Chapter 7-
7.1 ForΙ=10μA:
1
τ
'"
l0
25
ιΑ
=
mν
0.4
*: δΞ*π
=
,:Vn:Ι0V:tvΩ
ο | l0μΑ
A-_ρ-r'_η_
υ '"' o Vr
'"
mΑ/v
l0ν _
v
400
ν/v
ForΙ=lΦμΑ:
n-: lΦμΑ
25 mV
= 4mA/ν
__ιφ- :
rr, =
=
+,=
**
, o : β : lΦ :.rsιο
g. 40 mΑ/ν
:
R1"
r,, : 2.5 kΩ
λν6 : _g.ro : _40 mΑ/V(lΦ
Rr: rr: lΦkΩ
Rιn can be
25 kΩ
.,ιl = 250μΑ,g,: ##=l0mΑ/V
r'τ : __-!!L- : lo kΩ
}ιl
This makes
R;n =
rO :
μΑ
l00
250
kΩ
rzr = Ι0 kΩ
__
40ο kΩ
: 'g r[': -10
Ro: ro: 40o kΩ
mΑ:
1'Α :40mΑ/V
25 mV
:
__-.!!q_
:
.o :
.]9j( =
1 mA
lο
r,,
Α, :
.1o
Vo =
ιο
=
κ) : 4ω
0.4 mΑ/v
μΑ
lmA
4.Ο
10o
mΑ/v
40 mΑ,iν
25ο
25
'ε.V otoll Rι)
lo kΩ \rlο mΑt
ι5 kΩ + lo κ',ι t, .,
(4ω kΩ || lω kΩ) : 533 v/v
π
ιΩ
I Md)
4ω
kο
|0ο kΩ
400
lο kf)
400
2.5 kΩ
kΩ)= 4'0Φ v/v
-(o#*.)*"'',"'.'
Υo
Ι
l0 μΑ
mΑ,lV(400
ν'1τ= V'', R^ -Vo
R.,, * R'" - ζ,*
2.5 kΩ
mA/V (10
kΩ)= _4,0Φ v/V
raisω by a [acιor of 4 by decreasing
Α,,o
n':
lrnιΩ
/. to
rι, : l00lOv = 1Φ kΩ
μΑ
Α, : 4mΑ/V(IΦ kΩ) : 4ω
ForΙ=l
: ω,oπ
,ι) :V^: lΦV:
ι'
tmA
= 25okrt
0.025
I
7.2
.
--l
_ /
7J s,, = lq,so
Υov
2
, - E.Voy
'-\"'_- v)
"-::Ξ!!. _ 2_ mΑ
""',/ v(o.25
lD
"
From chapt.5, k'' : ιι,Co*
since g,, : uσncJπl υΓη'
,0.25 mΑ
chΔpΙe.7_2
: rτQΦ μ^7?'υr/L)(r5o μA'
2 mA/Υ
yielding
ι|/L :
pm)
:
20 μm
11'
/
L)
(l mΑ/γ)'2
A / Υ' 1(2l
(387 μ
:ωμΑ
'5 )
Note that many ansιγers are possible.
7S
V2ρ
'lτμρ.)(Wυ
V^'
:
1.8
V
v'4
Γ'o
is decreased ιo 25
μΑ,
Αois increased bv |
:
g.:
2 tLrtc-l
2
7.4 Αssuming that the MosFET is operating
above η,
ιf /D
'D
_
40(ο.5
λo :
2
40
so that
γ:
c.
z
Juo
β1s'"ρ*yιrlη.f,ιo
so, g,, is decreased by
Jυι : ιlz
Ιf
/,
Th€ edge ofthe satuτatio:n region is defined as
is increased to 40ο
μΑ,
1o 1. 6o."*"6 6r,'JτI : !
2
: 2
g, increases by
.'.
u'-:
Ιn this problem, we haγe two releνant
equations, bυt thΙee unkno',ir'ns. so, one
parameter can be choosen.
Ιf we multiply the equations for ,4, and 8π, we
ι:
γ:
l/λ
(2μ"C'' )'l/
λo'8- :
\' (2ν"c.')
'
since a must be Ξ ο.l8 μm, aδitrafily
choοse l = o.3 μm
τ"hen-
L
Ι
To find /D
=
-
:
g: :2pmc"'(W/L)ΙD
so,
2|.s
-
|Voλ
0.3
= l.5 v
:
0.36 μrη
ff:
7.A.l, V^'
From Tabιe
Io
:
5
*Ψ
ΥΙpm
/ν2)
_
5 V/μm (2) (0.l8 μm)
t0μΑ
:
A _ 2 (5 v/μm) (0.36 μm)
Vo"
o.25 ν
2V. Ι'
o
25(l mΑ/v)
5 ν / prn(2)(387 μA
l.8
h:
22
v. L
aτιd
= 6.46 ιιm
6'46 lΙ''τr.
0.3 μm
:
(0.l8 μιn)
,"" ' L
Γη
Jz<ιι'c*xwη.Γη so that,
:
:
so,
ιr^-ztν^c^ltwιl
Ao' g^
L:2
7.1
7J
|ro.] lv'l
The highest insιanιaπeous ouιpuι voιιage is
Voo_ |voλ
^Γ4
can eiiminate /D aηd
=
when |r,.,]
b) ιvith
1"
:
l0 μΑ
_ 2(loμΑJ _
k":+
v;" ι0.25 v)' 320 μΥlA2
solνing for yov with 1D
vov : ?b.-
= 1Φ μΑ:
k"
vou
: ΨiΓι
320
'tΙ
n_
_
μΑ/ν'
=
o.79 votts
Ιo _ lω μΑ
Vοvl2 o.19 ν 12 =
253
ιAlΥ
l80 kΩ
Cbapter 7-3
V,
L
μA/v( l8 kΩ)
253
c) Now, vr'ith a ne\p
/, :
:
5V/μm (0.36μm)
l00 μΑ
18
kΩ
= 4.56
V/V
w and Vov = 0.25ν,
ρ-
vt.
lΦ μA
2V, L
*r1
l0 μA,
d) /D iS now
(0.25
:
noιγ'ινith 12
so,
V
n-
16(0.4
μm)
0vι
'ι'
Vn'L
_
:
6.4 μm
80O
ιr]
l/0.25
20 v/μm(o.4
μΙn) _
l00 μΑ
ι : 0.8 μm,
Ι/ : 0.8 μπ( 16) :
μA/v
80 kc)
Ιf
uA
ro : Vι'L
Ιn
vr
l2.8
μπ
8Φ μΑ/V
20 V/μm(0.8
l00 μΑ
μm) =
160
7.9
l0 μΑ
'
4
{
_ Ιo _ l0 μΑ _
voνl2 o'o79ι2ν
{
5
v/ιΙm
)a
0.]6
10μA
e) The
loΨestΑo is 4.56
νlhen Von
:
0.79
0.36 μm
The highest Αo is 45.6
witb /D
:
|0
νA'
v/v
ν'
ΙD
:
uΑ/V
25_ι
ιm)
2u^ L
2(5v/μm)(O.]6
' _
^o_
0σq v
ω
: 45.6 V/V
Ιf
=
v)'
=Γ!_9=/Ξ1]!-EΦ-o.079v
k" 1.2 mA/v
Vn L
ro _
-Io
l :
V'
(ΨJ)
k'=+=2(100μ4):3mΑ/V2
v;
l0ο μΑ)
k'
8-:-:Ξ-.--=
fnd k,:
first,
2(
2ο0 μΑ/v'(o.25
l0O
νlν
l4'4
L
v
(2)(5v/μm)(0.36 μm)
ο25 v
OV
:
,,
2
2Ιo '
γ:
_
μm)
5 V/μm(0.36
:;k"(L) V
_
s- _ :9-= - lΦ μΑ
v\
ιιΑ/V
800
ΙD
so' ,r'
l00 μA,
ιo _ l00 μΑ _
Vονι2 o.25 ν 12
7'8
_
μm)
Since ,4^
"
100
μA'
:
ι
o.079
Υ
"
=
-:lVoν
g-
remains
constant, and ro is increased by L)
Each gain is increased hy a facιοr of l0.
Lo''l Ao= 45.6νN
High Αo = 456 V/V
L
:
anaιιhe cυrrent source is
ΑoVon
: ]Ψ!:-YΙ :
2Vo' 2(2o Υ lμm)
since 1o
γ:
(or since
_ 2v'Ι'
--L.
vou
ideal.
is increased
10 ιimes.
since ,4^
kΩ
v/v
Von
W/a is held constant. and
2v'ι'
18ο
L
=
0.5 μm
lι'"c,,'l(!)v-''
2Ιo
-----------_----Τ
(μ"c,,.)v;I
2 (50
μA)
(2ο0 μA/v':)(0.2 v)'z
:
12.5
kΩ
Chapter 7-4
7.11
7-10
Vrr:
Yon:2-5 Υ
+ 1.8
V
Η
Η
+
vi
V6=Vpp-V5p2
v
=vDD _lvιp|_lν.,1
=2.5-0.5-0.3=
since 1Dl =
(W\
2l
\ω"c"'l(f
,v.",
o'
2(100
μΑ)
(2ω μA / v'?)(o.3v)':
_
''.,
'-q"'(9':;#α
(
lω
since
μΑ /
.l,l
l)(o.3)'
,/r, =
|VA
:
ρ1
20 νlμrn
_ 20 v/μm (υ.5 μm)
:r:L
^L
02 o
|00μA
ι
o - Ιn, _ lΦμA _ 667 uA/V
vou 0.,1/2v
ol
2
100
kΩ
so.
v/ith eq. (7.18a),
Au =
:
Lε^,o:
-33.3 V/V
}{οοz
|voλ
0.2 : 1.1 V
'p\
-
0.5
_g^.c'l|
and
.l'"
ol"
/
02
ro2)
so ιve must nnd
_ A" _ _l _Φ
mA/v
9-ι
vΦ
ιD
v^,n
roι :
1; arιd ro":
5 v/ιιm
50Κ
ol 100 μA '
μm
6 V/ιm '
60K
02 lΦ μA
μm
since
: ηoιtι
-
'
so.
V
r, :
_
|v
1o lΦ μA ι mΑ/v
vov/2 o.2ν /2 =
ι':
r
\τ ),= 0"β.,ν",'
-
-
r.8
V
1.7
DD'
ιιzn){lωιο)
1ρ :
4ρ
or
L:
50
50
kΩ/μm(Φ kΩ/μm).L'1
kΩ/μm'L '60 kΩ/μm'L
|.41 μm
since ID
:
ξν"c"'(fiν'",'
ι'^ _ 2Ιo'
\τ), |'β
-f*
2(1Φ'μΑ)
'-
381 μ^/ ν2(o.2 ν)2
similady,
=
12.9
2Ιoι
ι4/\
)z μ"ρC oλ'
^|v
_ 2(l0o-μA) ;:58.ι
(
t-i
ιL
86 μA
z
ν'
(0.2
v)'
Chapter 7-5
This figure belongs with 7.12
o-+
+
vi
ν
Ι
7.12 Ιfboth MosFETs have ιhe same curreΠι:
all r νalνes
source νaΙues, and ]V
o
|ιι,',,| '
The small-signal model shows all
=
_
I
^rv r"r0
=
(iF
/F
")
are equaΙ.
vo = -
vos
+ Rl
:
)
?s
..
=
g.ιV nrQor1l ror\
+
2MΩ)
(b) small-signal model:
:
orll tor)
ι
μΑ(3MΩ
0.475
= 2'38 Υ=2.4 Υ
Rε=3 MΩ
18.zy ε,2ro
Vr,,
=
8.2' vEρ ro
since Ι/r,, : ri,
Α,:2:
A,
7
:
KcL
aι the oυtput node yields
vvv
J+ov
(\'*,.)(_)'^,")
'
ro
l1
siπce V",
λc.'ε-ι 'λ
:
+o
':o
ι1F
Vι
Vν
=
i''-r'+-9_Ξ
-13
/1
o
or
\
_
o_ (η c-J
V
ξ;t
vi
:
2Ι^
V
20ν '
:Yu:
200
ΙD
μA
0.89
mΑ/v
l00 kΩ
so,
(a) Ιf we negIect the current throυgh RF,
Ιo = 2Φ ιιA =
\ι,ιwl
V:
VoS
k|(w
:
Vι +
/
L)
Vo,
:
=
v;l,l
ιlξ"
E2Ψ.l*Δ..)
{
/o _
2,ne/v'
0.5 + 0.45
:
:
0.95
ο-ts
V
The cυrrenι ιhrough the feedback network is
:
"
'-
v:
RΙ
=
991J
2MΩ
Τhis is << 200
jυstified.
μA.
:
s.a75
,,4
so ιhis assumpιion
v
30ΦK
.l0oK
__86.] v./v
3ω0K
To find ιhe peak ofthe maximum sine\νave output
possib|e. we noιe ιhaι lhe currenι source iS
assumed ιo be ideal. Therefore, sinewave ampliιude ιγill be ιimiιed by !he negaιiνe eΙcursion.
Since ιhis happens Ψhen
Vo,
: Von:
0.45
V,
the maximυm peak amplifude
is
2.4
-
0.45
=
r.95
will be
V
(Thaι is, ιhe oυtput \νill vary between ο.45v and
2.4 + 1.95
:
4.35 V.)
chapιer 7_6
ftlc^
π*& Φ;.-τ;
to'ro
L-
:
R'n
L-
mΩ 3 mΩ !Ξ-!'ηJ
2
l
- ο.q9-ηΔ]J
* ]
.Ω i+h
-
33.9 kΩ
7.14 Refer to Fig. 7.4a and assυme Qz Φd Q'l
are matched:
,,
-':r+:
'
'neaι
lΑrl
(c) To find
Ri",
\rye
i" : 25o μA /Υ2 ,
V/":|Vλρ|:loν
i, :
The coπesponding input vollιge is
_
-]lΞ_Y86. ι v / v
23 mvΙ'eaι
apply a test voltage ,/r to the
inpυt
-
Roυi
since
:
1"'.
z.s
lωkΩ
u'.,= #,u #
- 200+/n' = /j:
= r,,
/", ι,=ΨΙoI
O.O5
mA
- 1r, :
0.05 mΑ
η.7.l8:
Av: -g^(roι || ro")+4o: _8.XlωK
+g-' :0.4m4/V
Kcι
,I4O
'R'
KCL
at node
A:
ν v -v
=(Υ),:
RF
at node B:
(Υ),:
_ν v
vxοο
R.ro'x
c.ι
,,(*-,.)
οl1
ro
\F Q2
uo
2Jq
Δrιd
|vovr|:
Rr
subsιituting inιo the frst φuation,
ι'|
\rre
geι
\
v, lη-ε'J
R, R- R.'ι;*R.,
/1 ι\
. V"'V'
'r
:
so that
l",:
=
I
-^.' * g.lRr
lrl
Rr RF l/ro+l/RF | + I
ro Rr
= v.r'
ρ3 have
=
Y
ξi,(Y)'v''"=( r),
(Υ)'=
"
and Q3 are matched)
0.05
ο.4
0.25ν
x2
(9,:
x
2
the same
0.05
;o , ro-' , or5''
(Q2
Vx
0.42
2x250xl0_3X0.ο5
l6
:
0.25
V
as O,. then
l
chaρte. 7_?
7.15 As discussed in Example 7.3, the ιransier
characteristic of the amplifier over the desi.ed
region (segment ιl[) is quite lineaι Therefore
Dc bias component of ιhe inpuι signal (for
maximum oυtput swing) should be chosen at the
ιhe
ylr
midpoint bet\γeen
and
y|,
ιhaι is:
:
o.88 + ο.93
0.905
v
Ιo,
7.16 Αs discussed in Example 7.3, the transfer
characιerisιic of ιhe amplifier oνer ιhe regiοn
labeled as segmenι ΙΙI, iS qυiιe linear.
4.47
v
Now to Find ιhe linear equation for segment ΙΙΙ'
iD1 : iD2
:
'''t(' -
p'.(l),ι'"'
ξ)
[*'(l),ω - l4r'(r
=
_
*
= zωιv, o.of(r
= 65 x 0.53'?
1rr,
"(t
65 X ο.5]'
1.3(ι,
_
ff)
:
_ 0.6;2
_ v'z lο
ι +!ρ
o'l|7
=l
ι6 : 8.51
Ιf we subsιitute
vre = 0.86
v
Y/,
:
62.51(ι1_ 0.6)'
= 4.47
= 0.6
+
_
ι}ou.> Voι
= 0.96 V
62.57
0.36
-
Therefore the linear region is
:
A I Uor
of poinι
=
vous
["'
..2
voνι='6:lr'rφ
= 0.1 V
25
2
l'
)''ι'ι.ι
= lVov.,l =
I
0.32
v
Vρ6=VΒ_V'
: iD2=)
=
=
o.36ν
$
_ 08)2
0.8|'z
:0.l1(Ι
(y,
=
","t(t. #.)
-rfff)
*',({),'ι"",(ι
(z,
Υ'then
1:
8'57
voo
υq
Vιs
Vo"
yos : coordinales
= so x
Tο determine coordinates of B. note ιhat
V"= V gg oτ Vη _ 0.6 = Vg'
substituιe in
and
{v1
for νoa
μι+
zs
By referτing to fig. 7.4d' you notice ιhat in seg_
menι ΙΙΙ, both Qt and Q2 are in sanΙration and the
ιransfer characteristic is qυiιe ιineaι The output
νoltage in ιhis segment is limited beι\reen yoΑ
rcox2iψI_
ι +!ρ
|+ττ';Ξ|
pA
o:(fl,t,
_ |_υo/|5
| _ o.o67 ι^
25
μl $ :
b)
Dι
20
=
l00
Now we find th€ ιransfe. equation for ιhe linear
section : (Rafer to Example 7.3)
yoy1)
(Note that yov2
i
20
0'6\1
:
At ρoinι B :
.'-t#)
1'5
W2
voA = 3'3 _ o.32 =2.98ν
."t:)
1r,;1'[r
__Ξ9-"(,
b?:)
*
:
Ιa99
:
L)2
(W /
=1,1
0.53
ΙD|
(Note that YsG2 = Vscr)
ouιput sinusoid amplitυde: l.4 V2
wecan ινrite
:
= l00 μΑ
_(W/L)3 _ W3
ΙD1
2.47 +0.33
Voι : Voo Vovι : 5
= l00 μA
a) ΙD2
The corresponding amplitude of the resυltiηg
output sinusoid is:
output sinusoid amplitude
=vootvou
22
7.17 Refer to fi9. 7.4a.
Νote that o,' 0, a.e ηat matched:
121
input dc bias
=Vr"ΙV,"
22 -
ο.86v<y,<0.96v or
0.36 ν
= vo= 4.47 ν
ο
εl'(r
x ο.:z'(r
:
+
+
&)
3'3*/")
0.127(ι.066 _
= 6.11(ι
ι
-
#)
/
(' r :"-)
0'0l9v,1
l+0.0lro,,
_ ο.03yo)
- 0.8)2 =0.l1(l -
No\ν if we solνe for
vbB+ o.oo33voι-
0.O3yo) Θ
yoB=vιR-o.8
o.ll : o+vo,
=
0.33
Thereforc the extreme νaiues of Vρ for which Q,
and 02 aιe in saturation 0.33
v = vo =
2'98
ν
v
Chapter 7-8
c) From (b) \γe can
vι
vΙB= voB +
Ιf we solve
1
fndvIA= vIB:
(Vu o.8)2: o.ll(1'0.03
= 1.16
7.18
= 0.33 +0.8 = 1.13 V
for yon = 2.98 v then '
+1.8v
Χ 2.98)+
V11
V
Large_signal νolιage gain
=
Διo : 2.98
_
Δo'
:
Φ
Δu,
l.13
- ιεq-,
=
"22
41 u^
0'33
1.1 16
vrv
!!!: Ξ =
v
t.65
"
Differenιiatinρ both sid"" nf ι ,
2(η
ο.8) =
:
- l?
dν,
0.ll
X
u"ol
δυ'l
=
(a) Ιf G and
either gate,
@
6.g
we have
or'(vc-(|'5v)_o.5η2=
(l.sv_vG_0.2ν)2
ιV- * l\2 = Ιl _ V^\2 +V^
m
g-r(ro, ll
'nr\
:
_210'6ν
o
t
becomes:
: ,,J
E*'oono ''?9r'ωr,o
l
:
: Ιoι = .ll v')to + )'
}{t
: 0.5 mA
(b) For r, : Φ, the small_signal model
Ιoι
: Υa: 50 : 500 kΩ
Io, 0.1 m
: 500ι || 1Mο
ΞR.*
.
R".t : 333 kΩ
A,':
V,)"
so,
"'
πA]Υ
_ V" _
1,
= ID2: :kρ\wtL)2lvDD_ve _ |v,|lι
'k'^(|yιL)'(ve
:
,^'
= o.632
D are open, and no currenι floψs to
ΙD|:
l00 _lMΩ
0.l
v
1
161ll 162
,^':y':
'' Ιo'
1.5
V2=V6 uιd Ιη= Io,
dυι
: 0.1ι(ι 0.03 X l.65)Ξri
: l. ι23 v
: _ 195.8 v/ ν
(v,_0.8)2
e'1 Ro,,
\avi'l
(_o.o3)Ψ
-6g6.11,,,
For zo = 1.65 V from
-
β)
D
^
+
u:
/ν
+
Vo:
V,-2(g,Vr"\R
: vi_2 g.Rνi
lo:Ψu:'!_2g^R
vo
l_
Chapter 7-9
substituting values,
Aν =
ι"l
1
- 2(l mA/v)(l MΩ) :
,.:Ι#
_1999
ν/v
20v :40kΩ
Substituting in numeΓical νalυes'
lMΩ
^"'
0.5 mΑ
Adding .oι and ro2
24.9
, tl_2(lmA/ν)(lΝ4Ω) l
'_liΞl]Χ l mΑ /vχ40 k(η
Κ
|
:25 kΩ
to the model, we get
(d)
Ifιhe
gate is driνen as shown:
ιi
G
l0ο ΚΩ
Γ*
ro
Rln Vi : Vμ
+
Vo
2
s
KCL
at D yields,
=
Ψ
2
8.V",+
lL
aιd since
Vo
R,n .tr
y.l"= lω kΩ + R,"
25 kΩ
.(
_
lοο kΩ + 25 kΩ
*'r.u,=Ψ-'+
λ,,:υ!:R
!
'ι"
- -"':
υ, L+2
R
so ιhat
|
ro
- 2s-R
1+2!
(e) |vr.| must be > |Voι'l1
:
:
with Yc
.'.lvoιλ
0,V65, = 1.5 V, V562
1.5 _ ο.5 : 1.0v
To Rnd Ri" , note that
7.|g
.
,Rl
a)
since
:
ιRF'F
:
Ιc1
: 3;#!
0.1 mA
Ιr' :
Ιcz: Ι :
Vo=
V
1-0.7
ι : -Ξ]-
.+
','u,λuo
1.5
a 1.5 v supplies'
_0.5V<νr<05V
I*nn
o =v,
:
-?.84 ν/ν
ciνenthe
Substituting numbers, we get:
. _ l_2(lmΑ/v)(Ι0οοkΩ)
κ'
'' -' ,οω
40 ι.Ω
: -39.2 Υ /ν
l_
39.2 v/ν
,,[ξl
5Ι
r'
0.5
mAΞ' :
3V
0.5 mA
3V
,1, f'-'*'^ll
ι ,-*,,]
t
to = --------
n-
so that,
o :v,=
,
-fr- rr.rl
t' '*'^,1
rO
b)
lVol
-
5OV .+ ro,
.rr: #:
:lvl:so=
/ 0.5
lωkΩ
100 kf)
Chapter 7-10
Toιal resistance aι ιhe colιector of 01 is
equal
r,",
r,",
rol
ιo
:
:
||
rrr,
kΩ
1σ)
||
thus:
lω kfi :
r-:
"
50 kct
50 kΩ
_ 0.5 _
cl ρ_' = Ι!!
vτ 0.025 2οmA/v
β
,_,
" =
=
E
20
c: r,, :
d) R'"
=
z.sιο
2.5
kΩ
,,
Iv'|ι
liJllol
=
μA(2) _
(0.5 ιnΑ /
v) -
mΑ/v
40kΩ
.o,
: Φ
lι-ξ0'05 mΑ)
5 v/μm
for Q2ιnd Q3'
162: lΦkΩ || lω kΩ = s0ιΩ
g.βo: _20x50 = l0ω v/v
0_5
ν
0.2
_ιοv/v
(ll2)
forQl. L,
Ro = roι ll
Aν:
Ιo, :50
'n_.
''' Vor,'
Lι : Lι :
aο k-Q(0'05
6ν
since we φant aI
=
:
o.4 μm
μΑ) _ ..JJ μm
/ μ'm
η =η
and a b€
an integer
multiple of 0.18 μm, we choose
7.?Λ
a=
Vρρ: Ι.8 V
3(0.18
μm):0.54
(Note: choosing 0.36
than l0V/V.)
μτn
μm resυlιs in slightΙy less
checking,
vι
,^' = -Δ!- _ 5V/μm (0.54 μm)
"'
ro2
:
Ιo
0.05 mA
For an ouΦut of 1.6
Vror.rn:
|yov|
V'o' :
ο.2
since 1o2
= Ιoι
and
ιv'Ι
\Ll,
_
For Q
: Ιoι :
: rιvj :
ιL/3
50
Ieast l0 v / vi
uιd Aγ
c^'σo'\l roz)
:
_
Ιf we want to make
So'
_
.
roι and ro2 equal,
)8.ιro
Α"
:
'o _ν2g.-,
kΩ
aΛd r62hadbeen eqυal, this ινould have
ι
at : 0.8 μm
Lι = 0.67 ιιτn
and ιv', makiΠg the area 4 times
/ν'
The closest integer multiple that saιisfies our
requiremenι is (0.18 μm)(5)
1,
Αy must be at
rα
For a gain of _20 ν
_ ,r.'
2(50 μA)
(387 μA / v'])(o.2 v)'1
64.8
g.,(161 || roz)
meaηt doubling
|μoCo,llVo12
μΑ)
(86 μΑ / v'?)(0.2 vf
_
greateι
2Ιo'
6.46
u'
: ο.5 mΑ/v(54 Kl164.8 K)
= 'l4.7ν/ν
Ιf ιhe gain
.L is to be douhled. and ιhe ζ raιios kepι
the same, rot ll r, must double.
lf
μA'
lL\v;ν
2(50
l'|
,l :
Λι
: 0'2ν'
_ l'6
v
)(μoC.,)(νv
(ιv'\ :
\L
l.8
l
Ιo =
Ι
Y
:
,o
/ ιm (0.54 um)
__nτ;Αro\ = 6V
=
Αγ:
-
:
0.9 μm.
so,ιvithι|=η=η,
(0.9 μrn) _
,^, : 5 V/μm
,n
"'
:
r-,
"'
0-05 mA
6 v / μm (0.9
0.05 mΑ
This resuιts in a gain of
μm) = l33 kΩ
: _φ.5 τnΑ / v)(90 kΩ
Αv: _26.8Υ/Υ
Αv
This represenιs
(#*Ι :
aΙΙ
*,-,
|| l33 kΩ)
increase in area of
2.78 (insιeadof 4)
Chapter 7-l I
(b) Foι / = 0.5 mA,
Vλ
7.2|Κ=40:gη7roz:
(o),:(9.:
ι/
ov12
so thaι
''u- ΚV""
4o(o'2Λ
=
R^
-
2
lfVo=5Y1"-,
--_^^r*th
Δ4η'
mΑl
|0.2 v|(0.5
:
:
250
]20 kΩ
7 -23
4v
:0.8,"m
1.=v^:
γ'λ 5Υ/μm
Voo
fαIι'
1
V* c*---{
7.22 γy'iιh both transistors being identical,
Yov, = Vovn = Vo, and
Ro =
"o
Vω4
Ξ''
lvnl
|Voλ
rL
2
Ro:G^ι
lvnl
t
'|Vλ
=Ψ
|voΛ ι
no
Vt
n)roι
(g^1r
Ε-ιroι : Αol :
,
'o4-
V1,: Vtn:
so
ιhat,
J
,l]?o
:
2|Vλ'
|v oλ
v6nr: v6uo:
v6v
La:\=L
V,=
V
'= γ
'
vDD
Vo.ι
o-''_l
VcιΗ
Now, if |t'r|
(a) for / :
ll
Ιρ :
i
=4v
w\
ιΖ /,,
t
\τ'l/\
)'
'
2(
and
Qι
sο that
Qι
/
ι_
=
(s-. r",)r".
|Voλ =
0.2
ν
/R,'
Finaf ly,
π"*,
"
/R^
2lo
ι Ι't
|γ
,
oΔ
: 5Υ/,,π'|v6/: oιΥ
|r,,|
_ 2(5 ν/μm)2(o.t8 μm)'7= 8.l ν
ιο.z ν)
R^:ιl9=
*"C^fo-"
lο0 μA)
2|V,|'
2l4 η'
" ' 1voλι= |0.2v |( l00 μΑ)
r r2
7lv
= 1:!
ForL:0.18μm:
L\v'6y
s.
loο μΑ / v'?(0.2 v)'?
n^
'|ι lγ|_!J_'|ι
'
|Voλ ι
z|v
_]_!-L
"_-
ρ^
&
= Αoι =
lγ ]ι'
: l-!-
v6o
0.l mΑ,
)ιμoC"'\ιW
ι
,
g.lroι
= l.6 MΩ
=
li
8'l
y
0-οι mΑ
0.0l mΑ
v.λπ_ ,'r''
2lyL:2(0.18
= 8ι0kΩ
''*'
= 0.l mAlV
:
0.065 n
Αssuming thaι ιhe driving NMos transisιors have
similar g, and Ro,
Α
ι,
=
μm)(0.18 μm)n
I
;c..Ro
Av- _'lo'l mA/v)(810κ)=
I
40.5
v/v
Chapaer
ForΙ
:
0'36
μm:
2(5 v/μm)'z(0.36 μm))
/R^ =
.
32.4
η
32.4ν
:
R^_
" ο_οΙ mA _].24okΩ
Φ2
v
υnchanged
_
..l,'2'ι _lιo.r rnA'/ ν'(3.240 Κ) = t62vlν
8.
remains
Αrca
For
=2
L-
LΨ = (0.36 μτn)'n(2)
0.54
:
0.26
μm:
1i^ _ 2(5 v/μm)2(0.54 μm)2 _
72o
-'-"
(o2
_ 7z'q
R^
" ι)
0-ο
l
ν Α-
,'.'
\η
7.29ο
nμm2
ν
kΩ
"' -
;
(lon)
(o.l mΑ)(2)
= lmA/V
(0.2 ν)
_ 8.| v _8!kΩ
R^:Ιl"
"
Ι
0-! mΑ
' ιnΑ,iν)(8Ι K) _ 40.5v/v
Α'' _ _1(|
L
ρ -
wl
νιillbJ'ιeπtimes larger
Areι : 2wL :
2(lo n)(0.18
μm)2
nμm2
ForL:0.36μm:
p _ 32'4 ν _ ι'1 ιο
o.ι mΑ
l''. : ! ιι
" 2 ' mA/v)(324 K) = 162Υ /Υ
Arca : 2 wL : 2(lo n)(ο.36 μm)'
: 2.59 nμmz
For L : 0.54 μm:
'12'g ν
_ 729 kΩ
R^ =
0.65
_lιr
a"" :
:
Αrea
No\λ/.
v/v
_364.5
(2')(lo n)(O.54 μm)'
l.o mA.
o.18 μm:
tοon
:
ρ-
:
= 5.8nμm2
for / :
Ψ=
:
wL
2
ι :
Fo,
roι/ν)(72g K)
2
Ι mΑ(2)
v)
(0.2
=
1o
mΑ/V
n"
Υ = ε.l ιο
" - 9'l
|mΑ
,η, _ _lι ιo mA / ν)(8.l
2
l (o.Ι mΑ/vl(7.290 kΩl
2'
: _364'5 νN
Αr€a : 2(0.54 n)(o.54) : 0.58n μm2
NoΨ, υse 1 : 0.1 mΑ:
a = 0.lE μm
|
'
sinc€ /D _ kP (w / LΙιro,,'
Α
7-12
V/V
: 2 wL : 2(lω
k)
_u1ο'5
=
Aτea
nX0.18 μ,ιn)2
= 6'5 n μm2
L=
For
:
R^
"
0.36 μm:
ν:
32'4
32-4
lmA
kΩ
A, : _lιlο ""'
Κ) - _Ιo2
2"" mA/ν)ι32'4
Area : 2wL = 2(Ι00 n)(0'36 μm)'?
v/v
=26nμm2
L:
For
o.54 μm:
_
.12'9k!ι
" ='72.9ν
lmΑ
l, _ _lιIο
mA/v)(72.9
2
R^
= _364.5 Υ
Area
:
2
Κ)
/Υ
wL :
2(1ω nxo.5 μm),
= 58 n μm2
The tabie below summa'izes the caιculations.
Comments:
(a)
Rρ
and
Αy
aΙe
increasω but εt the cost of
largerdeνice a'ea' Αs
ι
Av
by a factor of x2. Τhe
and
Ro
increas€
increases by a factor
ofx'
deνice area increases at this same rate.
(b)
1
lΙl'
3.
increas€s
with |Ι|, but Ro decreases \γith
Chapter 7-13
Table is for 7.23
L : L.', =
1R,
8^
Ι= 0.0l mA
ι
WlL=
1=
0.0l mΑ
WΙ'= lo
:
6.13
3.1
L : 2L.i" :
*.
γ
ΙRo
Ro
2WL
kΩ
μm2
8-
:
0.0l mΑ
WlΙ'= lff| ι
u
32.4
2WL
kΩ
μm2
o.t
810
-40.5
0.0,65 n
0.1
3,240
l_ο
8t
'ω.5
ο.65 n
t.0
324
!0.0
8.1
-44.5
6-5 n
lo.ο
32.4
g.
Vnu
, _= 8-ιVoν
ιo
-i:
Ro
:
_l
mA/ν(0.2 v)
_
2.6 n
t.0
729
-
t62
26n
lο.o
12.9
_3Φ'
l0Φ μΑ / V)'?
2(40Ο μΑ / v')( lω μΑ)
ifιve make 8^1 : 8-2 = 8.
and roι = ro? = ro' \νe can say
4ω kΩ = 1 mA/Υ.r2o
since
1
:
vi
: ,r.,
:
=
vov:
o.2ν.
vιη + voν + vov
+ο.2+0.2 =
0.9
V
minimum oυιput νolιage will be
2Vou = 0'4Υ
_
P-'
_ -,Ψ]tΑ,=
+
Vnu \o'25ν)/2
=
8ωμΑ/ν
Since all devices have the same yΑ and ID,
20 kΩ
kΩ) :
= lωμ-4(20
5 V/μm
58n
5
2
Γoι:fo2:fo!:foι
ν:l'
rο : _j_.
Ιo'o
5.8 n
.164, 5
For mΦ(imum negative excursion at the output,
we want the MosFETs ιo be biased so that each
= 0.5
l00 μΑ
(g^zroι)roι
,1 : lφ-!q = .^ :
" ImA/V
μm2
_.164, 5 O-58 n
162
-
Ηoνreνer.
that
2wι'
7,290
.'. set yc2
"o.
y
kΩ
tmnsisιor can rcach vD|
?b.
12'9
0.1
haνe been
ιrea,
0.54 μm
Ro
8-
(
increased, but aι ιhe expense of increased deνice
7.24 n-, =
:
:
o.26 π
-162
(c) Smallest area = 0.065 n μm2
Largest aΙea = 58 n μm2 Gain and
3L^r"
ΙRo
Ro
n
1=
L=
0.36 μm
:
0.4 μm
R-
lyol
/,
: av
=
0.1 mA
aοιο
= s. ,L = (o.8 mA / v)(4o
: 1.28 MΩ
kο,
n. : ε^ lo : l.28 Mο
Ro : R- ll R"n : 640 kΩ
Av : _ 8^ιRo
: _8Φ μΑ/V (640 kΩ)
: 5l2 Υ /Υ
7.26 Siπce A
y=
_g.' fo
Au _
R^_
" _
ε-'
8η
w:
L
: Jττq-(w/υ.Γη
2
g^
2μ'C^Ιa
ltall
so that
_2
have lhe same
_
(Refer ιo Figure 7. ι l )
20ο _ !ωk(!
fi^/ν
/ρand luo. and
iιo = R,,, || R"Ι,, and
8.ι=8.z=8.ι:8.ι=8.'
since
.
chaρteι
l2
ω'la:
k^*lll
Ro =
solving for ro, we get
"y'
' ελvoλ
'2-2
q--!:
2mA/ν(o'2ν)
:0.2mΑ = 2ΦμA
n-._
roΙ
ε^: ιΙΞμ'c-xwlυ.Γη
Since
(f),
ιγ\
/
sΙ
(4ω μA
2
Similarly,
Ξ
ν)'
=25
/ v'X2ω μA)
(2 mΑ
then
(2mA/Υ)1
2 (1ω μΑ / v'?)(2oo μΑ)
For
(ω μA / ν'))(0.2 v )2
vea' vea: vDD \vφ| _
^i"
:
= |voλ
ι
.
_
:
_
\V,o|
-
|Voλ
Fτom Fig.
fol
-
_
= 3.l
oλ = 3.l _
|vou|
7.l0, R6
Γoι
:
=,o _
_
:
4v
if Ψ
iS
L' is divided
)"
bν 4-
multipliω by 4, or φuivaΙently
: μ'C.'YLL*
Thus
8, for
Ψ"Ψ
Voν Iο
1_Yit.τι,..iε
Voν
is multipliω by 4, and Vρy is halνed, then Αρ is
0.8
16)
η.
vov
7.27 and
Auo: _AL: k-rδz
G.- 8'ι : 8. (same as circυit (a))
Νoιe that for the transisιor in (c), the 8,' and ro
are the same as those in ciΓcuit (a). Ιn summary
for circuit(b), yolis doubled,8. is haIνedΑ, is
v
doubιω.
_ o.2
3.3 _ 0.4
(g^1
=
:21
(b) Each transistor in circuit (c) has the same yoy
V
2.9
V
161
lv^l
sv
fr = δj;τ
= 50kΩ
mA 1mΑ/V
o.2Υ/2 =
0.1
1.
7.22:
|v oν1
o'2y
3.3 _ 0.2
|v
Ι,ψ =\_!ι
vλu" (η)
as the one in circυit (a). Referring to
The highest allowable oυtput will be
Voo
Forsame
(b) Each transistoι is circuit (c) has ιhe same
81.'l
. Yo3 must be
V oo
/ψ\
doubled for circuit(b).
Ιn summary' forcircuit (b), yov is doubled' gm is
so, YDa would be
voo _ |voλ
lι"Y, *"
Ao: 8-ro _
3.3v'0.8Υ o2Υ:23Υ
For maximum swing,
ro2
haΙνed, Αρ is doubled.
2(lωμΑ)
=
:
1
circuit (b) is halfofthe one for circuit(a).
4u
μ" C"'
v sD
+ (g^z + B^z)rozlrot
Vρy is doυbΙed 8^
=lΦ
ι - \ν"c.'(l)vλ'
Ιon'
-
tl
roι _
ιιen.L V"- g.rrorro, g
For same
w_
L
ro2 +
\L
7.27 RefeΓ to Fig. 7.1ο.
_
to,
rol
/
(y),: (fΙ
Ιo;
lLrx
R*,
7.29 a) I
\τ)'- 2 μ"c-h
'
ro, --
MΩ
2.5
Ιf \νe use equation 6. l 27 to approximate
I
'
: i"x
v
Vx
ro :
Since
k)' :
(1 mΑ)(5o
7.8 V,
,^=F=/ΣιmE)=rοιrl
" Nε.
2mAlv
_
:
Ro
ro
18^
7'l4
as the on; in
'Ι.22:
Avo =
circuit (a). Referring to Eq. 7.27 aηd
,A?,:
G' : 8-ι
= 8.
k^r.)z
(Same as
circuit (a))
Note ιhat for the transistoι in (c), the g. and ro
aΙe ιhe same as those is circuit (a).
Thus, the inιΙinsic gain for circuit (c),Α,, = _Α3
ιγhere Αo is the intrinsic gain for circuit (a).
chapιer 7_ι5
Ιn general' circuit (c) has a higher output resistance' and for the same yoy of ιransistors it has
lower ouιput swing. The output swing is limited
ιo 2 yoy on the low side for circuiιs (b) and (c)'
buι limited to only yov in circuit (a)
1.fr
Por Q,,
Vov: V,_ η,:0-8
ο.5 = 0.3
v
since all ιransisιors are identical. and
kn,=ko2:kρ:kρι
wiιh IDΙ = Ιρ, : Ιo:' :
ly.,l
v
o..r
(since
wiιh yG2 and yca nxed,
Vs| = Vc2
:
V51
-
1.2
/n _ ! ιιv"J'.l
2 ""
Vcs2
0.5
-
Ιoo,
0.3
:
0.4
Vn + Vov2 = 0.4
Yo is
The highest
Vr.,_ Vou.
:
2.1
so ιhe ouιput range
:
0.7
V
'ο.3 :
1.8
v
+
0.3
is 0.7
v ιo
Vρρ:
V6a:
|.7
Υ
:
1.3
V
V63
2.5
R.
TTiTn - s-,"''l s^
: i R1 and
V'i"
η,"
.
'_ R*-R- --;--Φ
,,ts 1+g_ro
(b) V6
ηic,
muΙtiplying and diνiding by
voRι-R.
V.* n.," , Jι]_j,
c)Ιfi. :
R"iε
:
'
I
ro
lK,
:
+
gnro
R.,"
l0 kΩ, Aρ
,. :
bro
:
:
we g€t
l Rι
8.ro
+
l
8.
20'
ft : u.nzv
]
R'" -10kΩ+
2o 2mA/ν = 1kΩ
Vo
R'
lOkΩ :5V/V
= Rsig+Rii: lkΩ+lkf}
ζi*
V
: V63* V653
= 1.3+0.5+0.3:2.1 V
The lo\νesι yo is
RιΙro
o _
-
^^
1.8
732
v
Υ
o--l
o--l
vo
Veι= Ι'2Υ
o--l
ιfl
is shoned ιo ground, ιhe currenι floΨing
ιhroυgh ιhe short is
a)
/'ω
.
'
α'l
R*+R. -
:0.8v
ξi"
i--
8^
+
ι
vo
^iι
"' V",e
R.," +
'
1
8-
From Fig. 713,
Ro = ro + R.ie + (8. ro)R.iε
b) lf ro + l0 kΩ. and
a) modifying eq. (7.34)' Ψe haνe
n-:b:-?9_:2.a7γ.
16 l0 kll
Chapter 7-16
^l!
"'' +1
g-
=
mΑ/v
l kΩ
0.67
Vovι = Vovι
lιο+ _LΙ
2 mA
n.,^
Ro + l0 kΩ +
+ (20Χ1
kΩ) =
-11
1ρ
Using the new model,
vo
: G.v"is?o
v^
-
= G-(Ro
= δ.u,
.,oru
ι| RL)
:
0'52
:
ν
gfi: 8.a:
#o' uνr=769
Ro^ ::
(8^ι'o)roι
mΑ/vx9 kcr)'
: (96ro.\rea
= (1.54
|| Rι')
(3l kΩ |ι
10
kΩ ) =
733
5.o4
v/v
Αssuming a cascode ampιifieΙ with cascoded current source. ιhe circuiι will be similar to
the one in Fig- 7.11.
voo
Ro"
:
= (0.769 mΑ/v)(9 kΩ)'
R, :
:
R"p
62.3
The valυe
R^ ll
μA/Υ
125 kΩ
:
62.3
kο
ll R""
kΩ l| ι25 kr' : 4l.6 kΩ
of R, ι| R. needed is
Aν _
s-ι
R,:
_lΦ
l.54 mΛ / ν
_*.o
kΩ
This is gτeateΙ than Ro!
Vcι
This can't be done \μith the present design.
One thing we could do is double cascode the
cuπent source to raise Rop:
Vcι
*voo
Vcz
vcs
vG4
o--{
Rop
rr
lv
l:
vcz
v'.
-
v62
2Φ μA
(w/L)
:
0.26 / 2
λ
o' L
Ro,
This rais€s Roρ to
Roo
2
o-J
vi
u: )ι"(l)v',
2Φ μA
o,Γ-*"'
+
ro:9kΩ
1or
vou
L_l
o-l
ΙD
(5 v / μmXο.36 μm)
μΑ/y,(;*)
Qι
avo
VL
tol=fοz:to3:ro
o-l
0'26
_
ν
,.ro
:
(g6r9)(g..rρ)ra5
: (Θ.769 mA/v)'?(9 kΩ)3:43l kΩ
we caπ no,,v find an R, that wiII allow a gain of
lω v/v:
since
43l kΩ |l l25 kΩ
:
96.9 kΩ
Rι(96'9) :
5"u;nn
65 kΩ.Ψe
- R, + (96.9)
Rι :
197 kΩ
To fiηd the gain of the
nr, =
ρet
cs
amplifieη Ψe calculate
(R* ll Rr)(g.rro)ll ro,
Ro,
:
chapteι ?_l7
:
Rr,
kΩ|]
[(431
(9 kΩ)] ]| (9 kΩ)
Rρ,:9kΩ
so, ΑvI :
=734
mA/v)
197 kΩ)(1.5
:
8.lRol
(1.54 mA/v)(9 kο)
V/V
13.9
a) Referring to both Figs. 7,l
R' =
From Fiρ 7.l1.
l
and p7.34,
ro+!+ro
8^
+ 8rro)
1.^
_+tro
also.
iιro
"_iΞn_
'o ''ι
V,g.r6 _
l_
rol:-+ro
8.
--
!_
8n
c)
:
Vι
V,g.rg
-!l
8.
v,P-rn
|s= ι1 :ι\:
Ιι:
+ 2r1
vtg-oo|| Rι)
t
- v,s-1,n1(\8-+ ,n\
,o+!+ro
'
8.
,,
V
ι
+
ι8.ro
--!ρ_+ro
I
8.
vιζ''{o
=
= !v'g.,o
| +l+Ι
v,
:
Snro
v,g,l{s
= jl gnrot
I
.2
v'() Ξ
-+ι
)v]s.lo
Ψiιh ηn""ι = 5 mv,
: -}(5 mv)(2o): -50
_ξ6
mV)(2o)'1=
_l
γ"^, : _\(5 mv)(2o) : _50
u't'(L *'o)
i,R.
,o*R,
_ i(l
d)
8-ro
mVr"uι
Vn*1
vdι),' _!rv,ιe.ro)
Vtg.
By current diνision'
ν
-\v,g^,o
vrn"^r:
8^
.
'
,'
Vιq^ro
ν74 = _!v'1g^r.''x
7,
:
l^t^
-tzrο
8-
v,r""*
ρ_r1
! +s:'o
n,'- 88-ro
R,= I +.,
b) i1
_V,8^ro
',
^,'")
ll
(s.ll
r,'?,)l
+ 2ro
mVpeaι
Chapter 7-18
735
Ro:
RS+ro+gnroRs
= (l *
g^rρ)R' * ro
V
Α'": # = _G-Ro
vi
Α"o -
:'' +: :8' |: ro 'l,l l Lhro'Rs ι rol
ι|
8n.o|RS+
g^ro: _λo
Α"o =
ι.νe starι
\rith the smaιι_signaΙ modeι:
736
t
Jr.
o_Ξ
+
vi
l"
Ξf-o
IΞ
*^.
From eq. (7.38),
R6: Rt*ro*g.roRt
: Rs(l * g^ro\ + ro
o-v^
io: s.vgr+;
Since
Yr, :
V;
-
io :
YS,
v^
i^:
ρ (V'_V
ν "m'
ι s'ι__!ro
Vs : i6Rs substituting,
_ ιoR|). i9L
io:
8^(Vi
;,(t +g,n,+
io
^
:
.Ro
fo
χ): '",
8^
ro
2
kΩ(l * 1Ξ41zo kΩ))
102
=
'^.("*
_""'(#τ)
,{,'"
"'
'__ι'"*
.=
- _4olo2 kΩl90+kΩ
I0ο kΩ
?J7since
L:
L
kf,
kO
o,o: +,
=
+ 2ο
1D
:
|ι',c*l(flιu-f
-l9.8 ν/ν
'
2Ιo
ι'ρc",'|v
oλ'z
this can also be expressed as
(
+
L:so
L
u:
(for all transistors)
o--
Ι
n
loo μΑ / v'?)(o.2
lo
v)'
= l0.8 kΩ
8^to
- (l+s-l"),R.+%
From eq. (7.38)
To permit the maximum sΨing, each ,/DJ'
should
φual |Vρr|.
n
So, refeιing to Fig' p7.37
'
Chapter 7-
Veι
:
Vez
:
Voo_|v"1_|voλ
= 1.8-0.5-0.2: l.l V
rοι.^"_
_
|v
'|
- 0.2) 0.5
= ( 1.8
Vcι :
Voz.n'-
|v
|v
"1
= ( 1.8 0.2 - 0.2\ - 0.5
Ro Ξ r o |(g.2l o)(g fil α)
-
0.2
d)
V
:
ιf R,
0.7
v
a)
Ro,
Ro: =
l1
v"
Ι
ε^,.(|'")= \ε-,''
=
R,,r
]l n., =
i"
=
li2
1r
18-ro
rι'
t
s-r.1s-rl)
Δ
: ':!
8-
=
2
mA/V
:
l0 kΩ
(2o)'?lo kΩ
lι lο kΩ}
l
_l29Υ/Υ
V,η
=
5
V, /=
2
2mA/ν
0.5 mΑ
Ιf ιhe base currenιs are ignored, \νe can use the
same r., and g. for each ιransisιoι
(|ε."1)ιι ι.ε-''"l
νcc
,r,<,
ro
8.ro
= aq^ro
u","_{
c) Ιf v,, is shorted to grοund,
*,",:
-
8^
a
18λ
1r,,+
v^7
ι
.1rι, + _
739 β = 5ο,
| ' 8^r: _ι
8.ro
ζtur.
8^
8.
8. -=
vi'-r"
-+ro
_(g.,.)2'"
\ _ |
v, 3r'' 2l g.
t l R,,
^ - ._
b) &,
r"
Calculating: r,,
R^:(g-' r"ι) ,", = g-r!,
,(,",
I
u*-(Lr)
Assuming that all transistors have the
&" = G,: r.l
-- *
8.
r,,/2f
8- and η,
o
.
η'=.^.L=^l
ρ1r] = (| mAzVl2(|0.8 kΩ)' = l.26 MΩ
same
6nt
= R.,.'
R,n, =
=
738
8-ι
--------8.r
oλ
oλ
|v
l"
i
n
0.2 = 0.9
'
l9
a,
-lL-.
-"*"::"
r*
Using current diνision,
I
8^
.
_
_
Ι,
ο.5 mΑ
:
vι 25 mΥ
β= 50 :
c.
2ο
lvnl
mA/v
5V
0.5 mA
20 mΑ./V
z.s
κο
:10
kc)
From Fig.7.19,
R,,
ι" :
R,, = (?sΞΔ)( l0 kΩ)( lο kΩ
L.
8-ι v''
:--_-_----Ξ
!' +!
2" ε-
= Q^2r,,)(r.all r*7)
= -----------
1r-L
8.r"
R":4ωkΩ
|| 2.5
kο)
chaσer 1_20
7.Δo
vcc
Δ
l
._{
r,
l
._-ι
ιL.'
r,
Ir
Ro
Ιf the traΙsistors are identical,
_- _lvJ
ιo|_ιoΣ_ιoτ
:
o
o
lΙ^l
:o
"
:
("8
Av
B
τ\')
BV"
8-
lrcl
R. 1Q^1t6)Q61|| r.2)
=
ff)(H ffi)
u
with 1. =
ι
nv
βvτ1
,*" = lvλγlvλ'
vτ L|ν +
No\r,
'
I
'r'' =\vτ/
|vΛ
|v^|l +
ιl
For |vo| :5v,β=50
0.l rnΑ.
5v
n_" 0.025 V + I
5V
/
|vι
ι
βVτ
I
w,- |η
l _-------------;
|vA|/vτ
;
'
Vτ / |vA| +
if β = 50
!
and |yΑ|
5V
L
πtr;
1
v'
=
5
_
_4oω ν/v
50
7.42
p\
l
0.1 mA
l-n,:
_:vο
e.
50
Ιf 1= 0.5 mΑ,
R,:,
τ
5ν/0'025v
ι',:
' _!2 0.ο25v * l
βvτl
," _ ly^l gv, lyrl
η '''pu,_τ
!-'π
βvτ
Ιf 1=
.
-llv^l
2 V, I
so that
A|
11η 1η + pv,
= s.(r,,ll n,,)
1', : _|Ιr| 11) lv,l . |vol ,
^ν_
= l-:.Ι
ftι:fτz:rτ:L:FΙom eq. (7.45),
Dll D _|lri
#βv'
_
Λοnll Λon
Mfr(*r-ηΔ)
:4ΦkΩ
n'
ΓR-=
{e^,.or) co2'llr1τ1)
Ιf 1= l.0 mA.
R.: rMc,(#f):20okΩ
7.4l lf
the base cuπents are neglible,
all transis-
ιors share apporoximateΙy the same |/c|.
. :lv^l: 0.lιωv
mΑ
_ 1MΩ
Ch^pLer 7-21
lιl
8.ι:8.z=8.:
ντ
mV
25
l00
β:
g^
rnl : rτ2: rτ=
with ιhe oυtput uηloaded' the small-signa!
model can be drawn as follo\γs:
7.44
ο.1 mA
4
mΑ/v
gC
= 25
Av = _8.'(R,,"
1|
100
MΩ
l MΩ)( 1 MΩ
R." : lω MΩ
so' Αv : -4mΑι,γ(ιωM|l lωM)
.
= (4 mΑ/V
:
2ω' o00 ν/v
-
'
=|Ιr1 =
r,,t
.
7.2o,
+s.(R"ll r")l
R,,
I
VI
0.5mΑ
25 mV
_ oo2A/ν
|Φ
:5kΩ
.'' =β:
B- 0.02 A/v
when R" : 0, R. : r.
a) For R" : 5 r.,
:
:
l0:
1ρ
'
kΩ)
..'
1 + (0.02 A /
v) (R.ll sk)
R''5 kΩ
sn,hr,
=
+
R"
5
k(}
.
q
0.ο2 A,iv
_ ηsοΙι
R,, = 50' η,,
50: l
R" =
4.8
49
o.o2 Α/v
kο
v
'
V.R
= _-:-:_ Λndv"
Vr
,''
r.+R"
:
.-*R.
sυbstitυιiηg, we get
,
nuo
Vo
- T,
Bnrota + R.
',+&
r",
, = _ Β-ro
Avo
since
8.r, :
R'
8-r.rn
ι'_
'
l+-l
rr
β,
. : _g-rο'
Aνo
,R.
βr"
--------^_
l+!
rτ
There aΙe severaι \γays to deriνe the €quaιion
for
G,.
Method
1:
c .--i.
+δ.o2ΑJv(R"l| 5 kΩ)
R".5kΩ
R"+5kΩ
since no current flows ouι the collecιoι
Vo :
8.v"ro + v" By voltag€ diνision,
Take the basic small-signal model:
solνing'&=495 Cl
c) For
I
R.
4 :
Rll
'" _5kΩ=0ο2 A/ν o.zιΩ
R"'5 kΩ
snluinn_
- R,+5k{) : 0.2 kΩ
kΩ) : 208 ο
R" _ 5 kΩ(0.2
4R kΩ
R, =
I
Re
dividing by
1 + O.02 Α,lV (R. || 5
ΞΞ
, : -8.ro:
8-ro
Λνo
,,+ &-
Il +s.(R"ll r.)]
b) For
Vo
o
Ε
kο)
|| 25
7.43 Refering ιo Fig
+
,o
vt
R"ρ)
R.ι : β,. : lο0(1 Mο) :
R,," = k-2r")(rdll r,7)
5
+
Fig.7.l9.
Frorn
s
kΩ
:
2.45
kο
Chapter 7-22
Note ιhat
y. :
io:
ov
--------
io=
v
_ -:
vi _ v,
+
+
g-V,
wiιh
16 :
i,
io:_ioR'+'frνi+ιoR"g-
:
8.η
0.2 mA'
o.2 mΑ
25 mV
=
_
k'Ω,
250 Ω,
8 mΑ/v
Iu-s
ιο
v
sothat,
Rρ
I
"._i,-
ι
since&<<ιusι,aιlu.
=
g.)
: lΦ
kΩ
+
(0.25 kο
|1
l2.5 kΩ)
tω kΩ)rs
'\ ηΔ)
ν ./
:
rο
io
t6 * (Re || r")(1 +
ro + rog.(Re ll r*)
Rρ
296 kΩ
t- R'
Avo : - 8-ro-4
8^ τ, =l+%R,
"^_
I+A
tπ
Method 2:
consider the model ofFig. p7..ι4Φ):
Ro
vo
^.-{-'T.l
P.o. eq. (z.slo),
Ro : ro + (Re ll r.)
.. g..
rorn
and R" :
β
,':
'' E- = -l!L
omΑ
^io8.
i
&
un6
β : 1ω, ro = lΦ
,_: |'d:
vτ
Then-
i,(r *&+ι"ε.)
f R"=ι
rτ
^8^ +ε-n.
"'=
l
8η(vι v,'
Assuming thaι io >>
Y - i^R
,in"" 'o
_
_ (8
: -784νN
8^ _
λvo
"'=ιιε.n.
:
+
mA/v)Ι
'
lω kΩ).
l
0.25 kΩ
Ιω(
,,
+
g.(ne
||
^
G-:
rn + ιζ.
i.'
-A'.
i,
%_
g^roro
-
2.61 ΙΔλ^l
vo
([rom pan
ηη
ro' 8-ro;ττ"
R,
R.
n
"-_ Φt'J.
foto
R1= 10 kΩ
l aboνe)
g.roro - R"
r, * R.
rooτ+R.)+8^roR.r.
Diνiding by forτ'weΕet
fofn
l2.5 kΩ
G.r)(R"ll r")
shoιting the outpuι removes Rι from the cKT.
_l"" =
)
s,"A/v(lrsτσ
r")]
g.'o'Ξ:-R,
kΩ
8 mA/v
+
or
Ro: ιofl
lω
ο.25 kΩ
A"
=
Aν
:3:
-G.(Roll R.)
2.67 E|ιAN (296 kΩ || lo
:
-25.9
ΥN
kο)
Note: Depending upon the appΙoximations taken,
ιhe νalues ofΑ, may νery slightly.
Chaptet
7
l-23
Ay6 = _G^R6 - _g.β9
= _(4 mΑ/v)(2.5 MΩ) :
-45
1ο
.
Vio--1
: r,' :
R'n
Ro
Rρ
=
:
:
Avo
25
-l
Ro
Fig. p7.45(c)
From pan (b),
8.ι=8.z:1mΑ/v
ο-lmΑ
kΩ
Ro = (g-7ro)rol
:
G.zro)toι|| r"ι)
Rρ
(4 mAΛr')(50 kΩ)(5ο kΩ l| 25 kfr)
Ayo:
3.33
:
MΩ
- G.Ro
-
Auo
-g.1Ro
= _(4 mA/v){ 3..ι3 MΩ)= _ l.ι.]
x
iοr
(1mΑ/V)(50kΩ1'
-Γ
\v
,--l
ol
Fig. p7.45 (d)
From aboνe.
g-,
4 mAN
-
0.|
o'2
mΑ= lmΑ/V
ν 12
2
: t-ι : 25kΩ
R6 : (g.216z)roι : (1 mΑ/v)(50
: 2.5 MΩ
Αgain, Rin
K)'?
Vιιιs
Ro
8.l = l
Fig. p7.45(b)
|ιo|
-2,5ω v/v
vo
-l
R6
"^=
:
rn
}-ο
o,
vo
i;n
2.5
νlν
}-__ o Vnl,ιs
-
:
G.Ro: g.ιRo
= '(1 mΑ/v)(2.5 Mο)
ζo___--t
g-, :
v/v
vo
Fig. p7-45(a)
/
1or
οyυtes
-,Γι"l o-I mA 4 mA/v
R.'_g.2 _τ:ffi
|ω _ zsιο
r-'_r-'= β_ 4 mA/V
B_ |yol - 5v _ sοιrl
X
:
4
mΑ/v
τοAΛΙ
' ,.a
=
25
kΩ
Ro: (g.rro)Qorll r"2'1
Ro - (4 mΑ/v)(50 kΩ)(50 kΩ || 25 kΩ)
: 3.33 Mο
Auo : -G-R6 ,
Aνo: g.ιRo: -l mΑ/v(3.33 MΩ)
3.33
X
101
v/ν
chapteτ 7_24
Voo-ves:1.8-0.7
Comments:
(l)ΑMosFETfor0,
makes Rin -+
οο.
(2) The output resistance when 02 is a
BJT is lim-
|tedby r.2. Ιn cases (a) and (d), Ro was higheΙ
due to the νalue of fo and 8π2.
(3) Ιn these four cases, Αyo was high€st
since
k'v2
with ι\ro
biasing.
7.46 Reffering to Fig 7.22' vDD =
: 1σ) μΑ,
Ι:0.5μm'lV:4μm,
yi : l0 v/μm, η : o.5 v,
k" : 40Ο μΑ/v'
=
2Φ μΑ
ξo'(Υ)":"'
2Ιn
2(2Φ
:
ψ1' :
+ W = 25L :
BJTs Α- was Ιowesι with t\ro MosFΕTs.
These results could be chaηged Ψith different
1oo.
1o :
5.5
kΩ
μΑ)
(40o μA/v'?Xο.2
25(3.2 μm)
:
v)'
_
25
8o μm
7.4
lvoo
* voo
|-8ν'
/ρ = ζ"1
Ιo:
Ιo =
ξo.(v)":'
: voo - vo^u
: 1.8 - r.6 : 0.2v
set lyovl
:
o.25
Ve = Voo_
ν
Vos= V6s= V,+ Voy
= 0.5 + 0.25 = 0.75
ρ = Voo. Ves :
Ι*,
l.8 _ 0.75
o.1 mA
:
V
l0.5 k(}
The lowesι yo ιviιl be Ψhen
v DS2 = vov = o.25
R^= r^:
:
50
ν
v 'Ι'
_-!_
ID
t0 V/μm(0.5 μιn)
=
Δ1o
Δη,
ΔΔ
Ιo
:
:
=
5 qo.
0.05(2ω
.o..^ =
"
1.8
"o
μΑ) :
_ 0.20
:
1.6
10
v
4&:
]-QΙ:
ΔΙn l0
μA
uλL-
Io L_'olo
v'λ =
L:3.2ψm
vο
noΙτiπa1 = vGS = vι
Ι,etιing yov = yDs
ι":
to'
μΑ
tεοιο
160kο(o.2mA)
(l0 v/μm)
+vov
nnn =
0.20 ν,
Ycs=0.5+0.2V=0.7V
-
-
0.5
_
0.2
1.1
80
|voλ
:
1.1
V =
V
l3.75kΩ
μΑ
)ν"c.'(fl|v"l,
z:
w
_
2Ιo
2(80
μρc"'|voλ2 ΙΦ
7.49 Referring to Fig. 7.23,
if W2 = 5
aηd wg |g1|, =
ξ
since
u:
since
,^ =
"
1.8
ρ=Vo:
Ιo,
1Φ μΑ
kο
o.5 v
^,' --'ΔVo
ro 5οκ :10μΑ
7.η
:
\v,o|
ξ{.(Y)ιv^_
lor: ,"uoffi,:
Vo = Vou
μΑ)
pAN2(o.2)1
ξ,
v,l''
20
μA(5)
:
1Φ μA
= o'2Υ
From eq. (7.59),
ι'=ffil",',(l'-Ψ)
vcs:
so,
vι+Vov = o.5Υ +0.2v+0.7v
for /D to equal51iεΙ.
orvo=vCS=o.7ν
lf ro,
'vo
: |!Δ : 20v
loι
ιhen' ΔΙo =
^-!9
ro
l00 μΑ
_ ιv
2ω kΩ
vGS=o
2ω kΩ,
_ 40
Chapter 7-25
7,50 Referring to Fig p7.5ο'
v-,' _ v-.'
ι o',
=
so
ιhaι
Ι
752 Refening
|!!4
/j? _
n'
(w/L)|
0ιhasW:
to the figure below, suppose that
l0 μm,
Q2 has
andQjhasW:Φμm.
^n6
w = 20
μrn '
Ι'','1Ψμψlι1,
(w/L\4 Ιoι Ιon
"n11r^1(!Β:Ιo.':Ι22
'
(w/L)Σ (WL\4
'Rn'|\w/υ,'\w/υ,
7.51 Refer to Fig. P7.51
vcs2
vos2& vGS2 vιp+vDS^σ:
(1.3
:
: v652 - (-0.6)a
0.8v: vofl
-
For P
ι^,
1.5)
v'
v.652
(1)
1:
2ο
μA =
].
2
' !08
εo
+W,:10μm
/r: l0oμΑ:5/"rn=%
1,:1*nn+ζ:
lV'
:
ι': ι'=Ywr = ι\Ptι=>W'
x(_ ο.8 '
= 5Wι =50μm
l0 μm
50
1x
μA = !2 x2ωz
=Ξw5: ι0
0.8
1.5
_ ο.8
ΙRΕts
μA
R:35kΩ
Ι.
l2i,ο'8
ι2
-'
t0 x 0.8
οο5
(2)
with o2 diode connected. and
:
1oo
rA(13) =
1ω μΑ(l3)
:
(3)
0.8
- o.οl
V
lf or ιγith ω
:
=
50
roο
μΑ
μA
ω = 20 μm.
μA
20ο
Φ μm
ιl(fi) :
/, = ιω μA(#) :
l,
400
20Ο
μΑ
is diode connected,
us
ιn
50 μΑ
so. \ir'ith only one ιransisιor diode connected, we
n= 200'7 ::stΩ
v.
= lω *Α(i3) =
:
can geι 25
Vo., = _0.8 ν .) vc2 =
vc2/
'*u'!##,
/. = 1ω μA(#) =
μm
Now ψe calculate R:
R:
ιο.ε
=
1.
,!
: 16a!Q = 4,,n'
2ω
For ρ5: yD55 : Vcss _ V,, For ιo\rest yo
( 1.3 - (-1.5) = V651 0.6 = V65s :
λ-
',
ο.6)']
with oι diode connected'
:
160
:
kΩ
oο
ιο,
:
0.7
v
μA, 5ο μA, 2Φ μA,
and
4Φ μΑ,
or
4
different cuπents
Νow, if2 ιransistors are diode connecιed' the
effective width is the sum ofthe two widths.
(4) lf Q I and Q2 are diωe connected,
ηtr :
20 + l0
/. :
1οo
:
30 μm, so ιhat
μΑ(*) :
133
μΑ
(5) Ιf 02 and 81 are diode connected,
weιΙ: 20 +4o 60 μm, so that
:
/,
:
(6)
ιf ol and o1are diode connecιed,
ηF :
1,
:
lω *A(:3)
10 +40
:
:
16.7
μA
50 μm, so thaι
lω μA(1:)
= 40 μΑ
so 3 differenι currents are obtained νr'ith doublediode connects.
chaρter 7-26
To find ιhe values of y.'G, since
l"
:
\i,(!)ιv,o*
Vo
v,,l''
for the case of a sing.le diode-connectω transistoη
we can use anyone. For €xample' for
ol
Γ
/
L\|
= li@
v.o
l00 μΑ
/
v'
(_ο.6l
-
v.. =
For
l.o5v
=
ο.6
:
0.86
'
Vo
v
o.7Bv
-
μA/v')(5ο)
-,*
,
,.rf
,_.,,11
Yi,
l
divide out ro:
: ε-ι g-ι(roιll
(+^*
t"')
l,
8'ι 8.ι Rι
:
Vovι
o.8v
*,*
Rι')
Assuming all rρ values arc >>
8-z
since /D =
50 μm'
@+0.6
γ( lω
V.r :
since
V'
μA/ν')(60)
n1|7:+:Ψ
11,
Vo
ii-]iffinA)+0.6
lω
η,Ι(
ηf, :
vro
+
c.,(,",l|
ε.,,,,)l
this would be the same for ysc2 and ysc3 with
multip1e diode connected ιransistors,
For w.tr = 30 μm'
(lω μA / v'?)(3o)
:
60 μm,
For ξ11
:
*.,[
}τ _ ε-,(.,,il
'-"''" ""[(,, * re''' ,"'
vr.: !t''D1 -vηkPlw
t
:
\ι,(V)u:"
Vovι
making g. α
' ιιso, ε.
Ι,
which is
and vcs2
:
α
:
Ycs3,
ffi
f
7,53 Refening to Fig. p7.53, the small-signal
mοdel can be dralpn as follows:
V,,,
+
+
vi
Η€re.
:
(Ι) Vo
: G
"2
(2) V'ι :
V
(3\
iD1
:
1ι1
C.1-*
: -Vrι r^,.o"-Υ#
V'z
:
so we coυld a|so express ιhe gain as
νoltage yx:
vrn
subsιitυting (3) inιo (2)' ιγe get
V"2
η.
No\γ' to nnd ιhe resistance looking inιo the
diode-connected drain of02, we apply a test
ro2_ 8^!vs"2ro2
-Vs"t
=
νo
^ /}yι\
ιΙL\i)
τ = 8il
8m3σω|| R)vι,1
oι _ g-2 Vr"2) roz
iD1
η
Vι"ι 8^t roz
(ι *'ιz,* ε-,
,o")
subsιituting (4) inιo (1), ιγe get
_g.2Vr"2roz
+
-l
*
v,
Rt,
chΔpter 7_2'7
ι,
=
!+
755
g., Vn,
v,,
vx, ix = --!
since Vr*z =
+
B
vx
['"
i"
:'::-+8^ι
I
R^,:Ψ:,o,l!
ιx
E
Q,ι'
2
The CS gain is
:
?
7.54
(a)
ε.,(χll,-l1,",)
lf Ιs :
ιO
16
A,
l-Εt
and we ignore base
cuπents'
vD.lντ
/q6s:11 e"'
y,, =
"'
:
:
:
:
ln[lo Ι6''|
') :
\l0-
r,(ffi) :
so for ιhe range of
lο μA = /RrF = 10 mA,
0.633v<ysrΞο.8ο6v
(b) Accounting for
finite β,
I
ιo = ιιtι'TΞΞΠ
For /*.',
:
10
: ]9-μΔ = 9.62
ι+!
50
:
0.l mΑ,
For ,RFΓ
0''
Ψ
l+--a
v
b:l'':.nt-:bι2ι
'
l. Ι',
o.εoο
v
Ι''''=lo*!ρtbo,
mιηββ
Ιo :
I
/ιl,ι, 1*_L*1
mmp β
ΞΙ^n:
/*..
t-l+m
Ιo
μΑ
Ι l
'f
-50
β
β
β
had been identical. rn ψould haνe
1
Now, if
β-l. =
consider that
50 aπd \re ψanι to limit
=
ffi
if β
mA
>>
l
rn Dropρin, l0%. ινouιd mean thaι
l0 9οl
O.98
which is eο. (7.69)
ιhe ιransfer error ιo 10%'
I f
r, :
t+l+.
eo- ι /. /Ul
/*,.= 1+2/p
Forl^u,,: I mA,
to
m
sυbstifuιing (2) into (1).
-
r(a
-1ω
For /*,,,, = 10
'ββ
been one and
= o'o98 mΑ
/o : _ι!Δ :
Ιo
yRΙ,2' so
lfιhe transistors
μA,
/o
1,:
ο-οl:l
ι0 mΑ,
o.o25
:
!! :
b
v-h
' Ιr, v-h
' ls,
Since /s2 : ,n /sr,
l0 μΑ,
0.025
For IRt,|,
Vιε
sothat
Ιc2
t'''' = Ι-+l, *Io ,l,
ysε| =
v.h(β)
For /RDι_
V-_
'
Ια : Ιc'
rΙ
:
m
-
1+m
β
Solving for π' the cuπeηt transfer mtio' we
geι
mA,
= e,62 mA
ιο.qot(t+l+"'):
ι
50
./
so that
m
:
4.56
l
Chapter 7*28
This figure belongs with 7.59
+5V
+1ον
lι,,
Rl=5 kΩ
Rs:3 kΩ
R. =20
Rι=2 kΩ
}ι,
'
ΙR4
kΩ
+5V
lη,
/
Ι-
''ι
R3:3,6 kΩ
}ι,
',,
Qz
_Ιov
l0v
7.56
= l^(l +
1μgι.
:
v",:
['"
V.
:0.91
757Ιo:Ixg1':2mΑ
=
7J8 /s
o'2
2
0.2mΑ
lo
IJ
n, ρ :
* = 1-'^":
β
59
u(t.i)
ι.z νa
o'l94 X
lΦ
1
= |9.4%
change in ,o
ιrith
: l0
R=
R,I"o, +
Δ/'r, lω _
7β9
= l09o chanρe
:
V
isS-0.3:4-7V
_ (_s' _
_ Δ!!ι
,o"n
" ΔΙo-ι"l Δlo
'^
+ Δ/o : ο.194 mΑ
20
Ιo
v'1n;7114-*γrJη1
yora* occurs \rhen 02 is on the edgΘ of
saιuration ot V"o : 0.3 V. Therefore yom.x
is the same as eq. (7.69):
Δlo
50)
mΑ
ft=
,^:\=E=soιο
"Ιol
For identical transisιors. ιhe traπsfer ratio
,^':Vo'=Ψ=asιο
"'Ιo2
Δ:9_ 19 |_ ηs=al^_
,^' =
"' ΔΙ" ΔI"
1.02
Ζ\
: 9.ρ25 l0-3 : 0.689
3)
ιο ,.(l
ι + 5ο,'
Vc = Vι:5_0.689 = 4.31 v
Qι
Ιo:
l
: t
/".. l+2/9 ,-2
"\
β:
Φ,
Ιq=Ι62:Ιx1
Vιι : Vcc' Vει : |o' 0.7 = 9.3
vE2: vΕE+vBΕ: _|o+0.7:
V''_V"'
ι :_ -_i;'''
-
9.J _
20
(:9.])
kΩ
V
9.3
V
chapteτ 7_29
This figure beloηgs with 7.62
+
v,
vι:
= 0.93 mΑ
Ιcι: Ιcz: Ιrr: Ι6a: Ιgs:
= 0.93 mA
Vrι :
V6 :
o.93 mA (5
Vgρ5
:
0.7
kΩ)
Υ
:
+ιο.οs mall'|Φ kΩ)
: 5 V _ 0.93 mA (3.6 kΩ) : l.65 v
Ιc9 : Ιcx : Ιρ1 : Ιa7 : 0.93 mA
I"o : 2 (0.93 mA) : Ι.86 mΑ
Vcl: Vce:0 1ρa R1
_3.72 ν
= o - 1.86 mΑ (2 kΩ) :
Icι = Ιcyl: 169 : o.93 mΑ
Vcg : Vcιιι : Vεto : 5_0.7 = 4.3v
V.,, : 0.93 mA (3 kΩ) = 2.79 v
Vι: o_Vιrι:
Since
β=
_o.'l Υ
R
:
10
\2)
3.2
ν
implied in Fig. 7.31.
+5V
x8
ι
ο.2 mA
lo.,
ο.4
ι
mA
m,ι |}r
ι
0.8
mn
}z.,ι
Χ2o
Χ5
ω,
V' _ Vεε
R
V, = Vrt + lru(R/2)
_5.7 + (ο.5 mA)(5 kΩ)
=
(b) For R : 1ω kΩ,
V, = O-0.7: -0.7V
:
_3.2
= 0.05 mΑ
2{lc1) = 2 (0.05 mΑ)
:
-5V
Ψiιh this schem€,
R = 5_0.7_0.7_(_5): s6kΩ
0.l mΑ
v
and each ιΓansistor has EBJ aΙeas proportional
ιo ιhe cυrrent reqυired. Multiple' parallel
transistars are ok.
Note: This laτge value of R is not desirable in
integraιed form, oιher designs may be moνe
suitable.
Eνen ινithoυι knovring exact circuitry ofthis
/,.' ψe can find the toιal power dissipaιion as
approximately,
P7 = Pqg * P66
, ' Vr,_Vεε _ -0.7 ( 5'7\
ιcl
R
!ωτΩ ο.l mΑ
: O) Vps = 0+0.7:0.7V
Vz: Vcc vΕB:3'7 _0.7:3V
V1
:
7.61 There are νaΙious ψays this design could be
achieνω, bυι the mosι_straighιforward is the one
kΩ.
_0"| ( 5.7)
=05mΑ
10 kΩ
Vz = Vcc- Vuu : 31- 0-7 : 3 v
since /c] : Ι6a = Ι62 : Ιr1
Ι:2(Ι61):2(0.5mΑ) = 1.0 mΑ
y] : ys5+ Vεss : o+0.7 : 0.7v
va: vBE+ΙrrR: _5.7
+(0.5 mΑ)(l0 kΩ) : -0.7 v
Ι:
_5'7
+(0.05 mΑ)(1Φ
V.o
7.60 (a) Refeπing ιo Fig. P7.60,
Ιc.R:
kΩ) : _0.7 v
v5 = vΕΕ+ Ιρoβ/2) = 5.1
Ιcι
V
4.65
vEΕ+
Pr : 5 V (ο.l +0.2 + ο.4+ο.8) mA
+5V(0.1 +0.5+ l +2)mA
P7
:
7.5 mW + 18
mW
:
25.5 mW
7.62 First, we draw the small-signal model
ofthe circuit in Fig' F7.62, and apρly a tesι
νolιage
ζ:
(see figure (1))
Chapter 7-30
This figure belongs with 7.62
ir: b*s-ιv-ι+_!!foι
r"ιll r"ι
y.Ι : Voι' =
since
ι,
: !!
+
foι
7.63
s^ιVx+
iι, Γ&
+5V
Vx'
Δ
-!!r"ιll r"z
)
It
I
\
i,:vΙl\loI +c-,+-_+-)
rnι|| rτ)l
o -Vt,ιin_-_
ι"
"
fοl
-+8_r
+
l
-------:r"ι ll r"ι
so thaι
R,. =
Ψ
lf t6
>>
R'" =
*
(,,ι
=
r.,
}ι
,",ll
,.,)
Siιcevo2
tP
:
,^"
V"z
Ro
-
(7.5ο) or sιfuce 8^
ro + g^ rr(R.
ro
>>
|'
ll r")
Ιn this case.
n-
: v"|
Ι, :
vτ
:
-1_!Δ
25 mV
-_ ο.οη
ezv
.-:β:]Φ:z.sιο
'' 8^ .04
,^:\=lΦV:lfflkΩ
" Ι,
lmΑ
Rρ : 1Φ K+ (.04 Α/v)(1ωkΩ)
R; ii
(4.3
^^
ξ: '*('-ι frl
..,ll .,,)
lf ro,
>>
nr:
t^"(*L,ll ,",11
,.,)
rn,
η.
can υse
ll o"1ll r"z)
so,
i6 : g'2
_ 5_o.7 _
Ι.:5-Vιε
' 4.] kΩ 4.3 kΩ lmA
since β >>1'l : Ιc: ΙΕ-| 1|'^
To find the output resistance, ιve
Next, we apply il and determine i": (see fiμre (2))
io: 8'ι
t"
Ξ
|
I
and ro,
kο
R6 :
kΩ)
ΜΩ
|| 2.5
6;42
ιoY
l.56ιΑ
Ιfthe coιIector volt ge changes by
ι0v :
=6'42lνldι
Rο
^Ι=ψ
Chapter 7-3
7.64 Referring to Fig. 7.32, we see thaι the
t)
1
Ι V, and Va51 : V6γ.
Vo-π:2vov+vι:2(0'25)
+ 0.6 : l.l V
Vor, =
t" Vrr.
/D
:
0.893 mΑ /
0'224
ν
:
- 96
=
16r
Using
8.93
the νalues
o'95
Vo :
:
2 V6s
2(o.5
V, :
Ψhen
:
of
2 V6s
ν
+ 0.224
5
v,
(7.77),
:
v) =
2(Vh+
l.45
Ιι :
25
=
!v
Vov)
ν
2
2γ)
5_l.45:04ιΑ
" Ro 8.9.1 m
^ι^:ΔVo
soat Yo : 5 V,
16 : lΦ + 0.4 : 1Φ.4 μA
(Y\ ν'^',.
/: l ι|"\L)
"'
2
: Vovι :
=
0.25
V
:
:
Vοι
:
:
0.85
V
since Or and Oa are diode-connected.
aηd /t)a
Vοl
:
1Dl'
: Vcι :
2Vesι
Qx and Q1 haνe lV :
times wt and η.
:
2(0.85) =
y-,, _
""Ι
2
μrι
o.J54 v
\W/L)2
(Y\
\L),
0.954
V
Vη1
:
Vργ1 or
: Vor' Vesι * Voul
since Ycsr : Ycs2 and /2 : 1r then
Vo
v|+ vov
0.6 + 0.25
:
The lοwest possibΙe voltage for the ouιpuι is
when 0. has
Vcz
Wa
.+Ι'=25xΦ:50oιΑ
'2
/r:0.5mΑ:11
1o : 0.5 mΑ
Vοsι : Vovι + V' : n354 + 0.6 :
Vcι : 0'954 V
Vo. : V63r l V654,
Since /, : /, and lV, : ζ ιhen
Vor, = Voro=+ V6o : 2V6s1
:1.191 V : Vc:
7.65 Refeπing to Fig. 7.32'
Ιoι : Ιoι: /η11, : 20 μΑ
Vovι
'6 mΑ / v)(50 kΩ)'?
"'?l v'^'
""|=
,t2
t'
the.e is a small
amount of channel modulaιion.
since
(|
ιιλ = Ιι Wι :
W':Wr:Φum
1 W.,
I' = ;k"l Vov|Ξ25
ν
yDJ will be equal on all transistors-
So, V6
η.
7.66 /RΓ|, = 25 μA, RefertoFig.7-32
vo6": vLh + 2 νoν
: 0.5 v + 2(o.22A ν) :
= /,",'
mA/ V
R6 1
kΩ)'?
The minimυm allowable % is
When 1o
1.6
ξ6 rρι rρ2 :
= 4Mο
rot
mΑ/v)( ιω
MΩ
ο25v
,^:Vo:t0V:50kΩ
" lo 0.2 mΑ
V
(ο.893
(2ω μΑ)(2)
2
:
σn4 ν
Using eq. (7.77),
R6
^ _ ID
6r
ζ_
(o.l mΑ )(2)
'^_ u*n-
Vou
Fol Q' Δnd Q''
v"" _ Γ_Jρ _ Ε:ρ]_ηη _
γ ,(" N4mA/ν'
= 2Φ μΑ
1o /trι_9
The lowest Y" will be when
: lffl kΩ
,^:Vo: ο_l|0v
mΑ
/, :
Ιo:
".
pammeters 8n and r. \νiΙΙ be identical ior aΙl
t.ansistors:
since
I
17v
40 μm, which is l0
:
'rn
Ξyonin: l.9l 0.954+0.354:
2ι^ 2 X 0-5
Vn, ο.:ι54
2.82
rολ/Υ
1.3l V
chΔρter 7-32
roι=roι:-;Ξ
v.
ιD
:
Eq. 6.189:
Ro
:
Ro =
ro1+ ΙΙ +
40 kΩ +
[l
?9
0.5
(8
+
+ 2.82
xΦ kΩ :4.6MΩ
:
7.67 V,
Vιε.ι *
:
+oιο
8.Β\rΦlro2
kΩ X.ω kΩ X 40 kο]
Vιει :
|.4
Υ
ΙF IRrΙ, in increased to l mΑ or ηuivalently mul_
tiplied by l0, ιhen:
Ιoι
: Ιoz:
ξ"
,o: ... =
= ,o
The emitter of OJ supplies the base cunents for
aΙ! ιransistor so
, _
,.,
(n + I )ln
ρ
/πυε: Ιε:.l":
Ιo :
/*..
1
(n+l)
t!
β(β + ι)
for deviation of
Ψ!:
|οο
ffi+ro
.
l
ι
% from unity:
(r+|) =n=g
'' ιω(l0l)
7.69 Referring to Fig. 7.33, the small-signal
φuivalent can be draινn as folioψs:
/Fl o
Ιf i,, is small,
-
-4.3 K
ν"",'"
|Vsrι uιr]ιΝ
lC' Ξ l.e "" 'τ .^
e
vBει'
/C,
ν'
lsP
Vιεz _ Vsεl : ΔVιε : V. Ιn l0 : 0.058 V
-9|ιr:
ΔVr, : 0.058 v + Δyχ _- 2ΔV'' : g]16y
No\μ we calculate 1o for Vo : Vr:
/x61,
316
:
1Φ μA
The acιual value οf /^
"
=
/,
Δ1o
ιo
:
0.Φ5
1o0
2ω
: 5X lo J
v'9
u"*
Ιncrcmental cuΓrent
ix
:
8'Vυ,ι
: 'f
,r",
iJ!
-v.^
".r:
/"..
: i''
ΙRΓF
I | 2/\9'1t 9)
___!ΦJΔ-_ _ 99.995 μΑ
2uJ1 +
Iρτ.'
, _2Ι, - _vl_v'β_
.Fι
β,.,.''G_2ιc
r-=r^:v':v'
'l '2 Ιr' /onn
Ιf igγ
_
I'zo
v,
8nV6.,
--->
Ξ0.0o57o
/πετ,
v*z
7-64
{
reι
vb.|
Qι
,"'1"': !fl.ixl.2\
2Ιor'
yυ", + yι",
Υlι
so that
*'ιL
β
ντix
/*.,
Chapter 7-33
V'
^
Λln
τ
For 1xg1'
Vr'V'
_
,-'.-λ...
:
1σ)
2Vτ
This makes 1ε] and 1r4 each
π
: ,il
2
μΑ'
R"' = 2(25 mν)
tΦ μΑ
+
I
Αccοunting for base cuffents 1rl aπd 1sa, ιhe
ξο0 Ω
output cuπents are
7.70 Referring to Fig. 7.34 and the analysis ιhat
follo\νs.
β
: Ι^': !B+2.
2 β β+| =!β+2'Ι
2β +
Ι^'
ιhaι Ro Ξ;p\ro\.
l
:
ΙRF|,
ιι,e kno\λ,
2/ttt'.= 1C--:-:.
29
/+1s] +/r1
I
+wΙ'
From eq. (7.80),
Jι - -J/".'-I+Ξ
=04mΑ
.o
ι
11,u1
βz
16:
:
Ι6:
Ιpp'ρ
;El
r"'^[
=
-1mΑ
:
v^
l0o kΩ
_
so,
vΕΕ
'vEE
: Ι+ β+2 .l:9'*29+z.,
(β+l)β
(β+l)β
Ro
ΔΙo
so,
ΔΙo
10,
Ιo
/"r,,,
731
:
lβ+2.,
19+l'
/ιυε= β2+2β+2.Ι
Ιn,
(β + l)β
:!
β(β + 2)
2p2+28+2
so, /οl
ψ
Ι02
Qn
β+2 :1ιεε.
- rREΙ.
2 Bt2-2/g
2
l1ι +f l
$r
a,
E
Assυming
eachΙ/p
1.' :
Ιcz
:
I, base currents aae
2
β(β + 2)
=,",=(!
\ u ---!t+2/8./"l
(b) Siηce the factor
___-]-
1+2/p'
is independΘnt of
how many outputs ιhere aΙe, multiple transistors
of different areas can be combined ιo geι other
current νalues. so for
/Rι,,
'V"ε
Vtt
l
so thaι
,,,,
,|,
: ηr:
γL' p2+2β+2
:
0.7 mA, and
β
:
50. \γe can obιain
cuπent outputs of 0.l mA. 0.2 mA and 0.4 mΑ
with 3 transistors haνing relaιiνe areas of l,2' and
4. Note ιhat eνen though /( 1 + 2 /
β)
is spliι up
between 3 transisto.s' the reqυired base cuπent
(being subtracted trom IRιF ) remains ιhe same.
Chapter 7-34
: /x6g'l
' '
t I +2/8.
0.7 mΑ
'--_:I , = ο'0999
7.73 a) see the aηalysis on the circuit.
So'Ιρ1
:
1REF: /+
ι' = l.:
(so)'?
Ιοι :
o.1 mA, ideaΙly.
Simitarly,
1*.ε
?.
1
/n, = (0.7 mΑ)
ρffi'
β(β+2)
292+2p+2
I
t+2/(p2/2p\
I
/"r,
2
observ€
that the deviation
I + 2/92
facιor 1 ,
+2/g'
I
The resulting circuit can be draιγn as figure aι lefι.
I
is
indepeΙdent of the numbel of outputs or the value
of each output, i.e.:
The current /REF can be spΙit into any number
7.72 one pοssibility is shown belo\γ:
+2.5ν
B2+29+2
β(β + l)
:1
/"on
6,:!
t+zt15o1t
: 0.1998 mΑ - 0.2 mΑ, ideally.
Ι- =Ιo'7.,ηl' .1.
7 l 2tl5Φ,
'
= 0.3997 mΑ - 0.4 mΑ Ιdeally.
l
\{|]ι
Ιo'
: !x
2
I
=
of
ouιputs through an appropriate combinations of
parallel-connecιω tΙansistors. (o3 and oa in this
a1 < \Ι
+2'5
ν
case) The reason the error factoa remains
unchanged
at
l
I+
2/p''
is ιhaι ιhe base cuσenι
thaι ne€d to be sυpplied by 1REF (subιract from
ΙRFF ) remains unchanged.
b) The 1mΑ reference current can b€ used to geneΙate thre€ output currents of 1, 2, 4 mA by usiηg
3 transisιors in parailel haνiηg relaιiνe aΙea ratios
of l, 2,4 as shovr'n:
Ξ/n, = _1t
Ι*.. 71+2lg'
: 0.998 mΑ (1 mA ideaIly)
-Ξr^'
vD
|
β+2 Ι
2 β+l
-2.5 V
: 0.2 mΑ,
^ vcc - v r",1πιrvr"\ - vff
1 β+2
2
β+|
Ιo'
Ι
v,
To obιain 1REF.
_
2.5
v_
0.7
ν
_ 0.7 v_( _2.5
0.2 mΑ
v) _ lr ιο
t B+2
τβΙ[
[Ι
Αlternatively, ,REF couΙd be obιainω from
another reference source.
Ιπ this case, maximum yo is when
Vo^^'
= V,,
Vεsl _ Vεc"^ι
':2'5_o'7_o'2 = 1'6Υ
0, is safurated:
o+Lil=
ιι Ι ρ!2
ΙΙ
ββ
Chapter 7-35
(e)
Ιfa small-signal model is addω to Fig. 7.35 Φ)
to accounι for o., ιhe circuit is changed to
lοz
Ιιιι
_2
l
ι l+2/B' -,
: ( 1.996) mΑ (2 mΑ ideally)
/,,. 4
Ιιl,ι ι l+2/β'
since yD"a = Vesι
=
(οo currenι into gaιe 3)
I
3.992 mΑ (4 mA ideaιly)
7.74 (a) First, we need an estimate for y.,
and η,.. since ιhe cuπents are all approximately
ιhe same. and
ιo -
uv},
|ιμ.c^'xw
t
Γ 2ι'
''
"o": ]uρ-1w l ιl
y^'' = / 2( l00 μΑ)
_
"" ,/(4ω μΑ / v)(l2.5)
6.2
.
γ
since no vaIue is given for y,n, Ψe haνe ιo esιimaιe
ιhis Ψith μ,coχ
:
40O μΑ / V2. from
Vcsι
:
Vοsι = 0
=
so
g.aV",ιr"ι
thaι vD,2
:
vc\
and
there is no effect.
Ru- (8.1,r.)r,2
Ιo _ o'l mA = lmA/V
n _ v.'/2
0'2ν/2
R,, - (1 mΑ/v)(2Φ kΩ)'? : 40 lιηΩ
7.75 Firsι, we draw the small_Signal mοdel of
Fig. 7.35(a) using the simp1ifiω T-model. liy'e
apply a ιesι νolιage η and determine ζ.
Table 7.Α.l, this fabrication process is similar to
the 0.18
μm
technology. we will therefore
approximaιe η,as apprοximately 0.5 v.
8^ι
ves: v.+v..,: o.5 v+0.2v : o.7v
(b) Vos:. : Vcs, * Vcs,r : 1.4 V, which is
=
(2 '
Vε'ι
v"")
20V :2ΦkΩ
," :V^:
Ι οlmA
_
ιι = ΔV", _ l'4 o'1 _ 0..]5
2Φ
kΩ
η,
Dz
uΑ
/.-1RεF-Δ1
so that.
l, * lω _ 3.5
(c) 1, - /Rl'ι_
:
8nιVμ1
96.5 μΑ
η'i, : vh+2v.,,
:os + 2(O,2 V) : 0.9 V
(d)
Chapter 7-36
:
Since ica
0,
i, = gr2yssι
V*2
:
9,3
:
g.1,
vμι,
t'
:
=
v
lJ
Ιfβ =
o.8
Vr, :
-
0.8
V
@1
r.,,(ffi)
vaεr_ vιε,
o ____Γ^, =5.7
:
o.742Υ
0.685
v
0.742 _ 0.685
lο μΑ
& - l0μA _ o.4mA/ν
vτ 25 mV
200
,-.:L:
'" s^ 0.4 mΑ/v :5ff)kΩ
=
ιvhich is >> R.
,^.:Vo:
"' Ι"
50v :
l0 μΑ
s ι.,tΩ
so, from eq. (7.98),
: [
g,(Rull r.)]162
+ 0.4 mA / v(5.7 kΩ)] : 16.4MΩ
change in cυrrent vr'ith a 5 v change in volt-
R"",
+
Ξ [l
The
age is
= 16.45Y
'
^:v"
R.
MΩ
^t-=
-- o.l,-.a
7.77 Refe.ing to Fig. 7.36,
Fora
/*..
:
mΑ _
οo,5ρ
+8.iε]ro
1 + (3.6 mΑ /
= 0.9./ =
From φ. (7.95),
v)(0.0293 kΩ)](556
:
:
Vρ2
: ιι25,ν)|"(#)
57'56
mV
0.οI mΑ
:
2.63 mV
57.56 mV
5.?6 kΩ
MΩ
MΩ
compared to 5
Fo. Ι' = 00l.
,Rtiμ
v"
'Ε :
t2s
ιrith l?ε
/^
:
kΩ)1(5 MΩ)
0.
l ιιA
=
mvr rn(.!!Q) = o.u5 v
ι l
R.:0.ll5ν:l15kΩ
' 0.(Πl mΑ
.,
:50MΩ
r'' = 50v
|μΑ
n_ : 0'Φl mΑ : o_()ι mA/V
25 mV
R.υι:[1 + (0.α4 mA/vX1Ι5 kΩ)}
(50 MΩ) : 280 MΩ
as compaΙω to 50 MΩ with R, = 9
: V, - V',
"R,
_ - 0.7 - (-5) : 0.41
1-'18
Ι-
n- =
v"": v'h(l)
0
50V :svΩ
r':
" 0.010 mΑ
mA : 04 mΑ/V
,' : 0'0Ι
25 mA
R" : Il + (0.4 mΑ/v)(5.76
= 16.52
kΩ) : 6l5 kΩ
ο.ι0,ζ: loμA
|ιεε
l0 kΩ
9o,A
Vnε
Vιε,_
1=
ρ._ =
mA
kο
ιbls-'
29.] Ω
Vρ6: (25ιnηh(Ψ) :
:0.8*v.l"(#*) :
VΜ':
[
εo.
7.76 Referring to Fig. 7.36,
Vgg,
_
compared to 556 Κ Ψith RE
if all3π" aτe equal,
R'^:Ψ:2
So,
2.6J mv
90 μΑ
o.ο9
οο,r.
Rouι* [Ι
:
substituιing,
'*(+)
we know ιhat
_ Ι.
"-
V,: V"z+Vrg :2Vrx
so, ysl2
Ι"
vτ= 25 mv 3.6 mΑ/ν
50V
,-=Υι=
' l" 0.Φ mA :556ιΩ
V'2 = Vg"ι: *,,r.rL-,Ι)
since
V', _
R.' =
Ι,
V. -
0.4J mA
25 mΥ
,'=Vn:50V
" ι
04]
.-=r:
''
8-
mA
mA
_
= I1ο.3kΩ
1Φ
l7'2mA/Υ
From Fig. 7.20,
Π'2 fi^l ν
:5.8kΩ
Chapter 7-37
Ro = r.[1 +s.(Re]l r")l
Ro: (l16.3kΩ)
|l
7.at
a) using a hybrid-π model:
+ (17.2 mΑ / v)(10 kΩ
V
7.79 \aJ ι o
ιf β is high,
Ιρ, : Ι6":
For /o
:
"''
|1
5.8
Ι V"r-
kΩ) l= 7.46 MΩ
V"r^
%i"
Ip.ργ
1R[|,, make
yrr. : Vιε, :
V"'a
ιγhere
V"6:0.8u*ηΙ"(*)
= 0.8
V . ι0.025
ν,tn(|ο uA) _
'
\l mΑ./
o.6E5
v
R,n
,4,
ρ:Vuu:0'685v _ 68.5 kΩ
/, 0.010 mA
(b) ιf β : l00 and y,a : 50 v'
50v :5MΩ
,:Vo=
" Ι" 0'ol mA
Ι" :0.οι mΑ
n = v,
- o.4 mΑ/ν
25 τnν
l00 :250kΩ
.:β:
'' t^ 0.4 mΑ/V
For 1o
=
- tl +s.(R.ll
R""l-t1
(5
:
R _
""
R,,
:
+,,'): p*
|1 250kΩ)]
+ l0l t25 k |] ι M]
MΩ
:
_
g.rΥ.r. ro,
using resistance reflection rule:
:
ιzt'".2.5
(.83
V., :
vo =
,,,)
6.9
x l0'|r mΑ
v_l,rΙ)
β}]
Ιfi1o.εzv',r1{r
--1280
6.42 kΩ
ι,ll
:
ffi1
ε2v.lg)(1 M)
v/v
(b) By incιeasing bΙas cuπenιs by factor of l0:
y'l
*β)Γly'll
r,.: Ψ*(l
Icι
ι rc2 ιclJ
βyτ+ y'η l
: β-L +,l * o,[
'
vA\J
.rhere
mΑ
24,:19o
μ
η,s)(.99)
lc,
ι
Ξτ mA
9 Xlο
l
0.82V.i,
G' : v.vo:
Ι
:
Vιεtv-
0.l mΑ
&
is
1[
l
'.,_ 11'Ξ;,5οοk]ιπlΨ]
I
0.1
'1
25'25ο + 10l (24, 390)
,,
and
Y,r"f+)
v)ln (;
25, 250
l0ο
16x11[J!9G5
lιrulLliοψml '|, 1,;]
.r.,) *
_-_l1L
'o,
mΑ:/se
lmΑ
ι _ e(0.7 V/0.025 V) :
a
r.'- R," ]Γ ι.,,ll
, "'' = ΓlRi"
l R.', l|ι.",ll
: ,, ," (i)
To calculaιe R, ιγe find Is
11
i#,,r.;+(l+
oνeraΙΙ Voltage gain
MΩ) = l l2.5 MΩ
Vreo
:
aa)
lα)(25 m) ..
Ri" Ξ 2.5
Vo
+ (ο.4 mΑ / v)(68.5 kΩ
and R can he deιerminert as
(o.ο25
ro, + (1 +β)(r,,
*ι, *ρl(Hil
r")1r,,
7J0 Referring to Fig. P7.80,
ιf oΙ and o2 are matched and
Ιoι : Ιoι : Ι' Vesι = Vor'
So, 1"
:
19
From eq. (7.98),
R".,
using resistance reffecιion rule:
/cι :
Lιcιgvτ+
/c2
:
chapιer 7_38
n," =
'.
Ιc VoΓ
ζ ι
γ
|
Rl"
ι/
l
βVτVι
1pv, +
'ι
ι
οr,',"
('ι'fiLfu_,,
^ _tωl frn"
'
ll i'*'
.'.Power will irrreas€
by
'
l'32
}κ
1
|'32o ν /ν
Α decrease in 6, of Φ9o
P : Ι*V
a
l
:
ιυl
ε^,
:
.64
frρ-Yrη
:
^lτ.2m.Ιo3 μ
mAN
n-' : Ιr': -J-JL : οο+arv
vτ 25m
,"': L = 2Φl0.o4 = 5ko
c.
frtor of l0
(c) Negιecting
1.82
R6
and draιving the hybrid-π
ciicuiι:
+5V
3kΩ
Ιe=o
: e.zv Κ|| 1 Κ
"ι3
V-, : g.,V."1(6.8 Kll r,u)
V"ι _ Vι ' Voz
V., : 9.,(6.8 K ll t,)(viVo
V,'(l
ι
ξ(g.r(6.8 K
1 +
vi
vo-
a) DC bias circuit:
+ 0.7 1D(6.8 K) :
o'7 = 10J uΑ
= 68K
This RguΙe is for 7.82 (c)
:
0
vi
||
9.,(6.8 Κ
-s.,(3 Kll
vo
v,2)
g.'(6'8 K|| r-,11 = ζ(g-l(6-8 κl| r.l))
+
1+
Ι^
-
+ v,1'
Ι-=J------Ξ:|m^
|
^*|ιF*,";ffi]
=
\ι,1vo,
Ve = Vc:1.32+O.7=2
]
v
2m
V6:
Ln.+n*]
^'1
*.
l'r
1'
tΓι(βy.+v^), ιι+ωι]
!
Πτ]Πδ ,
ηn decreases by a factoΙ of 10.
c v_
φ
1D=l03μ:
|[οv.*(1-β)[ffi]]
1
r"r))
|{
r",)
K)(g-,(6.8
Kll
r-u))
s,,(6.8ll r.r)
_0.α4(750)(.64 m)(6.8 K 5 K)
|l
r +.64m(6.8Kll sK)
-
19.5
V/V
.vi
Chapter 7-39
al
t.- -
(Ro)
- lo M
νo
Ι+l9.5
Ι
=
488
krl
vo
vo.-]L
Ri,
e)
v."
Rin
] |00 Κ -
:
R1n
vi
v*,= v,
Neglecιing Early effect
using Γesisιance reflecιion rule:
= _,r.,
488 k
te.sf
)
\488k+ lωk./
There wiιl no longer be a signal feedback.
r., * r.r(βl +
1) + 1
Κ(βr+ 1)(β + l)
ρ'''' : θ.5*βfi{lοll+
!kl!0!)'l
Ι.' Ι.' '
_ l0Ο(25 m) + |ω(25 m){ lοl ) , lo.2 M
98μ
9.9 m
:
+
Rin
25.5 K
25.5 Κ + l0.2 M : |o.252lνl
fπ2 _ιβ,
lωΚ1l' | \
Γ r,ι _β,Ro
ll
l
| jιβr+ l/
β, l
1ω(25 m) , t25.5 K , lω K"]/ |
Ro _
(l0l)(qr.,)'L lol - ια lιlol/
'\
:
R^
2.s + (2s3 +
=
Av
eeo)(#)
:
r4.8 ct
Φ0 V/V
1.
_ 1X1'0ω :0.985V/V
l4.8 + ι, 0ω
7.U
9:
t20
neglect ro
Rι:
Ri" ηow l0 M
|0M )=
c.' - l9.5r
ιlοM+ lο0κ.,
l9.3
7.E3
v/ν
R.ig
:
l0
ko
20
kΩ
Vo
νv
^
R,n
ζ
n'-*
Ri":(ρt+lX2γ")
,:L:25m:25Ω
'
ιΕ
lm
:
:
(|2l\(2.25\
&^
lω
kΩ ηn
c,.'
αzRι
η* η,-
6' 05ο Ω
(129). lο κ
6.050 .\'2',l
= 6' 05ο + 20 K
2(25)
7.Es
: 5 mΑ,/v
R"'. : R. = 20 kΩ
8.
Neglect ro
Find
Gv
vo
'
v'
Ι"z
Iu,
β+ι
lom
ι01
:99μA
99 μA
Ιu'
Ιr,
_
αΙει
:
.!!9ι qq
ιοl
= !Φ1
lοl'
1ρ
μ):98μA
m) =
9.9 mΑ
ΙRL
vou
^Vovi
""_τ
Since
R', =
'
v^
ζ
οc
v,
8-:5πΑ/V:k"(γ)v*
-
oouru
I
r
ξ
t
')
I
Chapter 7-40
1
'i
Ι
fl
ι
v,,:|-m:1γ
)m
:
ιo
\t,1v*1'
:
6'''v,!
Vo
d)
:
}ιs.lιrf
2.5 m (2o
=
:
k)
=
5ο
^
"'_
r.' :
c,.' _
2.5 mΑ
v/v
746
β = lω
i
CC - CE Αmplifier:
(β + l )β,R.
R."+r"+(β,+ι)r"2
r.r:5kΩ
_(ιοI)(lω)(l0k) : _l94v/v
t0k+5k+(l0l)5k
e) Folded cascode:
and neglect
a) using ιhe hybrid
ro
vo
+
vi
BmVL
10
k
yτ
,.
:
1Φ(25 m)
(
\
!!g),0.s .n,
= 5,050 Ω
l0l ./'
-s.Y.(10 K)
v.t"(r")
r.+ l0K
vo
ξ,.
_
0 CC - CB cascode:
(βt + l )α2Rc
K)
ln : _ω.ινπ
= _β1Ξ-g-L =
5, O5O
c)
Also assume
387
μΑlV2
Ψ
:
10
:lx38uxοls'xtο
Ι" :
2
43.5
pA
Ιa:
3Φ.6 μA
43.5 μΑ S ID Ξ 309.6
Ω
.
25m
_66.4
=
_
5,050
5,050 + l0 Κ
standard high-νoltage process: (refer to Table 7.A.2)
/s:5xlor5A
:5 X l0-9μΑ
νιε, u.
ιΙρ: Ι'e
ΞvDΕ: vτ]ι|I:|
lc : 43.5 pA= V"u
43'5 :
, o'572 Υ
= 0.ο2s tn
9
.ιIoκl
ΥN
CC _ CB Amplifieι
9Rc
^ - R.**
"'
2'"
r.:5k
c,, _ lω(l0 k) :
' l0k+2(5k)
μA
Now if ιγe considef the saιne range of cuτrent for
16 ofa BJT and we assume an npn transistor in a
o': _η*;',.ι*.l
:
μ"C.,:
process and
0.l5v<yov=0.4vΞ
(fffi)os,"
Gv
cMos
Thercfore:
Ri, : r.
.\10l/
l0 k + 101(2.s0)
7,87 Αssυme 0.18 μm
refer ιo Table 7,Α.1:
V6n = o.4 Υ ,+
ξ'.
l'J!9]ο-s
Gv:
ιror(i#)ιroιl
Vou: o'Ι5ν= lo:
\i"Yrv'u
-s,(r")(10 K)
r.+ 10K
5,050+10K
..
Rsig+(βl+l)2r"
50 v/v
_
Assumiηg operatioπ in saturation mode:
-(iff)(zi*),".s0)(io
b)
_ _66 v/ν
r-r = 5k
Gv:
β'
ι
- lrn(]!Q)rο
\l0l,,
_ _ _βozRc
.,v
lok_5k
π;Ξτ:
z model:
5 Χ l0
/c = 3Φ.ο μA+ vBΕ
:
0.62|
Therefore:
5o
V/V
43.5 μΑ < Ι6 < 3Φ.6
Ξ
0.572
ν
=
μA
vBF=0.62|ν
ν
Chapter 7--4
Fiμre
I
is for 7.88
ο.5 μm
η,,(v)
(v)
η'.
ΝMos
ΡMoS
NMOS
o.32
-0.54
o.27
o.'Ι
t.o2
1.34
PMOS
0.46
-
0.l3 μm
0.18 μm
0.25 μm
l.08
NMOS
PMOS
o.23
0.7l
- 0.48
NMOS
o.2
0.6
-ο.93
PMOS
-o.42
-o.42
Figure is for 7.88
g-
7J8
0.l8 μm
ο.25 μm
0.5 μm
NMos ΡMos NMOS PMOS NMOS
o,73 0.43 0.88
(mΑ/V 0.62 0.37
ιf the area ofthe emitter-basejuπction is
of 10, then /s is changed by
changed by a factor
ιhe saνe factor. [f ysE is kept constant' then
/.
is
15e
/5αΑ,
16'α
: lο ΑI
Ιr /. is ιepι conslanι.lhen
,42
/., = 10/.,=Ι.e
vιει
V
ν-
=Vr ln10
7.Sg
chaΙtges:
Vnι, ν-
t':134=,(y)
2
X
1ω :14'4 -14
93 Χ 0.252
7.9| iD,
:
ιo"-)ν"c.,(fl'vo'"'
ιιι
\νoc",(!vn'u,
'"'ρ
'
we also haνe g^' : 8.P
ε. :
Y = t0. /^ = 100 uA.
L"
vo"" = Voro
ξ+
rΨ\
-ιtl,=
ιvl
ail.o
Ι
/]
^
\t/,
lα)
7.92
]ζ ''=
ι^
"
:
0.25 V, 1o
:
: ! ι'Y γ'ov=!:
2 "L
L
Foι NMOS:
ι'':
0.48
=
: l0Ξ ysfl V"r,
: 0.058 v or 58 mv
7,90 1voλ
t.oz
Ιc|
ysε
Bε2
'
e
0.41
0.25x
For PMOS:
-
/r+/.α A
+ Ιc2 : lo
NMos PMοs
x
267
vεε,u'
=
PMOS
t:l μm
_ 2X lο0 :1|.98 -l2
also changed by the same factor:
Ι..-
o.
zο
4 -(l)'
lΦ
2Ιo
k,vt
g.:
10
(2)
1Φ:
ψP lω
!:o =
z.εε
mA/v, Vov:
o.2ν
ε.:ξ-l':Ψ:r#
μA
_lmΑ
1
!
t^:
γz
*Y = 2Ιo
oν
' 2ι'Ψ
L k'v'
nL
nov
2xlrΧrc3 _
387
Χ o.22
ng.2
Ch^pLet142
:
Ψ
L
ψ:8.xL
ιzs.z
for an npn tΓansistor:
ε^_
7.Β
_
I-
!=l, -
W=6.3μm
l0X0.025 =
0.25 mA
7.94 Vaγ
"/2
:
x-loX lω
o'25
9.,
i.e.
c-"Ψ:0.5
,
=
!
8'
.-
_-
Yr:
ι'
]zν'c",Y,
1Or
r : --Ξ-Ξ:::ΞΞΞΞ:
Jz zωx 1ο Χ 0.1 x
:
ιo'
ν^
=
:2Φ
7.96
2x |ωx lo
ο.5
τ'ν
l00ο
25
':
0.4
mΑ/V
ι':
ν
1ο0
0
=β8.
Voν = o'2
1.26
4oοο 4Φ
Rj (kΩ)
o'3
o.l
I
440
r, (kΩ)
L=
ΙD
0.t
A.(ν/ν)
,o - vo :vnL - 25" I - 250 k(!
0.1
lo ,D
A6 : 8^r. = 0.4 X 250: ιο0 v/v
^ w.. oν
8. : lL.c.'
=
Ιc
3. (mΑ / V)
1.99
MOSFET
BJT
(mA)
=o.25kr)
ψA/Υ2'
ιoν
Bias cυrrent
1.58 k(}
lo1
"
For case (b) ιηe have
Ι:r\|-'' ,,' l (β +B
ι)s.
8Pu,
'. _ (β + 1)/c_L_0.025
/. 0.l
r:250Ω
Vou=
β : lΦ Vr : 1Φ V
MOSFET: μ,C",
mΑ
15V/V
t x 0-3
387x1o-3Χo.2
oν
Deνice
case (a) Ψe haνe
2Ιo
^V
126
For BJT
μm,
w:
6 μm
\μ.c.'Yrv}u
rssι,n
1o = 0.155 mΑ
:
1.55
mΑ/V
I
4
100 l0
2.5
:}r:ετxflxo.z':
'"
15 k(}
3.88 μm
7.98 BJT:
7.95 Assuming large ro For both tmnsistors, for
:
μ"C
:
W
:
ε.: ξ-l":
1o : 0.5 mΑ
Υ
ψ= 8.xL :
o:Ιr=_9L=4.μ,7γ
νT 0.025
will have
o.2
l00 μΑ,
: 8^r.: I x 15 :
: 1"nc,,Y, vou+
Υ
8. : 4 mA/ν
:
o.3 μm, /2
vou
o.2
: vA - vλx L _ 5 Χ 0.3 _
0.1
Ιo
Ιn
= 44.?i-μr^Ψ
For an NMOS with the same
:
o _2Ιo _2XlωΧlol_lmΑ/ν
foΙ an npn transisιoΙ:
we
7.η L
Vou :
y =10 1o:1ΦμA
L
rΕc",
0.4x1
127xtO-3xo.5
=
μ'C*Voν
'-:
c,: ξYrc*+ cou:!wιc^
40
Chapter 7-43
wLoνc.'
+
c :
Ιn that case:
|
_
ι^"
" vou'!_+,q^ 15x!:7.5V/V
? x6Xο_] Χ 8.6+6Xο.]?
= 12.54 fT
Cr[ : C6yW = 0.37 X 6
:
2.22
'ΓG
8.
Jτ _
"fp
ΣaeJcλ
3
l.55 X Ιo
Χ
21r{12'54 + 2'22)
lo
|5
:
:
|'5 μ"V6u
Cr.
-
2zτI"
>> Cra, c *
C,l.
c^
23.9
wLov Co, ol Co| in cr*
"'τ
Ci"+
Cι"o
_
0.35
x lo-r
2'ιτ(214o +
Cι:
vnu
- 2"5"o'3 _
o,2
2"Loν
2
15
v./ν
r:vAL
'
Io
I,
J" -
0.t55
4
o'3
10
9.7 kΩ
2ncrr. - 2τtx9'7 kΩ x
10ο
Χ l0-
= ι64.2 NιHz
n :
r-
"'
2Ιo
: 82τC
= l.55 mΑ/ V
= 2.5 GHz
g.,
f,ξ
/r
or equivalently double
has to be doubled
multiplied by
4: g-
fτ" s." "[η
fF
lo
ι5)
=
26'1 lνlHz
Χ l0-]
|2
μΑ,
Χ l0'15
_
or /, has ιo be
:
/o = 4 X 0.155 = 0.62 mΑ
/. :
ΜΗz
ιr\ry_voltage process
lo':0.4mΑ/V
ρ = 10/
ο.ο25
C7. : l0Χ 1ο-I2 X ο.4X 10_]
C1":2x5lF=10fF
C":Cι"+cj":14fF
Cμ3Cμρ : 5 ΙF
Jτ
172.|
8.
,;(c|,+
= 4fF
r
c)
o.4X 1ο
2?r(5+14)x10-15
= 3.35 GΗz
1
Ιn order to double
3(n Χ
21τβ4{n + 3ω)
= 0.155 mΑ
_5Yo.3
20{n fF
16
ι^ = !ι'Yv, : l x rεzx 9ο.z'
"
1
: l00 μΑ. High-voιtage process:
8.: 10Χ0.4:4mΑ,rV.
Ca" = l0 x 140 : l{D fF
c": 3400 ff s
ΙV= 6μm
1Φ fF
2VoL
A^
"
l0-9 X o.4 X 1ο
1 gμ)
0.4
voν: o.2 ν
x
Cj" : 2l4O fF
: 0.3 pF = 3Φ
,",6"
and Cr7 calcuΙation.
7.|ω L:0.3μm
-.>
0.ο25
: τe8-:
C.:
Cμ :
GΗz
The approximaιion formula over estimates
because il ignores
4
:l40X l0 ''F: lωfF
C', = 2Cρ = 2xl = 2pF :
1.5Χ45oΧl0'4X0.2 :
f.=
2x"-xo.3'x1o-t2
/Γ
x
MΗz
vτ
ιvnen
=2rYr,
656.8
Ι64.2
7.'0l Ιc = l0 μΑ, Ηigh-νolιage process:
t
n _l, _ lOx l0 - g.4.ρn,
|6:t cHz'
Ιf \γe use the approximaιion formula:
JΙ =
:
1n !.lo-7"
fF
l0o μΑ' Low_voltage process
n:]Φ,lo-':4mΑ/V
0.025
Ca" : l0x4 = 401F
C.:40+ 10: 50fF' Cr: 5ΙF
o"to'
fr.
.,, - Π.6cHz
27Ι(5ο + 5) \ l0 ''
Chapter 7-44
ln Summary:
Standard
Ηigh-voιtage
low_volιage
npn
πpn
Ιc:
l0 μΑ
26.1
MΗz
'fτ
L:
7.102
a)
Standard
1
Ιc:
10Ο
μA
3.35
GΗz
μΑ
10
t'72.1
MΗz
For non
Ca.
l"= ! ι"Ψ 4u= w
- 2X1X1ω
:
MΗz
W
:
Φ mΑ]ry
" = 2τ( l6 + 0.3) pF
7.104
Ao: 8-ro
|ι"C ο rVoν
fot L
_
:
Crι:
2.3 + 25.2
=
x
'
: s.Φtr
x
10-'5
= 2.6 GΗz
25.2]4
:
μm, ro, cs,
n_=ψ=o.4mΑ/v.
6.3
2ΙDΙvoν
/t\
2τ|1wLc.. + Cou
+
Cou)
>>
cs, and
C. :
1
Ξ'WLC.;.
Lo:"Υt""l
vov
2τxlwLC
J
3 V"u
:4rr".V
3ιrn,,
k'w/L
aπ
',z
Αs ιve can se€Jfr caι be determined after knowing
yoy and a, it is noι dependent on either /D or w.
g^ ''' vov ιherefore in order to double,
vov has ιo be doubIed: yov: ο.5v. consequently,
:
39l MHz
Vov.
Jτ=._-_=-_--voνιL
o.8xlo3
W
__
: t^"+: tt^: t+#
(replace ID \ρiιh
o.2
fF :,M ΙF
WLovC* : 25.2 Χ 0.2
-
pF
τ;c;ττλ
lenιly c*,
43.68
fτ
14
ιf ιve assume ιhaι coy is very smal| or ηuiva-
2τ(,14 + 5.04)
b)
^od
_
"8^
Jτ
0.8 mA/ν
ΞwLC"'+wLoνc.,
C",
=
Therefore/4o is only determined by seιting νalues
25.2 μm
x zs'z.'. l x
40 mΑ,/V
pF
16
+ r,
2L ID-
Υ
o.25
"-_2Ιo_2ΧlωXlo-1
""
,'=u^'':25Xl:zsιΩ
" In
ο.1
:
:
Ao g.fo
20 ΥN
Cs' :
= ΖI
χ
2 pF
C.: Ι4+2:
l2't x 0.252
."+
:
τr8^ = 0.35 ns
C.. = -0..loF
1.6
μm' 12= t00 μΑ,0.8 μm - NMoS
V"": 0.25Υ
:
C* = 2x I :
l00 μA
I
4ο ιnΑ/v
27(Ι2ω.6 + Ι)pF
8^
ΞJτ
+ cul
Σ}d-c"
f7 : 5.3 ΜΗz
7.105
csd unchaΙΙged,
vov = 0-2ν'
L:
2
0.2 ψm, 0.3 μm, 0.4 μm
2v^
2v^'L
_ 2xsxL
vov vou
a^: ?9: rovrv
=
o.2
= 50 LVΛr'
7.r03
Ir:lmA+9.
For ono
:k:
vτ
ω
1, =l'Sιr'V'ou
2τL'
^ι,v
:
30 X 10 x 40 mΑ/v = 12Φ
"r8^:
C1 : 2Ci," : 2 x 0.3 : 0.6 pF
c" : 12ω.6 pF
Co,
C*:1pF
1.5
Χ 45o X lo-1 Χ o.2
2Χ3.l4xLxx],o2
={cH.
:
9
_
pF
ι( μm)
Ao (ν
/ν)
f, (cΗz)
o.2
0.3
o.4
10
15
20
53.75
23.9
13.4
chapιeΙ 7-45
7.106
L: o.5μ'π'V6γ =o.3Ycι:lpη
: lΦ MΗz
"fr
f,: βξ=ε-:2πC,!,
2'τ\L
= 21τx1ρFxlα)MΗz
=
62a
μAΝ
ε.=ξ+l"= g-xvovl2=ο'zεrξ
Ι9
:
94.2
μA
:
ι': !k''Ψ4"
z L -'y
2'LΙ.o
k"vλu
_2Χo'5x94'2
ltz
:
:
5'51 μm
5.5l μm
VoL_ 20 - 0.5 _
lo lo 94.2x 1o
,^_V^_
"
3
Ao:8.ro
=
Ι,'ιι : {,
-o
ft2f;u
x rω'z
:
lο6-2 kΩ
66'7 V/V
= l.5 MHz