. + RLΙ+ 1LRL
i.,. = r- ll (R.ie + r,) η"-ll ro,n, : 36kO
We haνe
ιi
+
3 4.3
kΩ
Εxercise 9-5
Thus:
to : |6x +5x36 τo =
36)4.3 + 36]
2.12 nsec
f, : :L I
:75.1 kHz
(d) Using equations (9.l02)' (9.103), and (9. Ι 04) Ψe have:
18. I η4 v 2πCu 2π0.3 pF - -,.-
,
40
CP(
+
g.R.)]R.',
" 75.1 kι{z = f pt =
+
(cι
+ cμ)R;
l [c, + c,(l + 8.R;)]f;iρ + GL+ c.)RL 2zτ IC"(CL+ Cμ) + c.cμ]R.iεRι f ρι : 25.2 ΜHz
Sincefμ <
/,
MHz
remains ιhe same as that of the example' to place J, at 2 GHz νιe ιeed
_
+ cL
:
l2.5
2πΧlοkΩΧ2GH7 94.5
Ex: 9.21 For
a
_s vιε :
f
,
f'
Ι'
5ω kΩ,
Example 9.12:
g-
:
'l.25
ιιλ/Υ
28o K
6.75
179.5
kΩ
(ci,, + cι) 'R*,'
κ
+ (5
f+
15
f) Χ
179.5
K
0.135 ns + 3.59 ns = 3.72 ns
Thus. iΓ,,
: _L
:
2πT g
42; γμ.'
Ex: 9.23 a) Low-fιequency gain
Ay, g,ro40, R,.: _
ro
: -g,(Rrll
Amplifier: Au
Since:Rr:rr-+ Av= \ιε.,ol: _!xιo:
re)
_20ν /ν
cΑscoDE Αmplifier: Au: _ g-(RolΙ Rr) = -g.(R'l1| 16) where Rq : roz+ rot + (.g^2ro"1ro, since ro2 _ ro| _ ro !Ιιd 8.z _ g.ι _ 8. Ro : 2ro + (B.ro\. ro : ro(2+ g^ro) 4216
:
'8-(42tol|
Αvcoscιrrι
,δ
=
_c^.ro
:_4o:z -2O
:
CS'Amρlifer:
C, + r : C"o =4 :
:
x
20 f
+
include R.is; also C"1
c.l(zoc") : CιΙ C"a Cμι 8^/Ι2τ(cL+ c'ι)Ι
Εxι 9.22 R1
0
:
:6.75 kΩ
b)J, : Neglecι componenιs of τ,, ιhat do not
Therefore.
fι
Cg,' Ri"
Avcs
"'t-=- 2π(Cr+C"ol 8' and f : "' 2τC"1
:
τρ : : τa :
+ Aν
\νe know that:
fu
K
20.8
(8,ro) . R.ie
+
Rsl :5ωK|1 28οK =
:
fF
CS amplifier fed with Rsie
l0 K||
(same as in Eq. 9.12)
CS
Ex: 9.20 Referring to the solution ofExample 9.l 1 we have /. : |Αr| ./, ' since |Αrt
lAul =s,= ' 2ιτRl 'f , -r,,
:
Ri,
: R. ll t?o Ra : ra * R.iε
'fρz =
: ' l3.l
R.ie ||
R,,1
I
+
To obtainjl:
Rε. _
.;Hz
fρι :.
2τΙc
: 20 fF, csd : 5 fF C1 : 15 ΙF, R.,* : l0 kΩ, n. : 2ρ 1f,}. , _ |' Rι _ l '' 500K % R^r" ll5 rn 12ξ ' 20 : 20.8 kΩ R' 50ο c.'' = b.2ν/ν R't" + Rl. ιo + 20.8 Cr"
x
4.3 + 0.3[(1 + ,ω
'
r,,
r+
a C"a
and from
:
20
kΩ'
τ4
:
+ where:
O.25 C r.
C",'Rsig + c s,ιf(l
+
g.'n.1R",*
+
π.1
(cL+ cd)RL
Rr' =
Ξ, τ,,
:
since Cr,r
ro
ll
r. :
ro 2
css.R.j.+a,[(t -
:
o.25
C' aιd
s)n.',]
g^ro = 40
Exercise 9-6
+1H:
css'
x cs,X
Rsig + ο.25
To calculate, we use the method of open-circuit time constants. FroΙn Figυre 9.30 ψe have:
21 XRsiε
=RsigΧcs,X6.25
R',,. = r.1||(r,r
R'.,. : R., :
cΑscoDE - Amplifier:
η
9.l ι8 and neglecιing the terms that do not inclυde Rsis Using
:
τa
R.iε[Cε, +
Cr/l
+
2'o (t'+Rtl : R,,: "' r^ll "'' \ g^ro l ,^11 "' ε-ro
Γ
τ, :
---l
R.;r|Cr'
X
+ 0.25
+
= Rslg ' cs, Χ ι.75 "fa
cιsconε -/ρ
_
cs
:
'""
_
t, : r,':
3.6
ι1η-
2x3.6
.,
'ι /,,,-
J
f
s.:40Ψ
and
r.
:
=|ιu1' 1o l l3.5 MΗz
,^ mA : 8- : ,ι'J -Γ Rιn : r.t + rΙ : 5 kr, + 0.2 kΩ = 5.2 kf} A. : 8-.'. : Φx 130 = 52ων /ν Roι : roι: ro : ]30kΩ 5κ ^ ^Ξr 5.- r^1+ Rι ,(. R' 20ο+ l "' F+1
8-z
._ 130+50
Ro
A..
lq
x
242
l 339 nS
x
_- 469ΚHz
469 Κ1-lz
Ω
;*
:
- azy
2ω X
130
kΩ =
r'* η,_L"'c'(β'" ll
V/V
to
MΗz
"
26
R.,)
MΩ
159 ns_ ιhus
x 10.6 K + 16 p x 35 +(C.+0.3p).(50Kll 26M): 159 ns +CL: l.4pf x 4.4 K
Ex: 9.25
R'.
+ 0.3 p
: =
Rι ll
l0K
ι,
=
2ο K
|1
20
K
From Eq. 9.121 we have:
, _
^"
8^R'ι
ΤΞi^R,
"8- ;qττ"'' Jτ :8GΗz
20Ι
=p2r92
o' :
:
: -_L : 2of
.,,"
16 p
tτ|:to2=r.:5kΩ
35
2τ
has increased from 175
l l3.5
this exercise:
:
| _ 2ττ'
To incτease, to ι MΗz we need:
= 5kΩ
Noιe that for the cascode amplifieι considered in
R1"2
16p)
242 Υ / Υ iιιcreased from 75.1 KHz to 'J"has 469 ΚΗz and, has inαeased from 13.1 MΗz ιo
Exe.cise 9.19 we have:
13ο +
', {. R,. Ιll (",*ffiJl L
compared ιo the cE amplifier in Exercise 9.l9,
Ex: 9.24 Referring to ιhe solution of
:
-
- ,,ll - [,"J I -l
339 ns
'
l.4rl
8nι
kO
X 35 + (5 p +0+0.3 p)(50 Κ|ι 26 M)
τ = |Au|' f ιι .fr.n..,,,n. /lΑvlc,ιscooι\,lt'"fac.ηscooι\ =
4.4
C"). (nι ll Ro) r, = i6px4.4K+0.3 px 10.6K+(0+
c) f
Ι,-
:
Ro| + (CL + C,", +
τcAscoDE
R.ia.Cs, x 6.25 R*ip.Cz, )/ L75
R'",.
Rol : 130 Κ I| 35 Ω:35 Ω Rnl : R',r*(l +g-,Rot\*Rot Rr1 : 10.6 kΩ τn: C.ι.Rot * C*l .Rn1 +(C""r+C.:)
/'#1
cs,ιl
4.4 kΩ
Ro, _ ,o,l R,",
8^Rιι\l
: ,"ll'# ='fi
+ R.1.)= 5 K|ι(0.2 + 36)
1.25 x 10 : Y 1+1.25Χl0 6_93 v l _- 1-25 Ιn 2τ (2of +5f)
| .8. _ 1 . !Ξ-9 _- 196112 -'r' = 2τ C"" 2τ 2ot
: 10 Κ R.i"+R',. 10Κ+l0K t-s-.R! t+t25xto
R3a = R,i3
:
1.48
kO
Exercise 9-?
:
R..
R.ll R.
:20 K
:
R..
||
20
:
n"1 (."U
K
1|
1.l kΩl| [2.5 + Rμ = 1.Φ kΩ
;)
= R'7ΧC3a : 10kΩX5fF : 5οps τ., : R., X Cr. : 1.48 kΩ Χ 20 fF:30 ps 7cL: RcLΧCι = o'74 kΩX 15 fF:ll ps 'rH: τ'ιι* τr,+τca: 9l ps 79ι
ρ = R''iρ*R'ι _ l.l + 0.99 - 1.1 . o-99 Ι+R''ip+R'ι 2.5 2.5 rτ r.
τΙ, associaιed ιγiιh c"i, c.", and cι
we haνe:
= 55ga: 19
9t
x lΦ
l
_
l.75cΗz
Χ 25
mv -
2.5 kΩ
n :L:ντ J-ΙΔ: ηοn,'llν 25 mv noting that in this case
R, -+.ο
(β + l )(r"
R'.ig + (β +
"
β+
R'*iρ :
l)tr.
, \νe haνe:
|| R1)
+ (r,1| R.)1
l
R.lc * r, Αssume r,
:
Ιω
ο
+c - 0.965 Yv
lΦ x
β+ l 2zrC -r-
1
2πC,r,
zoc 'β-!:+ l
2nlc
8"
+
'ρe haνe:
R''
R'.', ll tr" + (β + l )R'.]
: R.,.*r,: l kΩ+ο.l :
:
l.1 kΩ R,||r,,=
l.l
-
A,t:
(c) For a
r. :
mΑ/V2
0.2
a o8m n - -L Voν o'2- o'v
g-(Rr|| r")
where
sοιΩ
=
4 m(5 K l! 50
K)
:
18.2
Y
CS amplifier, when R.i8 is low:
2π(C,
I + C
ινhere
l
Eο-9-
"nlR'1
'
l
α5l
R'. = Roll r, : 5 K|l 50 Κ : 4.5 kΩ and c. = lω fF + CDB : |lo fF
kο]|
l0o
kΩ
f.
2?r( |1ο
:
Σ 4.5 K
295 MHz
: l0 kΩ _L(Eq e.8s)
R.
τ, : Cr,' i, *
+
Cι'R'ι :50f
csa[R,(l
+
8.f'.)
+
R'ι]
Χ l0K+... 10fuo K(l +4 m x 4.5 K) +4.5 Kl +...
thus' τ11
(lοοf+lοf)X4.5K
:
0.5 ns +
1.96 ns +
0.5 0.99 kΩ
f.)
where, from Eq. 9. 84:
fΙ, kΩ
f + l0
(d) Using the open-circuiι ιime-consιanιs method
/, :
C!\
ΦmΑ
R'.,*
=
Vo 20ν ,-: " (Ι/2\: 0.4mΑ
for
c"'' :4" -c,,n - 2π,4ΦΜHz ^---=:- -2pF 2τ l. c" = 14 ρF ThusF : β+l :459MHz ' 2ιτC.r. R, =
:
(b) Α7
Voν
ιhυs.
ψe have:
r: "
1a1
n2Yt
_ 9V' _ lω ,'.-ΙlmΑ
C.R,)
MHz
55
Εx':9.27
2π Χ 9|ρ
+
= ]_37ο:
Ex: 9.26 ινe have
G"=
2n(cβp
/ρ :
are:
lοΙ
R.:5ι fl
The percentage contΓibution of ιime consιants to
!!'ιul:lzco 9l | |- - 2πτ'
+ l)0.99]
we haνe:
:
E x tω 9t
(lω
8οο
0.74 kΩ
To find τr,
:
Rμ
ns
fH:
:
2.96 ns
1/(2τ Χ2.96ns) :53.-Ι ΜΗz
Exercise 9-8
Using
Ex: 9.28
,:z
n'.:
I
J
2τ' C.5'
Ad
:
,
ro
tgn
0.5 mΑ 25 mV
_
R._ ! =
lωv -
ι/2
0.5 mΑ
20
mA/V
2ωkΩ
τ" :
:Σ7
Ex: 930 (a)
AM Aιι
To calcuιate
l1
vo :
.,)
γ
using the meιhod of o1εn-circuit
η
vo : ι(--.19_*)v.π.
ysis
: r. aπd
2βV
1:j-Ξ'
:
lΟ kΩ
= 6l'2ΜΗz
z"'{}
+ c)
20X 6|.2Μ = |'22 GΗz of Rs
Ro: r6fl + g^Rsl : 3ro Ro:3x20kΩ:6οkο
$
: ;
'4Ψ-2eF)(10Kll
10K)
:ι6.4Μ\1z
rl
Ιeι = !Ln6 s '', :8MΗz
pδ |οπ
Therefore, the transfer function of this CC-CB amplifier is:
Α(s)
:
-s^roxl;L** = -2xzo
20 =-toY , 2ο+ω γ o-: Τf-r-η= fh:'ξ : ?^A 3ν
z,"t
1
f pt =
we haνe:
The gain-bandψidth p.oduct is:
(b) With source degeneration
ω
',νe haνe:
+8DR'ι)+R'.l
cL.R'L +20)+10 Kl τΙ1 = 20Ιx2oK+5fι2ο K(l
+5fΧ1οK 1Η : 2.6 nS+ f E = *-
:
(9'84)
+
AM=
rr'ι
i(πk;),".
V,,,
kfr) = -'o
|l 20
time-constants we εan employ τρ = C.,'R.ig + csdlR.ig(l
GBΡ :
'* '"" GΗz
we haνe:
_c,(Rι
τρ
r.' :
βι:βz:βlvehave _ 2r, ^ : ,l?,n = ro ] r.
MΗz
= '1lx R'ι : : _2gΔ x 1zο kο
X lo9 Mιlz:1.ι
Since iπ this case
x7pFΞ ιzω κl1lω κlο 0.8
}a:
R,.= (β' + l)(r.| +..2)
ι(roι|| roι\
:
10
=
Exι 931
l
= 0.?96 MHz
2ο+ l ι kΩΞRs. _'"j'-tο'skΩ
1.46 ns+ f 11
GBP =
The dominant pole is seι by the output load capacitance C
=
T, = Cρl?g, + Cρ' Rr1Ι CιR'ι :2o f(1o.5K + 5 f Χ 235 K + 5 f Χ 15κ)
=A,:''X20ΨX2ωkο :2ω0v/v l'ι_ 2τ'
'δ:Ξ_o
R.ie * Rs ').
where,
Ι /2 8^:τ_
= i5 kΩ
R"i" + Rs R +R. -----_---------_ "' l u 8.R, |] 2Χ 20 ";u, η
ampli6er: 1
Ro:2oKll ΦK
Λ..:
loaded bipolar differential
a
(9.l53) to (9.157) ιve have:
R,.ll
Rr2 = R.ig(l +G.R'L\+ R'. = 235kΩ
R55
I - 2zτ ' (0.4 p)25 Κ :15.9MH2 Ex: 9,29 For
η
lΑ(s)1"
=
a
^Lι
('* 2of ρ' χ '*#Γ,") δ
ρ,
1!λ 11
oΙ
le(,)l'
_
A:M
2
Exercise 9-9
Ex: 9.33 To obtain Rφ: A1
A,M '-
(, ' ιz"1"1'](, , ιzτf uf \ 'G;})
ι' π;Λ'
('*&\('*/i):,
\ ri,lt
r'",)
: R,l! roz \l ,"s R::20kΩ l0Ο ν ,^':Vo "' Ι"' - 0.25 mΑ = 4ΦkΩ r-. = (Β + t,V' : lοl ,25'V ,Ι lmA Req
=2525 Ω
solνing this eqυaιion for/r/ ψe haνe:
Thus.
fυ =4.6ΜΗz
Using the approximate formula, we haνe:
| r-Ε-Ξ t-2 ηlΛ
_-sMHz -2
tp'ι
η (9.l78) G'' -'' : cc from ΕxamoIe 8.4 G-ι:c-ιz=0.3mA/V thus, forf : l0 MHz: c. : _!.!_!Δ/J- : 4.g ,p 2τr
ηn.
Ι, - :'+ι1τ'Le
G-z
_
8-o
: 91,:0'6mAzv ' 27,\4.8pF From
ηn.
'..:4 c,.
_
ο.6
Ceq 0.6
mΑ/v
: =
lma =*u ,^ mA ν
=
25,nv
=
=ξ zτJ
τ
40m --------jj-__:::2π x
4Ιl[| m
48 MHZ
'f '"
:
^ 2
D
14
pF
2oρ * 14pF+2pF(l +Φx3) 258 pF
FiοaΙly.
2oΜΗz
pF
c.. '
+
thus.
mA/v :
2τx2
/.
''''
(9.l77)
r - C.z trz lo.g,
:
^ Ξ Ξ ('-ξ
X 10 uι I0'' (9.l73)
|1 2525 = 2.2kΩ
Cη Ceq = C"2Ι c"s + cμ5(1 * 8.sRιs) Cn': C", = 2ρF R6:R1 :3 kΩ To obtain
Ex: 9.32 From
From
'Req:20κ|] 4ωK
I
2τ. Req' Ceq
2τx2'2Κx258ρ = 280
KHz
Exercise
Ex:10.l (c)Α = l00 v/v andΑ.,": 10v/v
Α l Α.= , llβΑ =g_| AJ Α
β:
l l l0 ]ω
= 0.Φ
I
=&: Rr -l-0.09
.t:
Ex: 10.4 For Example
o.09
Rlβ ";n""'11&:
l0-l 8.1
Ao: σ{ω' β = 1o ιl ,Αβ): (l +(6Χ lor) x 10-1.1 : 7 :' fιιl : fι'(| +Αβ) : l x7 : 7 ktlz 3
Ex:10.5 1 :1o.rl
(d) The amounι of feed-back is:
v^: v.' +A',A'.' = * γ "--+ lrι = v'| l " 'l A1A2p "l ΑlΑ2β |XyI _ (| xlω)ΧyJ + l+(lωXl)Xl l+(ιωx1)Χ1
l +Αβ:1+ 100X0.09= l0whichis20dB ys = 1Υ''Vq: ηV5:10 X l:10V vr: 9.vo = 0.Φx l0: 0.9V
= 0.99 + 0.0Φ9 Tttus, V,, - I V and Y"l
γ.=Vo:-]9:ο.lv ιAlω
an improνement
(e) FoΓ
(f)
IfΑ
decreases by 20%:
,4:ο.8X1ω:80v/v
8o
A.: ,
:
1+80xο-ο9 : ΔΑ, 10 _ 9.756l +
:
Er:10.2 (c)Α
New s/N ratio
-1ω/ι
-
of 20 log(
Ex: 10.6 Replacing
0.01
lω
V
/l
)
:
,ιο dB
the amplifier by its small
signal model
9.1561 2.44% of .47: 1o
1Φv/v andΑr=1Φ
A' _ A +g _ l _ l = ι _ ' llβΑ Ar A lor l04 : 9 Χ lο |
: !-1:110.r .-E R1 β
Φenloop
(d) The amount of feed-back is:
1+Αβ = 1+lo4X9x10_a:
is 20 dB (e) For ys = 0.0ι
Vo
:
v'
Ar. Vs = l0r x 0.01
Vr: F.Vo: 9X
:
10 ιγhich
1o_aX 10 = O.Φ9 V
^n6Φ
:
_
=_Δ ! + Αβ
and ιhe largesι closeJoop gaiπ
Ιfthree of these ampliRers aιe cascaded:
: λ1lΧ AρΧ Aμ
the total
dA'
A1
= lω0 v/v
variability is:
+Φ
λ2
*
4Δ A1
=
Rl
factor:
r: fi-v,:
cιosωJoop
gain ΑJ.
g.Ro
rh
vo'+ρ
:
_ Jr
'
possible occurs when Α : 1Φo V/V 9 Α/ : 0.01 X 1000: loV/V
A1.'1
the gate
6.396.*i.u,
: _g-Vr,Χ {(Rι +Rr) || RD} ., (R, + Rι )RD = _8αvss' RrJ. R' + f" i >> butRz Rl Ro Rz+ Rl + Rρ:R2 Vo
κ1"/o
| |.44=ο.ol . l lL_( A| \lJΑβ/ Α ι|Αβ A,, _
Rl ang
A.=b vt
lo4
0'1o1"
Vo = -g-Vr,Χ
Vr": -Vs+ A : Fe€d-back
l0 V
γ'=Vo=]9:ο.oοlv ' A ι"rιο.s Φ : AtA
gain: Ψiιhout R2 and
groυnded:
and
'->vo:
(_g.v,,.ξ#)
= g^V Ro ",, -v :v"-v.:v.-
RtVo R, +R)
+ RI
vlΙ
Εxercise l(}-2
^.Ι'' " -
v,
{r
+
β: \ιo
R,yo l
Rf+ R,I
Ιo
: ,.*,,,
ffi}
8.Ro
I + 8-RDRr /(Rr + R2)
ι Ξ (8-RD) 'Rt /(Rt + R2) >> g.Ro , _ _ R, - R2 ^' 8-R"R/ιR' -R, &
if
,4
' β >>
1
R.
Rι Ex: 10'7 From Example l0.2 we know ιhat:
_ζ'zRρ o,' _
' '
8"'ιΚ
ιhus:
\νhere 8-2RD isιheopen-
A arA.: , ι+βΑ Ιf Αβ >> l + RFΑl8.2 A ιg-ι . n,=F,-Ι/,],
'
--c Ξgdg
Αg >> l
>>
t*& RM
_s'lRo -- (' * Α'= , g.zRD \
Ex: 10.9 The equiνalent small-signal circuit for |
Fig l0.l I b)
Ro1
vo
RΔ,/
AV,
Fig 10.10 b) is:
Vo
=
AV,
>v,: _!ι
Ι": Ι,+Ι, ,':
, '
OpenJoop gain: V, = V,
_g-2V""' and V..
R,
:-"(#;.
-
AR..
RF
+)
: -"{*-.;.
#.
*1*,l =-'oΙ' RF |ARi,ι A
AtVt
:
ti,
=
'tvo: -Ι.
o,r-, if
(l)
vo
]
:
A
ξ: #
v,-v" -v.,
α)+(2): /s
+ Ι,, = g^2(A|vi)
and
Rr
Ex: 10.E The eqυiνalenι smaΙl'signal model for
:
>> l
-->ι., _-!
l+& RM
ΠlΞ1 \ RM)
Ι"
R7
Α, : } ' vs Ι6: λ'g^2V, and y, : yJ vF Ι6 : A'g^2lV5_ V7} : Aι9'zfVs_ ΙoRι} + Ι6(1 + Rρλ1g.r) = Aι8.z'Vs Aιg'ι A. _ Io which is lhc samc ' V, I +RFAfi.2
ι>
Ioop gain i[ looο-ρain
R7=β:
=
ClosedJooρ gain:
Thus:
tr.:vo: ' V,
.r,-\h
-R, f.) (l*]* \ Α ARta)
Α >> l and ΑRin >> RF
=l+1+jι-trnaA.-_R' A AR;.] _
+}
Exeτcise
Ex:10.10.
Αo-60Φ,β:10r (l | Αβ) = (l+(6X ιo1) > |0-1.1 : 7 :'fιιt:.fιι(1 +Αβ) = l x'ι :7 kllz Ex: 10.11
15':162:0.5mΑ V67: 10.1 -0.5x20= V.= O.7-V"r^:O fel:
o_:]
:
Riπ
From Example 8.1
1ε1 =
1
5mΑ fc2: Vo/l:
20t l0 =
R,l _ Rs =
n"l = (R*,ll
Ξ R.,ι :
i.)
=
191
kΩ
#π = j$
= rε.εο
l9.1 Ω
Er: 10.12
The feed_back network is composω ιhe voltage-driνer resistors Rl and R2
of
a) The loading effect of the feed-back network at the inpυt
+0.7v
is: R| |ι R2
b) The loadiηg effect of the feed-back netιιork at the output is:
5oΩ' r"l : 5Ω
Τhe
Rι
+ R2
Α circuit is:
Α-circuit:
Foτ the CG amplifier:
,η
To
A:\:
vι
2
|l
loι|
[,",
*
,,
*-jg1]
:
rεr
ο
B_circυit:
.__'____n^λ
u'}r' :
.
vλ
+
Rr)
]
obtain β:
vΙ'/v,'
= _L :
6.1γ
s:5= ' vι,
A SubstihΙιinρ: ι,, : |+Αβ , 8^Ro Rr+i2
if Rt + R2 >> RD \re obιaiπ the same result as in Exercise 10.6
R, 7
R,
Rl +R:
From ιheΑ circuit:
I
9+1
(Ri
^'_,*4rιl-lLη
R2
β
ll
/V
R, = RJ + (β + 1)(r.Ι + r.r) + Rει| R4 = 10 + l0l(50 + 50) + (1l| 9) = 21 kο
:
ε,tRr
t20 || (β, + 1χr,1 + 2 || 10)]
= 85.7 V
Rο
:
γ
A _ 85.7 _ 8.96ν,i v l,_V._V, l+Αβ tl85.7X0.l ' R,, : Rll + A9\ : 2| X 9.57 : 201 kΩ
= ι/g-Ξil. : l1l *,ιρ1
R, : R,
8^
ll (Rι
+ R2)
+ R."ι
_
(RD
l|
l
Rt 1 R,)
+
Αβ
Exercise
Ex: 10.13 Referring
Eι:
to example 10.5
| Λ'-J'+R. ' =5x !o-r '-RF
_
1(μ 10.15
zoοΩ
A'8'ι A.: ] l + A |ζα'RF -
:
Rιn
2οoΧ2Xlo ---------------l+2α)λ2Xlο'X2οο R,a + RF +
=
4.a4 mΑ,/V
(A'gn2RF)Rid
=l0o K + 20o+ (2ω X 2 Χ lo'] X 2ω). 1ω K
:8.l MΩ since ro2
>> R,.
R"u,:roz(l :
+
'
i.e 20
K
>>
net\νork:
20ο
ArB^zRr)
:20K(l +2ωΧ2X lo rΧ2ω) l.62
MΩ
lf 8.1
I
Δ!\ A1
Εx:
:
mΑ+Α/ :
4.878 _ 4.94
-
4.94
: ι0.ι4 B ' Ι
1=!:V, '
:
4.878
mA/v
_1.25%
Rε:'Rεl
RF2+RFt+RF
11
A.-! '- β |Φ-lt" _ 8oοlι ΞB' _ !o _ |Φ\ 2ω+RF
with .4/ =
v
Fe€d_back
-j: v
lω
lα)mA/v
enι1
mr'ι,/v
_lωmX6ω: _60v/v
θ:lε:l ' vο (a)if ΑB (b)
RF
>>
t =l.
- 1 s_n. β
Α-circuit:
vo:
-8^vr"(ro ll R.)-and
v.. = /r(Rs ll v
Rr)
+A: J = /s
(R.
'- = l
-(Rs ll Rr)s.(ro il RF) I + (Rs ll R.)s.(ro 11 RF) / RF
'ι.
A +
Αβ
This figure is fo. 10.15(b)
11
Rr)g.(ro
l1 Rr)
Exercise l0-5
:
R,
(c)
R5 ll
Following the procedure used in Example Ι0.7
R, " = l+Αβ =l:1*4β R,, R, R, I l . {Rs || no1g.{r"|| ". R7ι r!.^ \r/e cal (Rs 1l R.) R, R,r R, 1ι : g^('r' |l Rr) R,.
]+Εorρ-. =!Rir= R, Ro substifuιing fo.
t,
Ri =
||
_
R. ll
& μ
nr:
R,l = RJ|| RF || Ξ μ ,, R. :R" ll
----
(1 + μ
)
:R, t
Rin
:
-_:-
μ' = 8.(Rs
l t*4β = Ror= Ro Ro
||
-:
:
n.,
:
'o ll
Ro:
.^ Ιl
:
r,
jζ
.+Rn,, = ro (e)
R",
" l+μ l?"r : R.,,, ll R,
Since:
For
8. =
ιl
5$
||
Rρ
||
\
_(l K
l+Αβ Ro R-': _ l+Αβ
ι| lο 30.3 kΩ
4.0J
=
-,-.. 99l-! = t.οοιο 4.ο3
-τ;##τ , *l
t.66 kΩ
μ : tω
A
: lΦK,
unchanged
_ ιrn. : _ι' -_!ω 'R, {., R, = lο 90
ll
=
ι.ι1 X
and
R, :
o.
1o3
_l.ιΙX
m(20
κ
lo3
ΦK t00
=
9.9|
= 9Φ Ω
Ro = 9ω KΩ, unchanged R.", : (l + 111). 9ω K :
Αll of
/,
Α/A
lω MΩ
Rl - 0=B _ Rι+R) :
_
Ι
is fed_back. >>
1-+
ideal
l. = _! = ι ιt ι 'β R,:Rsll R,, ll R,=οo|| cο|| n':n1 Ro : ro2+ (Rt l| ο) + 8-ro1(\ l| o) : ro2
rr:2okΩ κ).5
Α/Α unchanged
I+ III
.=c if /β
*;foπ
||
β : _0.1 Α/Α, Αβ:111
Εr: t0.l7 Ιf R,
Rs:1kΩ
: :
"
μ
Rr:10kΩ ,4
n,.
R.- = '"
substituting foι
Ro
l0Κ l+5m(2οκ l| lo K} =2slΩ R, : 9Φ = 225.6 Ω =
R..:
Α., =
RF)
R.
,,
R", = R, ll
n. ll R,: l 1l t0 = 9Φ f,) Ro= roιl Rr:20 |l lo:6.67kΩ
R,:
R,
(R. ll RF)s-(r- ll RF) -.:- + " ifινecall Ro Rr. (ra ll RF)
R.r -
_7.52k!t
Refer to Example 10.8:
Folιowing the procedure used in Example 10.7
R-.: l+Αβ
mΑ/V
Rs=-rol:1KΩRl:loKΩ Rr=90kΩg.:5mA/V ro : 2O kdl
:
ro ll Rr
Ro
0.1
_30.3
Ex: 10.16
g.(r, ll Rr)
+
Rρ
(d)
J
Rs ll Rin
Αβ :
R",, _ 20 * ll ,
Since
i,, :
= 1/l0K= X'0.1 : 3.03 _.1.0.1 A.= A _ + κ _ ' l+AB I 3.0:l 9:_l/RF
R,
l| l0 Κ)
Replacing R2 for 0 in Eq 10.69
Exercise
ιR.r o'
o = i'
'''
' Λι
A
From
R",,
-
1+Αf
_
μ8.R,
-
1Μ
β8-Rι
_β'8-'Rl l +},s-.R'
η 10.77: R, _ μ'τ'''.ror'
R'
-
μg.ror'R1
To obtain,/?in:
^R,Rr l+Αβ l + ι,€-R' n,,: '" I R,, Rs _ ρ"= x_lR = R' l + μg-R, I
Thus
I
Since
μ8-
i
ι
π+μ8-
= l/ρ8^ ', l=π," ,ll
rο
:(
Ex:lo.2ο A(l&,)
\l + j,o/ ι1./ 1oo1 β : -3 ιan
Αt ,,.,f'Φ
:
(ωtsο/loa)
: .E-r'r, if
=
-v. : vι
g^z'Roι Roι+ Rz* R]|| 1/ 8-'
(o**.) Aβ:
toa rad|s
= l: β.,
: --:J JωtxοrI
loω/ (l
{
= 0.Φ8
+ ("ξ)'?)3/'?
Εx: 10.21 Pole is shiftω by facιor (1 +,4,,β)
^,
4 m Χ 1ο Κ (l0Κ+9κ+lκ l|
r\1 K ]κl/4 mr) ll
.'. β". =
x
"6
6ο"
|Αβ| < 1 at ω,*,,.
J,1
Er:10.1E Small-signal equivalent circuit:
:
l80" =s ιan-l(ωl80./lo4)
ΑmpΙifier stable
\νhen |Αβ|
,l'
10"'
=
l/4m)
10K=|666
Compared to 17.39 obtained in Example 10,4
Ι1:
l+lO5x0.0l:
l0Ol
+Α,β) = lωX !ω! = ιοο.l*Ηz 'f u(|
l' β : r _ r'ιι lΑ B)= lot{lο0l) _ lo'Hz -Γt
For closed loop gain
=
1
Ex:10.22 From Eq. lο.92 Poles will coincide when
(ι,l"' +
ω")2
Using
Αn
: lΦ.
ωρ1
4(
:
l
+
Α"β)
\oa
'
ωpι
ι,l"rιιo"
:
(1ο4 + 106)2 _
t+
4(l + l0o β) Χ (l.ol)'? X Iω/4
corresponding
8:
l0oβ : Ξ β : ο.245
------Ξ---Ξ=- '
ro+R.+RJ
Rξ'vl
_ν 8- r^ + vι ro Rr + RJ -r=Αθ=--_-a-.R. 5mX2οΚXlK Al1 :-:t.22 ' ,n κ + lο κ + ] κ as compared to 3.03 obtained in
|_u(|
radls
10'o = 0
o
= ο.707
and
π-ffiΞβ-ο'5" _ " * looβ)^1o|ο -
Corresponding gain is
A: Exercise 10.15
o
ο.5
For maximalIy flat resρonse
Ex:l0.l9 V.' =
106
:
A : t+Aop
lω
I + l0ox0.5
= l.96 ν/v
Exercise 1O-?
Ex: l0.2J Closed looρ poles are foυnd using l +Α(J)β : ο
&:μ!-
'" ;;β:0
l+
(l+ Si l0')(l+s/4)3+10]β=o
4-q*_{*1,+lωβ):o 0'' 10^ t0'
20ΙoglΑI
t
=s;
+
3ξ
+ 3s, +
:
(1+ 1ooβ)
o for
s"=+
to'
Roots of this cubic equation are:
( 1
-l
loβt/r),
+5βl/3
t j5J3β1'1
Amplifier becomes unstable Ψhen complex poles are onjιo axis ie. when β = β*
1oβ;.
:
Εx:1o.A λ
:
-- 2+ β.l
;;; ι
+
j l9
j
1+
Φp
l + /2/1ω Aιf = rc4Ηz
thus
f lf
Φ
=
tan
_ lo5Χ0.ol _
:
^t;;τπ
lo6
!( ιo4/ 10)
+ f
:
making phase maΙgin 180
Ex: 10.25 From
ηn
ffi θ:
ιAJυω/)l = quency gain
-
P
:.fn:
l
=
_Φ =
iΓ;l;ΖRi
1ο'
respons€
_2o
80
A=A^-2O dB
Φ'
l0.105
lo 1α
Ιr'
-o}-ro*r."-
1o3
lΦ 1σ 1Φ/ f"+
Ex: 10.2E The φle must be moved /Pt to
/, ,
Ι20'
R + l/SC
106
origina|
_9ο"
|At0ω)|/ (|/β) = l'o For PM :9o". θ : 90' IAt0ι,ι.")| l (ι lβ) = o.lm β
_
l0O l4z
l8o" _ PhaS€ margin
PM: ω'.Θ :
20 dΒlde.
loa Ηz
PM = 30'.θ = l50" IA/jω)1/ (| /p) : |.g3
Ex: 10.26
Α
lω
For For
_fρ
" !D
to' : |+jf/Ιo
B _ o.ol /4sl
:
Ex: 10.27 Must place neιν dominant pole at
ο.ω8 A
=
Rate of cιosure
wheιe
Freqυencγ of 2nd oole
Ao* A'
_ ιox
_t
I
: +
SCR
tσu
1o4
<_
(1ω dB _
20 db)
to3 Ηz
The capacitance at the contτoιling node must be increased by same factor as/is lowered. .'.
c"e* = coιd X
lω0
Exercise
Ex 11.1
vcc _ vcu,,u,
'
l-l
Ex 11.3
Foτ Q,
,
I
_
15
&
u'
0.2
rkn
Ι = l4.8 mΑ
_Vo_(_Vcc) ρ: 14.8
_
ι0",rt'ι' : (8z 'Dl' o.r, * ρ,'Rllω = P': 2v',, x l:2 x 10 X 10οΧ lο r
:2w
0.7 _ (_
15)
n = &,
Efficiencν
14.8
= 0.97 kΩ
q'..":y.,._ηr'"
= QΞ?
: 15 0.2 : 14.8 V
τι.,"
=
: :
2
=
_
y., +ηE-,
-
15 +0.2 14.8
Er
output signai swing iS from l4.8 ν to _ l4.8 Maximυm emiιιer currenl: 2Ι : 2 x |4.8
:
Εx 1l.2 Αt
Uo:
10Y
is
:
= ο.64
Thυs,
Thυs,
l4.8
ο.ε + o.ozs
v
-1ο mΑ and l0 : 4.8 mΑ.
0.67
: :
(b) P,
0.64
{.|)
.
PeAk innιlt
= 24.8 mΑ
0.6 + 0.025 ln (24.8) 0.68
(e)
and,
=+l0Yiεl Thus,
:
|
ο.995
1+
0.Φ17
:
|
+ o.00l
=
62 : ntx4
ν2^
#
^
τ'Rι. o-9t
w
sistor:1qXη.'
v/ν
Toιal power dissipatω in the two transistors : 2la x v,,
l'1 Ω
0.998
= 24.8mΑandre|
oo= ιi l
=
(a) The quiescent powe. dissipated in each tran-
,ξ 4. = ο v, r.ι _ ffi
uo: v,
: P"p-,'
Ex 11.5
4,a
similar|y. al
lιlιl
Usiηg Eq. (l2.22)
R,
-
2'5]
2,2'15
_ I x4.5 51 4 : 22.1 rοA
Po"-,.
I + 0.0052
w
οlrre.,. : -Ι- Ι' β+lR/
V
rr-,omΑandir|
2.53
ι0 =P :V^^x:__ τRl
:59Ψο
υi = RL+ rε\ Αι υ,, : l0 V' i61 :4.8 mΑ and , ' : Ε : s'zο ''
1: v,
:
_eι!xΦ=z.tsw
To calcυlaιe ιhe incremental voΙtage gain we use
Thυs.
(4.s)'?
24
,", ''_&ytοο_ ' PΙ
V vr = 10'68 V uo
!
V
= +0.67
At v.,:+16γ Thus, v6g1
:
I
: v,
+
!Y ι^' P': ' 2R'
π4
\l/
: 9.36 V = 0V' iι:0andi61 : 14.8 mA v"., : 0.6 + 0.025 lnΦ l0
16%
ιnf|a8)
v
x lιn
11.4
mA
29.6
ιhe load current is
the emiιter current of Q1
Thus. l/--,
tοο
P.
:
0.999
v/v 1
Ω
v/v
:2Χ2Xl0 : ΦmW
(b)
/ρ
1X
15
is increased to l0 mΑ
Αtη:0,,tr:i:10mA From eqυation l
l.3l
V' _ 25 _ l.:sιl iρ+iπ l0+l0 R, : l0ο '!: ιi Rι + R,,uι lω + l.25 R'.-
Exercise
1: Αt
o.sεε
:
τr.
at
A:
= 0.l
,*,
irν
:
{
'
Io- "'1
isbecausebiasinediodeshave
I
areaof
the output devices.
1α:0
_ lωi"_
Ξ
This
r'"
,,' _ λι,. _ /or: 0
I
l.i
lOomΑ
φuation 11.27 to calculaιe
use
-2
ζ:/,'._r*:3_2:ΙrιA /\ V*= 2V'lπl , l0 '
v.: ov
16γ
,,' _ ιοv l0ο {)
I I
99.99 mΑ
using equation 11.26
-ι-ι
ι" =
&
R
_ ν'
''''
i^/
ν" ,i
+
25 iP = 99.99+ l--
Rι Rooι lω
Ιn example l l-5, In =
V,
,i
Ω
_
+ o.2A75
l _ο'988X rco _
-t =
cbaΙρe
!::
1ω
RΖ +
R.:
0'2475
l.2%
η{' ηd fq 4 :
]
L:ε'zsa
6
ii/+iP = 2+2
R, :
Rι + Rnul 1ω 10v
4: lOv : ,,: ' lωΩ
lΦ
:
+ 6.25
o.q+
-vcc
lοo mA
Again οalculate ii (for
i": 99'96 mA' ' Ιλ 22 ' N 99.96 R,, : _yj= :
10
:
2 mA) usingequation
Ι|.27
25 99.6 + ο.04
Rι
,r
R,
+
-t =
ι% Chanρe
:
R",,
=
0.25
Ω
:3ΦmW
1
x
l0 X 10 1X
15
15ιR.: 1ωΩ and ( : 10 " Α and β :
ι.4' y." =
on and o, matched
Forz,: lOv. l, : 19 = l0ο
Αs a firsι approximation
91Δ _-2.1 50+ ι
in
o.t
50,
I
: 0.l A'
i"
L
(l)
ηΗ)
Ιι
1 and
jl)]
ιzl
J
2
l = ,,l" |.'l--t'1 \ li ' |---ι9:]l!xlο''| \.]
:
11.6
v'F':
v.l"(fl)+r.'(
r\ 2v,|n
l_0'94X lΦ:6Ψο
From example
+
ηuating φuations
-
Total poνr'er dissipatω = 2
Ei
vιF^
= y,1n ' [i"(i,-_
l0 mΑ, change is l.2% 2 mA, chanΕe'S 6qο (c) The quiesent poψer dissipated in each transisFor 1o Fot Ι a
:
0.04 mA
I
τo
Βνlvs'
:
0,
λι
:
l' = iυ(i, - o.1) ιΓ,. "l τloΞ]P l.
'n-,
i,(i,-0.1)=9x106 Ιf λ is in mΑ, then
,Ι(,,_1{0)=9 ,,1 _ lωi,_ 9:0
+in = 1φ.1 .o ι:i, i,:0.l mA _I0 _ -ο.l Fοrι,,- = lOVandi': lα) : -lΦmΑ Αs
a first approximation assume
λ:0
siηce
i,
:
e
i, = 1Φ mA,
0, current through diodes
:
3
mA
Εxercise 1l_f
(\
3X l0 l Ι Ij' lο '''1
'.'van 2v But
: ,,,, (;ξ)+ v.L (;h)
ys,
(ffi)*
_ ,.,,
:
i'
Here
0.1
v,l,
(fi)ι+,
A
ηuating ηuations
/\ ?y-|n|'rΧl0ι| ' l1^16,, '1 v,l,
,.,,
Γln|
3 and
4
,]'
8l Χ lo
where
*
mA
*#
Fοr
Δys,
V39
:
= +2ω mv:
r"^:
1.4Υ
1
Xe
0.48 +
O.4
V66
/cr
:
:
0.55
For
Refer to Fig. 12.14, (a) To obtain a terminal νoltage of 1.2Y and since y/el = Vnu : 0'6 V βt is νery laηge' iι folιows' that I mA Thus /c.1
:
Ι-" = l'2ν : !Ξ : Rr +R, 2.4
:
: v""'2
0.625 V
: ι
1'25
=
ΔνnFlντ
xe
2.'Ι2 mA 2''72 + o.52:
:
For
ΔV;g =
+
= o.sε
,e
l" :
V
,ιβ
:
e
=
0.52 mΑ
ltll21/οι|25
3.24m^ l0Ο mV
161
0.85 mA
I xe-oo5/oo?5
Δyr, : _2ω l^
o.κ
!): :
o.37mΑ
-
0.59
nι
O.l1mA mA
mv:
:
{o
=
o.ιtl mι
: l Xe
/:
"'""'n2l = o.o18mΑ 0.43 mΑ
Ex 11.9 Using φuation 1l.43 ιw / L'''
'ο_'"'',@tD,
+50 mV:
Ι^
I
rl
/cl +/R = l.5 mΑ
Vug:1.25Υ
,al :
o.s
0.37 :
/ = 0.46 + 0.13
mΑ/mV
:
Ι\'ο231\J'015
ΔVr, = 1φ,nγ'
Vg6: 0.5Υ
Δyr, :
2.4
l.l5
Ex 11.8
(b) For
!J
l.15V Ι":#:0.48mΑ
V3g:l.0V
1:
eο-o5/0'ο25
.39
7
Thus'Δ/.: o.4x2Χ 5 :4mΑ
Thυs,
X
7.39 + 0.54 = 7.93 mΑ
vBR:'|.lΥ
X2mv,/'cX5"c,mA
:
:
For
g, is in mA / mV
=
1
:
Icι : 1:
L=0.8mΑ
ΔΙc = g.
6.54 rna
= o.os v
= 0.575
"
i,(i" ιω):8l i",-1οο,"_8l:0
Ex 11.7
: lJ :
Ιrr: 17"ινιε]vτ: ι
Ι/ "'2
Expressing cuπenιs in mΑ
i,:η
ll
: v". "'2
V33:
'o ,,1 ,lo ,'l lt ,^ \.1 '
Ξ iP = |ω.8
v 1-
1..'r
,ct : 1 x e')l'0ο25 : 54.ω mΑ 1: 54.ω +0.58: 55.18 mΑ For Δyss : _50 mv
',
iΡ(iP _ ο.l)
i. (i,, _ 0.1) =
=
Vοε: o.'1Υ
1
(ξ-}1)+,.'(;a)
|., ,.
v..
|W / LΙ' ,' : n., "'(w/L)P
\w / L)" (w / L)\
Q':
Ιsι^,
:
o.z
:
=(f),
lo'(Υ),,u"'
:,
V,),
)xo.2so(y),(0.2r
Exercise 1l-4
Qι:
Ιιl^.
:
\ι,,,(Yr)"ιv^_1v,1l'
o'z
=
lro.Iωr(Ψ) \Lι1 2'
=lΙ) \ι,,,ι o",'n : l
Lι''(l)"ιv.,
V,)2
x (14.14 + :3.5Ω
o'z'
Ex 11.12
l x o.zsοx (Y\ 2 \ L lΝ
:
ι
see
lv,|l,
l xο.trnxfΨ] x \ IP
=
10
zoo
λ*",(l),ιv",_ 2
(Ψ) :
L
ο.z,
srn
\Lrρ
No\, %o = %s, + %.,
:
(V.,r
+ V,) + (Yo", + 1Y,l)
: (ο.2 + 0'5) + (0.2 + 0.5) :1.4V Ex 11.10
1":η-,:10mA .. Io
: lι.'[Ψ) t \L
tn
v:'
- l, 2"'ο.:sοx zrnx + yον : 0.63 v
iο
Using equation V
: o^^' : :
I
ya
1.46
Voo_ y""|Bi".
y,,
-
0.63
2.5 1.t7
Ex 11.11 New γalues of
-
O.2
V
WL
-
0.5
Vo"ι
are
(Ψ) =?9Φ:lαn \ L)P 2
(U'l =!Φ:arn \L)Ν 2
ι":
ξι"'(l)"va'
1Χ l0 + Y., Gain
:
3: l rο.t r, to'rr 2 :
Error :
4μΙρR1 0.ο35
: .--l-+ : μl8Δρ 8-t) I
=
'n
R.,,,
= Iιn
: = lΨ) \ L lΝ a",
,19.21,
ξ':::':1:' "="r^ro"i*u"' 0'i4
]"'"]',':=
0.14
lοοo X v;"
V
4Xl0ΧιX10
3Xlω
soluιio
14.14)
x l0
on ηext page
3
Exercise l1-5
Εx |l.|2 Νeed to prove Ψhen
yo':
Αssume Qγ off (V65,
:
4 Ιqβlthen Vcsιιz y/i) so
i
2 =
:
vιn
ο and
i..':i,':--'!!!:4Ι^
Rr"
|ι(Viz_
: :k;(Ψ),r,"", "f
v
lV,u|\'
2(V sepa ' |v ,o]) : (v 56rz - |v η,]) V ro"2 : 2V 36"q _ 21v ιP| + |v ι?| = 2V,orn - lV,ul
V6p2 :
the gate νohΔεe' v cP2'.
(Vρ11- V'o'n) +
-
Vo) :
+v 01
)
(v
:
: _
V
Ι
μ(V
or'
seρo + 2v scPa
|v
V
tcPa
56rn
lv ,"1)
μμ
-
_
v o7
V '2') ιρ|
+ Y:-9!9
Ρlug ιhis vaΙue for y,2 into the νalue for ycrr2
Vorr, = V,n (_uΑ+ vesNd+ p(Voz_vi2': vaN2
lv,J)')
Find yi2 for
,2
ιρ|
μ(Voz Vn) _ Vro"ρ* μ(Voz_ Viz)
Vscpρ+
IyGsP2 οR] _ Vscρz
using (l): _ 2v scPo + \v
v^"
4Ι0
(Vο"'_Voo) =
μ(Vo'_ V'r)
and show
u.*o.Αul % where
(l)
VoνQ v
:
y'ι'o
(V esu, _ v ι') v y'πo +
V,,
:
Same proof for p ιransistoι
ff)
u".''
:
-
(v scPa
ιν
Vo.r, Q.E.D.
ιΙ,|)
(
uy's)
Exercise
Ex 11.13 T1
_ Ta:
Ξ
θs,t
:
:
lo.7
Υ
and,
/, : 15
Θ,':175=].s'c/\ry "5ο θr,a
l-6 V5'
Θ4 Pp
200_25:Θrex50
But,
I
-
10.7
:
θr. + θ.. + ΘΙ,
0.86 mA
+
l5v
3.5:1.4+0.6+θr,
T1'T6:
:
1.5"C/w
ΘrcΧPp
Tg= T1_ Θ16ΧPp : 2ω _ 1.4 X50
:
ι3σ
c
Ex 11.14
(a) From symmetry we s€e that all transistor$ wiιι conduct equaΙ currents and have equal ysε's Thus.
l5v
+
- 15V Neglecting ,s3,
Ιg11Ι6
=J1=
0.86
mΑ
Buι at this current
:
vrr, _
x lo t) \]_] x lο 'o,'
o.o25lnro.86
:0.6v Thus,
V36:0.7 V Theπ
V., _
0.7
V and 1,
_
ι5 _ 0'7 5
-
2.86 mA
Ιf ιve neglect Ιaj then
: 2.86 mΑ Αι this cuπenι, yrf is given by
--
Thus
x
|0
]') , \-1.3, to'"/ yεl : 0.63 v and Ι' = 2.31
o.ousln(2's6
(b) For ι,,
:
o.οr
^o
i6a12.87 mA
+10v'
To start the iterations Thυs,
v
and
|o_o''7: +9.3ν
,-
11:-.(:]Σ
letV3
=o.7Υ
v
:
"5
:
Νo more iteraιions are required and
icι : ia: ig3:
+ 1ο.6
vn:
V
161
y," _
:
Vr'
η
:
0.88 mA No fur-
ther iterations are required and /61 3 0.88 mΑ. To find 162 we use an identicaι procedure :
- t5v Ιf
h(?)
v.
jg2
=
0.ω3
vn:
Ιz:
:
4.86 mΑ
o'o25ln(4'86 \ ]-]
v
Χ lq-]) Χ l0 '-,
lo -.643 : +9.351
4.87
πA
Ι62=4.87 mA Finally,
Ιcι
: Ιcι
= 33 71n-ιι
γy'here
Vu'_V'' : γ"-: "2
Thus, lca
:
/6a = 1.95
n'62
mA
"vιεlντ y
Exercise l1-7
The symmetry of the circuit enables us to find the νalues for r, = _l0vasfollows: Ι
ρ1 = 4.81 mA /c2: 0.88 mA
: /ca:
/ca
ι1:
For
+
1.95
V For ι1 = _ 10 V ι, :Vει_Vsεj =-9.351 - 0.62 : 9.98 V (c)For UΙ = +|0v' l:_- l0ν
:
10Ο
,lω_ ','
:
:
Thus,
-
V51
V363
: 10.58 ιo Vr2
V664:
+9.98
-
9.86
:
I.n
:
0.72
9.36 = 0.5
3.3
x 10
V
+9.86
V
l4eosl0025
:0.02 mA
For symmeιry we Ιind the value for the case
[: lΦmA /6j
0.72Υ
Thus, ι,,
|oΥ' ι,,:Vg1_V'Ε3
x t0 ''l
-t.J
:
mA
10.6 0.62:
=
Vur' = o.o25ln( Ιω x lor:) \
Ur: -lοvas. : 4.87 mA /cr : 0.02 mΑ u., = -9.86 V
mA
Ιcz :0.38 mΑ
Ιρ1
]δΤ
:0.5 mA
=
1c+
lΦ mA
Ex 11.15 For Or
:
"rgIVτ
λ,f
i.
β,+l
Ξ:
Ι )
i6
ιc-
ρ
"ειΙντ
nΓD/ντ
=βμ
lsp e "
Thus' Effectiνe scaιe cuπent
.Qz
ffi+lc
+
ΙD
Αssuming that ydε| has not changed much from
:
V61
10.6
V
V
:0.38 mΑ
:
:
"e' 0.58
:
V
o'o25ln(o':ε ^ lo ]1 ι }.1 X lο ',"
,',5
15
10.58
Thus,
/6.1
:
0.88 mΑ
vn:
l0
-
l
Y
βn(β**,)
l;-
lPμ*l]
ι'1.
0.643
:
9.357
Ι94=o
:2οΧ50:iοω lω Χ
1o_]
:
Jρ
><
1ρ'l'ι
ι* = ιa
βρ
βι
"''εε10125
:0.651V
100
V--, = Vrl,
:
/c2:4.87 mΑ (as in (b))
/61
*
se€ FiguΙe t 1.34 when yBF : 150 X 10 ' Χ Rll, then
4.87 rοA
Αssuming ιhat
mΑ
Ex l1-16
I
:0.30 mΑ
Now for Q2 we haνe: V 3g2 : o.Φ3 Υ
Iz:
lllJΝ
,c
τ,Eu: o.o25 ln (2 X 1oll)
V
_
-jc
Jlthi*
(b) Effecιiνe curτenι gain
10.88
Vεl
p*-,
βoβ'
/,: ,5 15 106 = 0.88 mA Ιu: Ι|_ΙΒ = 0-88 _ 0.5 :0.38 /61
β]v /sP
-ieρn εi,. Ξ v,,"
0.6 V. then
:
mΑ'
:
25
x
0.651
(f
)
10 ,Ι"
V
(%Ψ)
/cs
:
/F'.
Ιη'
Rxercise
ι5ο X 1ο rRε, = 0.651
fi.'
:
ο
4.34 lf peak ouιput current
"ν
: |o-|4 eo'434125 3 0.35 μΑ
x
4.34
X
l-8
P,' _
= 1ω mΑ
_ %^-R.,> toomA : 0.434 V ia' = 1, ιεslvτ
I
lω X |o'
(v. /_J2) RΙ
: JiΞvΤx
Τhυs ι7,
Φ
Er 11.20 Volιagegain:2K
3
rryherer
:
Thus,Α":
υo _ _ 2υ, _- L --: Toιal cυrreπι ouι of mode B
=
ft 3
:
Peak-to_Ρeakτ,o
(i:. ft)^ : -1
P, =
2R -'
=,"ι;-πJ
π"'
b-_ ι. lR ΑR' _ - 2R2/ R\ + (R, /
:
Q.E.D
ΑR)
Ψ
+*
η.
6oV
:30v =:.zsn 8Ω
=('-#^)Ψ -6
= (,
:
.
fr), -,
z
Εx|L.22
Ιoυ:
1
Ιoρ =
1ωXlo 3:
For ΑR >> R2
o.l:
μ - _2R' Rl ιi
(lvcsl
1.44
}ι,,c..}ιlv.rl }xz11vor1
'v,l'
_:1,
-v,)?
Ξν(,S = 3'32ν
vω:zvοs:6.64ν o ' Voo = 6'Φ r = ,. 20mA 20x l0
Ex 11,18
Ex 11.19
kΩ
:
(12.58)
Refer to Figure
1l
'JrΔx
θro
-
10
20
_ 3 _ 3 = _6 mV/'C
ΘT but From
=ξ: R4
35
=
x
βo /rt)2 = 56.25 W
δVce
rhus
150
3
\νe wish to vaιue
R,
_
= t's
Ex \1.21
_2R
D 'ρm,i -
R1
:
Ρeak load current
R\_
t.ft
V/V
Ιnput resistance
Thus
l
82ν
or 16.4 V peak-to-Peak
Ex 11.17
/ι
8=
50
-
Using equation
,"o
2.9W
to be less
For Fig. 12.32 we see that for Ρ66"1rr1;on 2οv is than 2.9 Ι , a maximum supply νoιtage of called for. The 2o'v-suppIy curνe inteΓS€cts the the oυιput 3% distoration line at a point foΙ which Ψ. Since poιιer is 4.2
= (, *
ft)r-.
*
_ (, * \)v ",, ιv
6.64: (l+2)Χo.7*(r
R, -_-R2---
.l.:z Ω
*}),
01
u,
_4x0'1
Exercise
Ex:12.1 V
ιcιιι'^"l Ξ
V
oo
-
|V
orr1
_
Ξ+ l-65 o.:1 ο.5_ο3 s+ο.55v _ yJs
V
+ yoy'1 + v |"
Ξ'cl,ι^ιnl2 l.65 + 0.3 + 0.5 > 1.35 V v
o@iΦ<
v DD
-
<+l.65_ο.5 =
1v
|v uλ
_
_
l2-l Fιom ιhe above we can write
|v
ou'l
|v
ν
l.u' = G.2V'2 + --.!μ
Ξ Εxι
\iJhere
V,, = -G.rVi,rRiand Vιnι: vo +νi2 = c.ιRlvnul ιherefore:
ιP1
ο.5
ovλ
v
/,,,|
:
8-^8-|RIy"uι + --9η
/.,,
:
y.,,(s..8.,R,
+
+ 1.15V
8-Ι,8^lr",lΙ r "ι) + r"οl] r"l
l.65 + 0.5 _
Since
1.l5v
30v' /u =
:
Κ (v
0.5
r,, is huge,
\νe can negte.t
_|
r""ll r"t
:
R = ----------i'' g-68-|( r"r|| I
0.5 mΑ,
Vou, = o.2Υ. Voro
ι :
r.,o ||
and have
12.2
|yΑ| =
i)__!-
Vnι.,nlΞ -..vss+vov6
Ξ
:
V
Exι
oiz
r,,Δ)
12.4
: r(0.5)'?Ξr : 2 mA'/Υ1 For Q2 Ι, = 2(o.2\2 + Ι r: 0.8 mA e-: J_ = ε^,: ν',ι' = 4 ,nn,' γ V
Forρ6:o.5 ι
oν
-9Ξ
=n'.:Φ:ImΑ/v 0.5
,.' = \-,'': "' Ι
29 = zs
οR
ιο
:Ξr':r'=29:ηοιο ο.5
Aι: _8.zroz: _4x25 = '1ω v/v Az: _8.οroa: _l Xω: _40ν/ν A : A'A, = (_lω)(_4ο): + 400oV/V R : ιr.ll .-1 :19!:26Lρ 2
Ex:
l2J
The smaΙl-signal equiνaΙenι circuit for ιhe op-amp in Fig. I2.l on page of ιhe Text is rωιawn b€low for a uniιy-gain buffeι From Eq. 12.8, |2'7
Text:
'12'14'12.15 c.| : 8.ι G-z: 8.a
R, = ,u\l
r,,a
R1 = r"oll r,,,
oΛ page of
ιhe
G.ι:ImA/ΥG.ι=2mA/Υ Cι : lP Γoz: foι : 1Φ K,r,,o : r,', = 4oK |'Α/V _ lωMH. Q| ''|' _ !-!L 2τC6 2τC6 Ξcc: l.6 pF A1 : -G-1R1: (l x lO t)(.,,,11 ,.,) = (1 X 10 '11lω κzz1 : _50 V/V A'2 : _G.rR : (2 X 1o ')(r,,o ll η,.,) = (2 X 1ο ')(ao( κ)zz) : _40 V/V A: A'. A2 = ('50)(-ω) : + 2000 lα) Χ lο6,/2 X l03 = 50 KHz f ,,: f,ll: (b) ιo moνe zeΙo to s : ο" l R=l-: G^ι 2x lο =5ωΩ 1
ΕxeΙcise l2_2
t ^- G.z _ o.2 Σ to 2τC2
1
2,Ιτlo-9
: taι_1!!: fρ
Θ
1,n
-l
]ι8 Y
=
lΦ Χ
106
3l8 X
106
ιoο
:
Hz
l7.4'
PΜ:90_Θ:'Ι2.6" Ex:12.5 Find sR for
(c) |.2ν
MΗz
(d)
2τ Χ 1ω X
Χ o.2
106
= |25.67=126Υ/μS
sπ=|+1=SRXCc ιc 126
= 2Φ
μΑ
X 106Χ 1.6X 1o'l2
_ 0.3 +0.5
+ 1.65
:
+ vιn
+ Ι.85V
yn{.in) >
-
V
_
oo
|v ounl
1.65 _0.3 _0.3
Ξ+
>
:
1.65 + ο.3 + 0.3 + 0.5
V oι^^'l =
_
:
0.55
|v ou|
+
1.05
v
:
-0.55 V
' ou : 0.2V, Ι = 1ΦμA 2Ι - 2 Χ lω Χ 10-6 : l.o mΑ/ν G-"' 0.2 "' , :V^ = 20Xιωι06+2οο kΩ Ro : |8-4r.4t.2|| r"lo)] !l [ε,ο..οr,sl 20 Υ
V
V
c^,""Γ'ιι| η]
= 1.OX
2002
A : G.R": =
13.33
x
X 1/3 X ι06= 1.0
103
Ex:12.t Given : all Vey
Vη: V55:
x
lo-3
13.33
13.33
MΩ
x
106
0.7
V
v/v :
0.3
V, lVrl
2'5Υ
(a) Yrc(.",) for NMOS
v ιcuι^^in=
x
v DD
_ voν + vτ
:
v
ξκ1w
l
ι11vor, _ v -1'
l r11v
or,
-
V')"
Vr)t V7)
Ex:12.10
: pnp: /s : npn:/s
lo'l4Α,ρ = 2ω'vA = |25ν lo-raA,ρ : 50,V, : 56 γ
v
Εxι t2.7
=
ι,: ι" :
-yss + vovr + vovt + vn
1.66 + 0.3 + 0.3 + 0.5
: |v Λ|
V
Εx: 12.9
- V)7 = 4(V6s2 i.e.,ycsι V7 = 2(Vo'r_ oτ Vor, : 2ve9_ vτ
V,cuι^in'2 -Vss } Vorι, +voν|+vιn
Ξ
(v ιcM)BoτΗ < + ι.2
=
(V65,
_ lcιιι^^')Ξ v DD vovg
s
1.2
_2.9 v < (v ιcr)o'e|Δ < + 2.9 Υ
\κ1wι Foτ I| : Ι2|
Εx:, |2.6 V
Ξ - 2.5 + 0.3 + 0.3 + 0.7 : _ + 2'9ν .'. -l.2v =(vICM)NΞ
vτ
+
_2-9ν<(vιcΜ)P<+|-2ν
Vou' = o.2 Υ
...Ι:
vιcνι-ιn1"> _ vss+ vov + vov
(b) By Sym.
/l = lω
SR:2τf ,V6y:
s+2.5_0.3+o.7:+2.9Υ
=.C V
"r:
V'Ιπ!
ι"
V-" = 25.u,n-1!]:633 mν lο-'o
8^: β / rε : 2ω/5k:40 mΑ/v rn- βη - 200 Χ 25 + 5 kΩ ' 25ο ,' :V,Ι :25mν lmA ,' =Vn: l25v - l25kO Ι. lmA Ex:12.11
Ι:
/ Ι5) ι""u"'u' -v =V 'ln(Ι ", : and Ι, = Ia,11 Ι2 From ect: Yrrr *V"rr= V j4* VsBa
v.rnL*l+ r.'"[*] -
r.,"[*ji]
/i ] .-l"lLΙsl ,i. Ιszl l:hΓL/sl/sl] .'.
l'' :
I
1,[/sl/sηl) 'L/sr
/J2.1
Εxercise l 2_3
Εxι |2.12 Vs6 : O.7 Vforlc
Ic V
:
l0 μA
for
:
*
"6u,
o.7
:
Εxι
mAforoll
1
QlO
v,l"|Ψ#] :
0'585 V
"'
v V ιcuι.inl : _
: ιΞ /-.+R r".+R+ΔR
ι^.
"
-ΔR ror -
+v
v s5
+0.6+0.6
ΒΕs
+ 0.3
+ v Bfi + ν 3"aι + v
-
^re^
o.25Ιo,',,
R
kΩ
0.02
ε-"':
* = 3 normal
: l80μF.iJ
Ιrι
: l80μx3:540μA
Ex:12.15 Assume 1c7
-
1c5
-
/c6
:
kΩ
and r" = 2.63
for 74l
t.oz +
}Φ
=
5._ξ
x Ιo
1
9.5
μΑ +2.63)
= 3.63/1
2ιε _ 3'63ιε R1' β+1 50 '2ιL 2o1
vs6
= ο.ο8rε
"' - ο.ο81. - ο.oοοal. Eji ]δlv
Vgu+
Ιgr",
= .].6]1. + ο.oεr. = 3.84kο X
(β +
β
l)Ιη
=
Rω llRolo
R.r: t,,r:
? :
(. ''ιsx ιo
6/2ω X 25 Χ lo-1
R,'
:
x
10
:
6
|9x|ol
Σ5X1Γ1ι
R.ro = 31.l
19
MΩ 3l.1
2.63 ||
:
2.63
ll s
,
ΜΩ
Iο'))
2.42l\1Ω
Ex:12.18
ιιllr,= ?+ΙΒ5+Ι86
{c) ,s? =
12.17
From Fig. 12.23 on page ofthe Text:
ιlf
V"6: Ι'(R2+t.6\: i.(l
_
Εxι R.
lΙ:[J-'Δ
Ι',n
'''
lο2+a.sj
I
Q|a' Q2o hΔνe
:
o'o2
οp_amp, we have
Assume Q1s, Q19 (now diode connected transistors) have normal aΙea
R-'
:
sυbstifuιing R = l
+0.6
l2.9 Υ
(d)Vρ:
0.02
BΕ|
Εxι L2,t4
(e}
ΔR
r.ξlRlΔR
o.ο2R €-"'=- R +0.02R+ r"l
14.?
(neglecting R, & R2 drops)
(a)
= l?,
r.5+R+ΔR
vιcMlnui': Voo V"ε^* viol+ vBΕ = + 15-0.6_ο.3+ο.6
= =
+
r"ιlR ^ r..5+R+ΔR
12.13
15
ι€ι Rl
ΔR
ΑSsumo ρ >>l and re5 = rΦ, then Vg5 = Vg6 = i(r.5 + R') = i(rω + R:)
Rι : Vιει, Vsειι : 0.115 v = 0.7 0.585 O'||5 ν-1 =>l-'-Υ3!-n'= "' RΙ ' l0μA tt.s ι(!
=+
i
Rz =
VoItage across
Εxι
12.16
see Fig. 12.22 on page oftheText,
x
ΞΞΙ
9.5
μA
2Κ"
Ex. 12.l6 and l2.17 in ιhe Texι. we
€, =
5.5
:
/ l0-'. f,,
2.4]
p.qη
haνe:
MΩ
Hence:
Χ tο 2x2.43 t
c... -
5'5
' - = t.|.] Σ lο o mA/ν
lO"
FΙom Eq. 12.95 we haνe:
CMRR
: 29''(R.'l| 2(9.5
1E
kΩ ιΕ -.].84
(i... _ Ξι
From Eq. 12.93 we haνe:
CMRR
=
\2s
Χ 1ο
x
R"lo) /
€.
6Y2.63
l
_r.ι.ι μ] lο '/\2.63 μ+3I.l μ,/ 5.5x 10 r μ
Exercis€ l2-4
1.68
x
(A).t"
105 or 104.5 dB
Ιf the common-mωe fe€dback is not present, as exp|ained in the text, common-mode t aηsconductance and common-mode gain are both reduced by a factor of ρr. Hence,
CMRR
:
1'68
CMRR
:
7O.5
:
Χ 1o5 5ο
:
:
-G^,
R.
8.ι :
_6.5 Χ
_526.5
v/v
Εxι lL23
3360 oi
dB
Exι 12.|9 βιa=
βιl=2Φ
, '. = Ξ!-: 16'2 μA
,''. :
l.54 kΩ
mΥ :
25
45'5
0-55 mA
Rε:1ΦΩ,Rs
-
Ω
25 mV
ρΑ Κ
16
= 1.56
:2N 4ο X 0.165 : 3o.3 kΩ
= 50kΩ
substituting iηto Eq. (9.77)
Riz = 201[1.54 + 50|| (20ι X 0.0455)]
=4MΩ Ex,12.2O
Vιι'ι β l' Rs ι r"n 1 β
Vι,ιl 45.5 + lω
Vιr, ι45.5
(Re
ll RirT) vδ!7 _ v'2ιR9ΙΠi-;Ξ;; needs Ril7
:
:
l 6.6 X0.917] .' _ '' -"Γ"LτEττ _' κ ] : Vr[0.05 + 6.05] '8J
"
v-
+R.=
(ρ + l)(r€!?
+ Rs)
2o1(45.5 +
l00)
:
29.2
kΩ
= 163Ω
"
Εxι |2.A
..Vbn-V12xO.92
[R,*,,
*'.r,.
Εx; \2.21
:
i,2
β'o+1
R,r3D
\rhere Roι3B
Ror,
:
ll R,r7
:
f,l3,
η.9 1ρ
roρ(l +g,,r(R3|| r,17))
:
0.55 mΑ
ol7
: -l9!- : 227 '1
kΩ
0.55 mA 0.025 ιΥιv
:
22
.'1
=
9.09 kΩ
θ2ω
ο
mA/Υ
Rε: 1ΦΩ Thus Rn,,
+
722
kΩ
Hence Ro2 = 9α.9ll '122k F,xz
- U kA
1L22
I' I yι : 18.8 = 6.11
+ 6.6
x 0.917 =
open-circuit νoltage gain
o.o5 +6.05
0.025
- * 0.0ο5
R^
f ,o, n
\ +
2ΣΙ * r, *'1 l80μ 5l / 201
Αssuming β23 = 50 and β,o 1.'3 n :
Ro
-
Ro-
180
μA
5 + ( 163 + 14.4
dl
#e] :
2ω
and
from Table
l?9 + 1588)
-
5 +94
Exercise I 2-5
Exι
Εx'ι 12.28
12.25
SR
:
0.63
using eq. (12.129)
ν/μS
using eq.
0.63
"sRlavo-nt :
Εxι
2π(l0) X
1
X 10'6
|'2.26
:
,:
10
kHz
νr'here yr : 25 mv H,,(f) R::Rn:ffi=zο,ωoο
n''
:
π,: A,,-243147x10s
c_,:_]-xtο1 "" 5-26 .'. Αn : G', R .'.
R=
+
A''/ G^ι
Εxι
12.27
SR
: L
: 2L n,, c_,
o = 9=t=sn
2l
R,
ΜHz
ιιhere
a
with
1279
inserted in emiιιers of
lΙ _ 4rJ 2R' 2* mv-r ι\ε ____^
^
:9.5xl0
now
6Α
0.050 _
g.5
SR
5.26
X lο_6
kΩ
: 2J!!y12γr'γ
= 4ιvτ
RΕ
2r,l
z' ^..0.025
for I
oj' oa
+ Ι RΕt2]"'
1pr1
QΕD
n.*c.Φ ' c,.= 9'ι 2c, .'. c. must
be reduce
Χ
factor of 2
c,".-:Ψ:f:rson
GainΑαG.t .'. Α alsohalνed Αn"*: Αοlυ 6dB: l01.7dB f
= f ,lA =J
has been halνed
'' f ,'.. : 2x f o,,tu: 8.2H2
ΙRρ
Ll"l'Ιe) Ι
\Ιζ1)
ffilnιzl:
l,7]3Ω
Exercise 12-6
Εxι
12.29
: R!:20kο and 1 : l0 μA from Exercise 12.28 For /r _ l0 μΑ = /. ιhen (f), (#), useR1
/'\ IJ
:
For 1,
For/,o:5 since ys
μΑ
20
:
"ι:
21, then
(r')"
:
!,then(Υ),":
Find
$ ιBl
for (V,r)
Assume iβP
-
iE2 aΛd ιc1
i"
.iP
-
iB2
: ^ι tl-.ιt-^
Ιει lrρ\/l)/l ) /\β, iιβ", ιβP
'(Υ)' i(yr)'
has to equaΙ the originaΙ
(ycc_ /*R4)
: Vrr_o.2
soR3,R9,andR1ρ
Qρ
can be found by
0'2 R,: ' loμ : j'2 : R": zυ
n,,,:
|''
{Φ
20kΩ |0 kΩ
Ψ )μ = ωιο (b)
Ex:12.30 (a)
Find
β,
:,,
βr
(;tχri;)
β;
(40)'
='9
iN
: ξ
ir. = |01
!ι83 for 1V,*)
Αssume
;--
ial
^2
(Αssume
:
6.25
i",
: 4
i,,
: --!Φ- : (10)'.40
β;β"
β, - 40)
μΑ
(Αssume βP 2.5
-
μΑ
Io )
Εxercise l3_1
Ex:13.1
Noise margins sιay unchanged, because
Ιn ιhe Ιow-oυpυt sιate' the transistor is on and
v
7,. Since V.
ν"c.'|{vo, _
v,)
μA
Αlso:
Pο"*."," =
yol : v oo;!ξL fo + fos
and
hence:R,
To obtain
w
7
:
48
3 rr, : 219
2x fi! Χ
|25
.Ψ-" "L
Κ
w L
v,)
2.5 X 50
:
l25μιv
Ex: 13.2 when input is lo\ι, ιhe ouιpuι is high and equaι ιo
yorr. Ιn ιhis case. ιhe switch is connected to
Rcι
Vo, _- Vcc_RcJΕΕ:5
*. ur", (-R^ "" L .
o.o89v, RD
:
R.ι
-2X l
Eι:13.3
V'"
RD
_1 ν
.
Roι
:
= l0 kΩ
and
K,
Ηence
o'zzν
lo
:
"43
><
0.22 _ 0.22
V, ,,voo' v,
_
= 1.3lΥ
1.8
I + 1.8-0.5
**η" voo
0.22
and sub-
:
\ro: \vooιo':
lrv
r"
- vo,
RD
P,,'..,,"
: ]' 2
Νoιe ιhaι keeping
l.ε w
7
"
ξ;,0:Ξ Iο kΩ
= l3s μ\ν
unchanged resulιed in Ioινer
noise margins and higher power dissipaιion.
l0*ο'
L
0'26
Voo- V,o : 1,8 - 1.31 : 0.49 V NM1: V Voι : o.12' 0.26:0.46v NM"
=
= 3v
_ /_ loy |oι -L=[ 0.089
y4 _
V,Ι l.ωrryΙ'Dv, v'
= ο.5 + 1.63
φual
3Φ μΑ/V2:
}()0, t0 ^x Ψ x
and wilh
frιdVoo'V'n'V'r'Vor'' Voιl: Voo = l.8 v unchanged vlL: vι+ v' = o.5 + o'22 = 0'72 ν
Pa
to yo. . The switch is connectω ιo
!
=
use:
The power dissiption becomes:
llence,Vo!:V66:5V.
t,' :
unchanged resulιed in
'r-
is zero.
\iy'hen the input is high, ιhe ouφut is Iow aπd
yΙ :
μW
To caιcuΙate ιhe ne'ιγ noise margins, ψe have to
when the s\ιiιch is open' no cuπent is draιrn from ιhe suppΙy: PDD : 0
sιitute
y.
=V_:0.o89 v,t = R,),
μΑ.
X 10_6
Rc2. |herefοre lhe cυrrenι ιhrough
'R"
V'z
ιh€ cυrrent dra\ιn from the sυpply is 50
To deιermine Ψ
302.4
unchanged:
ι X l0 6Χ(2.5 _ο.5)
: V2]ρρ:
μΑ :
|68 μA
: l5l μw
Y, , we
To determine
when the switch is cιosed oΙ in lo\γ-ouιpuι sιate'
P16
;Ppp
168
_
Ex:13.4
kΩ
{Vo,
^Λd
higher power consumpιion. bul noise maηgins stayed ιhe same.
. ι,νe use:
p"c.,Υ,
I
Note that keeping
Subsιituιing for 2.5_Ι9!=_
l.8 _ o.Ι2 lo kΩ
lDD'
= 56Lρ
Rr}rrr:0.l :
ι'
has not changed. noise margins
ιDD V"" V"' R" Poo = Vooloo: 1.8 x
Voo :50ιΑ=R^*r Ι^^: RD+ rD\ 50
v DD' v
stay the same. Ιηo.der to calculate ιhe power dissipation, ψe need to first recalcuΙate
Therefore' ιhe cuΓrent drawn from ιhe sυpply in this state can be calculated as:
= 2JΙ
oL'v oι]'ν tL'ν τH only depend on
_ :.ls
ν
Exercise l3-2
EΙl
. '
Ex: 13.7
13.5
t(w/L\,: ι/l : L\' \Ιll3 -' Fιom E4. 13.20: Voιl : Voo_ y, :
From Eq. 13.35 ιηe have'.
4(w /
l.3
P7r,: loox
v
=
unchanged From Εq. 13.28:
ν"o'_- (vDD v')ι _ (l.s_ο.5l 2k',(voo- 2v) z :1ιι _ 2 Y ι).5) ' : 0.12 V From Eq. 13.22: vιL : vi = o'5 ν unchangω.
Voo I lk, _'|)v' _ ,, '" _ --τ:+l :0.7V
].8 - (3___]lρl
-.r-l
From Eq. ι3.26 ιogether lι,ιh setting
zι'x|ιuι
=
",)'o
oo
:
_ qΨ '''','] aυI
_ v, _
υιιl *
1lι
(V
aa
ι,
-
ι6)2
Φ :
_
_1
dι'
'υoL"!
:
+v DD
1γun'
_ vι _
Ex: 13.6 The inverter area is approximately
A = W,L, l w'ι'
since
= a.,r"1,uu" |
Κ.
'
l, = l.lV, . Αssuming Lt: dandw2: L.
= kd2 + kd1
k. >
W"krWz
:
2k"d2
Ll
= J(-
anο
= k'L,.Δnd
Thus:
A : k,LrL, *
L
:
l,
v oL
: l.8_o.5_ο.12 : 0.61V
NMs: V6"-V,":1.3 -.61 :.69V NML: vιL_ Voι = .5 -.12 : .38 v
L7
_
o.5
32'4 μw
P-!!! '
: crv'oo, C
dvn2
x52 _
0.13
v.lι' _
66.7j
L22
"vt
oo"
:66.8
v.(ΦΙ _
ιy.ι..) _
V
"(o''\"-''' l#)' v^^U"o-(Ve6-O\e '"'" ,Voι -t:
.'.
1
+o'5 + 0'121
".32Ιo.l2_vιΗ 9[0.74 _ v Π] : 1.18+ Vrrr
Ι!
:
Pu..'C,v2 oo +
for
aι,I J '"Ψf
odΦarη
Now ifwe substitute for
k,2fιoι
lo_6
1.82
l]
_(_'Pι'l)
\ 2) = Vrr'C
2k,ι|υ6+ @1_ L _2(V
loox lo-'t x
106x
13.8
lnr _
-t,
drr
x
fCV oo2
Ex: 13.9
Fιom Eq. 13.23:
Φ:
32'4
Ei:
P* :
we have
1r"12, + k.W2z
ι"'-
:
tpp:
Voo'C ' Ι
o'un
l0 psec. with C = 10ffandyDD:1.8V
18 10/.0.69: ιο Ρ
t.2mA
Exercise 13-3
Ex:lf.10
For rPrγ the output starts at yoa and goes ιo yol, ιhrough the Pu which is 20 kΩ
:
voQ): vor-) (o@) - voQ\)e tt. * vor) : Voι_ (Voll_voL)"'PL"τ
\tvon
(- \v""
ιhrough Po which is t0 kΩ
* \tv oo ()v
*
\v.,)
(Vog) + Voι
.(
R.C = 0.69 (20 K) (100 = 138 pSec For ,Ρalι, the ouιput staΙts at yol, and goes to yoι ,p/-fl = 0.69
"(i))
:
,-
'"
,pιιl/τ
*
v
or1
: _
v 61- (v 61- v on)" '"""t
",) vor+v* = lv "'
"(-,"(i))
:
:
',Pιry'τ
}v
"-:
,P.,.ι = 0.69.R.c
t"
.
)φ""r+ '"r"1
0.69X
:
"
10
){ιιz
KX l0/ : p
*
69
p)
:
69psec 16ι
rr".
Εxercise
Ex: 13.11
: v oQ) : v o(t)
,is
vo@' _ tyo(-) _ v o(o')]e 'ι' o _ [V aa _ o'!e ιl(7Κ' |α'ι' = r
ιγhen voltage
'|Vnο: ln(.1) =
V
is
ooe
-
-(0
.1
yDD
ι.'la
rr
2-3..t=tΙ .2 Κ. 1Φ t J:2.3
.,c
:
6.46 n.""
1
3-_4
Εxercise I3_5
ι' : r"'(l).|ιv' - v,")v o _ :v "']
Ex: Ι3.12
η. l3.58: v., _ r(vpp-lv,,,lt+v-
a) From
'"
r+l
+0.2
r(12 04)+o'4+0.6+0.6r
0.6 = =
or
1+r
0.8r+0.4=r:
_ι-Υι_ι
,= n!μ"." E{"_ 44 Γ'.ι
|νa
.1(w|:
μm: ο.52 μm) Vpo = |.2Υ'Voι: oΥ lt _ vιH λ(svDD_2y,) i(5Χ l.2_2
b\ V6g :
v', =
From above:
4 X 0.13
= 0.65v
f,1ιvoo+
2v,)
:
:(3
. 0.4)
0.55
v
.,*,
_
o,: " =
since oN and oP are matched, the outpυt resis-
.:
/]xl=οs
hence:
V-- :
= rry) _ \ L lμ \ L lh
0.5ι
:
0.4
IPHL g.5.1
γ
Ex:13.13 Using Eq. 13.58 and 13.59:
u.. -
lv,ol+ : 'tvoor+1
v.\,-zs
/{-5-ll+l
l.99
)4.7 ps
ol v
"'
k'(γ\ '' \L l" gl-,,, - 24.6'
* !pμι)
\{,","
η.
:
4
x L! .ι
:
l{zι.l
* ιo'ι)
:
ι'lν"
l3.68 \γe haνe: and if ιve substitute for RN
o.69RNc
from Eq. l].7ο. i.e.R,ν
in ιriode region and hence ιhe circuiι is giνen as|
rιv]
(τΙ"
Χ 10
t2
^
O.2v,oNoperaιes
= -J2Ξ 1ρ
11'"n'
\L)' ι-''' = 0'69. lΞ x Io' ι6 _ ΦΞ! (w\ (w\ \τ1' 50
r+1
Ιι)n I Fπ l/2 ni μ"r" y, : \νhen vDDΔndvo :
L'ι,
=
Ex:13,15
0.4) +
l+ο5
ι.
αD
\Τ )"
FΙom
l.2
α',C
H = L
we have
Γ'
4
:
|ριιι *o ,tw\ ,o
l.0.
an
lζ
= 49.4 ps
2.9 kΩ
(Ψ)
|v ,n|' then
\lo\lo ]α)x1ο''Χl5X1R
ιPa _
tance in ιhe high-ouιput staιe is ιhe same:
a,rn.
lο
lqq
l.g9
'
= 29 kΩ
:
-
:r_ = [Ψ] \ L I,
:
'"
ι,,,'(Ψ)
I
rρ"x
.ρ-
z]-.l'o.s*ΙΦ)': 4 18 ιl8.'
v,"l
43oXl06Χ1(l.2-ο.4)
:
= s
"oz']*(f),
z1 -tv'. +(!:!!\' 4 Vno \VDDI
' lpιlι ιpι
r,rr"
L
qYλη;
Noting that v
ofthe inνerter in the lovr'-
ν"c".(l)'ιv""_
.
l
lο ^xfΨ) \ t"
αnC
output staιe is:
':I
sο,<
Using Eqs. l3.63 to 13.67 ψe have:
= 0.55 V
c) The ouιpuι resisιance
l
I
Ex:1f,.14
x|.2+2xo.4)
ΝMg: V6g_Vη1 : 1.2 _ 0.65 : NM1 : V1 Voι : 0.55 V
_
- llοz
[ιs
I
l l0 ' :
.-ΓΨ) = \ I, L
'
8625 X 20
q1
X lo-Ι5
rq) \Li"
ls
similarly, υsing Eqs. 13.69 and 13.7l, we obιain:
Εxercise 13_6
Ex:13.19
,Ριr:0.69RPc: 0.69X30x1ο3-Ξ-
a)
(Υ),
or
ιPLιr
:
2o.'l Χ
lgr-!-
(Υ),
=e
59;
ΑS mentioned oο page of theText,
Γ
2
2Cr7z
Χ 0.|125
Ι Cιι
+ 2
b) From
c
:
parι of
η.
ο.69(R"qC"l + rρ
}n"oc"',)
is: o.69lR"qc",, and in order to reduce
increased by a factor ot z. πote ιιrat
: :
Cord
C".* C""*
(-
+l: L
exrinsic
+.lp
orequivalenιly
t" = ilpιlι t tρι'g) : 'ι'ιοld
/,.r, n *
, :
476 Dsec
2BD
*
10625P 6.25f
-
ΞΞ
:
0.69(R"qoci"'Φ + 1n"orc",,). π"n""'
Cin, + C",,
ρold
ιvP is reducω from l.|25 μmtoo.3.l5 μrn
ForS _
...9ΞΞ x
= 20.5 ps
Cgι: = Cgal =
33Eo redυction
d)Α -
0.3375 fF
C6: Cgι:0'7875ΙF
C = (4x0.3375f)+ lf
+
lf
+
(2x.7875f) +.2f
:
24.6Χ rc']tr
ιρη1
:
3l.5
tl ι" _ ^(ι"''
x ιo " +
16.6 psec
x(#):
2t.3 psec
:ιl6.6p
+
2l.3p)
:
Ex: 13.18
l =---------------ι : ,|ι.gcRz " ]-^'_ -llo 2ι28 X l0 '')
The minimum period at ψhich the inverteΙ can
reliability op€rate is T'n" : tPEι'+ rPrr. Thus,
-r1l =
17.96
Ton Hz
ιPHL+
}Y
=
:'6ff\28ps 'o!?ιξ?,'
Σ ιandsince (Ψ) isaouυlωunοι
\L)
is constant. then
Using
Χ(Υfl):
ι"''\ _
2:,ρ*
Α or
aΙea is aΙso doubled.
Ex: 13.20
4.225 fF
ιPHL
(f ) -o(f), ι'"*,.υ"
1!!Ι _ ciiι+cιlls ι
ιω :
.
increased by factor of 2. c) From Eq. 13.79:
Ex:13.17
|.Ι25
s: $ R"o
Therefore, R* has to be reduced by a facιοr of 2
.10625 pF
lΓ
:
part by a factor of 2, S has to be
οιd
+ .lp = 6.25f
2'9 fF
τι'e e*t.in.;"
and
*
l :
13.79, we have:
the
_original
+
= 3.35fF
s.r
L οld
|pιH
l
0.7875 + 2.3625 + 0.2
:' ιPH] : ιPHι _ orgina| * :-!l:
Ipllι =
is the
* Caιz
Χ 0-3375 +
c. :
C"u= Cgι+ cga+
,P
tPιlL Δαd t PLH aτe ploportioΙΙal to
:
.'. Cinι
\ι),
Ex:13.16
Ι
C'n, = 2egιι
19-t2
Ii+|/ιy) _ =2o.7XlotY20Xlo \Llρ (w\
q
conιribution ofintrinsic capacitances of 0N and ρP. Therefore,
ιPul
I
2|o
2Χ2s
η.
ι3.35
Po, : fCV
oo
2
:
t x loe
x6.2sxlo
= 39.1 μψ
t5
The maximum possible operating frequency is:
ι9 ρsec
I ..nence' " 'J*: η' PDP : Paι"x to : f *CV2ooΧ = Ι x cx Vzooxιo:9!3.2 6.25xlo-tsx2-s2
:
19.5
to
fF /J
x2.52
Exercise 13 7
Ex:13.21 a) For NMOS devices:
W ι
ο.ι8.-.0.l8
ΡMos
For
The Dower densiιν. i.e.. o.21
w:
4D : 0.18 ._ , ..
b) For
NMOS devices:
-Ι
Y: ι For
olR
I
2.t6
Using Eq. 13.94, we have:
0.l8
ο.54
-
Since dynamic power dissipation is scaled by
-
l
l
s'
-s
bγ : . hence. PDP I
and proDasaιion delaγ is sca|ed
.s',
8
Ex:13-24 lf yDD and vt aιe kept constant,
-
unchanρed_ \phile
o, jL.ιhus v""
c
{
α
scaled by s. ιherefote ιn is scaled by
will
,('
:
43ox 10
6
X
1.5
ΧΜ4(1'2- ο.+
}x
o.:+)
would be:
io '2- ]ν4.16. l0''\ l.5(l.2
(ι
is
+s ,i
ο.6
_ I
,."''ar'r, ',r'*r'"o
CY- oo and ιhυs is scaled
thus Pdyi increases.
l3.lΦ
and thus iρ is givon by Eq.
o.4)'?
l by 'ξ 'l/s'1
:
|y,ρ|
:
0.8
v
and |yDs|
:
1.2
v
are
Ι s
+ 0.1
Ξ
X l.2)
yDs < l.2
:
55.4
v
ο.o)
μΑ
wiιhout νelocity safu ratioπ
ι" ' !l ιιo\ = 59.1 μΑ v
' 2tn
are satisfied and the NMos transisιor be operaιing in ιhe velociιy-saιuraιion region
6λ l.5 xo'6(|.2 o'4 1x i^= "\2)ll0Xl0
I s
p'''-
η. l3.l0l
boιh larger than |v o-λ = 0.6 v ιhe deνice will be op€raιing in velociιy saιuraιion and
any-
remains
is scaled bν -.s1 . and
"""orrr*n"n,"* "r","
l.2 v results
conditions in
|yorl
more. They are kept constant!
" k'v""
:
:
in the absence of νelociιy saturation. For the PMos transistor. ψe see that sinc€
the entries in
obνiουs|y, vDD and ν, do noι scale by
,ince
2
|o
Vos = 0.34v ιo l.2v cοmpared to Vos : Vov : (1.2 0.4) : 0.8 V to 1.2V
Table 13.2 thaι change are as foΙΙows:
oC
lo')
X (l + 0.1 Χ l.2) = 21|.2 μA saιuration is obtained oνer the Γange
: I so pop
decreases by a factoι of 8.
,,
-
(1 + 0.l X 1.2) : 154'7 1ιA if velociιy-saιuration_were absent, ιhe current
Ex:13.2f
ιx!:
Χ l0
1.2 - o.4 : ο.8 v which is greater than yλd : 0.34 V. Αlso' Vo': l.2 v iS greater than yDsaι' thus boιh
in vcs _ v ιo
io
l.
's2 S
400
NMos transistoι ycs
For ιhe
cuπenι aνailable to charage a ιoad capacitance is thaι provided by a single ΡMos device. The maximum current aνailable to charge a Ιoad capacitance is that provided by four PMoS transistors. Thus. ιhe ratio is 4. (b) There is only one possible configuration (or paιh ) for capaciιor discharge. Thus ιhe minimum and maximum cuπents are the same
isscaΙedbν
X lo : x
o'25
Ex: 13-26
0.l8
13.22
ratio is
ι y.", ' μ,
= 0.63ν
0.l8.-_ 1Ξ 0.l8 -x
(a) The minimum
Ξ
osu,
V
l'ο8
deνices:
ιry ι' '
/.S'
Εx:13.25
οlR
ιn:9JΞxax1.5: 0.l8
ΡMos
Εxι
.
1, ,"u1"4
,S-l
0.l8
deνices:
_fu_
deiνe area
"ν=
v Ds= Ι.2
I.5
v
Χ0.6(l.2
ο.4)](t
+o.l..
or 0.8v < v Ds= 1.2ν
Note thaι the νelocity saturation reduces the NMοs currenι by 33% and ιhe PMos cuπent by
-
1qo.
|.2.)
Exercise l3-8
Exι
13.27
Eq. 13.102, ινe have
a) Using
io: Ι",Vo'/nV,
iD = log/s
v^-
+ -i+log(e) 11V τ TheΙefore, the slope of the straight line representing subιhΙeshold conducιion is giνen by: Thus' Ιog
!!ι los(e)
= z'znν-
: 25 mv atV 65 : .2|Υ
b)V7
|ωn :
Ιs
for iD
:
1Φ nΑ
eι|ll '2x(25r)
'=.lnΑ c) For Ycs : 0, iD : .tnA Ι,o'o1 = 5Φ x lo6 X .1 X 1O-9 = 50 mA Pai,,: ΙιoιoιΧ Vrr: 50mX 1.2 : ΦmW ir:.lne
Εxercise l4_l
14.1
(w/L)": ,Pι, = ,P., :
('v/L)P:
Ι.5
14.3
v,
0.32
0.5 ns
v,.
1
o" + zφ' uσ-v
1
^[_z6,1
= νDD vι'
v,_ v'''-1aσoo
0.ο3 ns
v,
+
zq _ JΣo,ι
substituting νaΙues, we get
: 0.5 + o.3vt/': ιJιεv_v,+ο:sv_^Fasvl Y,
14.2
Using φ. l4.l l) (
Γ ιvDD _ v,)| l_
voL -
tl _]|
[
Jl
(2.5 0.5,x
=
Vor=
o'27
Υ
NM"
l l=ο.zv
,ν 4 |l'ΓΞ_
ο'5-,:,[
:
ο.5 +
|Voo _
ι v,)ιl _
./ηιsll
): ιrl"
2
,o
"(η: ..(Ψ\ : ι'ιι
(:)"
ρractical νalue for
y, :
ο.648
: Vρo
1.15
86.4
0.s1'z
μ2.5
:
=
ro
36.4
""(")
=
jιsο
ι""ιο
0.22 mW
:
", 'lΓl_,(t)- (Η)'] = '* I' lr., _ l'68 \7 /x l0 - o.lι lO '
|.44
/
=
I
t5
x t0
i)
\{tror+
.'.
t"ri:
l.8
v
ο.648
v
:
oo
"(Υ)'(
_
V
u,)'
r,ιrv'{ffi)ιs
v
:
V.)'
tι"(l)
oo "ιv
\tzo"ι,ν\(|)ιsν
ι5
25
./
,
}(.11n
0.03 nsec
:.. + o.o3n)
l Vl' : 3φ ro
!v)' :
= 1120 μA
)(Η-(tΙ]
β
l..37s)
\
;k
Fig' l4.l2(a).
j20 μA
=
ο.07 nsec
+ 320
μA
,D/ν{ lPLγl' we no|e ιhaι lhis siιuaιion is idenιical ιo that in ExampΙe l4.2 and we can υse
To obιain
t.9Χ7X1ο 6
:
v,
ic(o) = iDN(o) + ιDPω) = 80ΟμA
ns€c
2.5
1.9
_
:
capacitor cuπent is
using eq. (14.17) and ( 14.18):
'
ζ
V
using eq. (14.15) and (14.16):
-i(,
v
(b) Referring ιo t
: jι:ο ιι)(l.49Q.5
z/|ι
o
solνing this quadratiο equation, yieΙds one
vo!)5ν voL -'oν
using eq. (14.12):
"":
:
y,)
(a) Referring to Fig.14.12, without loading,
\ L IP
>\
-
ο.09(2.65
14.4
'=
Po = Ι.,n,Va6 =
:
o.o5
sothaι,yi _ ο.356y, _ ο.189
=
rrs,,(ffi)
-.,".(Υ)
v: _ o.446v'+
Voι.
'51
2 /r_ . J.',1.4',)=ο.εsν
lO
o.3vl'2
squaring both sides yieΙds
V, ΙVηn v,)Γ| /, t_Ξ] L γ "y Jγ(γ l trl
ΝM, .
:
^lri5Υ
using eq. (14.13) and (14.14):
NM1 =
y,
- v'_ o.3vl "'σ.ε5v vl_ 0'223 = u rσ.65 _ v,
_ /TJl t { 4J Γ'
:
+
Since v oH
THE NoΙsE MΑRGΙNS wΙLL NοT cΗΑNGΕ
l.,,,
:
ιhe result of part (c)
:
iDΝ\PLΗ\:
50
in"\t",') =
ι"(Ψ\ ,
μA
\LιP
|ιu"'_ u,.lξ
_
'
i(Ψ)'1
Exercise 14-2
= ιzo
ι'ιlν'1(!)|,,'_
=
μA
2'15
:
Thus,i.(rP.r)
iρ|"' _
ι
= 0.24 ns
ι""o
=
722.5(
ro
oo
ι""ιο : \ι,:"(l) =
oo
"ιv
iDλ(o) + iDP(o)
:
4l;v;
2Φ
_
J
J_2Φ
i
|y,.|
_ 0.613ν _ o.svl
lv+o.svr"
Jrs vl "./ξτν -
_
|yφ|-
_ o.g76|v,P| _ l.o24Υ2 = o
solviηg, \re get |v,r|
ιo
8Φ μA
V.\'
8ΦρA
+
,
:
(d)
:
lV,ol
squaring both sides and setting one side equal to zero, we haνe the quadra.tic equation,
V.)' v)'?
:
"
t{ξτ _l';Γτoj v
=
tt
:
)|ι"'ι
1.6
v ns { 0.19 ns)
1(0'24
0.22 ns
R-_ '"Aν=
+ 320μΑ
=
+ ιPHL\ _
14.5
_ ιv12 :32n"s
ξιzo"ι,ν\(ξ)ιSV
,c(o):
_
"||
V
lv,,l
|v,ρ|'
v_l
l v"l(ξ)15
+
Solνing for |V,r|
Fig. l4. ι2(b)'
l|'(Yr)
ξοoυι
μA
6)A
^ιv
=
325
: ν'
|v'|
7ο(lo_'5)F(T)
tcl"
(c) Referring to
:
\γhere
| 325μA' _ 722'5 μA
4Ψ) ---:---
PΙ'H
',(Ψ)_ i(Ψ)']
'
50μΑ + 275μA
1lll20μA
O, will tum off when
Rτcι
* Rτez 2
:
ιΩ
4.5
+ 6.5
kf)
= 5.5kf)
ιe6 : 0'69RC:
= 1120μλ
ι""ιt""'1 : *"(l),x
t
pLH
:
0.69(5.5ko)(70)(
lo
O.27ns which is close to the value
'5)F
of
0.24ns obιained in Exercise 14.l4
l,u''_u,;Ψ_i(Ψ)"] _,,,(Ψ) = so llzv'?(|)[ιs,
14.6
_
\L)"=(Ψ):ls \LrΡ
l'Ψ''l
i(T)]
Using
= 688μA To fiιd iρ/ιpg)' ψe firsι determine yφ ινhen
,, :
}rr1r1"1''"oo"soonosιo
: tv,"t =r
v
:
^,1ffi, - lra,1 o.s v"'[F. oi v -
+
}
)ιzoνι t
"(Υ),|?'
;k:
v"l(!)|ξ _ l.ινv]" :
14.1
ic(ι pιι ι) = i DN(ι
PH
)
r.ao
,"ι
irl-,
'zoιro'5lr(}) 9ι4(lo
η.
6)A
(14.36) we se€ thaι
l2'5 '" = (ιv\ kΩ
R--
l2'5 = (l) kΩ
0.19 ns
_
ι2.5 k!!
\ L ),
Using Eq.(i4.38), 0.69[(C,,",1 +
cIcl)Rl
+
(ci"2+ cτc2'
+ R2)
= 0.69[(1OfF + 5fF)(i5kΩ) x ( ι5 kο + 12.5 kΩ)] ,/, = 0.64 ns
So,
15kΩ
\L)ρ
with
x (Rr
W*Μ:9l4μA
,(Ψ)
v
l, :
+ i DP1 Pι!L'
=688ρΑ+20μΑ:708ρΑ
',ι.'':
J0 '' = ι w\ kΩ-lqkΩ_ (2}
R-,
λ]' zo
ε.:ιο
Usingη.(l3.7l)'
: ^,f.o v] |v
(ι4.36),
r,.:ffi": ff:
v,..
Thιs, ιDP(,P,.) = =
v"o
η.
+ (10fF + 5fF)
Εxercise 1,ι_3
=
]^sοrt,_s !l' 2
= 4ΦμA i6,(ιl'1
Y=A+B
OR
_ V,) _
= 175μA
i^,l -
NOR
(c)iDI
Y=AE+AB
xoR
XNOR
14.9
(f
so rD(yDD)
τιis new(f
wirl οo
)
:2(76.1 ).,
μA):
'"(Ψ):2(68.9ρA) This doubles
/a',
ι
"and
"
152.2
o(
137.8
o)
145
:
"",
-
=
μΑ
c(v^^ Y2!\
\"")t -'-τ:-
30(10
'5)F(l.8v o.gv) _ 0.l9 6)Α 145(lo
Πs
14.10
:1(w)=!x1:r ,,,(Ψ') \L)"q1 2\Lt 2 2 2
(b)i,,(ι1, =
v
"'s
:
\ι"(l)
_
i,,,ιl,.| _
8αi
:vl _ lι,|Ψ) ι: - ll' ι \Lι'c|
i^'l : !X50Χll] 2 =
lΦμA
(e)
Δv ''' =
It'
i^'l .Δl -:jj.]jj: Ctt 6X0.56X lo'9 _ _t0οΧ10 40Χ lο 15 Uγ2
decrease to 3.6
Y
14.11
oιl = o Voι : _0'88 V sHoULD BE SHΙFΤΕD BY _ ο.88 v Voιl: 0.88v AFΓER SΗΙFπNG Voι : _ 1.76V ΑFΙER SΗΙFTΙNG L4.12 Refer to Fig. E 14.l2 Neglecting ιhe base current of ρΙ, ιhe currenι through Rt'
is
,
5,2 Vo,
_
5.2 - o.75 - 0.75 0.907 + 4.98
_ v,l'
Vo,
RΙ+R2
Vn: Vι_V"'':
""
Χ lο 15 Χ 4 6 288 X ι0
40
= ο.56 ns
Thus,Y, = -IRr
l
",,ιv
288uA
Δυu1
i,l,,,
D1, D2 and R2
Refer to Fig. E14.10
: lrg):.L,.1: rη) \Ll,,]ι 2\LΙ 2
ι'iJ
V
the new ,psa is
ι
:
(d) Follo\νing ιhe hint we assume that
Thus'
ρΑ
:
to 2('lz.5 μA'\
o
μA
io(})
*lII also οoυυle
=
oo- V')2'
""ιu
ιae(!)
_ v,lv, _
remains saιuratω during Δ, '
Υ=AB+AB
aoυυrinε
2
Cιι(Voo- V,)
i/,.,|.. '"'
|ι""r,,'l(l)
'ι00 + l75
l,',Δ, -* C11
6, =
:
""
=50xtΓιs t2J |)t 1,'rl
Y=AB
since iD(y,D)
ι.(!) \ L l.qtL|ιv
:
:
o.6285 mΑ
-0.57 V _ o'57 o.15: l.32
ν
Exercise 14-4
t4.13
ιoR
Refer ιo Fig. ι4.26
' 'u -
|
vR_ vBΕ|oR-
_______τ
ιΔo,
v εΕ''
o.7't9
V.Irrrr= 0V
4xo'245 =
'|
(because the cu.rent throυgh Rct
V
:
",|n,
14.14 Refer to Fig. 14.28
:
_1.24 Υ l5.2 mΑ
o.orsh(Ψ)
o.75 +
= 0.818 V
'o.49
0.818
_
_1-3t V
t4.15
ForV'
: V'r'
:
' ι.=+
_ ι..12_Vιεοn+5'2
Ιρx
:
REFER To FΙG. 14.32 for
Vι:Vιιι:_1.205
99 Ιρ1'
0.779
Αssume yrε|oR
:
Thus a better estimate foΙ voi is
ιon =
is zero)
IρR-
: _#-
0.'15 + 5.2
Vc|ρ": 1x 4x Rc1=
'
_ 0'49' 0.15
A better esιimaι€ for yrr|o2 is
"
- 1.32
:
o.75Υ
0.99 X 4.018 :
The νalue of 1ε ιγe found in Exeτcise Ι4.14 ιo be
4.l2 mΑ. The vc|aR: _0.22 x 4.l2
Ιε:
'
4.018 mΑ
= -0.906 V
υΝoR-'0906 - 0.75 : _ 1.656 v _ l'656+ 2 I|^. = = 6.88mΑ ο.ο5
3.98 mA
Thus a beιter estimate of Vru ,^ is |
= 0.15+ o.or5ιn(Ψ)
A betιer estimate forysEIca is
= 0.785
v
V
,-' _
t.32
V
"'|n,
and corespondingly,
V' :
For
r 5.2 _
0.785
0.179
_1.32y
r-' _ ' l.32
'
Ι9*
:
o.'Ι5 + 5.2
0.119
-1.97
Ι9o
:
:
mA
Ι6/2'
4.018 mA
Thus a ωfter estimate for Vrr|r* is
V96|'^
:
= o.16'Ι
ν
o.15 + o.or5
h(Ψ)
:
"'|n'
o.75 +
= 0.798 V and correspondingly
Vιρa
:
(b)Fol
τ,
_0-906 _ 0.798 = _1.704
: Vor:
:
= 4.58mΑ
:
o.75 +
V'": _1.205 v, Ιq:99Ιρρ' l'2o5'o''15+5.2 = 4.Ι66mA Ι-_ ' 0.779 bett€r estimaιe for yrE]ρΙ is
vBε|oA = o.?88 V
o.15 +
oozsl"[Φ}-1JΦ)
1.
'
\1V
l.2ο5 _ ο.788 + 5.2 0.799
__
0.788 + 5.2 o-7't9
Vρ.lρp:
1
_ 4.l2 πA
2ιε:2mA ' : x Thus, Yc]oi -2 0.245 : -0.49 V
-
0.788
v
4.53 mA
: _0'22x4'53: _1 V Vιon: - ι 0.75: _1.75 v /t-_ : --_l:ji--l : 5 mA vc|aΛ
ο05
Ve6lnr:0.75+OOr5h(i) = 0.79 V
-
and correSpoπdingly
_
0.or5h(Ψ) :
ο.88
4.00 mΑ
v
Α b€tter estimaιe for yrEloΑ is Vgg|'^
FoτV' =
Thus a
_0.88V,
,,-_#
and correspodingly,
1ε
o.oδh(Ψ)
"or (c) The
V __
_ | _ o.79
:
_|.'19
ν
input resistance into the base of O3
(B + 1)[r.3 + R.]
= IοIΓΞ Ls
+
5οl =
l
5.55 kΩ
is
Exercise l ,ι-5
u_:)rΔ
vi
_
:
o_ν
(5.55 kΩ || o.22 kΩ) r.|nn + Rε
Thus, --------
-5.s5ll 0.22 _ _0.269 [25 * uzl)r. Iο' \4
5t
5οΩ+5f'
ιc|ρι 1μ5.
J1Ψi orlo^
:
:
+Vr:
Αssume ysE
:
_o.24ν
/ν
-
o.79v
)
ο.7?9 kΩ
2 X 0'75 4.98 + 0.9ο7
:
V
"lρ' Vι = _
:
0.629 mΑ
v _ο.75:'1.32v o.57 0.57
_
|'32 + 5'2
6.l
:
-5.2 V (because ιhe cυrrenι \νilI be 4 to 5 mΑ). Αt the range ofsaιuration'
= o.99Ιu
Ο.6]6 mA
Thus ιhe reference circuit draws a cuπenι of (0.629 + 0.636) = l.265 mΑ from the 5.2 v supply' Ιt follo\η,s that the ρo\rer dissipated in the reference circuit is l.265 Χ5.2 = 6.6 mw. since ιhe reference circυit supplies four gates, the dissi_ Daιion aιlribu|cd ιo a ρuι.
Ι, : αΙ'
x
6.59γ
5'2
Ι.|^ - |ν|
(η-0.79{.3)
ξ_0.79
0.99
14.16
o.22 kdt
(
:
Refer to Fig. 14.26. For the reference circuit, the cu.rent ιhrough Rl, D ι' D2' Δnd R2
is
d) see figuΓe beloψ.
0.3
0.779
0.9Φ
0269 X ο.9o9
0.79
o.22
vι_o"19+5.2
l
5οΩ
ιNoR
+
i.
Φ _ Δ
l.o5mw
Ιn addition, the gaιe dm\rr's a cuπent
/ε
:
4 mΑ from
the
5.2v sυpply. Thus ιhe ιoιal
po\reι dissipaιion / gaιe is
PD:4Χ
5.2 + 1.65
:
22.4 mW
Exercise 15-1
Ex 15.1
5+Ω
Refer to Fig. 15.5 (a)
:
when vΦ
"': Ψ v2_ ΙΞ9
satumtιed ι/^
ur6
ρ' *1l μ;n
---
(Υ)
|ι"
""'
- )' "
""(Ψ
vlξ
:
:
"
- ;(ξ1'1
ro*{1)(Ψ).(Ψ _ ε_
;(Ψχffi[ιr
,lz, lο-ulfΨ) . = .\ L,/.
l?2
x
(v),: ιιz=;*ffi 15.2 Bits for row address:
2Μ:
:
logr( l,024)
μ ,i Iog2(2) :
logr(1,024)
,,=ιogr(|'ο24) ' Ιog.,(2 )
lο
Bits for column address:
2Ν
:
Ι28
:
32
logl32) log?(2)
:
(}Ιl."._,,"[a5,l Ex 15.5
ξ:vooιι_"'
:
C
" ξi^u
'
ξι"'c,,'l(ξ)ιvoo-
144
Δ,
)
v,.-
vu)"
x lo-o
1o
ux l.5 x1l.ε_o.s _o.s1'
μA 144
μA
a1
2-8 ns
-J_ Ιs61 lWι
th6ss161g.
Δr: l.7nsxΞ1-5
= 2.8ns
or: 11
ι '1cR
To find Ι5, we us€
al(ξ)": zs l. : 1* 3r, lo-u x 2.5 x 1t.ε - o.s - o.s1' : 24o μΑ l, : 1:-]C1Ιj? : t.z n,
=
o":Voo(l-""c")
_,,
=(f).= zs
s
Ex 153 |/
ol']
\'
vDD_vι')
ωl(f).: ls r. : 1 x:ωx
Bits for bloοk address:
Bit"
10-6
}ι
\
2Ao
π:!99:Ι!?8):7 10E2(2) 2ttt :
*)'
os{f)-
(ι_ v^
Θl..
',
Ι'o24
logr(2M)
6.9 ns
ι, :
Et
u
(fl"1
k; : 4k; : 3Φ μΑ/V2; lY,l : 0.5; Vaa: 1.8
:
,
Ex 15.4
FΙom Εxample 15.l:
}ι:ω "
-7:ζJz
ιd: CRln2 :2Χ|o'|2x5x1o3xο.69
; (Υ)
)o
tLz"' ,Ι:i
-
1.1o6"
n*'(Υ): L(Υ)'" \-'
or,.-_τ_<η
',
oφ ινillbe
then
\τ)"
'
Εxercise l5_2
Ex 15.6
y. )'| v"'_u.1 ]
(Ψ)
\τ)",t"Γι_(ι\τ],-=tU} τ"L' \' (Υ),=(Υ).,+,,[r
-(, πΨrΙ]
(Υ),=''(Υ).* = ,,
(f),,
"zs-([),=
: Wp:
lV.
:
0.18μm
CB"!ρρ 2 ΔVφ1
.lο
ι
o,
: "}.rij' :
Ψ
P : VoρΙ :
M2M(NMos)
:
'
CV'oo
1.8
Χ
18ο
l8OμΑ
μA :
324
μw
Refer to Fig. l Ι.26
From E4s. 15.14 and l5.15' ιve have:
9ι
ι:
Δ, _
our decoder is an extension of ιhaι sho\ν; We have M bits is the address (as opposed to 3) and correspondingly ιhere Ψill be 2M word lines. Now' each ofthe 2M word lines is connecιed to M NMos deνices and to one ΡMos transisιoι Thus the toιal number ofdevices reqυired is
Ex 15.7 ΔV111
From Eq. l5.l8:
Ex 15.12 e.zs
Foι minimum area: selecι
W'
Ex l5.t I
x lo '' , !Ξ _
0_lrΙο ''
2
o.υ
:
.v
+
z"1ιι,Ios1
2M(M + 1)
Ex 15.13
=
_ _3ox lo_'' : _οrn v _" yV ^ Τ C "y29 2 oJ lο "^ " mV
_9}
= -60
Refer to Fig_ 1 l.28: Our ιree decoder will haνe 2Ν bit liηes' Thus it wilι haνe N ιeνels: Αt the first leveιs ιhere will be 2 ιransistors' aι ιhe second
22,...,
at the Nth leνeΙ ιhere
will be
2N transis-
Ex 15.8
tors. Thus ιhe total numbet oftmnsistors. can be find as
Αrea of ιhe storage array
Number
=
:
64'loAΧ lo24x2
μm')
134.2 rnrn2 or equivalently
= ll.6
X
mm
1.3
x
134.2
=
l'14.46 mrn2
:
X
13.2
13.2 rnmz
Number
Refer ιo Examp|e l5.2.
Δl
,
=
S. υ-
ψe
by a facιor of 2 by decreasing
same facιor.
Δι o
"
o
can
τ byιhe
L
G.
Hence' cn has ιo be doubΙed.
G. = g.'
+ 8-b
8,, and 9,1, have to be increased by a iacιar of 2. The increase in 8- can be achieved by and both
'Ι
increasinρ ιhe corresοondnρ Ψ_ ιhus:
ο.54 Ι" z^ ο. ι8 6 (Ψ\ = z x 2-.J9 : zι
(Y\ : \L
\L)P
: .'_| Ξ 2Ν_| r I 2-l =2Ν l
ThυS,
since ΔΙ is pfoρoniona| ιo
0.t8
:2
^ :tum
Ex 15.9
reduce
!-
Geοmetric series .
11.6 mm
Total chip area
=
134217728
: 2+22 +23 + ''' +2Ν 2 (l+2+2'ι'' ι2n')
:
2(2Ν |)
Ex l5.l4 l J" - 2x
'o
I 2 x 5 x
tO
e
:
l00 MHz
Εxercise l5_3
Ex 15.t5 (a\
(c) Ιn one time-constant the volιage reached is
Ι." = k; (y)"[,, _ |)2.5 -'2'517 = zo x
4Uo
:
mΑ
1.65
ι.lzs1 '--> ιr=2.2 <_ +5V
Voo1Ι'e
ΓrΕ =
11
1
'"ι
: o.632V pρ : 3.16 V 5
V-->
4.5
V
+2.5V
t----'
τ* : Thυs, ,ch'sins (b)
tt : 2x 10 x2 : r.65 x ro '
6.1 ns
sahΙrated
ι":
\ι"(l)'<ι.ιe η" : !xsοx9xz.lε'
:
22
0.35
^'
mΑ
cBΔv ΙD
- 2x
ιο-ι2 x 0.5
0.35
= 2.2Χ
:
19.8 ns
3 Χ
ro
t'x3xrol
x to
3
_
2_9n"
Exercise l6-1
Ex:16.Ι Α : 20 Ιog |T| tdB] Tl : r 0.99 0.9 0.8
Α: 0 0.l 1
0.7 0.5 ο.1
1 620
2
Ex:16.2 Α.', : 20log l.05 _ 20log 0.95
e-.:
zologΓl |0.ω||
Ex:16.3
T(J)
: ι =f
(s +
=
:
0.9
dB
ηoοs
7(0)
Thus:f(r)
=
since
i2\ (s-
i2\
t t-J
J i,r
44
I ----------------
(s+ l) (r'+ r + l) 1
^,Γ .,./
1
-
--
=κ-_ r'+s+l 1! :
s+1)(s2+i+s+1)
Jτ ;1]|
(s2+4)
,
(
:
(,-l",f)(, -\_'β)
,
: t
.. T(s)
T(j,r):
, (sl +41
7161 =
ο
1_ι 2Q2 Q:I
f
1
I
ll = ____j ,=r,υ, Α2 (l + ιυiaι )
: _2οloρ -fi l
.4ι_])
(" *4) .-rr.t:1 ' 4.'+.+!
.' +.' +.' +.'
o.ε.o
-. Thus:
1:! 4
;T;;τl;
:28.6dB
*Y
Ex:16.6
Γ!
Jlo"' _
Er;16.4
Γ(s) = t
(r
,("' + 4) j8) (s + 0.1 j8) j1.2) (s + 0.1- jl.2) s1s2 + 41
Α(ιo.)
(.l'?+ 0.23 + ο.65) (J'?+ o.23 + L45)
Ex:16.5
l:
o.sοεε
1rU.)1
+ 0.1 +
(., + 0.1 +
lradzs
:
_20 IogIT(iω.)l
_t *.'(9L)'nl ιωP,/
=
10 lοsΓ I
.]
Thus. I0log[Ι + 0.58882 l{ : l0 LΗs : 29.35 dB l{ :1 I LHS : 32.87 dB .
Use
.
N : 1l
X
I.52N]
Ξ 30
and obtain
A^''':
32.87 dΒ Fo.Α.," ιo be exactly 30 1o €
ι]ol2Q = 1cos60:1Ι2
:
logtl +.' x 1.5"1 =
9
o..t654
0.54
Ξ Α.,' -
ω,,: 1andQ:1
ω6/zQ:1cosΦ:
a
log.[ , ο_,ωl
:
The real pole is aι J -l The complex conjugate poles are aι
J:2cosω'a
1
zο
36
dB
Ex:16.7 Αs sho\νn, ιhe paiι ofcomputex poles has
dB
:
0.s
t
jJi 6
jsin60"
Exercise Ι 6_2
valleys are obtained Ψhen
:
co.'Γπco.''f9)l L
oh(+)'o
\ιor,/J
ι
scos'Γ9):kτ'k=o'1'2 '.;
T(r.)
: (s + 1)(s + o.5
-rf)β
for
(s+l)(s'+J+ι)
DC,-
-
:
o., _
;f)
k: o'l'2
= .","",(Ψ}
dr1
:
ιυ"Cos0
:
ω,
ιλ,
:
ωρCosξ
:
0.81ιoe
δ3
: ιorCosf :
Εr:
16.9
I
O.3l ωe
_L
\
Α(ω,) = ΙoιonΓι t .'cosh'(,νcosh
Ex:16.8
: :
lo ΙogIl
+ 0.34932cosh
'''\l δJ]
-'2)]
2(7cosh
64.9 dB
1og,. = 'ηdL1: Α(ω3) = 10 logIl + ο.5o882cosh
ForΑ.,,=
:
o.soεε 2(7cosh-|2)]
68.2 dB
This is an increase of3.3 dB
lr(jω)1
Exr 16.10
=
+.'cos'[πcos '(9)]
r for ιo
{
ιoρ
[π"""
"..'s [s"* scos ...δ
:
Jroto
-
1=
0.5088
(a) For the Chebyshev Filter:
.
Ρeaks aΙe olrtained ιryhen
*"s
Γ!
.:
''(-e)]
:
''(:ι)] :
'(Jι) :
Qk +
N :
o
:
ι:
;,:,"co'(fi,,) :o.so',
:o
log[1 + 0.50882cosh21/vcosh -ll.511
7.4... choose λΙ
:
8
Excess Attenuation =
;,:."co,(ff) :o.ss."
;,:",,co,(rt",)
10
Σ50dB
o
')ξ,k
.,cos[QtJ_Dz],
:
.A(ω3)
1O
o,|,2 o,
ι'z
_'l.5)Ι _
log[1 + O.5O88'cosh2(8cosh
:55-50:5dB
50
(b) For a Butterιγonh Filter
e
:
0.5088
Α(ω.)
:
:
lo ωε[t
lo log[1
N: ι5.9
+;(:ι)'t
+ o.5o88'(l.s)'?N]
.'. choose N :
Ξ 50
l6
Excess aιtenuation =
l0 log(ιl t 0.5ο88'( l.5)"] _ 50) =
ο.5 dB
Exercise I 6-3
Ex: Ι6.1l Rx
"..-*Ξ:pΞ-.
9 :
l0rradzs
O
selected to yeiΙd a centre. Fιφυency gaiη of 10.
Ex:16.15 (a)
lοo =
_L π'
cRΙ = 0.0l μF
C
μ.p 6u1n = R2
:
10o
l0 kΩ
=
:& :
- l0
Rl
kΩ
r(sl
Ex:16.12 Refer to Fig. 16.14
'' : lcR :
ω,.
:
t'+ ta sz + s ωο/ Q + ωλ
: .__:!_Ξ_ l'
lr(l.)l
t0'radzs
' ω','
lok()c:
__τ_: . I
l
l0'
lο"
:0.l
r
dιωi
For R arbitrarily selected ιo be
_
\''2\nλ
I
μF
The two resistors laωΙled R, can also be selected to
be l0 kC) each.
for any two frequencies ω, and ο)' at whiοh
EE:16.13 11τ1 =
2
-------J9n
(fordc gain =
s+s'/2rυ,,+<υ']"
lr(jω)] :
ll
.?-b - .1 .b 'i)' @b -7)'
(.3 -
: ω2@?6 : ω,ω, ω$ (l)
',1.}
2
.=>
+ 1zω|
ω')
ωj1
L
t
ιυlωο 1
-b.,lQ OlΦo
ιυ,,,
||= ]
J2
vaΙue at de(unity)
which is 3 dB he|ow ιhe
Q.E.D.
Ex: lιl-14
'δ'iιο|;':Q'ΣJ]=l
1ιυ}
> ^Λιμ,
u,
|-"/ηοΑ/ll,-
1- "πo4'l|1-
',
This 7(s) ιoο
=
:
1Oas
.2+|ο]"+lοlο
l05 rad / s
_ l sι;g 1
ω1ι,l, ιυf Q .,, 1> ωl_ωl o ^/ηδΑ/l,,
9ρ(b)
vn _ oR
-9ι ιl
BW.^tloιl ForΑ = 3dB
Φo
BιvJd'
BW':
_
ωJQ
l
I
1
BW,, Q
/
ι,ι|)
Now to obtain aιιenuaιion > Α dB at ω, and ω: wh9re (δ _ ω, : B\r,
toloρ_ Γl
At ιo =
|r| is
the same
ι
q.ε.ο.
ιno
B|η' Q.E.D.
ιtl
Exercise
Ex:16.16 From Fig 12.16 (e)
1μ From ExercisΘ 16.l6 above 3dB bandlridth
= <ιn/Q
2rrωlQΞQ = 6 : Q ω"CR 6 2τΦ Χcλ |o4Ξc _ 2τ10 =
ηj''' /ιυ.'l2+1 2QI radls, ,'tn
dc gain
= lo
1.2
radis Q
ω1
a:+
2
I
Ι.6μF
R : |oa : ι.42H ω.Q 2τω x 6
ι:
) '[;) ,2 2
2
:
l
Ex: 16.19
1.44
f" : lo k{z Δ/1u, : 5Φ J :|oo:20 ο: _ Δ"/',,n 500
l.ηall_l)_l ι 2m/
Ηz
Using the data at the top of Table l6.ι:
= 0.986 rad/s
:
lr(jω,-)l = 3.17
lr(j-,
rπ)1r}"Τ
o,
=
l. 7 2 -l a2|( ω l _ ω mu)l
: ,L*:
flaι Ξ0:
Q: ω"CR+C = ll25 pF
κ
I
=
J1
10
Ex:16.20
I
rtx2τ105x|o!
η
- ωρ
(16.25)
ιιs|:+x
L
**
ιn"L
lot z",lο5 x
tu
Ex; 16.18
265
(s'
:
kΩ thenrz =
8.lω8(J
90
lιΩ
2τloa ω^
--Ξ------------_
l0
5
+ 0.2895ωρ)
+ o-42g3ωe2)
1
+ s0.1789ωn + ο.9883ω,,')
The circuit consists of 3 sξctions in cascade: (a) First order section
_&
R: ..r: ωoQ
:
Ιo
r| =
selecting
*. :
kΩ
20
centre-frφuency gain =
(s2 + so.4684ωP
ρ-
13.26
2τ104 Χ 1.2 x 10 the data in Table l6.l for the bandpass
'ι+12/r|=
*--Ι-Γ
1;.o
=
ω.. :
Q/
Now υsing
R : lkΩ
=
l.2 Χ 'ιo-9
Φnζ
'ψ
ωo:2τXlωX1or Αrbitrarily selecting
x
2τ104
Ro:
o.eo
Εx:16.17 MaximalΙy
cΑ:c6=1.2nF Rι=R:=Rt=fξ:_l
!
= 2.25mH
6
L
ΞΓ
*-
th€
number co€fficient
11.1'
:
was set so thaι the dc gain =1.
-0',2895""
s + 0.2895ω,
kf)
Exeιcisc 16_5
Let
n, =
dcgain= as
19
1ρ
RrlR, :
Εx: |6.21
1 = R, : lokΩ
C=lnF
jω '+ ..
lr( n,rl
ι'_ ^t
,
2985t,
0
ιo
: -_L ωCR'
: l5.9kο
2τlo4 X Io_9
Rt = 10kΩ
Usiηg Eq. (16.62) and selecting
Rr: ρ, :
--------
Using
(b) second_order secιion wiιh t.ansfer fuηcιion:
R1
0.4295ω1
s'+
l : ΦnC
π:
0.2895Χ2π.104X1o4
r(r):
l6.Ζ choosing
Refer to the KHN cifcuit in Fig.
coefficient was selected to yield a dc gain of unity \rr'here the numerator
1ρ
: t0 k!l R2QQ -l): ι0(2X2'1):30kfl freuencν ρain= K _ 2 1 l.svrν '0
:
Ηish
0.4684ιυ,, + o.42%ιo?"
16
η.
(16.63) and setιing
R2
The ιransfer function to the output ofthe firsι inιe_ gτator is
v^. v"' l y,scRv.'ω^2
sΚ / ιCR) r_ + s--_'+ ω-ο
O
Thυs the cenιre-frequency gain
=ΚQ-Κo_l'5\2=.lν/v CRιυ.' Ε\| l'6.22
:
Select Rr
Cι =
Cc
R: =
"σ.ιzN
: C
R:: l?5:
10kΩ 2.43 nF
x 2'r l04 X l04
= 2.43nF
Q : ιι
a ='9:!:-" 0.4684ω,
κa
ω,,C
(c) second_order secιion with Transfer-fυηcιion:
r(s):
0.9883ω' s2 + so.1789Φρ + 0.9883ιo2,
The circuiι is similar ιo that in (b) aboνe but Ψiιh
Rl : Rz:
R1
1 ,: ωu ' l0"
cr=c6= =
1.6
: Rξ: IokΩ
nF
Jσ.98&r .
giνen
2r, too
, lo'
,=ffi=r.ro _ ο 1789
Thus R6
:
Q /
ι''οC =
55.6
v. V,
kΩ
Placiοg ιhe three secιions in cascade, i.e. connecιin8 ιhe ouιput ofιhe firsι-orde. section to the inpuι of the second-order secιion in (b) and the oυιput of of section (b) to ιhe input of(c) resulιs in the oνerall transfer function in φ. (l6.25).
*K(RF/
RΗ)S2 +
(RF/ R)ιn2n
,' *cιo,,/o}ιo2n C : lnF R. = 10 kΩ
I _,.il.83 kΩ . lο'l l0 " Rι: l0 kΩΞRF : iokf) R, = l0kΩ+Rl: RΛ28-1) = 10(10_ l) = φkΩ : 25.6 kΩ ._
I ω.c
2π5
*&t,,'=','-',:to( )'
DC ρain .R, =
κ& :
l'10: R.: ' 2 t/5
?-
RF b) R,
l6.7 kΩ
Exercise
Εx'ι 16.23 Refer to Fig. 16.25 (b)
1
:
ΚQ
l.59nF
,'+
r13
, : -:
I
x
104)
:
+ ro8
x tο',t,',6 x
I
| l
o
to* _ 4
x
ld
2
= _0.382 X 10a and _2.618 X Ιo4
: Q20 R":R/Κ:20R:2ωkΩ ..r( =
2 I _)* +,( + \lο-'xzx lο5 tο 9Χ5X κfl
,t
1 : Cκ:!+C ωo - 2τ104 x 104 Rr:QR=20xl0:200kΩ
centre frequency gain =
lffi
@ s
Ex:.16.27
Ε\| |6.A
Refer ιο Fig Ι6.26 Δnd'lΔbl'e 16.2
c
C = 10nF
{*'*"*^, }ι'ιl_"
1 : : l6kΙ) Φ"c 104 X 10 Χ l0-9 QR:5x l0: 50kΩ C1 : CΧ flaιgain= l0Χ 1 : 10 nF n:1
Rr:+= galn r:
Vo v,: ' o- scrRl
Σ1at(Α)
R.zl = lokΩ
Υι * sc,1vo-
lOkΩ
n
5
galn
x l0 =
50
kΩ
From eq (16.76)
cR:29: ?JJ:2y1sos
+
lο4
ForC: Cι : C2: lnF ρ=2xlo-a :2ωkΩ Thus
R]
:
α
v
vi
:
_α/Ra
sc,+t,/R,+
-
I
I
This is a bandpasδ function whose poles are idenιica| ιo ιhe zefos of (s) in Fig. ( 16.28 a). FoΙ c1 = 10 'oF,&:2 X l05Ω&R.:5 X
sοιο
The tΙansfe. fυnctiοn of the feedback network giνen in Fig. (16.28a). The poles aΙe the roots
is
of
|), Ι _o t I .'+.l._L \cIR] CrRjnc|R4l cΙc2R!R4 ForCt: C2: l0 eF,R3: 2X lO5(t, 5X,o4Ω,
|
scrRlR4
\CrRr+-J-). C2RJ CtCtRtR4
.s)+.sf
Εxι |6.i26 the denominator polynomiaΙ,
c,+
cΣR]
_Sα.z (C1Ra)
269 111
m:4Q2:4 Thus. R. : E:1Φ: 'M4
o
=ο
scrRlR4
vo
From eq. (ι6.75)
R4 =
Rr/
Rπhl
Vo ,, , sc|yo - (l αlVρ , αξ τ - ' Jι". ιyo ' scrR, ΞζΞ]F| ' κo
Εxι 16.25
l0
vt- vo :
+
1
v,4) +
c,: lαΩ Vo: SX2XlOaxα yi s2+sx104+tos
For uniqi centre-frηuency gain
2XlOaXα = lOa+α:0.5
τιr*
&: rιlα
ιm kΩ.
-4'i :
tω ιr)
ο-
t8
|οln
=ο
Exercise 16-7
|-Δ.o
_
Ji
ΞΞ-
tJro
_'
_ 1=4Qs9 "42Q2 vn
Δ'r:_1_1:
9! (vn-
scavoΙ
y.,)
C3
Ι
Δ'Q
= ι' 'ι =
a
: η,(l Σ,
: ''c, + s::AQ "c, ωa sl:44
sα
4Q:
R,)
at (.4)
l
sc'vn ''" !,},}sc,π' Rι R' R,
!R, _ y^Γ.s,c.c.n'. "t vo _
sC.'R, ,
R,
yοsc4{sc\f2)
,,nsι(ι+&)* " "c"R,ι' RJ
ω, = ":
I
as ln Eq.
^!c{4R1R2
(
l
D.C.gain:
asinΕq(l6.7El
R,:R,=R:l0kο
C'2Q ωοR'
2π14
2/rt X
|01) lOa
_
5.6_ι
nF
m=4ο2=ι(!\=z \2.,, C.
:
η.
From Eq (16.l ο0)
Co
:
Exι
16.32
5',
Ro
:
Ex:
I 6--1-1
(b) ΔR'
l R':
ρ.j14
:
I
,p
: 2τ1o6 x3.2Χ l0 6X l5o : R: R.|] rρ ll Rr: 2kΩΞR.: l5kΩ
Bw
ωoΙ'Qo
(R, 1l R,")
.,r^O/O 2%
2
=!2 Χ2
1%
: flQ
+Cr":
Cr
+2o/ο
ωo
: R\2 -l
:
c5
/
ωoL
lοι
: _l /2 .aΔ-o : _ lx 2: s'" R'] .So
7
ll
|οt
(2,ττx 455 X l01)X5 X
RefeΓ to ιhe results in Example l6.3
=
(16.99)
6.:Co=9Ξ!i:ο.;lanε '420
0=
Ex: 16.30 ^R'lR'
x2o
6.283 pF
2.81 nF
(Δ)
10r
Cenιre-freoυencν ρain = ]-9
Ι Q.Ε.D.
2Qlω.
|
2ω X
From
From Fig 16.34 (c)
:
s9,99+s9^ΔC,+ο
Fromη(l6.96)&(l6.97) C1 = Ca = ω6TρC
:
Ιxl. 16.29
cR
0%
:2,ιτl0aΧ
lo. / /)
|\ '-crμε [l 1Rr* Rr)
2
''C "C, Q :o(-2)+(o)(-2\+o:osa
' C'qR&
Cι''
a- _ +''
_!ι _2l _
2
Ex:16.31
.sc,. f,I
I / C 1C 4R1R2
''
_.1ι _2\ +
1) :2-2 =
+ SC,V" R. +
oq.
49s :
V,,
2qo
(d) using ιhe resυlts in (c) for boιh resisιors being 2Ψo high we haνe:
Ι u-'o Rr
V^:
|Ψo
Combining the results in (a) & (b)
(c)
R1
]rz_
= 455/35
:
13
l0
6
Kιiz
+
ttbL
_t : c1
(2ιτx455;19:12x5xΙ0
24.4'1 nF
: 24.47
0.2
:
24.21 nF
6
:15
3kΩ
Exercise 16-8
163
Ex: To
^l
jusι meet specifi cations
: fo :1Ξ : 6'Βwlο . Rrll n']R," = ΦoL
R,ll r2R," =
45.5
45.5
x
π2R', :
Γ!Φ: 2A.36
10
6
= zι.ι'l
V
:to.ε
aol?'6s = ro.rf,
η
13''7
x
.'.
I
lo.1x
106)2
X 3 X 10
x lo
|2
X 2π x 310.8 Χ
1O3
=6.9s kf)
Ex:16.36
B|
= fo+Ψ
:
l0.7 MHz Bl
Jn,= =
{2
:
2ΦΙ
|0.77 MHz
{2 :'!4|.4ΚΗz ηι|6.|16}
MHz _ 2ωι2f,2
n.:2Φ 'J1 : For stage I
η(!6.l15)
, 2Ψ ιπ. 111
Ιo Ψ
1O.7
α
(16.110)
CR
:
CBt
74.7
x lo
12
x !4l.4 x
lo3
2zτ lo3
15.l kf)
: !xts.sx15.l:117
ιιl, (2ιτ
6
(2π10.63Χιon)2X3Xιo6
ωλzL
rl +R.,,,.I x +R"ι,"",
g- ΙRln 'here
_1
Jo,
x 2τ
tional to
13;Ι ρF
ο
141.4
Gain of synchronous-ιuned amplifier at to
c: ,nbL -L: ωo
ro_ζ
"
lo
Gain of stagge.-tuned amplifier atlo is propor-
o, = ΙR/n &
2Φ:t/alt'''ι o
τzε
kο
Ex:16.37
Ex: 16J5
&:
15.5
:
the νolιage deνeloped across R ι
t^"^:
:
106)2X3 X
For stage 2
r.-,o
Ic: g.Vo,:
12.8 pF
x
= 14.1 pF
1(R, || n'Ri") . Thus,
:
x
5
nF
Αι resonance,
:
-^l
1 ..*9r: -' n2 ,nb L
T
x 103x
1ρ
1.36
Λ/|
Cι:
455
(21τ lo.'Ι'Ι
n: CBιI :
ο
650
,:
:
as.s
ωλιL
|4|.4ΚHz
:
10.63
MΗz
6
R.,,a"1
Χ
: 6.95 :48.3 Ratio
R,,uu"2
Χ 6.95
: β7 : 48.3
2.42
Εxercise 17_l
Ex:17.1 Ρole freqυencyi = 1kHz
:
cenιre frequency gain
+.s
AMΡLΙFΙER GAΙN
lv;v
=
:5+0.93 =
L:
,fr
ι:
V
v"
30
:_2Υ /Υ
:
Rr
R.
60
(}1 +ξ _'"-"
ll
vb+15
3'
:
!(o.t
=9o
nxι
sιope in ιhe Ιimiting regions
R,.ll
*
ξ)-
.l
-o-ο95
30
ιl.ι
:
a) for oscillaιiοns to sιart, R, /R' 2 thυs the potentioΙneter should be set so thaι iιs resistance
y
ιo ground is 2ο
z' lal ,-{Jl' : (ι * !ι1 \ RttzP+zs
:i
:('-ft)('-λ)
-+{
k(!
1ο
kΩ
16
x
1ο
kΗz
Vo Vo , ,-Vo, ' η scRrR- sCRlR scR (vo, vo I I
μ.--.1;-ηl andc:
l+516x10s+
16nF
,, |
s
x
16
x lo
s
The closed loop poles are found by seιting a(J) = ι' that is, they are ιhe νalues ofs, saιifiying
]+S/l6y l0 t , ____Ι .s
Χ
16
X lο
&
''
Thus
ι(J):
x lor x
ιJy'orking from ιhe output back to the inpυt and conιinuing ιhe equaιions \γe get Ι
3156'ρ1_! .scR :
27'lο
Ex:17.5
ι
\νhere R
s
t0.68v
Ex:17.3
ι
ιs)
from symmetry' \rr'e see ιhaι ιhe negatiνe peak is equal to the posiιiνe p€ak. Thus the ouιput peakto-peak νοιιage is 2l.36 v
Υ
'! 2.9'l
3 . Now if we neglect ιhe cur-
l?6
νo_vb 13
_ω
Thus limiιinρ occurs aι
\νhich is about l/3 of %; thus
Τhus.
,,(,-ft)
τ:
?lo /
Rξ
+5.93V
V/V Llmltergaln : -R.
o.7 +
Ι
( io vu-vo _
. j)
-5.93
Αι ιhe positine p€ak to ' the will be one diode drop (0-7 v)
rent through D, in compaΙison wiιh ιhe currents throυgh R' and R" we find ιhat
=_ts,]_ο.?(l+;) =
j)
(b) The frφuency of oscillaιion is (105/16) rad/s or I kHz
volιage at node b above the νoltage
L' =yRι]Rs+v"(l-ft) rs(l)+ o.u(,
i_:_(0.015 :t
(c) Refer to fig. l7.5.
2
Εx: l7.2
:
:
.
= l.or
_ =
.scRiR,
vo
.scRJ-
l(Vo'
Vo \
sΓιη scRl /
ff(,.*,*)
_VoΓι, | , l
l(I*
scR,ι scR scR scR\ :-vo(ι+ -l- + | scR,\ SCR 1ι6]ρ2./ ')
ι sc
|Ι
scR/l
9
Exercise l7-2
t( ;*&._ *
(#
."l_"^J
].-*
ι;
Ei:17.8
ι
ν.' v"l+ L= v"(l ,l l "\ r s'cι'' sι, sc
Thus:
:
vo
-.scR
Node €qυatioη
^4t 'r+rar+r1:*:
: v'",-,
jωC
Υ9,
+
ι) *;(l-cn '\ οιCR)
The circuit ψilΙ oscillate at the νalue of4, thaι
Ι9a
i-t
"
Thus.
/.
: --l_=-^ : ωnCR
"/__1CR
Hz
For oscillations to begin, the magnitude of
v^
;:(jω) that
shou|d
ηua| to (or
gτeatef than) uniιy.
is
4 Thιis the minimum νalue ofR/is
l
12ο kΩ
1+Cn: .,^: " CR ) πlo3 : ForC:16nFi l0kf) I
as large as the νoltage
across the resonatoζ ιhe peak{o_peak amplitude
is
4V
4(2\1.4\ -Ξ-:'.oν
+
D,
(oscillations have staΙted) it ca'Ι
s'ιrιrc(ε^+ 1) +.sL+ R
l:
+
s'1ι'c
+
ιrc1
ο
Substituting S = j@
Ιl-ιn2c(L1 +L)!+jω
[?
_
t
β"
Rε:
!r)x.,ι,ι'c]:
o
a.E.D. + lc(Lt + L2\
o=rΦο:
Lt+
L2
L2
+B.R :
Lt/
L2
for oscillations to start
g-R > L'
l7.7
.'. ιhe output is ιwice
y.
ο
ι'ι|'L7C
4 4R 4R ^ = -i_-;--fil ' ιoiC'R= _-;---;--Ξ: ιυiC'R' l :12 R or
Since
s2c Lr)
/-:O+3.R+Ι:_J-
-lct " nn.'ΞΙ
Εx|.
*__L):
l
2τJax|6Xlο-9XιoXlο] = 574.1
o
be eliminated resulιing in
a reat ntrmher
Ιι follows that ω" is obtained from -1.u^6'ρ
:
-! #,* '.u-.+('. #") *L( l so \
Ex:17.6
.rk".
t^u.* rL *
#,t
R'
at collector:
/
L7 Q.Ε.D'
Εxz |7.9
R:
q|l R.||
:
lω
Γo
11 2X lοrll lωX X |ο 3' :l0|l 2|l 1ω: ι.64kΩ
lο6
Ξ='.r:Φx1.64:65.6
lo1
Exercise l7-3
cr:65.6Χο.0l :
.t
EΙ:17.13
0.66 μF
1 ι .l . ω; '' c2 ''cr + I
o.o1 Χ 0.66 Χ 10 1012 X 0.01 + 0.66
6
Ξ lω μA
Ex:17.10
η
from
_
''
:
from
(l7.24)
ι_ ;;τΞ,
_ zn
Ji.sz
2.0I5 MιΙz
η
"
αon
''
lo "
(17.25)
JP: ---!-C'C z.n
lι'CD+
4
"CP Time delay =
I
o.0l2x4xlο
:
---!-
ms = 0.125 ms
Ex:17.14
2.018 MHz
ι'ι Ι'
O:
-i
ι'l" t'
=
-!-
_rΙ"r.oir"ro""o.r, 120
^= 55,ω0 Ex: 17,11
Vτι=Vτι:βiιtl 5= R' x13 R,+R,
&:l.ο Rι R:
:
Ειι
17.12
16
Α comparatoΓ \γith
kΩ
:t 12
EΙ:17,15
lyJ
R,,
Vτι vτL=;|Ll ,ι2
s:
of
Ιeνels
R2
Possible choice
Rι :
]!9 2
:
ιhreshold of 3
sο mv
50X1ο]:10&R2
&xro
R2 = 2Rl
:
ν
a
l0 kΩ
Rl = 20kΩ
Rz: Rι R2 :
to 0.06 200 R,
forR' = 1kΩ
ι, :
2φLρ
v
and oυtput
Ex€rcise
1
7-4
Ex:17.16
R, y too Β= ' R,+Rr: ιω+ lω0 :0.091 v 7 : 2,r1!!-P l_β 6 l'09l ) 2 / o.o| / lo Χ ι06 X lnr :0.Φ365
1
=
\t -
s
! : T
0.09t,,
274μ,
k----t<-----+1
T'
η
During
Εxz 11.17
T,
vΒ(t): Ι2_(|2+V)e-ι/5 vR : vD att: TΙ2 vD: 12-{t2+vo)" t'"
τ:
:
2
+v
o\ z,ιn(|2 \12_νD)
xo.r x
10-6
x rox ro' x
h(]]+)
"T
^(ry._η)
Tt: T2: T/2 vA(v\
5ω
'"(i#)
:428l Ηz Aι 0"c , yD
_":
:
0.7 + .05
iω, = l,rJ2Ξ) \ 25 .'
.fl-.
Αι 50'c. VD
.
=
\l ι.]5./
: 0.7 5ω :
Αt l00"c' yD
:
.lnt-t /t255\
ιι
v
0.65
= 4.6|l Ηz
l,(1?Φ)
fl._^-^
V
0.75
],995Hz
0.7 _ 0.05
= 5ω
rl
:
0.15
:
ο.55
5.451 Hz.
!.45-'
Ex:17.18 To obιain a ιriaηgulaΙ ψaveform with to-peak ampΙifute we should have
vτE:_vτι:5ν
BuιV-, '' Thus
,5
: _t+&
R,
= ,ro x
Rr: 20kΩ
v
!! R2
l
αv
peak-
Exercise l7-5
For l kΗz frφuency, T = lms
To obtain
Thυs. T
/2 : o'5" rn l _ rρVτι-Vτι
:0.0l X l0 6XiXΙο,/lo R:50kΩ
4
Rz =
η
lωX lo R]
:
_ nI X
10
"XR, |"(i#)
+ R, =
Αι ι :
_
0.6q,/
lο'\ |0
''(Rn + 2Rr)
Now. subsιifuting into (2)
Use 7.2
kο
v'
3
1.25 the
1ρ
circuiι provides
5-
l -Rr l2Rs ο69X lο "' l4.4q kΩ(l) Using η (13.45) A+Ru o15 = RΛ + 2RB (2) Re * Rs : o.15 x |4.ψ : 10.88 kο (1) (2)+R, : 3.61 kf) '1.2"Ι
8ywe must have
i _ 0.l' 9 _ o.gmA.Thusιheeπoris
o.69C(Rλ+ 2RB)
Rι :
=
; = 3 = ο.οrnn ινhileideallv
Exz 17,21
lο''
1,
8.8-.1 . 8.7 5 ι.25 Rl
T : l.lcRΞR : Τ/1.1C : 9.l kΩ
tοο X
(i.e. to obtain
ιo select l?, so thaι i = 6.4 mA,
Εxι |7.2o
'
4V
kΩ
To obιain a perfect match aι
6171 Ω
T:
1.25
y:
υ z__] υ_7 Rt R.2 R]
'
(17.37)
o
perfect match at
for υ>_'7v
Ex:17.19 Using
a
i : 1.6 mA) t.6:!+4-3 5R,
kΩ and 3.6 kΩ
.
standard 5% resistors.
Exι 17.22
-0.3 mA.
*
Αt
v:
5ι4 ιhe circuit proνides
, =:_= 5 125 - 2.6 mΑ.whiιeideally i - o'l / 25 _ 2.5 mΑ Thus ιhe eπor is .
+ο.] mA
*Αt υ
: 7ι
the
circuiι ρrovides
i - !5 *1.1.25 ^:\ ιhe error x
Aι
i
=
τ, :
4.6 mΑ. ινhi|e ideally. Thus
is _0.3 mA 10ι4 the circuit pΙovides,
Ψ+19-3+19..7: 6 1.25 t.25
ideal|y i
_ l0
lomA'ψhile
mA. Thus ιhe error is 0 A.
i (mA)
i
8
R3
i =o'l,1ρ
Tru Slope = 17ρ,
{
i:o.l; At ι: 2ν' i = o.4 mA Thυs
f,, = -1 ' 0.4
:5kΩ
For3yΞy=7V
'
υ υ-3 l?t R2
Exercise 17-6
Bxj \7.Δ
Thus,
υ, :
0.5
V
so
1ra : 1to + vp : 0.51 V ,, = 1V=9/o: 1V ip:1mΑ, υp: o.1 Υι1 : |.1 Υ Ur : '1V - The negatiνe feedback loop is noι
operative.
ι": oν
ιA =
_l2ν
-1V
Ι"r: =
Ι +(2.42V7)/R
*2-Ξ:ιf lιLrRl Γ L
2'42 v 2.5
-'1
vτJ
=
lι\
/,l
/ 1 Δ''\ Ξ /ιl +2J-J
+24\ 25.'
Exr 17.25
ι.2= Ιl'2-4\ \ 15ι ιo=
(V,'"_ Ι,ιR,)_ (y.._r"ΙR")
: ("1_ Ι,)R,
:1ρ'γ272-! ' 2.5 _ o.25x |o x zx?-! _ ι'ειν 2.5 :
*
lt
For the diode to conduct and cιose the negatiνe feedback loop, Uo must be negatiνe, in \ρhich
F'xι 17.A υA
|υ"
γR
case, the negative feωback caυses a νirtuaΙ short ciτcuiι to appear between the inpuι terminals of
rr
the op amp and thus U, =
For positiνe ,r, the
op amp saturaιes in ιhe posiiiνe saturatioπ levei. The diode will be cut off and vD
R= lkΩ
.'. The opamp is
ideal υo
ι1 : lOmV ιρ:
i, :
!.9ΞY
Given
=+
:
16
:
ι, for
ι:
>o'
l0 mV
i, : 1 mΑ 0-1 mΑ ιn:0.7V 0.6V
10
μA
o.5
v
0.
Ιn summary
τ,:0 for τ1>0 ιo: υι fol 1]τ=o
Εxι 17.26 Refer ιo Fig(17.34)
For
,a
:
U/
:
+ι v:
D, Ψill conduct and close the negative feedback loop arοund the op amp. U_ : 0, the currenι through R| and D, q,ill be at the op amp output' v,ι
l
:
mA. Thυs the voltage 0.7
V ιγhich ιγill set
Exercise 17-7
ιIζ0-D,on - τ]ο goes the & forms D| off
D, off and no cυrrenι ψiιl floΨ ιhrough R,. Thus
For
ι, :
Ι0 mV
D \νiιl conduct
-
Rr: v1 :0V -v^:0"Ιν
R,& R,to vr. The negaιiνe feedback Ιoop ofthe op amp will ιhus be th.oυgh
closed and a virtual ground Ψill appear aι the inνerting input ι€rminal. D,will be cυιofi The cuπent through R,, R,and D, wiιι be
l9,ζ' _,o IkΩ
will be 0.5
μA.
Thus lhediode. D,.νolιage
V
υ,,: oΙ l0μAX10kΩ: 1]A
=
υD. + z,,
:
0.5 a
o.Ι
:
+0.l
Forvr=_1γ This is similar ιo the case ψhen The current thrουgh R|. R?' D,
7: -L = 1.a 1kΩ .'. ιlr, : 0.7 V 1,o : 0 + 1mA X l0 kΩ : vΑ
no currenι flo',νs ιhrough
= 10+ιa1 :1o.7Υ
vl
0.6
:
v v l0 mv
.
ιγill be
10
Εxι
17.28
V
Exι 17'27
V/> 0 - Equiνalant Circuit
-D,on,D! off
-J
υι}o current floΨs from U, through R,, R,, D, into ιhe outpυι ιerminal of the opamp. vο goes negativ€ and is thus
ofi The following circuit resulιs:
l
Ε +!! 3R R
=
Αs ,, ιhat
goes negaιiνe, the aboνe circuit holds so
ιο = 0.
This occurs as the
15
v
sυpply
sources ιhe cuπenι l eνen for smalΙ negaιive U,
This situation remains the case until
....!Ξ+1:o 3RR 'o: vr
Rz Rr
υιζ 5v
D, offD|
-on
1
:
0
.
Exercise ι7_8
b)
z,
:
1
γ
-similartotheciΙcuiιin(a)buι
with all of the υndergroυnded opamp input termι
,Ι : l v
naιs at
Ι : l/lo kΩ:
0.l mA
υA: l+υD2 =l uo
=
0-1R
(QJ)
l0 V - similar
to (ε)
"\ l/
:1.6v (c)
=ο_Γ]l+1)ρ ι]R Rl =45
υl:
& (b)
- all inpu( ιerminals (noι gΙoundω) of opamps is equal ιo l0 v'
: _ιι_4'3>o :. τD2=o'Vn
note υΛ
ιoρ
ο.l
+ ο.? +
= 7]0+o.7
_l , _ l0 1ο
mΑ - diode νoltaρes = o.7
ν^: νo+Uor=
lo
v
+0.7: l0.7V
d)vr:0.1 V Vs: Vr*V^ = o.l + 0.7 + = 0.63
o.
t ton f0
02)
-\ 1
./
v vA=_12ν
lmA
+ 12=
=0.1V
{o.or.ne
o
Upι
Vs=Vo+
>Dι o
0.1v
ι0
0.lV
=
η
{o.οr..ι.
(e) τ1
/:0.1
+ 0-7 +
for all circuits, cuπents aΙe given in mA, resis_ νoltages in
V
/ \
0.l ιοg-\, / o'o2 \
=0.63V
: -1
v
_
use circuit in (d)
mΑ
: 0.1 mA Ισ: Ι + Ι,:
NB
kΩ &
l
0.l
1,
a)vr:6.1 γ tance in
=
=
Vo:_ψr:1γ
vB _l2ν
o.02 ΙnA
o.2
mA
= 1+o.7+o.ιloc(Ψ)
Εxercise l 7_9
ιr : _ 10 V /:.0mΑ (0
,() :
η:
use
circυit (d)
10V
ζ Ι^=2rι^
:
lo+0.'Ι +o.l loc(?)
v
is a sinusoid of
^D
be
vB = v0+ vD| 10.73
τl, 5
= 1.0mA
:
Ex:17J1 )
The average curΓenι ιhrough ιhe meter
:'t '
cuπenl mυ5ι 1
.
peak' i.e.
1 lo I mA. Thus Ξ \,srt '1τR -
be equal
obιain
υ,
:
maximum
zo
will
To ohlain fuΙI.sca|e leading. This
4.5
=
: 5^Dv.ΑιthisvalueofvA,we
vD| + yM + yD1 + yR
Voι = Voι = 0.7v
νr'here
and
: J-!: x005 : 008 V V,, 4.5
For
ι,
v2
=-
=
ο
,
|ι1| and |z,]
For υ2
=
i.e. v1
'2
Thus |vr|
x _|ι,|
< 0, i.e.
+
|ι'
y.l.," =
,
:
v, = |r,1 _|, _ : vn lηll
ι,
: 0'
=
+ ]ι,|
Thus, ιhe block diagram implemenιs ιhe absolute value operation. Using the circuits of Fig(l7.34 a), wiιh the diωes reνersed, to implement ιhe half'waνe fecιifier' and a \νeighιed summeι results in ιhe cifcuit sho\γn below.
UseR =
Rr:
l0 kΩ
ο.7 + 0.8 + 0.7 +
Similarly we can calcuΙate
νl
:
8.55
ν
_
kΩ yΑ when is at iιs posiιive
mΑ' which Ieads ιo R
V. willbe
Ex:17J0
5-v rms (peak voltage of
:
5J2
:
8.55
v
