Chapter 8-1

8.1 (b) Current source requires 0.5 V

(a) Referring to Fig. 8.2, v cu1^a*1 : 2.5Υ - Ο.5 v - 0.8 v- 0.354 v
note that ol : οz : Ι - 0.2mA : 0.846 V
Ι
22
Ι

: 0.1 mΑ Vcιιι-inι - - 2.5 - (qr#)(4 kΩ) o.8 v

"
u: )t'l(vr)t, : -2.3 Υ

ΙΙ
"'"-
Γrh 2(0.1m4) 8.3
,l["Mrυ- O.4 mΑ / ν2 (12.5) Refer to Fig.8.2

: 0.2ν l] - lΙ-_;-2Ιo Γ λ0JmΑ)

l-#:
"ov - U.2v
Vcr:Vn,*Vr,:0.5 +0.2:0.7Υ Νt<"ιw l tl η/0.4mΑ /V'(l2.5)
(b)Ιf y.-:0' (a) Vοs : Vov Ι V, : 0.2v + 0.5 V : 0.7 V

V,ι : V,z - vc Vcs:0-0.7 _ -0.7v Vs _ Vc- Vcs : Ο-ο.7 V : -Ο'7 V

0.2 mΑ Vοι _ Vυz: Vοο- iorRo: 1.0 V - 0'1 mA
Ιοι:loz -Ο. l mΑ
(10kΩ):0v
Vοι : Vnz: Vυυ- ΙoιR, Vor-vor':qY
: 1V (0.1 mΑ)(10 K) : 0 v (b) For i21 = 0'15 mΑ' ioz = 0.05 mΑ,
(c) Νoη if v ΙCM : 0.1 v ,

Vsι _ Vsz - vc V65:0.1v-0.7v

ι", _
l'* ξ-o,o:(+ ').ro,
: -0.6v z,"' _ Γ2(0'15 mΑ)
Since 1 is a constant currΘnt souΙοe, L 0.2mΑ- ι]ιo.z
]
V) - ο.ι V

,csI : i zισ'υΞλj +o.5v:0.745V

Ι,,1and Ιr2remain at Ο.1 mΑ
This means that
Vμ and Vr2 are still V η/ο.+ ma /ν'(l2.5)
: Vc-
0

(d) Vlclι : -0.1 V, Vs Vcs' : Ο.1 V - 0.745 v_-0.645 v

: * Vc Vcs: -0.1V-0.7 V oι _ V oο - ioιRo: 1.0 V - 0.15 mΑ(10 kΩ)
Vst Vsz
: -Ο.8 v = -0.5 V
Sti1l, 1Dl : Ιoz : Ο.1 mΑ
Vo' : Ι.0 v- (0.05 mA)(10 kΩ)= + 0.5 V

Vo' : Vo2: 0Υ Voz-Vot:1.0V

(e) V..(max) : Von ΙoRo- Vo,* Vcs
(c) iοr : ior: 0:
0.2 mΑ,

= 1 - (0.1 mΑ)(1Ο K) - 0.2 + o.7 -- + 0.5v Vcι : υia : J2.vov: 1.414(ο.2 v) : 0.283 v
(Γ) V5(min) : Vr, + V.r(min) |]
ι Gs - / u1ο.z*η + 0.5V : 0.78.1
- η/0.4
]---------------
: -1+Ο.2:*0.8V -
mΑ / ν'\|2'5)
Vrr(min): Vr(min) * Vcs: - 0.8 V + 0.7 V Vs : Vc- Vcs : 0.283v - 0.783v : -Ο.5V
: -0.1 v V o' : |.0 ν - (o.2 mA)(10 kΩ) : - l.0 v

Vo, : + |.0Υ
8.2
Voz-Vor:2.0V
y,_
|r - -0.8ν k'"J!:4mA/Υ2
ι[,
l^' : mΑ]
(d) "' 0.Ο5

|"
οf (b)
iοz :
(a)Vcι:V62_0Υ }opposiιecase
0.05 mΑ

|V
: Jo.s ,n / 4 mA / Υ2 : 0.354 V For example,
"'|
ly"rl : lv + lv,,1 : 1.1s4 v
tpl
'" : Γ2(9'95
,,., 1Α) r]ιο.z
L 0.2 mΑ l'
V) : -O.l V
Vs : Vcι + iyorl : 1'154 v
v-. : i zιο.σ5 mλl + o.5v - Ο.64ι
ycs-]-----Ξ- V
Vοι _ Vnz: _ 2.5r-(ψ#)(4kΩ) Λl0.4 mΑ /ν'(|2.5)
Vs : Vc - Vcs : - v- 0.641Υ : -0.741 Υ
- -1.5 V
0.1
 Chapter 8-2

Vo' _ v - (Ο.Ο5 mA)(10 kΩ) : + 0.5 v

7.0
ΞVcsl :vt+ 2Ι
V o2 : l.0 v - (ο.ο5 mA)(10 kΩ) _ - ο.5 v knw / Ι.

Voz-Vot: -1.0V = vt+ "f2vou

(e) lpι : 0 mA, ir, _ 0.2 mA is the opposite and V., is reduced to ζ thus Vr - -v t'
oΙ'(c): Then z,,,, : τ.1551 * ι,t5

0iι: -L2ιvou1 - ,f,zιo.zν1__ -o.283v

- v, * vou v, : ,[2 v""
Forio, : 0.2 mΑ, Vcsz : 0.783 V, Sothat
^f2
Ιn a similar mannef as fοr the NMOS Differential
ys : -0.783 V
Amplifier, as υ reaches - .J-2 V ou Q1 turns off
Vr1 : 1.Ο V,
1

and Q2 οn. Thus ιhe sιeerine range is

V oz : -1.0 V -+ Vo'- Vo' :'2Υ
''f2Vor-ViΞ -..f,2Vo,
The ιesults are summarized in the tbllowing tabΙe: For this particular case

Cα vi,l ιοι ιD2 v,(v) Voι( Vοz Vυz V"ν: 0.25 mΑ : 0.25 V
(v) 4 mA/Υ'
SC (v) (mA) (mA) v) -vD
Ι(v) fz x -o.zs = υιl ζ "fz x o.zs
(α) 0 0.r Ο.1 -o.'Ι 0 0 Ο -0.35<υ;7=0.35
(b) 0.15 1.0
when V,77:-0.35 Y
ip1 : Ο.5 mA, io2 : o
Ο.1 Ο.05 -0.5 Ο.5
0.645

(c) 0.28 o.2 0 _ο.5 _Ι.0 1.0 2.4

Vs: Vrz: *0.8V
3 Vol:4 kΩX0.5mΑ-2.5 _ -0.5v
(d) -0.1 0.0.5 Ο.1 5 Ο.5 -0.5 -1.0 Voz:0-25ν _ -25ν
0.74t when υ;,1 : +Ο.3_5 Y
(e) 0 0.2 1.C) -1.0 -2.0 ior:0;iυz:0.5mΑ
o.zε 0.783
Vs: υi,ι-tGSΙ: 1'i,ι V,ι
3
:0.35v+0.8V:1.15v
Vυι: 2.5ν
Vo: : -Ο'5 V
8.4

8.5
vc, : ιl'oio1 : 0.11 mΑ
vo, : 0 ioz : 0.09 mA

ΙD, :!l,'Ψ,v^._ν^f
2 ιιL ""
For Q,:

Ο.llιn - jsrιycsι
1

-Ο.5)2

J Vcs, : ο.71 v
For Q2:

O.O9m - jsrιycs2-O.5)2
Vcz.: O
iVcs,: 0.69 V
Vcl : υiι
Vs : -Vcsz : -0.69 V
When all the current is οn Q,
υia: Vsf Vor,: -0.69+0'71
:

l: )(l,,ar)ιV",,-V,)' : 0.02 v
 Chapter 8-3

V or- V o, : 10 kΩ (io' - ioz) 8.8

: 10 kV (0.11 2 0.09) m (o'a^u*/2\2 : γ
:0.2ν \V,,)
thus
+2vorJ=Κ : 1lidmι"
voz-vot :0-2:lο
Q.E.D.
o.o2
when ir' : 0.09 mΑ and ιo,: Ι-(*)Ψ'-_ r
ir, : 0.11 mA
: ζt !-
is the reverse cοnditiοn from the case we just
io,
L YOV
?'vouJk JT-κ
studied, thus υ,, : -0.02 v
+iο,ι: ξtlJκιι-κl
ιhus Δ1 : zιJ-κl _ κl
8.6
Vcs : V,nl Vou : ο'5 v + 0'2 V : Ο'7 V
Q.E.D.
Voq:Vcι - -VsstVcs: -1'2Υ +0'7v For : 0.01 K
: -Ο.5 V Δ1 : 21ωτl(l -0.ου
Voo-Voι - (-0.5 v) : 17 kΩ : ο.198 X 1
R_ - 1.2v 0.1
0.1 mΑ mA : 2v : 0.2Vou
V i,t-u* ovJσ.0|
Voo-Vo' _o.2ν
ρ^: o'4 mA /2 - l.2νΟ.2 mΑ : 5 kΩ
-'υ ForK:0.1

^Ι_2ΙJo.1(|-0.η:Ο.81
Vi,]'^u' : 2vovJσ1
(f),: (r,),:ψτilo"u j"7
: o.894.Vοv
: 0.2 mΑ[(0.25 mAzv2xο.zv)'z]
ι: 2o

(y), : mΑ] 1 :
8.9
0'4 mΑ[O'01 49
Ι : : ι"'
c"" (l)v o, - V,)'
: "
rΨ) : O.l mΑlO.Ol mΑ]_l lO
\L)4
: V,n * v DD - σ / 2)RD
1Ψ
2 - -""'\0.5/" " _
1ιzοοl[2Q]rv.,
2'
O.5)2
V c^(^u*)

: 0.5 v + 1.2ν - (0.2 mΑ)(5 kΩ) : 0.7 v (Vcs- Ο.5)': |/20Υ

Vc'1^ιn'1 - - Vss 1 Voul+ vιn+ vov| Vr'5 : o.724Υ
: - 1.2ν +o'2v+0.5 v +0.2v : -0.3 v - - 2Ιo - 2.40Ο μΑ
8' v(;\-Vh- αn4ν -O.5V
8.7 : 3.57 mΑ/V
lv,rlro, : 160 mv V7 fοr full current switching

(V : ο'ι : "Γz (v.r- V,") : 0.317 v

'ι/2\2
\ vοv )
Tο double this νa|ιe,V^γ, so quadruple 1ρ to l.6 mΑ
,,2
uou - -
(8OmV)2
8.t0
o.l
Vov _ 0.25Υ g^: 1 mΑ /ν vfu:0'8v
v^.,:@:25.tmv
' ov η/ ο.l k -- 100 pA /Υ2
/: 0.4 mA
n
Ι: g^Vov : Ο.25 mΑ
k; : 0.2 mA /Υ2
Ι" : Ι1:
Lo;Yru'o,
W : ι.Γκ'v2]'
L Ln oνJ W : τl(t'v'ov)\
Ι' \n
: o.4' t(0.2xO.253)'zl Ι : 31'.2
: /Υ2' (0.25
o : l - Ο'4mΑ
o.25 mA / (0.1 mA v)'?)
om V ou ο.25_] v
- l.58mA/V : 0'25 mΑ / 0.00625 mA : 40
 Chapter 8-4

8.11

i" : :ιr'f σo, V,)'

I ---.-- ,
-:Υ4o0(vc\ lΓ -'o"a
)
50
2 (?#)'
=V., - 1.5V
: Ucz:0, r, : -1.5v
Fοr tln,
(b) see plot
slοpe οf linear portion
Fοr υ67: υcz:2Y υs: +0.5v
The drain currents are equal in both cases.
: +('!".y,,,)"') : ιRD/Vov
dv i,t\v ov
ForV62:0:
(c) see pΙot
To reduce iρ2by 10Ψο,
when the bias current is doubled, Vρy so
io2:o.9x50:45μΑ
ipt _ 55 μΑ 2ΙRD
voo/v,,, - -(v'o/2 Ιι \'
6- Jl vou{ \"f2 vou)
Urcr - l''"ι l - Ι.47Υ increases by a factor of J1 the slope ofthe linear
Λl400
part has increased by a factor of
Tcst: 1 _ 1.52ν ^D,

(d) see plot

Τhus, V., = ΥGSl - υ652 : 0.05 V
Ιf WiL is doubled, Vρy reduces by afactor at ^f1
To increase i92by l0oΛ
l22 _ -55 μA 2ΙRD
|, γvιtl "Γz1z
so Von/V,o -
iot : 45 pA {zvorΝ \ Vr' l
ιl,352 : 1.52 Υ
The slope of the linear part has increased by f'ac-
of 'f2 compared tο
υa;r1 : 1'47 Υ
tor (b)

ΞVcι : -Ο.05 v 8.13

1 : O.4 mΑ wlL _ 32 kn : Ιιn Cn,

iD2/ iD| ιD2 ιD] vcsz vcst vc- vc' : 200 ψA/Υ2
(μΑ) (μA) (v) (v) (v) VΑ: 10Υ RD _ 5 kΩ
1 5Ο _5ο 1.5 1.5 0

0.5 33.3 66.7 1.4ο8 l.577 o.t7

vou'-
Γ'"ιn - JM/ φ2.32)
ο.8 4"7.4 52.6 1.48'7 Ι .5 13 Ο.026
: 0.25 V

o.99 47.75 5ο.25 1.4886 1.50t2 -0.ο13 n'..- Ι -0'4mΑ-l.6mΑ/V

V ou Ο.25 V

For io' io, : 20 * iy : 4.76 ψA ,^-Vo l0ν

" Ιo - 0.2mΑ - 50kΩ
/

ip1 - 95.24 p"A

Ycs2: 1.154V V651 : 1.690 Αa _ g. (Rrll .r) : 1.6 (5 |l 50)
Thus V61 -Vcz: 0.536 V : 1.6 (4.54) : 7.3 Υ / Υ

8.Ι2 8.t4
(a)v"a: vvD2 v / 212
_
γv 'o 0.05
zR o);-- (r
(Voo- io ,R (Vr (iDι - iD)RD \ VοV ]

USΙΝG 8.2}33rΑΙ\ND) 8'2Δ

8. (0'1/2\ _
\ vοv ) ",η'ΙL5

v.o : ll,v Γ' /) id

(?#)'
t' )(r?)^
^.,) y^,, - ο'05 - 0.224 ν
νb]r5
Ι
8", - Vov
 Chapter 8-5

The figure belongs to 8.1 2

Voa Qno)

1.8

t.6

1.4

1.2

1.0

0.8

ο.6
(d)
0.4 \

0.2 ι
I
V;a (V"")
_1.4 _1.2_1.o -{.8_0.6 -_ο.4 --ο. 0.2 0.4 ο.6 0.8 1.ο 1.2 1.4
\ 4.2
4.4
I
-0.6

-0.8

-1.0

-1.2

-1.4

-1.6

-1.8
Ι : g.Vou : (1 mΑ/ν)(0.224ν) : 4ι lzl/(k;
: 0.224 mA ξ vbv)

Aa: 8^Rο : (1 mΑ/V)(10 kΩ) : 10 ; 1 mA / (0.4 mΑ/V2 . (0.2 v)2) : 62.5

Vol: A7V1: (10X0.1 V):1V BECAUSEwE PΙCKED mΑTHΙS ΙS TΗE

Ι=1
SoLUTΙoΝ wΙTΗ TΗE HΙGΗEST ΑLLow-
W : ll(ι' v'\ ΑBLE PowER. TΗΙs SoLUTΙoΝ wΙLL ALSO
L \n ov1 TΗEREFoRE HAVE TrΙE wΙDEsT RANGE oF
: 0'224 / (0.2 Χ 0.2242)
DΙFFΕRENTΙΑL MODE oΡERATΙoΝ. AN
ΙNFΙNΙTE ΝUMBER oF oTHER SOLUTΙoNS
:22.3
EΧΙST.

8.1s 8.16
* 1 V supplies nοt more than 2 mW Α, : V/V
l:.
5
'""'
2mW : lm.A
%: O.5v Κ/ : ψnCo" : 0.4mA/Υ2 1V *(-1 V)
Ι- 2mW _ lmΑ 0.4 V : 2[2 vov Vov : 0.141 v
rv-(-rv) V\ :
0.5V:lkΩ n^ : l'Υρι: .vzν[Ο.l41 uοs ο
ρ^:lV-0.5V-
" l/2Ι / \ lmA)
0.5mA
o A" 5 V/V : SmΑ/V
w:tt(k'v')-rzs
:--_j--_
6η - RD - 1kΩ
---: L \n ov)

Voν:;:#ffi:0.2V
l7 - Ι _ 1mA
 Chapter 8-6

8.17 8.20
Au 20 - 0.426 mΑ /V ΙVοο
la'] 8m- -" - ;td
(b) 1 : 8.Vοv _ Φ.426 mA/v)(0.2v)
:85μΑ
(c) Vn, : (85 μA /2)(4'7 k{ι)
!n":
:2Υ
(d)Viaι"ιxl: Vcιι + 10 mV : 0.51 V
Voo2V,aψo"\-vt+ ΙDRD
F.7
: O.51 Υ - V, + (85 μΑ/ 2)(47 kΩ)
: 2.51Υ - Vt

8.18
For a CS amplifiΘr A, : - 8.R o - V,,
Fοr a differential amplifier Aa : 8'Rυ with Ι : 2Ιp

So the differential pair requires trvice the bias cur- (a) Differenlial half-circuit :

rent as the CS amplifier.

Τhe power dissipat'ion at the diffamp is also twice
as high.

8.19
To keep the total power dissipatiοn at the diff amp
equal to the CS amp, the bias current 1- 1e16's,
where /p1qs1 is the quiescent current ofthe CS
amp. Τo keep equa1 Aα(olpp) : Aγ1q5) then

8m(DΙFF) R, _ |-8.ιcslRrl
sο- Ι
2Ι o'cs'
V nu,o,rr, V οurcs,
To find the equivalent value ofR| ,consider the
lo,rr, 2Ι orcs,
' lru ou'"' small-signal model of Q3
-
V ον(olrr\ V ou(cs)
1 ',| ov\Dι|Ι)
:

Alsο1r,rroo, : :
| \ιrrrn
: )ι"Yrιvoul' - )ι"Υ(!,
",,,,,)'
so Wottt: 2Wcs
the width is tννice is wide for the diff amp transistors.

Αpplying a tΘst voΙtage τl,,

i.: j!+grrursl
rol
since : υ'r,
τlr"_:

. (1
ι.. _ u.[;+8..lJ
\

υ" | -
so.R^ - i. - 1 * g,njr61
.t _I r g,,l
roι
 Chapter 8-7

since Α, : 8-ι'zσoιll R), Voo

nn-Π-
^- ζ-Ro
|+%ηJ,
Aι: 8.ι z[,,, ,ll (-;^11 when R5 : 0 A,ι : g.R, (agrees with Eqtn. 8.35)

when R, : 2 tι"differential gain is reduced

(b)Ιf r9 is ignoreζ

n'r:=J-
Ι =ι8.
by half
+,
' 6m
ro 8.22
sο that (q) Vcι : Vcz : oV
_ Vνassuming matching components
Aa : 8^ι,' rl-)
\8mι.ι/
Vsl
Vsι' : Vct' - Vcsl : oV - (vt+ vov)
: lτ μ C"Λw / L) η'
since g,,
: -(4 + Voγ)

1 - 8nl.2 - Jz μ" CoX (W / L)|'2 lD (b) zero current flows through Q3

8m3.4 J2 μu Co' lw / L)'\'4 l D Vovι : Vc - V st
* V' : Vc*t(V, + vov)) - vι

F" jw / L\,) : VrΙ V,

- (w / η.' : V6+V6γ
^/;
(c)Ιf μ, : 4ιιp and,L1 : Lz : Ll : Lι : Ι', (c) Vcι : *Vcz : V'o/2

Α." : lΟ V/V : (Eb.@ Γ5' is now more negative than in (a) and Ι/ρ is
now Ιess negative than in (a) so there is a voltage
μ. (w L)3'4 /
across Q3. Ιf this voΙtage is sma1l and if Ι/,_ is such
10 16
τ - ,l\w'^)
that Von > V1 tben Q j νli|| operate in triode.

,o,. : u'".f_'
(Y.2\ : 2s
Γk'"Y,
\w14)
l lz ι|,Υ ν}'
ζtl : 8-z - :1/2k" wν ou
8.21 Vou Ι'
ΗΑLF-CΙRcιΠΤ
: :
b^e\'
Vou
so /rs3
V oνι 8nι

(d) rrsr : !ρι- l

V ovι 8'ι
(l)Rs : J- :.vorr: vou
8mr

From(b)Vrr, : Vc+Vou so Yc: 0V

ξ"ι
(rr) R. : ;]- i.V ouι - 2 Vou
/ 8-ι
so V, : Vo,

8.23
Vcι : Vcz : 0Υ
small-signal analysis
{a)
ν''
tf
vr,: - B.vr,2
R
Vsι: Vsz: -(V'+Vou)
Zero current flows through Q3 and Qa
Viιl2
V'':
ΧJ Q1and' Qahaνe the same overdrive voltage as Q1
R_

'*ε^i
and Q2

Vοι : -q^Vg, Rο : *
r V''/2 ι rοs3 : rοSι :
ε'|77ξ η12]R"
|o"(Υ)'
^r""r,1'
 Chapter 8*8

: 8.24
Γr"(o,).
^'ou'''f_' +2.5ν
8 ιtl'2 : o",',
)o,'(lr),' ru

V ov ι, z :'',,l)o
"(Yr|),',]'
fοSz: fosq: fnsz-ι
_']'
=
|n: "(Υ),'''.,',Γlo "(Yr)'',]

= (Υ)''f ' ζ"*

|'' ^',(Υ).'
^,

- (o'),,

(Υ). 8''l'z
,'2'
rΙl'\
tt!t*t
Rr:2rorr..:
*-,,t1).,,, -2.5ν

For 1xgρ : μA,

1Ο0

(b)Aι: vo,ι/vid:
##kl R-
Vou- Vo'
Ι *r,
1.5
- - (-1.5) :
Ο.1 mΑ
30 kΩ
(See solution tο 8.21)
Vcsl : Vcsι _ Vcss : -1'5-(-2'5) : 1v
8mι'z Rυ
Vovl : Vovι: Vovs : Vcs-V,n: |'o.1 :0.3 V
Vcsο : Vcsι : |.5 - 2.5 : -1 V
Vovο: Vovz: Vcs-V,o: - 1'(-0'7): -Ο-3v
From section 8.23, we know that
8nι'z' Ro A,1 : g,r(rolll rno)
Since p, and Q2 circuits are s1τnmetical

, *(Υ),', With/: Ιnεr:100μΑ

,(Υ),, Ι
lo: λ:50μΑ
fol:fo2:to4:fo5 : ιy41 : 20ν : 40Ο kΩ
ΙD 50 μA
Sο,
80V/V: 9,,,(400Kll 40ο K)
and
8'ι:4Φ μA/V
 Chapter 8-9

Qt o. Or Qι O. Qο Qι
ΙLCo' 30 30 30 9Ο 90 30 90 ψAN2
ΙΙ) 5ο 5ο 1ΟΟ .50 50 ι0Ο 1Ο0 μΑ
Voν -0.25 0.3 0.3 _Ο.3 0.3
-0.25 -0.3
w 53.3 53.3 74.1 12.3 12.3 74-1 24.7
L
vcs -0.95 -0.95 -1 1 I -t 1 v

g. : t, I
Ιo': Ι/3
v#π
Since
Ιor:2Ι/3
ΙVou'\
: |vourl
: |Vouo| -- |vou'| : 2Ιo (b)Voν -- Vcs*V,
Vovl':Vovz:Vοv
- 2(5Ο μA) : O.25 V
400 μA/V ForQl:
so, ζ:\o"(Υ)rι,
Vcst: Vcsz: Vov+V"p: -0.25-0.'7 . ι/
: -0.95 V -"ov-irζπ;
po. [Ψ).^tin.
\L) 2Ι^ : 2Ι
(c) 8. - τ#
vou-+ 8-l :-
Ι" : 3vou

So that
:νc"'(Yr){voul'
8m2 - 41
Ξv*
W 2Iο
L - υoι taid.o l\D
6m1 '\ τ
ι" Co"V\"
For Qr,
= -:*'Rο'υιa
rΨ)
\ L ]1
- 2(100-μΑ) -:24.7
90 μA / v2{0.3 V;2 νoz : l c.rxΨ.Ro
For Qaand Q5,
=:*.Rο.υia
(n),: (Ψ),:,trffiΞr :123
For Ql and Q2'
_u# = (Ξ'.,); o"

(y)': (Υ)'_,#&r:533 =zxξ.no

For Q6and, Q3,
8.26
(f).: (#),: ωf#ffi
Ιn summary the results are as follovrs:
_ z+l By equation 8.38 Α,,
= 8.ιf(9.ιr o), or|l
:
(8
8.ι(R,, ιl R,r)
-r, o), or]
Ιf all transistors have the same οhannel length and the

8.25 same |Vrr|and |Vr|Sinοe g. :

**o
(a)Ι oι : )lr''Yr{v or, - v,)'
,' : with g, and r, the same for alΙ deviοes,
l", : )κ''(zx!)rvo,,- v,)' ξand
:
Since Z65 - ζis equal for both transistors :
^' *((#, Z)t) t ((#, 2)2)
*Ι#: );Ι,, : 2Ιo, 2tDΓ 2vi zvλl
= v."lv.rΙ"||', v*Ι"J
but1: ΙD|+ΙD2:3Ιoι
 Chapter 8-10

o yr' ι^ 4 .1o/'
l- RD .^RD-2Χ_rΟ
(2Ι 'l
ιι|CM' ""
=
\V ou v ov t Ι.)) π.r, ζ
_ 2vi: 7( lYol 1'
= 1.33 mV/V
v'ou -\lv
ovl ) CMRR: μ4 - 5.68 : 4.27 x !03
|A.'| -'|
X
For Α, : l0οο v / V and |Vrr| :
l.]_] tΟ
0.2 ν
= 72.6 dΒ
lv.l
lο00 : 2' n'.
2

ΙV ouΙ' 8.29

vo: Jωo.0'2ν :4.47Υ (a) Ιr' _ Ιa2 _1mΑ_O.5mΑ

2
rrIvo| : |0V/μA
:
γ : -!!J-- : o.M1 μm
=
\rr'' mιtν21v2o,
ξκ''Yrv'ou
10V/μM yo, - ΓΖ J jl_'Δ 1' ' - frro : 0.632 v
For high g, the bias current should be high, but L2.5 mΑ / V'J
V Supplies Vοs : V,Ι Vou:0.7 v + 0.632ν : 1.332 v
with -ι Ο.9 the bias current must not

.^"..d lΨ _ 0.556 mΑ to keep power dissi- 7s : (1 mΑ)(1 kΩ) _ 1v

1.8 V
pation at 1 mW V,u : Vs t Vcs: 1ν + 1.332ν :2.332ν

(b) Α1 : 1'Ro: !- r"

8.27 Vοv
Ι_ 0.2 mAR5": 100 kΩRD: 10kΩ
- A.'Vou 8 Vi v.0'632ν -
1 -
R^ 5.06 kΩ
k Ψ -:ξ rπ mismatch in drain resistances
lmΑ
Lv' ('c)Vo:vDD-ΙDRD
: mΑ- : _ 5v - (0.5 mΑ)(5.06 kΩ)
Vov - Ι /k-l
" t.
Ο'2
0.258 V
_ 2.41ν
3 mΑ/V_
Ι 0.2 mA
o:-:--0.7-55mΑiV (d) From Equation. 8.43
δm V nu 0.258 ν ΔVοι R,
: : -5-06 kΩ
Equation 8.35 ΔVcu l ,rp^^
| Ζrι{{
ο.632V+2ιlkΩl
8n,
- l mΑ
: 8'Rο: 0.775Ψ' 10 kΩ _ 7.75ν /ν
|A,1| : -1'92ν /ν
RD . ΔRl) (e)Triode whenVo : V, - ο
Equation. 8.49a|Ar'| - 2Rs.s RD
V

(Vc'ut + ΔVr') - (Vο- Acν'LVr"): V,

10kΩ 1o/ (2.332ν + ΔV r') - (2.47 ν - l.92' ΔV r')
7lΤδΤο' ' 'o
:0.7v
= 0.0Ο05 _ 0.5 mV/V
Equation 8.50a 2.92 ΔvCM: 0.7V - 2.332Υ + 2.41ν

7'75 - 0.838V
CMRR - )Aλ- _ ο.οο05 _ 155OO - 83.8 dB
: v : ν
1A,
'l
ΔV c. ο.838 / 2.92 0.287

8.28 8.30

'Y
γ μL - 4 mA / Υ2 Currrent source resistance (a)RDl - R"rΨ Ror- Ro Δ-!2

30 kΩ Δ8'
o - δt
δrtl o *Δ8n, bη2
o ' - δm
o
lv oul - Ι /kl,v-L : frΞ;Αi 4.A77 2 2

; -- ζu,tv irt ; ItzV

: ο.353 v
i,,n
"'' grRr, "' - 2grRu
8.: Ι/|Vou| : ## :1.42mΑlV io, - ior: (8.t - 8-r);#;,
|A,1|: 8,,,Rυ_ (1.42)(4 k) _ 5.68v/v
 Chapter 8-1 1

v. s
=
^8'ffi(rt
ι^':ι^":4
β+1
: i-' : l0ο.Ο. 25 mΑ : o.2475 mA
idΙ + i,r2 (8,nt * ε.s
ffi 10Ι
Vcι : Υcz _ Vcc- icRc
V',^ - Vi,.
= (2 o''2grR*
'- ρ^) R*
(2) : 2.5 ν - (0.2475 mΑ)(8 kΩ) : o.52Υ
Vnι : Voz- Vol _ - iarRor+ idlRD.
8.33
1 : mΑ Ιcι : Ιcz:
= - ιo,(Ro- Ψ) * ιo,(no- Ψ) 0.5 So 0.25 mA

:
Vu: Vr-Vru Vrr:0.7 +0.025 lrfT)
V oo Ro(io1 - id2) + r \1/
ξr'o, :
'o,1 for ig -- 0.5 mA ,Vιε 0.683v
Now substitute (1) and (2)
if vB|:0.5Vand V32: oΥ,V'o: 05Υ
Vοa: Ro(o'-;k)-Ψ(};) .Ι Ο.5 mA
Ιεl _ _

o - Vr.- - !! 'Δg- , ΔR,

_
ΛcM
Voo 7+e "id'"r
--------:;_--;; I +e -0.5 v/-025 v

R* π ,R* mΑ :
- 0.5
0.5 mA
Rο ΓL8^ , ΔRrl
=rR*l*- RrJ 1 + e-20
j
rι --
/
-----------77--- -0.5mΑ
(b)R":5KΩRss:25KΩ
'

,*avia/vr I+eo's"o2s
lf A,^ : 0.002 V/V, use the result of (a)
Ro Χ
-lxIο|2A
A.''' - o.oo2 - |Δg' * A&.l
g-
4.85 1Ο"

;., :
2RrrL Ro )
J!Q6.5 mΑ : 0.495 mΑ
So, ΔR, can compensate for Δg. 1ο1

r' : 2.5 v - (ο.495 mΑ)(8 kΩ)

O.Oo2: 5kΩ.ΔRο
V

2.25 kΩ 5 kΩ : -1.46ν
ΔRD : 0.002(5ο kΩ) : 100 Ω icz-0 Vrr: 2.5Υ
so a 100 ohm compersation in Rp(a 2Ψo adjιst- Vε : 0.5v - 0.683 V : -0.183 V
ment) is sufficient. if Vil: -0.5vandV'2:0Υ
8.31
V'o : -0.5Υ izl- 0 irz- 0.5 mΑ
Ιf Αo : 100 (40dB) R55 and therefοre CMRR (Same equations as above)
Vr' : 2.5Υ Vcz: -1.46ν
,

will increase by 40 dB.

v^ Vε_o-ο.683v_-0.683v
no - u*u
8.34
y, _ -Vou : v)
^ lO0
2 ΙΟΟr0'2
\2 )
= l0 V Using eqn.8.66

V,^,* : - .!r*,+ 0.4 v

forV'o:10ν,L _ cc
V
1μm
μm
= 2.5 V
_ 10ο[0'5 mA)ε
uιl * Ο.4ν : 0.92 ν
8.32
1Ο1ι 2 )
1:0.5mAVr. - -Vr,:2.5ν Using eqn 8.67
VC'^in: -Vεε+vCS+vBΕ
Vclη:-1vRc:8kΩβ:100
Vu':0.7 v&iC:1mΑ = - 2.5 V + 0.3 V + yuF

ι-:!:Ο.25mΑ. V,, : o'7v - o.ozsrn(ξ}β) - O665 V

'2
Vr"^rn: -2.2 V+0.665V: -1.53V
vBΕ - o.7 V + 25 rV.l.rΨ'):
\1/ ο.665 v
So-1.53 Υ ζVr"<o.92Υ
Vε _ Vcu- Vaε: - 1v - Ο.665 v _ _l.665 v
 Chapter 8-12

8.35 Ιn this analysis we assume all other parameters of

thο differential circuit are matched
5-0.4 (.I?cr :
Rcz, β1 β2' etc.) :
,},: :4.6mΑ
1
1kΩ
8.38
4.6 mA
;:-0.3+0.7
-0.3 v : +0.4 v (Δl υ.' n,u,
: ι)Γl'2 _ V ι-c - !' n,
+.ο me (b) Ιf the currΘnt is steered to Q l , then
Vcι: -5 + 4.6 X {
ιcι _ Vcc-
1

: _0.4v I kΩ ΙRr,achange
-5V οf : !2'n-
8.36
: : Vcc.achange or +ζ n.
Vuu 69omV ati. : 1 mΑβ : 5Ο
z,cz

:
vCΕ(SAT) 0.3V (C}r.u,n.* - Ι : zs-!n,
Rc : 82 kΩ ycc - -Vr, _ 1.2Υ
=1R. 3 V
:
Ι:20γιA
(a) (d' Ι/2 <2ιA
β+l
V uu _ 69o mV + 25 mV ln r_lqμΔ) - _szs .v +Ι<4(β+1)μΑ
\ 1000μΑ i
_ Vι-Vιε: -575mV Thus,1 : 4X 101 μΑ _ 0.404 μΑ
Vε
1_ mΑ
Vcι _ Vcz _ 1.2ν * (10 μΑ)(82
select 0.4
kΩ) _ 0.38 v 3ν -7.5k{)
(b) from eqtns 8.66 and 8.67
R._3V-
' 1 Ο.4mA
- V-
Ι
V,.,vn* o)f, +0.4V
8.39

= l.2ν 1Iο ,,.R .82 k(Ι + O.4 v .Ι

lεl--
5l 'a\1:υ31-υg2
-urt
vτ
= 0.8V I].e
V.*r," : Vιε*vCS+vBΕ Δi.| iΓ|_ Ι/2: 1r,
= 1.2 v+0.3V+0.575V: -0.325V ΙΙΙ - -.'.
So -0.325 V ( Vc, < Ο.8ο v Define normalized Gain
(c)
Un -
^iΕ|Ι
1. =l.l(Ι) l_ι*r'ou' υn

\21 vi,t .vr0.55

-
I *e τ;d (mV) 5 10 20 3ο 40
'il / v- Gn 9.9-Ι 8.3Ο
0.82:e '
9.8'7 9.-50 8.9-5

-7.1n(Ο.82):Via:5mV observe that the gain stays relatively cοnstant

if Vg2: o,Vιt : + 5mV υpto V,7 near1y 20 mV. Then it decreases signifi-
cantly with the increase in signal level. Whenever
8.37 gain depends on signal leveΙ, nonlinear, distoπion
With only common-mode at the inputs occurs.

vr, - vcz - vrr-.lnr, v, 8.40

DesireVn,1 :1VwhenVd : |0 mΑ1: l mA
therefore the ripple voltage directly appears at the
single-ended output V61 and Va,
Vgq: 5Υ
However, because the differential output .Ι
ιΕl: - 0.599mΑ
l0 mV/V-
Vua _ Vcz - V.' doesnotincΙudethe l*c I

cοmmon-mode output, the ripple νoltage does iεz_ Ι-iεl'_ 0.401 mΑ

V,,,1 : RC Q6i ic) - Rc (Ο.599 mΑ -
not appeaι on the di11'erential output.
0.401 mΑ)
This is an advantage of using the differential
output compared to using the single ended output.
 Chapter 8*13

soR-- lv -5.05k{}
For example, if Vcc -- 10 ! a gain of 200 can be
' 0.198 mA achived by increasing R6' to 10 k Ω, the maximum

Vcvιη,ηx: 'Ι
Vι-e -"λo, +Ο.4v
common-mode input voltage would be Vcc - 5.5
: 4.5 \1 Ιfa gain of30Ο is requireζ ifcan be
achieved by changing RC to 15 kΩ. Ηorverver this
: 5 V _ lΓl TA)s.os k() + ο.4 V : - _ l.75ν.
\2 )
means that Vrcv1ma"1 Vcc 8.25

:2.875Υ 8.42

Voa
8.41 Av: u
ξ7(mV)
Referring to Fig. 8. 15,
'

(a) Ιf we use equations (8.73) and (8.74), we find 40Ο0

that 3500
i.' - Ι
via/vτandi.. 3000
1t l 1
: " aιt- "Via'v7 250Ο
wιth V,,, υaz : 5 mV, andα _ l, 2ΟΟ0

icι - iu' : Ι
--5^ν/r5rv : 0.55 / 1500
' ,
l-te
100Ο

in - i"ι Ι _ 0.45 Ι 500

5mν / 25m v
l+e
Vcz- Vc, : (Vcc -
0
icz Rd - (Vcc- icl Rc) 5 10152025303540
: - Ο.45 IRc + 0.55 1R6 : 0.1 1R.
Αs shown in the text for F ig. 8. 1 5,
uo (0.1
, : ::=
' v,o: 0.oo-5
) 1R. : (20/RC)V/V
A, : ::' Eqs. (8.7Ο) and (8.71) can be simplified by noting
v that i'' * i', -- 1, vrith α: 1' Ιc: Ιε
so that,
(b) Each collector is biased at Vcc -t Rc lει
:
ιεl
:
iΕ| + iΕ2 Ι (AgZ Ag1),/Vt
Ιf we want to maintain the same differential input, 1Ιc
each col1ector should be allοlνed to fall by with V id= as1 - os2t
o' I
./Rc below its bias value.
2
i.' : Ι
I V ,/V-,
so,
l*e
V g@iΦ : V 66 0.5 /Rc - 0.05 1Rc
Similarly,

: cc - ο.55 1Rc
V ,,,: --*--u^,
1+e'n ι
Ιfthisispermitteduntil υ., : 0'
vrith α : l andi6 : ξandnotingthat
Vtcu{^u^1 : Vc(.in) : Vcc - 1Rc
υoa : Vcz - Vcl' : (Vcc - icrRc) - (vcc -
0.-55
ic\Rc)
Ιf the gain is 20 1R,_
Uοιt - ι'"
τ,'':ΙR-( ι | )
1R- - &
'20 sothat
\u "-*'
-
;;7;1)
V,rr,ru,, - ο.55 Α : Vcc-
- Vcc 0.0275 A,
For.1x6:5Ywehave
tr Plotting Vo6 ν s V'1 and' the aοtua1 gain (Vo6l V'), we

seΘ that aS the amplitude inοreases, gain becomes

so, for a giνenV6g, Α, reduces the maximum less rvith non linearity increasing as well.
allowedVl6y,

A"(νN) 100

v ιcιι1^a*1 vcc - vcc - vcc -

(v) 2.75 5.5 8.25

lRc (V) 5 10

RcGΩ) 5 IO
 Chapter 8-14

This table belongs to 8.42

V1(mΥ) 5 IC) Ι5 20 25 3ο 35 40

%a (mv) 498 981 t457 l 8οο 2311 2685 3022 3320

98.7 98.7 9"7.1 95.Ο 92.4 89.5 86.3 83.0

ξcu,ul
8.43 18Ο mΑ _ 76 μΑ'R,
_
1 mΑ The current'μrill divide between the two
3
RΕ _ 2.3"Ι kΩ
transistors in proportion to their emitter areas.
Τhus with no differential input,
Ιil - 2ΙΕ2 and 1r, + Ιεz: 3 mΑ so 1r' : 2 mΑ,
Ιεz:I mΑForα:1' Ι61 : Ιεl:2mA
Ιcz _ Ιεz: 1 mA
Tο equalize the collector currΘnts we apply a dif-
frentialvoltage Va: Vnz * 7r, suchthat
(vBt- vrYvT
iu, - Ι5u, e "' and
(Vsz-v|) vι
'
Iει: lrz: Isεz€ (d) Withοut R6, a V',1of 20 mV causes a differen-

where /ru' : 2Ι suz tial currentof76 μΑ

lvR'_v|\'νT
Sο e "' '' ' _ 2e
lVB1 vLt'vT
"' - !_Ε+
G','
2Ο mV
- J.8 mΑzV : i26.1 Ω; Ι

(vsz Vν'\/vτ with Ru : 2.3'7 kdl a υ'a of 200 mV causes

"' '' ' a
2 _ , 2: or '

,(u
ιι _ VnΥ vτ "(vιz_vsι)/vT differential curΙent of 76 μA
: : G."' : Ξ+4': mΑ/V - kΩ) l

"r- r,
V V Vrl,n (.2) 17.3 mν 0.38 (2.6.]
20Ο mV
So G,, has been reduced by a factor of 1 0. This is
8.44
the same factor by which V,7 increased. So we
(a)
have traded differentiaΙ gain for a wider usable
V", _ 69o mV ' 25 mν h(92!2] - ο:z rnu input range. (see figure 8.18b)
\l)
R":0,Vιι_0 8.45
(b) Eqn 8.73
Ιει: Ιεz _Ι-:1ΟΟuA
Ιcι:σIcι=lΕ1:-
z99-μΔ_ : l38 uA
2

1 | e '"''' Ο. lmA -
ρ -- = Ξ:-J--JJJ-: 4 mΑ 1V
Eqn 8.74 ο.ο25 v
icz : α iιz=iεz - *L. _ 62 ψA
Eqn 8.81
_e'"'"
Ι'oo V, 2-5 mV
R, : 0, V'7: 20 mΥ
γ
' -
ι/2
----'!_
0.1 mA
(c) R'o : 2(.β * 1)r" : 2rτ : 2x 151 X 250 Ω

Vυει = 69Ο mV ]rΨ] : 64Ο mV

25 mV |, - 75.5 kΩ
\l/
8.46
Vrrz - 690 mV 1 25 mV l^ |.Ψ) : 620 mV Rid>1οkΩ Αd: 1ΟΟV/Υ Vρ6:5Υ
\1/
Voεr-Vnεz:20mV Ria_lOa:2r,ττ:zxL
20Ο mV - Via _ V'' - Vu,
: (Vιεl + 138
μA RΕ+ vΕ)
n.._2β_20mA/V
R,a

-(Vuuz+62ψλRΕ+vΕ)
2Ο0mV : Vιι yB2 + (138 ψA- 62 μA)Rε
 Chapter 8-15

- 2(r"* R') -_ 20Ο

ΙΓ V,,, mV ,n
6m vT π"- η 5rnv - -"
Ιc _ 8.Vτ : 20 mΑ /y '25 mV : 0.5 mΑ r"lR":2Οso Rr: 19r"
Ι : 2Ιc : 1mΑ
F.qn8.93 Ao : g*Rc Ri": 2(β + 1)(r" + Rr) _ 2(|01)(2or") _ 5Ο kΩ

n -Aa-
Λa--- 100 :5kΩ '-
r_ 50kΩ : n.4Ω
20mΑ/V 4040

'' vτ 25mV
Ι,:
8.47 ," l2.4 {ι
Ιf Vi1 : 10 m! the corresponding half-circuit Ι :2Ι" _ 4.04mA
input is 5 mV
R" : 79r" -- 235 Ω
v- v- 25 mΥ :500Ω Rc : 25(12.4 Ω + 235 Ω) :
' 6.19 kf}
ΙC Ι/2 50 μA
Ηa1f-circuit gain
8.50
oRc kΩ :
: *R, : 10
20Υ /ν (a) Eqn. 8.93 Ad : 8.' Rc : 40 Υ / Υ
re re 5Ο0 Ω
At one coΙlector we expect a signal of + 100 mV Eqn.8.80 9", : + T,

and we expect a signal of - 100 mV at the other vT

cοllector. ΙcRc :
Between the two collectors is a signal of 200 mV 40ν /ν Ι-R- _ 4Ο.25 mV : IV
vr
Quiesοent power dissipation
8.48
1:lmΑ Rε:200V Rc:12kV : Ι(2.5 v-(-2.5V))=2mW
V6: 100 mΥ 1=?t!:0.4mA
5V
Vr 2-5 mV
γ^: For /: 0.4 mΑ, Ιcι : Ιcι: 0.2 mΑ and
' ΙΕ 1mA/2
--.L

1v
(a| i. :
V'a
- lΟO mV : Ο'2 mA ' 0.2 mΑ :5kΩ
R-:
2(r" - Rρ) 2(250 Ω)
Vcl : Vcz : +2.5ν - ΙcRc : 1.5 V
r, (Via\ : lΟ mν (b) Eqn. 8.86 Ri,, : (β + 1) 2r"
'vu" -
;1v]jJ V'
(b) Eοn.8.8l , : Ι/2
i'' - Ξι ι l" : ρ.5mΑ+02mA : Ο7mΑ
R'ι : (β - (β + .+ -o
,,. :;_2' i' : ο.5 mA - Ο.2 mΑ : 0.3 mΑ
'n(h): ')
: lοl . 25 mV : 25.25 k{L
(c) 0.4 mΑ
Vr' : -igR6: i,'Rc _ -Ο.2 mΑ.12 kΩ (c) For V,, : 20 mV
2.4 Υ v., : _oo.20 mV _ -40Ο mV
Vr, : +i6R6: i".Rc : 0.2 mΑ. 12 kΩ _ +2.4Υ 2

4'8v : 48V/v :
V., +400 mV
β\A" : V^"/V''-
'" 100 (d)
mV
VurΞVrr+0.4 v: 1.5 v -0.4 v + 0.4v
8.49 :1.5V
V6:200mΥ : 5Υ < mV
Vcιι : vΒ|- :
V1 V6" 5

Ri" > 5Ο kΩ
v;d/2 1.5 V - 1Ο mV
: 1.49 Υ
β=tω
By eqn. 8.94 8.51

Λι" - R^V :5v :25Υ/Υ (a) Vsc < ο.4 v

r"*R, - -V,o o-2 ν vr-vc<0.4v
R6: -
25(r" + Rs)
{V cιl + v idi 2) - (V cc - lc'Rc) < 0.4 v
 Chapter 8-16

So V., .u, -_ Vr, + 0'4ν -Ψ i.,R.

Vr, - *5

_- V,-c+ o.4 V Ψ (ι, * ε''ξ)n,

RC Rc
using Eqn. 8.88
+3V Vcι Vcz +3V
Eqn.8.93 Aι : 1'Rc and Εqn.8.8O g, = !
VT

Ιc _ 8'Vτ V"t
/ \
VC'*": 1V
Vcc+o.4ν -Ψ 1V

+(s.
|ικ'v,R.l Ψr,)1
-Vεo
_ Vcc | 04ν _ o,ΨJ
Ψ |o,
u, *

_ Vcc+ 0.4v -Ψ - ι,|v, +ξ) (a) τ,ει - l + 0.()05 sin(ιυt)

υs2'_ | -0.005sin(ωt)
From Fig. 8.17, we see
(b)
that since
V;a _ 10'Ο mV
- o.o'
Vc],'.u, - Vr,, o.4V -ξ _
u("r'Ψ) vΤ 25 mV
the output will be fairly linear. With the informa-
: 5 v+O.4v η#- lΟ0(25.v+l!'v) tion given, we use Eqs. (8.73) and (8.74) and note
thatsince ic : iε
: 5 V+0.4 V - 5 mV - 100 (30 mV) .l Ι
: 2.395 ν 161 - ______;_1.,_; anα,.' : il, ιl
Vnι : Aι'?iι : 100' 10 mV : 1 V
|)eι'1ι|e'''
υ()i: 7'C1 '7ΙC|
ΙRC - 2ΙCRC Εqn 8.80 8," - f, - (Vcc - lc:Rc) - (Vcc - i6-1R6)

: g,nV τ ΙR6 - 2(g. Vr) Rc Eqn 8.93

c
or
Ι
,,,,'- ΙRc ΙRc
Aa: 8.Rc ν''l-
ιιι ι v-ν- ι
l l e |+e'n
ΙR, : )|7rAr - (2)(25 mv)(Ι00) : 5 V : :
with 1". 4 V and lV,ol 10 mV,
/ : quiescent power
- 510mW : Ο.5 mΑ l
ι',.. - ( \',r) V a,,,! ιn,,r - 5Vr____]- \
\l+e '''δ | ' e""5)
R__5\':lOkΩ
'Ι : 989 mV

tcr For l'.* : Α," : V,,,

-
792 mΥ :
-. OV sο_ 'o''
lo mV
79.2

V,o ι,(zs rV y!]

V,,,
'-,
5 \ - Γ).-+ v - 2 "\ 2)
Vc'ι 0)

ι') ι : 5.J \u 5 mV - Α, (30 mV) 3.396

_ξ.395 v :
3.0
_1 , 18Ov/v 2.604
l0 mV
l iιlr 1-,., - 10 mV)

t*ξ2
ιιith 1R. : 4V, andassumingthatα: t,
Ι
|'cι - cz Vr, - )'
V R,
_ 5-2 _ 3ν

I
 Chapter 8-17

(b)υgl:1+0.1 sin(ωr) 8.s4

ιsz:7 0.1sin(a-ll) -Uoιl
: iaβRc
2
arrr.p: :
Vr Ψ-ηy
25 mV
8 FromFig.8.l7.we υιa
2
see that this will οlearly represΘnt large-signal ιb
rτ+(β+1)R"
operation with significant distortion.
So.
Using the same equation,

7'o,!^._-4ν(__ J- -
l ) ξ vn.^
\1 -μ n_2οο'25 , * -uor1 :
"zoι'zs) 2 rτ + (B + 1)R"
-4.0v
5Υ
'' V,ι _ 0.2 v :zs
A,-Vno
waveform is distorted; upper excursions are 1im-
ited tο 5 V
VcιQ)
5.0

3.0

-vΕΕ

ΙVcc

v,a Differenιiαl
2 Hαlf _ Circuiι

via
2
Dffirentiαl
Ηαlf Circuit
notlng
v-
'
|Aλ : Voa
_ g,nι(roι|| rr,) Assumingthat
IF

- r"*RcR"
v,a
ιl'ι : |V.o| which is identical to
Ιc : Ιε : |ν'ol_
Eq. 8.94 The half circuit has

roι: lV^l 1Οv R,: rτ +(β+1)Re

'03 ΙC EC
R'7:2rτ+(β+1)(2R")
lΔ: 25 mΥ
vT
ι, Τhis is equivaIent to Eq. (8.87) :

R;a : (β + 1)(2r"+ 2R,)

: ι
Aι: *(i)(Ψ) 5V
25 mV
Vr.:v'r+λR"+ys
 Chapter 8-1 8

(b)

D iffe rentiαl H αlf _ C ircuit:

vir
2

NegΙecting rρ,

-Ψ:,-(*.ll +)Ψ

^'=l?,1: '-(o' +)
ll

Dffirentiαl
v,a Hαlf _Circuit is the 8.56
Sαme αS the circuit οf
2
Pαrt (a)

So, using the same derivation,

RC
lιl'ι:|v"n|
ul _-
r" Ι Rn
\v,,Λ
R,,, : 2rτ + 2(β + 1)R" : (β + 1)(2r"+ 2R,)
Vc': vΒΕ+vS
Since the quiescent emitter currents do not pass
through the 2R" resistance, there is no drop so that
Vg,,,C&Ι1be lower in case (b) than case (a) -Vrt

8.55 Vo" : 4V, V*6: 40V7

From Eq. (8.94),
40v r

' Rc ΙC
"-,otR" Vr,4Vr
ΙΕ ΙΕ
-v,^ + ,l lf α- 1, Ιq1Ι5,and
Q' Qι
A," : 4ov'
-- 8
5V,

_ V,,
 Chapter 8-19

8.57 (a) Single-sided output: Ιgnoring rρ

^
ncM _ vo, -ic,Rc,
- Vιcl'ι ηlη+rR*)
: -αRct
2R*
r" *
Ιfα - 1 andrn<<2Rgg,

lΛ l R,^l I.Ol(2gk!υ =
|Aι-u| - ,or': - lMΩ
0.0202 οr

-33.9 dB
(b) For a diffeιential οuΦut :

-α(].οl)Rc.,v
'Ιv ol - _τl;;ττ Ι(-l'|

,,'οz-- α(0.99)Rc. ,,"ICM

From Example 8.3, 2ηt1
v'o : R,o "
V,ic R,;ε f R;a V'
'οι1
: V'_
'ο2 V'-
lοl -α(0'02)Rr u'r"
ru
2Rfft
From Eq. (8.94), "
'' α(2R^\ A." =| V
- :ι0 |
02/Rc
and if α :
A,:=+::'---"So,
v
" V,,t 21r"FR") |V ,r"|
"d
2R r, ι r"
Ι, and

r" << 2Rss,

7),,,Ι aia
\]V V nι
ο.ο2(20 k)
ηr- η v* o.^"
2Rr, -
-(0.02)Rc
2(0.5 m)
- O.O0o1 or

: R'o α(2R.)
-67.0 dB
R,;ε * R;a 2r" * 2R"
where R,, : (β + 1)(2r"+ 2R")
Rid :
Ιf
Rr;s * Rιl
O.5. Γ_
**'=
αRr'
ιr., :
" 2(r"* Rs\ Vl.ν
μ'
Ιf β is doubleζ α may increase slightly' but not {,
much if β is >> 1. **
*,
Rla
Ηowever' with
= + R,d
R.,s = - O.5. this imp|ies _l_
Ξ
that R,o : Rs,.8 .
γFF
Common'mode haξ ci rcuit

Sinοe R,, : (β + 1)(2r"+ 2R"),

8.59
will increase to approximately Referring to Fig. P 8.59,
=+^
R,iε * R;a υo αRcz
2Χ _?_o.ο.l υi - (2r" + 2R")
Χ+2Χ 3
vr: v, _ 25mV :125Ω
So. c,, increas., b,
"
0'67
- .34 or 34Ψο ΙΕ Ι/2 0.4mA/2
Ο.5c)
' Assuming α - l,
8.s8
2Rεε:2(ο.5MΩ):1MΩ
υ"- 25k :SOV/VS
q-(2)(125)+2(12s)
Ιf co1lectοr resistors are 20k with a tolerance of
Rι: (β +1)(2r"+2R")
:]

7o/o,|et R61 : 20 k(0.99) : 19.8 kΩ

;
: : : (1o1)t2(\25) + 2(Ι25)1 : 50.5 kΩ
R62 20 k(1.01) 2o.2 kΩ whichwill
result in the worst-case. ur.-
 Chapter 8-20

8.60 Since R" >> η,

Referring to Fig. P 8.60 R.
ly,,l _- 1.o
with each transistor operating with 1, : 0.2 mΑ'
|ν,n| 2 re Ιf o.- l.
,' -V, - 25mV - 125()
|'"| _- R.
lyl . : -:-:::-
2Κ
ΙF 0.2 mΑ - 20V/ν
|v,,,| 2r" 2(5Ο)
Y- - :-!!^ if r, is ignored αRc
Vi (2r.+ R"l (b)Α.-..-
'"'- 2Ro + r"
o: β -l.so
β+1-
v" _ z5k :5ov/v
Vi Q)(ι25) + 25o
Ri" : (β + 1)(2 r" + R,)
Ri, : (1Ο1)t2(125)+250] :50.5 kΩ

8.61
Refer to Fig Ρ.8.61
(a) Αs a dff'erential amplifier, ιhe gain is given by
4' αRc (l
υi
}

2r" cοmmοn-mοde half circuit

(b) Transister Q', canbe considered as a cοmmοn
Ιf α: 1.
coΙlector stage. Ιt is biased at1l2 and has a resis- A''- 2Κ
ιance re2in its emitter, thus.
-- 2(4.3 K) + -s0 -o.23
(c) CMRR (dB) : 20 log,"
i,--:-:+ (2)
t t"2 Γ"τ /Γ"
I l''Ι
-
|u,,"u'n| 2Ο lοg',, l4l -
'"|ο.z:|
38.8 dB

(d)
vΒ| - 0.l sin 2π'X 60r + Ο.005sin 2ιτx 1σJo t
Vu2 : 0.l sin 2π'Χ Φl - 0.ΟΟ_5sin 2z X 100Οt
V, : Ο.Ο1 sin 2τ X 10ΟΟ r
V',,,: 0.| sin 2π-X 60 r
Sο that
l Ι/
v,' - ι A,.'vt,'
L

|----|.vi,Ι
|v
v"(t) :
'λ
le - 20 [0.Ο1 sin 2π Χ 1σ]0 τ1 + 0'23
[0.l sin 2ττ x 6o t]
Νoιι-. Q' is connected in the commοn-base cοnfig- v,(t) _ 0.2 sin 2π X 1000 r + 0.Ο23 sin 2ιτ Χ 60 t
uΓatiοn. Ιt has an input signal current ie. Τhus its
cοlleι-tοr sigπal cuπent (in the direction indicated) 8.6J
ιτill be i.' - αi (3) Referring to Fig. Ρ8.63' the differential haΙf
Ttε output ι'oltage wilΙ be a, _ icz' R,' (4)
circuit becοmes:

Cοmbining equations (2) (3) and (4) provides:

iι' αRc.
i, - 2r"
'ιν-hich is identical to the result found

atrοι'e in paΓι (a)

8.62
Reler tο Fig. P8.62
u'ith Y- : 0,V,: Ο.7 v
(a| t _
Vε- Vεε _ -0.7-(-5):1mΑ
RΕ 4.3 K

r:: vτ 25 mV
= 50Ω
' ΙF 1mAl2
 Chapter 8-21

α: β -1ο0 8.64
β-1 - lO1 '''
with 1

(a)r.: V': V' : 25 mΥ :100Ω

v^ 100v :20ΟkΩ ΙΕ Ι/2 0.5 mΑ/2
'' - Ο.5
_
Ιc mΑ
r..:- vA 1Οo v 4ω kΩ
v, 25mV :5ΟΩ '' ΙC - 0.25 mA
' ιΕ
- 0.5 mΑ
with r. and R so large and α: : 1ο0
o
=e'
β+1 1oo
"
η is >> R- ll
'" &.
2
υrι will have some eΓfect. Ιf we
- 1, we can calculate the differential gain as

ignοre this, the differential part of V, and V" are o,-2(Rc || r) - & : 2Ok * 2οοv/v
2r" rp 0. lk
related as A"" : ι!-λ :
lV'ol (b) R,, : 2 (β + 1) r" _ 2(101) (Ο.1 k) : 20.2 kdι

Λ -:-Ξ αΔR,^ ΔRc

*,ll + ι^\ _4
|cl -Voa _
- 2RΕΕ + r, - 2Rμ
* 0K|l l0K:25νN V i,^

,'2+ξΞ .50 + 15Ο

- (0.ο1)(20 K) : 0.0005 V/V
2(200 Κ)
R,, - (β + 1) (2r" + R')
: (101) t2(50) + 3Ο0] _ 40.4kΩ (d) CMRR(dB) - 2ο log,.|lz-|
Ιf we ignore r", the common-mode half circuit lA,rl
becomes: : zo tog,"| 2Ο0 l
|0.0ο05Ι
:112d8
(e) Using equation (8.1ο3),

1+&
R,'-βRεε;e
ro

Γ lΙ 2οK
2R,,' 200 kΩ : tω(2oo Κ) 1ο0(40Ο K) : 9.76 mΩ
l+2οK+2(2ο0K)
40Ο K

8.6s
νlith +1% tolerance on R", 1: 100 μA, β : 5o,Vo: 2qy
Α.- - r.lΔR.
"'' _
_ O.O2(l0 K) :
Roo 50 + 2Οο Κ
o.οO1 v/V
For Q.,

Rεε: fnι : Vo : 20ν : 2OOkΩ

Αn estimate for Rr, is 1 0.1 mΑ
*(Αlternate method)
Rr.
= fo : fol : fo2:
2Οv : 4Ο0 kΩ
11
+ l)(R56) : l0lX200 Κ) : l0.l megΩ h: 0.05 mΑ
7ιο Using Eq. (8.103),
Ιfeq. (8.1Ο3) is used,'(

1+&
l+&
Ri.':βR66;e R
'-= βR'" 1 * !"fu to
r ,
Rc
ta 10K 1+(5ο)(aω
: 1OO(200 κ1-___--!!9G99Ξ)- : 6.56 megΩ
Ri,^ : 50(2OO t) ' k)
Rc + 2(200 k)
l l ι0Κ+2(200K) 200 K
1+
4ωk
Ιf Rc << REE
and R6 << ro,
R,". - 5Ο (200 K) (.5) : 5 MΩ
 Chapter 8-22

Τhis figure belongs to 8.65

Γ
Rir'.
vu o_{ et

8.66

Rc: t0k

Vtt
Equivalenι
-vΕΕ
^ v, |Ον :2okΩ
Κg5:Γul,:i_O-5mΑ From Eq. (7.88)
1
R"
r':r,:r 25mV -lOOΩ =
-V'Ι /2 - 0.5 mΑ/2 'β,r,,,
R_
"-- 1ι tοοlιzο ιl : ]MΩ
Sinceα: β -1Φ-t. 2'
β+1 1Ο1 ^
ΔR.
n''=2ηaγ,1
A.-R'- lOk
r| ο.1 k - lOΟν/v (0.02)( 10 k) _ Ο.οΟ01 v/V
rιΔR_ _ 2(l m) + 0.1.k
/L\ ^
|υι ^ιn]
(0.02)( Ι0 k)
2Rru1 r" ,(20L) l0'I k
cMRR(dB) : 20 : zο lnn,,,l t00
: -
vlv '*,,1*l "lΟ.ο001l
|

Ο.00499
:
cMRR(dB) - 2o ιoε,,|!ι1 = r0'"*,,'l##,l 12Ο dB

:86d8
 Chapter 8-23

8.67

R,,: (β + 1)(2r" + 2R,) : 161ρ

R.
Α'-10ο_
" 2(R"+ r")
ι,,: ξ-(ff)(i).'",, +R.: l0Κ
Α_--. : Ο. l
R'
: =
i,,
t- (#)(z)'^'' 2Ro* R, Ι r"
+ft,>50kΩ
νcι : Vcc - lf, nrjrin.r
ζn, - For+2swingξ,:V.,
:Vcc
:
T

_ nrjsinrl λ*r:'
ζn, -
υcz Vcc
ff *Vcc:2+|o-3x104:12ν
Vrr,vrr> 0
:0 Choose V,,: + 15 V although 12 V is ok.
=10-51*0.-51 2 R'"^: (β + 1) (2 R" + R,-l r")
1: l.8 mΑ
1Rι,-: MΩ
Vn : V.': 1Υ 5

Α," : 2q kΩ. where r-' : 25 : {ι

2ro 0.9
27.8 8.69

Thus' Α, : 360 V/V Τaken single-endedly Αcn. : :k

'lJcz
- 1)cι : 1.8 sin al r, V " 2R"
Let colΙector resistors be R,. & Rc + ΔR., then

Acm -- * (n. *
2R"' ' ΔR. -
Rc)

|i'ψr:'t"', ΔR.
u''tγ q.9_+ 1V
ι
2Ro
Which can be written as
.
Λcm':
ΔR,_: Acnι- ΔR.
αR7_
V,, "
2Ro Rc "Rc
CMRR -
Ad- 2.AS
A,rd Ar,
,^-l!
Rc

:Ar'2
8.68 Acm" ΔR.'
At Ι,': 7 mA, r. -- 25 Ω Rc
R,, : (β + l) 2r": 5.05 kΩ < 1Ο K 2 :
Ξ Thus.20 lon
" 40 dB
need emitter resistοrs ΔR.
Ιn this case: &
-+ ΔR, / R, : 2Ψο
 Chapter 8-24

8.70
The bias current Ι wiΙ1 split between the two di1'-
: (?)(ξr) : (o.15 V)(0.o2) : 3 mV
ferentiaΙ transistors according to theif base-emit-
"",
2!
So both Α, and ξ. increase at the Same Γatio Since
ιer areas. So. ιhe larger deνice will ,unν
boιh are prοpoπiοnal ιo u/ξ
Amperes,
8.72
while lhe secοnd ιransistor will carry {-l Α Worst-case Δ7, - l0 mV
Αssuming, for example, ιhat Q, has the larger
v' Worst-case 4& : o.οη

"' - 2Ι/3 undr,=vtl/3

RD
arex.r,
Worst-case
/ L) _ Ο.Ο4
Normally, we cοuΙd apply the common-mοde half (w/L)
^(W
circuit symmeιry. Buι here. ιhe amplifier is nοι
From Eq. (8'1ο8),
symmetrical.

'' r,
Ιf fl -
3V'rnd
""- ,'':'u'.
'"'
vos|(d,'"roΔR,) -Ψ ^# -+'{ο.ο4)
2Ι Ι
i.'Ι + i,2 - 4mV
From Εq. (8.ι 13),
so, (ri : -i.zR,, -(-i,t R,) : -
:Ψ
(ir1 ir.2) R,.
withα-1 vo,,(d'".^(Ψ))
v,
^#i3
i', l i , :: i"| l ii --JΙ.lΙ 15iηqe' R,, >> r
Κει
l
:Ψ.(O.04):4mV
!n-
A''.'|V"'1 - 3 12Κ From Eq. (8.116),
''|m Ruu- 3(50ο K
-0.Οο8V/ν Vοsz(d," tο ΔV,) _ ΔV, : 1Ο mV
",l
)
|v ,'
The absolute wοrst-case total offset would be
4 + 4 + 10 _ 18 mV Howeveη sincethese off-
8.71
For1: 160μA' set sources are not corre'lated a realistic observa-
tion might be [from Eq. (8.1 17)]
Iιo--Ι -160μΑ-,- 80μΑ
) Vnr: lvο.S_,l t vοS',2 I Vos- l
ν
)

8. : F;Ψ Jλ4 mΑ ' V')(s0 μΑ) : ι (l0)' :

":
Γlυ

^l(4f 141'+ l1.5 mV

: 0.8 mΑ/V The major contributiοn is from the variation in V,.
R, : l0 kΩ, so that Ιf we attempt to compensate for V65by Changing
Ao: 8,n R,,
: (0.8 mΑΛ/) (10 k) : 8 V/v RD,
using Eq. (8.108),
v ou oo'o.
(Ι.,)r4&) ΔR, we have. t 1.5 mv - 2RD '
v^. - \2i\Rn)
*h"r"
Ru
- c}.02

ΔR, : 11.5 mV :
(worst-case) 0.115 or 11Υo
RD 0.2ν / 2
. Ιn
slnce gn Ιf Ζρ5j is reduced by a factor of 1 0,

vo, ' Jt4\' t (4',' t |)' : 5.74 mv

',', Ι τ'
Jj:-Ξ-0.
V 80 μΑ
lv
2 g,n Ο.8 mΑ/V So that
Therefore, η, : (0.1 V) (0.02) _ 2mΥ ΔR :514mΥ:5.74%
360-μΑ _ RD 0-2ν / 2
For / - 16Ο μΑ. /, = I 80 μΑ
2
Ιψ 8.73
8,' - |2k"'+ιD - J2(4 mA/V))(0.l8 mΑ)
Ιο _ Ι μΑ :50μΑ
t -
lνL 1Ο0
: 1.2 mA/Υ
2

Au: 8mR,,_ (1Οk): l2vlv

(|.2 mΑ/V)
Ιο: 12 κ"(l)v3"-
V", l8OμΑ :
2 -Ιog'"- 1.2 mAlΥ O. 15V

sο ιhaι
Vov -
 Chapter 8-25

: 02ν 8.77
From Eq.(8.108)
ιor:o,ξnr_ orξR,
V ιo ΔRp) : : o
ιo.orl
651(due
Ψ Η. 1u = !Rr("', -
:5mV ",r)
β' β,
2'\β| +l - βr+l))
=!R-(
From Eq. (8.113),
W/L'l : V ou . / L')
v
^".( aun ιo ι wlΙ-)
""\ 2 ^(W(w / L) For β,, β, ,, 1

- o'2 v.(0.05) : 5 mv : !R-. β' - βz

2 ' 2 ' β,'β,
^,,^
From Eq. (8.116),
= Ιρ-(t -
t1
Vrr,(due to ΔV,) _ ΔV, : 5 mV 2'\β, β')
The worst-case offset is
: , - Rc
no-i-rν, ΙR-^
Vos Vosι * Vo''l Vosι: 5+5+5
:15mV
The root-sum-square value from Eq. (8.1 1 7) is u",:*:",(L-r|)ο""
vo,: lv3r, + vSrr+ v3r, For β' : 1Ο0and βz: 200

= ilΞΠ'πry:8.66mV vos:2s(#- *)
8.14
= -125 ψΥ
Τhe ouφut offset voItage is ΔV. - ιn,
ξ 8.78
: 8'Rc: ΙRc CΑSE 1: BJT Diff. Amp.
Aa ' Ψvr Rr: 2v.t From Eq. (8.121)

l|/,-lΔy.|
|vosl - Λ --ΔRcI
1
|v,,l : ,(Ψ) :25 mΥ (O.04) : 1 mV
"n
Ιη CASE 2: MOSFET Ditr Amp.
2V, From Eq. (8.108),

rΨ') :
- VT'\Rc/ 25 mv(0.08) ,"' : (Ψ)(t ) : Ψ (ΟO4) : 6 mv

|Vo'|
: 2mΥ
Ιf the MOSFET widths, are inοreased by a factor
of 4, and sinοe 1p must remain constant, we see
that sinca
8.75
From Eq. (8.126), ι": )κ'(ξ)v3,'
Ιv,,l : ,.(f)
The new V ov :
lvorl : 25 mV(O.10) - 2.5 mV

8.76 ofits orginaΙ value.

So, the new offset voltage is
Δυgg: ΔR6ξ
y^:(1ΞΨ)ιo.ωl:3mV
R. : R. : ΙR.c
" 2Vr-o 2vT+ιRΕ
rolR" _Τ
ΛJ

' ι\Ε 8.79

Since the two transistors are matched exοept for
-+)
-

Δo,
',vos: --;- : their V1 value, we οan Θxpress the colΙector cur-
A,Ι Ψ(,, rents'ιvhen the input terminals are grounded as,

ι,, : l,(l -Ψ^)

 Chapter 8-26

l,,: ι,(ι -ξ) t,,(n, Ψ) (j +)'"

Where 16 can be determined from
Ιr' * Ιr': Ι = /,,(R, Ψ) -(l-ξ)η
+Ιc Sublracting οuι the { r, lerms, we have
t -vr, -v- " 2'
Vo, vtz. o+)-
l,,(n, -
Νotethat for V *a4 7ll. V or. l r: j τnus.the
+
Ψ)- ξ,":1,,(*, ξ,"
Ιn terms of the emitter currents' this becomes
differential gainΑ1can still be written as (t
t---t Δ/\
nu-
^ -Rc - ,V,
Ι Rc \2 2)(o
Τ 'ΔRs)-ιl..
ιη υl"r-τ) 2',
The offset voltage at the output can be found from
r/ - Δ/)
ΔVc: υcz- Vct: (Ιr' - τ)(R., 4&), U,
-\'(β+l)\"'
Ιrr)R,

*,(ηy-,-y"^)
2 ) 2''
=
', o'
Subtracting ,I
" --2(β , , from each side,
+ l)
:'*(Ψ^-Υ"^)
IΔRs ΙΔRs ΙΔr"
: ΔV 4(β+1)- 2(β+1)- 2
Thus^ V^.
A,ι'' /ΔR. ΔIR. ' ΙΔr"
4(β+1) 2(β+1). 2
u":
'(|#-ξ)
Combining terms, we have

For
V66 _ 10V,Vιτ:.100V and
2(β+1) (β+1)
R' 1ΔR'
\(β-l) l r-)")- 2(β+l) .o,hr,
V12 : 300 V Δ1r
yos:2s(#-#) Λ, 1ΔRs 1

2(β + 1) Rs
*'"
= 1.7 mV (β *,)

8.80 rrfr - l.
^vC- ^ΙΓRΓ
|/ - /ΔRsRΓ .
AvC
l
^
2Φ'D
Gjτ'
η
'"
Since V,,. is always referred to the input and
Ao: I' Rr,
1ΔRsRc . I

2(B+l) R"
(β+1) '"
JΙ,-

1Ι
!/
',uo'- ΔV, :
η, 8- Rc
, * IΔRs

Vut
2(β + 1)
,-[ffi.'"1
ι/
νoS -
/ΔRs . 1

Consider only the incrementa1 currΘnts invoIved.

2 8. Rs * (β + l)regm
Assume the mismatch ΔR, is split between the Since (β + 1) η : rnandroB,:9,
two base (source) resistances. The emitter currents
wiIl be differenι, as shown.
(*) oo,
Equating the voΙtage drop from each grounded
Vos -
, ,8,,Rs
input to the common emitters, we have. β
 Chapter 8-27

8.81 Using Eq. (8.129),

(a) Referring to Fig. P8.81
o, : lΙ ιt - Ι : 11.244 - 3.086| μΑ
icr :5k(1.04):5.2kΩ
Ι
'z|
: 1.84 μΑ
Rcz:5k(Ο.96):4.8kΩ
Offset nulling will occur when x is such that
R61 + x(l k) : Rcz + (1 -.τ)(l k)
8.83
Assuming that Q, has a junction area twice that of Q,,
Ιn kΩ we have 11
'
5.2+x:4.8+1-x l'' : ΞΙ and Ι', : \Ι
2x: 5.8 - 5.2 - 1, so that
Ιf β >>1,1.
x: 0.3K.
(b) There are two ways to interpret this problem, ΔV,._V,,-V-=\ ιn,
but for this οircuit, the end result will be different.
One way: assume that the transistors are such that Nominally. Ao:8'ρ,: Ι!,2
VT
_- p,
20oλ of tbe nominal value of area is equa1ly split
between Q1 and Q2. For instance, with α - 1, ΙRc
Ιc, - 1|0E"(')- , ,ιΨ) : O.55 mΑ v., - lk: Ξ _- 1r,, : 21ιzs
Ad lR, J ^ν1
,V,
Ι,, - 90Ψo(i) : .r(Ψ) : o+s ,n :16.7 mΥ
Thus, small-signaΙ analysis prediοts that a 16.7 mV DC
Offset nulling will be achieved ifx is such that
voltage appΙied asVn * V":
mV would restore
16.'7
(0.55 mA)(x + 5) kΩ : (0.45 mA)(1 - x + 5) kΩ the current balance in the pain and reduce ΔV. to zero.
Ο.55'r + (0.55)(5) : (0.45X6) - 0.4_5 x Using Ιarge_si gnal analysis,
x: 0.45(6) -0.5s(s) : -0.05
vιl-Vε
So, perfect offset nulling is not possible under ir, : Ι"re ι

these conditions.
VΒ"_VE
Second Ιnterpretation: Vτ
Ιf the area of Q1 is 120% of thje area of Qr, tbe ιΕ2 : Ι.2e
οonstant current 1 is split such that Thus,
ιr,-#orι) : #ο mΑ) : o'545454mA
VΒl_VB,
it, 1.,
Ξ.e
vr
i''
!99ι l mΑ) :
-:
r., - 220 0.454545 mΑ
Ι r,
To restorebalance making irι : iεz,
Perfect offset nulling is achieved ifx is such that
VBi_Vι,
(0.545454 mΑ) (Χ + 5) kΩ : (Ο.454545 mΑ)
v-
(1 -Χ+5)kΩ |:2e ' οr
soΙving' V n, - V ο,'
x - 0 (the wiper would be all the way towards the _ lr(1') n.
vr \2)
R,., side.)
V", - Vur: -V.ln (2)
8.82 So that
From eq. (8.131), V", * Vo, : η.ln (2) : 17.3 mV so, results are
similar.
'
ΙB'a':' Ι l Ι/2
β-lΞ β.--I 8.84
The solution to this problem depends upon what
- 500 μA : 3.Ο86 ιιA is οausing the offset. The mοst-straightforward
2 approach is ιryhen the offset is due to the toler-
80+1 ances in R.. Ιf sο, wΘ can assume that the maxi-
FromEq. (8.130), mum offset of -ι3 mV occurs when R.., and R-
.
,B'i' - ,Ι l ι/2 are at opposite ΘΧtremΘs. That is,
p+;+: β.*+ l
R-, : R^- ΔR.. and R-" - R_ +
ΔR'
'2'2
- 500μA _ l.244 ιA Using Eq. (8.121) is apprοpriate to find ΔR :

2
200+1 tv,,t : ,.(1T) sorhar,
 Chapter 8*28

ΔR- : ΙVosl .
r-' : '1 mV . 2ο kΩ : 2.4k{ι _'μ# : 10Ο μΑ
' V, 25mV
Ιf this value is added to the smaΙler resistance (R.,
Ιοι _ Ιοz:
in this case). each collector resistor wilΙ now be \ν"c""(Yr)vi"
2Ο kΩ + : zι.z kΩ and the circuit ,*vill be so.
^_Β 2
ιv\ _ 2Ιo
ι7/'-, - ;i:ξ"
/ 2{ l00 μΑ)
balanced.
Ιf an offset nulling arrangement like Fig. p. 8.8 1 is
4Οπ;]τlα'τ;
used, the same 2.4 kΩ νalue could be used for the : 12.5
potentiometer, making it possible to adjust for offset For Q,and Q,,
of + 3 mV. The appropriate singΙe-digit potenιiοme-
(W\ - 2lo
ter value would haνe to be 3 kΩ to cover all cases.
ι7,,, . - LofilτΣ,Or"f
2( l00 μΑ)

Note: Ιf it is assumed that the off'set is from οther

sources ('eg., Vrn, β, or οther mismatches), then :50
the sοlution would be slightly different. (b) Frοnt Eq. (8.141),

o,, : 8*o"2l| r"4)

8.85
Referring to Fig. 8.32 (a),
ξ:
Ιo : ! Since r.,2 = fu4= Γuι Ao :
Each transistor has
2
Lg'r.
ΙD
lv o'Ξo^-o--o:o-
Γn) : Γn4: r,': ^l
δml 6m2 6nι' δ]D4 6m
vov/2
- ΙD
From Eq. (8.142), 100μΑ:1mA/V
.1 o.2 ν /2
Aa : aBntr,, The νalue of r., needed is

:2Ao _ (2)(5Ο v/V)

Since 9,, : ιD ,^
" l mΑJV
- lO0 kΩ
vοv/2' 8.
A,'' : !(!p\. lΙn : lv ol substituting, we Since r, : l# t,
2\Vovl ΙD vou
l00-$(0'l mΑ)
find that
L _- '!: - - O.5 μm
,",:Ψ: 20ν :0.2v Iyo'| 20 V/μm
100 v/v (c) Ιf y.,,, :
0, the maximum Vois Vo, Vu": -
1 * 0.2: +0.8 V with a single-NMoS cutτent
Ι _ 2Ι "2
n - Q\:k'(w/L\v2.,' transiStor, the lowest V, should gο is V* + 2 vou
Ι : 3.2mAl\ρ φ.2 V)2 : 128 μA
: - 1+ 2(0'2ν) : -0.6 V so, the range of V,, is
-0.6 v to +0.8 v
8.86
(d) Q, delivers Ι 2o0 μA, and L : : o.5 μm'
*Vrr- *l V V.- : 0.2 V. So,

,^.-|Vo'|.1- - (20V/μm)(O5μm) - 50kΩ

1 0.2 mΑ
fιlι: Γoι: r,, : 1Ο0 kΩ
Using Eq. (8.148),

zt"-Π:-rR*
^1'o-164
I

l+8^.ro,
. lοοk
"cm 2(_5ο k) l+(lmΑ/V)(1Ο0k)
ψ, : -0.0Ι
CMRR(dB) : 20 log,sf{
|Α.,l

-y"Ξ -l v
: 20 lοg,,, (ffi) : z+ αε
(a)Ins: Ι _ 200μ"A'

Ιo': Ιr2: Ιo. 'Ι

ιοι - 1
 Chaprer 8-29

8.87 8.88
RecaΙl from Equ. (8.15Ο)
:
CMRR (s, rJk,,, R*) *Vno
(a) For a simple current miπor
Rss : ros=(fοr ΙD--Ι/2)
CMRR : (9'r)(9, ru)

- (2Ιo .v o\ . (2lo ' V o1

\V* π) \π π)
: 2.!!.JlL
vou vo, vot4 t_-vo,
o\t
.\w) V
lVo,
- ^( v.D.ΙJ.

(b) for the modified WiΙson current source

Fig. p8.87
οf Ι
l' ι: z00 μι
RSS-g-7'Γο1'fos Qε ο-
+CMRR : (8^rd(8-. 8.l rοl rοs)
For Qt,6,r.s'.

llv Γu
ovs --
,]lτπ
while for Qι, z,:, +:

vov : - Yss

For transistors Q, to Qu
ΞVovs: "l1 Vor'
Thus, (forl: 21p) Ιo:!-200-μΑ:l00μΑ
"22
CMRR- Ι Vι Ι 2Ι Vo.vo 1.
vov u/2) vou ΙΙ ΙD - :Κ'(w
2 / L)V'"u
"' so.
,Dv ou
4 vA] ^ πVA3
lv oul
"l1vi" viu
: : :
Fork W/L: 10mΑ/V2 lror,_rl lyλ + lvoul 0.s + 0.2 0.7 v
1:1mΑ For transistors Q'to Q", Ιo: 1
: 200 μA so,
|VΑ|: 10ν
l'tv^.t: l7(2OOμΑ):0.28v
ovl -,----------------Ξ
ι,vou: Γ 1mA
i----- : 0.316V ηJ5Φ0 μΑ/V'
ι/ l0 mΑ/V'
lr"r.
: :
,l o., + Ο.28 0.78 V
= For the simple current mirror case:
For lVrrl : lVorl ineachtransistors,
CMRR-2rΙΟf:2000
\0..11 6/
(V οο + yss) : 2|yG"1 .l * ,|ror' rl
-
-r 66 dB
: 2 (o.7 v) + 2 (Ο.78 v) : 2.96Υ
For the Wi1son sourοe:

CMRR : 2il2 (to)1

-: 8e442
(0.3 I 6 )'
-+ 99dB
 Chapter 8-30

8.89 Ηoweveη if we use KVL,

V'ι
lu υi't-
1

: u,r-
-() u,
-J _
__ 18t 4
ι6 -
f,, Γr,

- !I -ε' V,n 4ro

Yil inconsistent

Qs 1 -. '''Ξ= 8^Viι
Roο i' - i, _ i6 : λ8",
V'ι ζn,Vi,t
- 4
(which is the same as i3)

ix - ξ.Uj'l :'.(2 A- 1
8'υιa

*Ξ Qι V",o,
i,) - iι :
1
^

oεmVιa

Ι1ι1 Ιγ-ιl:. 8'Viι _ 8.Vιι :0

4 4
il| + iΙ0 : i9 of

: (8-ιroι)roz ilι : ig- ilο : ,n :

'-d'o
(b) Ro+
- 8,,, 13 (which is the same as 17)

R," - (9,. r^) ru*

l12 - s,,Uq'ι _
I

: 8," r: oε.Via
A,ι : 8.(Ro41| Ro6) .ll
llr - l' ι - iιz '- oε'V',t _ 48",Vi,l : 0
: 8-'18m7 zzro
Note, through, that this is inconsistent \η,ith ΚVL.
2Ι^ v'λ Ιf i1j: 0,Vοs: 0,butVμ: Vcz: Vι,ιl4'
Δ _-----L :
6ιn 'υ
' v,,
lΓ11ρ - 0' v o' :
Vο, ιD
but this confficls with Ι/r',
thus, g,,,r,, : 2Vn I V,,u
t.
*Aa:2(vA/voλ2 b"in,'4- Ι4.
Q.E.D.
Ιt appears that the approximations for Ι/r., and
Fοrξ,:Ο.25V&V^:2gy υ.,

A': 2 (201 0.25)' : 12800 viv prevοnt a clean solution. Ιfthese were morΘ exact,
all current and voltage relutionships should be
8.90 consistent.
Referring to Fig P8.90,
8.91
!2k' Assuming a configuration similar to Fig. 8.32(a),
o 'o) _
tl :
v
ro
- ro
'"n 1

18nυiι Ιοl _ Ιοz: Ιοι ,r^: '1: ΨEΔ : 50 μΑ

: g'4 ιlrr4 _
8' υia
ΙD
i2
4 8'ι : : - 50 μΑ : 0.5 mΑ/V
8n2 v^n αzi lz
υi,ι υiι
i^:iι iz: 8. υiι - 8'4 - 8' 4 G,n _ 8*ι : 0.5 mΑ/V
2

r':r--Von 20ν
ιD - -4O0kΩ
ι'i,t
Lτ τ]
f υi,ι]
_
8ιn2 υgs2 6m 0.05 mΑ

-t r._'r':|!λ- l2ν -24OkΩ

1 8n' !i,t D 0.05 mΑ
4 I

.3
is: ιι_'8'
From Eq. (8.140),
: f uz]| ,"o : 400 k I| 240 k : 15Ο kΩ
-ui,t
4 R,,
11 Ao : G.Rn _ (Ο.5 mA/v)(150 k) : 75 v/v
le : - 48n,υiι: -rSn'υiι
-1
lι- Ll: -o 8n' υia
Gain will be reduced by a factor of 2 if
Rι _ Ro: 150kΩ
 Chapter 8-3 I

8.92

tVυυ -!2
^(Υ)^
(Υ)^
Referring this current back to the input, we have

Δ rΨJ
1 \L,1
V ,: ^I" -'(|η//L)o =You.fl
ο
6m
I l'.'
v
""n
' (o,)^

(b) To consider the miror transistors, assume Q

and Q2are matched. Therefore,

ΙτΙ
tot-tD2-j
- Yss
lf Qa is mismaιched by Δ ruj
\ι)-'
(,)Let (f)' : (or)^-)o(Υ)^ , Ι T, Δ(w/L).]
'o'-1L'- ψ/LkJ
since 1ρ3 must equaΙ /21
(o,),: (o,)^-L^(Υ)^ Ιo: Ιpa - Ιa2

ι : ll, -^(w
(w/L\,J- 5: !.^(W
L).]
l o, : {' - o (Υ) :' '"-tL'-
/
1 (wL),
Ι- / L)^

L l(or) L
^]u Refering this to the input,
-:
^
Ι
ι';' l(Υ) l (rr)
o }' Ι/2.Δ(w/L''' ^(w\
^- ^], l/ -ΔΙ" _ w/Ls -V',
'τ ^\τ).
since the nonminal value of1ρ1 : Ιυz: j unaer 'o"_-- _|J2- rψ)
baΙanced conditions, we can rewrite these as
v-ρ \τ)-
(c) The worst-case mismatch will occur when

Ιοι
ΙΙ
-τ-
- 22
[1^ (y)^] Δ rη) Δ rq)
\L)A _ \L)^ : o.o2
(Υ)^ (o,)'

ΙΙ V"'2: V"u
Ιo, - -τ-
22
tO.ο2 + O.o2l

worst-case

u',:Ψ'O.Ο4]:4*γ
Δ
LΙ o' :!4 ',vorst-case

(o,) 8.93
^
Referring to Fig. 8.37(a),

Ιεl : Ιεz: ψ#: 2ΟO μΑ

y - vτ--! : ::--:::-:-
25 mV :

The current mirror makes Ιοη : Ι ol

' ΙF 200 ': μΑ l25 Ω
since1ρ: Ιοι- Ιoz,
R;a : (β +1)(2r")
Δ'Ιo: ΔΙρa- ΔΙρ2 Ril : (1Ο0 + l)(2)(125 Ω) : 25.3 kΩ
ι6V:8οk()
ΙC - Ο.2 mA
,^:|vol
''
 Chapter 8-32

(So the common-mode input range is -4 V to

Ro: rn4ll ,"r: _ 80k :40kΩ +4.7 V)
7 2
Frοm eq.(8.1 67),
- - 8,n1 - 8n'z : Ιr- - 0.2 mΑ : 8 mΑ/V
U. r''o
V, 2.S mV A ''' _- where R." is ιhe resistance oΓ ιhe
'''' βιRuε
A4 : G,', Ro : ($ mΑ/V) (40 k) : 32Ο viv
current source
Ιf the next Stage has R, : 25.3 kΩ,
:

Aι : 8.(Ro l R,) : (s mA/v)(40kl| 25.3 k) ^ jVA

Ι(Γr: 1ο0V
- kΩ
/ Ο.2 mΑ
_50c}

: ι24Υ/ν so,

8.94
λ -
Λ'ιι (1 megΩ)
(1r5)ι5OO k)
+5V Α,,, : -0'0l6 V/V
CMRR : yλ _ 20OO l25.OοO
ιA, '| ο.ο l6
or CMRR(dB) : 20lοg.0 (125'000) : \02 dΒ

8.9s
+5V

Ιr'' I
{ l: z00 μι

-5V UseR_2kΩ
I--- Vεεa: (5-(--5)-0.7)v
For G,,,: 4 mΑ/Y we note that ι'κLr -Vrr-V'' R 2 kΩ
/: :4.65 mΑ
"'- Ψv,
G',,- sο ιhaι 2V 7G.: 2ι25 mVl
Vοεe - Veεs:
(4 : 200 μΑ
mΑ/V)
y-1n/ /nεε') : (2_5mV,'n(a'65mA)
-'o _Vcc-(Vtt)
Vruo _ 5 (-5)*0.7 ' \ 1,, \0.2mΑ/
Ι 0.2 mA : 78.7 mV
: 46.5 kΩ
p- -' Vuro_ V''' _ 78.7 mV : 393.5 Ω
' Ι Ο.2mΑ
r,-r._ V'-25mV-25OΩ The only amplifier parameter that is R,.,,. This is
ι/2 Ο. l mA because now R,,, is:
R,1 : (β + 1)2r" : (125 + 1) (2) (250) : 63kΩ
Rr5 - (1 Ι gr,5 R,) ru5
Vo 1οΟ v
Γ^ι 'ι:
Ι,^-':"'=|megΩ
Ο.1 mΑ ,.:VoΙ - lοον _50Οkt)
r, 0.2 mΑ
R,, - r"o1l .", : : 50Ο kΩ n.: VΤlll - 0.2mΑ 8mΑ/V
I

,r" 25 mV
Ad: Gn Ru_ kΩ;:296ρuru
R,s : (1 + 8 mΑ/v(0'3935 kΩ)X500
(4 mΑ/V)(500
kΩ) _ 2.07 MΩ
Ιnput bias cuπent
Ri,. _ (β + 1)tR", l| r,,']|
Ι.'
-6l - Ι".
'βz
_ Ι /2 _ |00 μA - O.79 μΑ
rnzJ

β*| 126
r,:r': Vo _ |00V -lMΩ
v,c'|'o."_ Vcι* 0.4v _ vCc'vΕΒ+0.4 Ο. l mΑ0. l mΑ

:5-0.7 +o.4_4.'7ν R;,," : (125 + 1)t2.0''t MΩ 11 1 MΩ ι| 1 MΩ] : 50.7 MΩ

V'r'|n,n - - V uε * V crs@rin\ l V sε
: -5+0.3+0.7:'4ν
 Chapter 8-33

8.96
Ro : fo2|1 ,"o : 60 kΩ
tvc,
ξοzoκl:
Aa: G.R, : (10 mA/V)(60 K) : 600 V/V
rι :Vr: 25mV :1OοΩ
ΙF 0.25 mA
Ria: (β+ 1)(2r") : (15ο + 1)(2)(l00)
: 30.2 kΩ
For a simple current-mirror cuΓrent Source,

Q' Qz
^"- Vo
Ι
30V
0._5mΑ - 60kΩ
From Eq. (8.167),

A : -rοι - - l2Οk -
'-Cm 0.OΙ:t3
β, Rr, ( 15ο)(60 k)

οr 13.3 mV/V

CMRR(dB\ : 20log,n, |Aλ

-vΕΕ ιA"|

:200μΑ, G-: 4 mAΛ/,

: zο roρ,^] |6oα
"'"
}: 93 dB
From prοblem 8.94' Ι ιl0.0l33ιJ
β:125 With source resistances οf 10 kΩ, vοltage divisiοn
r, : r. :V,Ιr, - l/2 vT _ 25mV : 25OΩ οf Α7 will occur:
0. l mA
R,.
since R7 : (β + 1) (2) (r" + R")' Aιtιι,νcrαlt\ - 'A,t
;Ξ-R
Ria
R' : 2(9tt) l2-5 kΩ _ :
' 2(t26) 250 246 Ω :l,( 30.2 k
,'"'-'l ,::l(\. 600v/Υ1:36! v1ν
\2(10 k) + 3O.2kt

r ' "": Vo - I00V : lmesΩ 8.98

Ι/2 Ο.lmA Refer to Fig. Ρ. 8.98
modifying eq. (8.157), we have (a) Ιf R.,1 and Ro2 can be ignored,
Rn2 : roz * roz g..z 0rrll ,,r) ii: Gr"-Vi..
But now, 2R" is included in the emitter circuit: υo i' - G
lA,n -r.VιrrrfRoη
Rn, : ro2f1 -| 8'z{(r"ι + 2R") ll .'r}]
=
substituting in for i,,
fnι : f (β + 1) : o.25 k(126) : 31.5 kΩ υo: lA.Grr"orVir* * Gr,r-vιc.7 Ro,n
"2
Ro2: (1megΩ) [1 + (4 mΑ/V)
: A,.: + - G.,^Ru.(A^- l\
{(0.25 k + 2|0.246 k]) || 31.5 ki] 3.9 megΩ i,^ v
Ro: RοllRoq: Roz||r"a: β9 megΩ) ||(l megΩ) (b) Refering to Fig. 8.32 (a) and relating it to Fig.
:796kΩ p8.98. i, :
i,6 * A,n i1
so.
Aa : G. Ro _ (4mΑlv) (796 k) : 3184 viv ^ -
^m
'_ii-iro3_
------7- -
iro3
t, ---:-- -,-Vrε]
- I -----
lι li fo\lι

8.97 8^*Vrcι: A-i'so,

Refeπing to Fig. 8.37 (a),
0.5 mΑ
^; , sincegr4:g.r.
A^: l- "'''
Ιcl:Ιcz_Ιει:ΙΕ2:!: 2 2
8.ιf"z lι

: 0.25 mΑ ,1
''m l
^ Ι. 0.25 mΑ 8n1 fο3
U'_8-ι:;: 10 mΑ/V
25 mV Continuing, we can substitute this into the equa-
tion of part (a):
fot : fo2 : fo\ : - lvol
''o - Τη - 30v
O25.Α () I
_ 120kΩ A,.:G.,.n..1
..(n -ηCm-'nm|l |
-ll
t )
' 8n] Γo3
 Chapter 8-34

Since V,..G.., -
v.
;;:,.G,,,
_t
2 R,, tA r-ligll
l (rη|
t^ rI
-'ll
)t

|2R,o l, ,
substituting,
l ι' d' ll
'|

l
.l -l
ιr Ζll
lΛ , : l*,,-
t"'n" ρ'll
l** |ι
|
t* 4
β']|
ll

1^ l _ Rn. 2lβ' Rnn'. l

ln'nιl - 2R* -
| +2lβ' ** β'-,

8.100
Using a base-curent cοmpensated current mirroη

t,",: ι,Ι, πiτl

βiυ so in a differential amplifieι this means that oηe
collector current is while the other is
$,
α/Γ- 2 1
2 L β(β+1)]
lι;l :{Γr rr- 2 |_ rιΙΓ 2 f
ii : i' :
_t-t
βiυ+ 2iu+ 2| (β+l)β] z Lβιβ + t,l-]
;* 2

A-i' : ioβ :*lΓ 1 l

L(β + l)β]
A,ii - (*r), : Ψ where G- : Ι'' - αΙ
vo, ^ V, zv,
G.
4:(β
- \β|2))-
l
αΙ
l12lβ
Now, substituting into the resulting equaιion of vos- β(β l υ - 2V,
,,
paπ (a), "1 β(β-D
,η
Ar- : G-r^Ron(An 1)
Fοr β" _ 58,
A,, - G^,;^(#- t) : _
ffi#
and since
|V,,l Α.6 pΥ

u.r. I

0"":P*
,R*' 8.101

-#, (#) Frοm eq.(7.80), for a wilson current mirror'

Ιo :
Ι*r, ι L 2
sinceβ >> 2,
β(β + 2)
o _
''' - -'no Αs an active load' this means that one collector
βrr" {.
current will be whiΙe the οther is
2
8.99
Φ(ι* 2
Ιf the Wilson mirror from Fig. 7.34(a) is used,
from eq. (7.8Ο),
2\ β(β-2)))
.tl lι;i -{Γl 2 _t):"1Γ | l
''' r+ ,- l+2 2\ ' β(β+2) ) lβ(β+2)]
(β + 2)β β' α1
Using the general equatiοn from Prοb. 8.98, -τα1 δm
ι'νnι
VT 2V,
|A,-ι _ to' - 'l|
|fu*
 Chapter 8*35

lll l-Δi αΙ 2Vr

|'osl - q- - uru-" Φ,d('u,,)
β(β + η
αΙ ' ι-, : -------------:-- _ )u,o - u,o
β,o-)F,o ]
^

π '*:
Fοr βo : 50,
: G'Ro:
A1
|yorl :
l|Ι l- 2(25rnΥ)
affi : l9.2 μV
Ψ
Ιf β:1Ο0andyl:30v,
Ι.
t^ -- ir' ro : -! so'
V^
8.102

: +5V
: :
^, |(*)(ff)ι'*l 4Ο,O0O

or Ao - 20Ιog
,o(40,
ω0) : 92 dB

8.103

Vcc : *5Υ

l
0.1 '

o.1 '

iVtt: -5Y
: G.R6
A1
:
G- 8-ι (a) υo.*r* - Vm : o.4Υ
Ro : Rorll Ro, bιt'. V', : 5 - 2 X ο.7 v
where Rr. is the output resistance of the cascode :3.6V
stage, and R* is the output resistance ofthe ΞUor*:3.6+0.4
Wilson current source. :4.0V
From Eq.(7.88), we can express Rry as (b) For a 1.5V max. positive swing the DC bias at
the output should be:
Row : I

Vorr:4-|.5:2.5Υ
)Ptrot
From Fig 7.19, it is clear that (c) Ιn the edge of saturating pa
Roc : (gmaro)Qorll r.o) VsIns = Vo r,n : 0.4 V -+ Vo .in : VBΙAS - 0.4 v
Since r, isusually 2t f τ4
Ιf vo .in is; o, -
V o 1.5

Roc:q-qro4roa:β4'164 2.5-1.5:1V
sο that =9 VsIas
: 1.4 Υ.
: (d) Ιf vBΙAs : 1.4 V
Ro Rocll Rr, : (βq ro)|ι
Qv,'",) ΞVcz: l.4-o.7 : 0.7v
Ιf β+ : β, : β and assuming that
r64a161:rg;
 Chapter 8-36

This figure belongs tο 8.1 03

Cg

t,
b7

: lbι
ilr(β + 2) : iυι and 163
The upper limit of Viclπ is sυch that brings Q2 to (β+2)
the edge of saturating. j6'3 : βj63, sο
V i,,min- V g2 : 0.4 Υ
. βin'
Ξ V i,-nin : ο.4 + 0.7 : 1.1 V
ιrυ_-
(β+2)
iι: iυι-icε :
8.104 'r,-#?Σ
(a) Referring to Fig. n8.104,
ιl. ΓBι2
- |-pΤj']ιυι Βl. : ιυιtβ
l 2 f
,]
'ΙΙ1
ιtι_14 lsι--,
(ilυ
compared withe just i6r and no base cuffent com-
Ι.'-Ι.,:s!
"- '' 2
pensation,

Αssuming that R,,i.β+2-βhinh".

'"22'
Ιcι: Ιcl: Ιιι: γ #T 8.105
The input bias cuΙrent is
To obtain maximum positiνe swing V6iu, must be
ι ι ι _ : Ι- l _Τ α/
Is - Ιsι-Ιc*
I
as low as possible.
r'(β- η (β { υ
To keep the top current soufces out of Saturatiοn:
rβ+l-β) V cc - 0.2 - 0.7 : V b,u,

_ / \ β+l / _ I. ^o*
1 -1(l-α)
'R
I
Vuior^o* : 4'1 V
2(β+1) 2 (β+1) 2 1B+1)2
Αnd: V, - Vo,u"nin : +0.4ν
without Q7 aιd Q''
SinceVρ-o +Vbιu,nin _ -0.4v
-Ιl
I.: -
" 2 (β+1) + Range of V,o. is:
So, the input bias current is reduced by a factor of V
(-0.4 << y/,i," < 4.1)
I Fοr: 1 : 0.4 mΑ, β, : 50' βN _ 150 &
(β+η
Vt: 120
withοut Q2andQg'
R;a: Rin: (β+1)(2r,) ^ - 8,nι: Ο.2mΑ- -mA
bn, δ
25.v v
ν- 2V- :
r. _ ---:-: ' soιhaι For the folded cascode'. Roo β'roo
'Ιl2Ι
. - 4VT(β + |'t For ιhe Wilson mirror: Rρ. - β'ξ
''' Ι
(b) Ιf we ignore the effects of ro in each transis- + Ro - [9,,..,o il
ο' }]
toι \iie See that io. - iu',
: : 12010'2 mΑ : 60Ο kΩ
f oq f os
Drawing a simplified equiνalent of the input and
Llιe Q, _ Qg current mirror, we see thaι -+ -U
R^ - Γsο ,, ooο k ll |5ο x Φ9 rl
ic.+ifi *ilι:inl-iι,'. L 2Ι
since ro, - ror. i61 : iot
: t3Ο M l| 4_5 M]
so that,
:18MΩ
irε(β) + iu'* i6' : i1,1 A,1 : G^Ro: 8 Ι8MΩ : 144ΟΟ0
ΨX
 Chapter 8-37

8.106
2(Ι
Referring to Fig. Ρ8.106, lvoul,.r: "ι)
ιo, : Ιor: - 0.4 mA : 0.2 mΑ *,r",(Υ),',
5 2
: 0.25 V
ιrr: Ιru-ξ : Ο.2 mΑ
l|t l ΓτΙ-
|v ovlι ι -- l_--;_ -
G^_ 8'z: βΚΙwlL)η \ βnL or

-Ali'ar^A):
4ΣβΑ 1.6 mΑ/V : Ο.25 V
fοz
lY"l
_ !---Ξ] to v : 50 kΩ :
ΙD 0.2 mΑ lv oulr.r,"
*rro"(Yr),.,.,
foι :|vo| - 30v :15OkΩ
Ιc 0.2 mΑ : 0.25 Υ
: rozl| roo : 50 k || 50 k : kΩ
lv"rl :
Ro 37.5
lv tl + lv ovl , so all are
_ G^Ro: :
lvorΙ : :
Ad (1.6mΑ/v)(37.5k) 60v/v o.75 + 0.25 1.0 V
Ι/2 - 225 ιιA : 0.9 mΑ/V
P- - |vovl12
8.107 "'''|_4 0.25ν
Referring to Fig. 8.41,
Ιo' - Ιo* _- Ιu _ Ιυo: Ι : Ιnεr : e* l
"5_8 - | VovlN -2(225ιιΑ):
225 ψA l.8mΑ/V
ο.25 ν
Ιοl:Ιοz:Ιοz:Ιυ, Ι 225 μA'
t* 2-- 9v -
"l 4 :u:
Γa
2 8ΟkΩ
: 112.5 pA Ιl2 o.225 mA

From Eq. (8.1 80), systemic balance will occur in fn :|vλ: 9v

this circuit when
"5 8 Ιl2 0.1125 mΑ -4OkΩ
Aι _ -7^t(ro'll ,oo)
ru:,ψ : *(0.9 mΑ/V)(80 k |l 80 k) : -36 V/V
Az : -Ε-ο(rouΙl ,or)
(o,)^ (y),
: - ( 1.8 mΑ/V)(40 k lι 4ο k) : - 36 v/v
(r\ Ao: Aι XA2 : (-36)(-36) : |296ΥN
(1). :,# :,,,H
16ο)
(o,-)^ (j:) : 201og rc(1296) : 62.25 dΒ
lΖJ, ι'πJ The input common-mοde range is determined as
fοlΙοws:
_20 Τhe lower limit is when the input is such that Q1
0.5
and Q2 leave the saturation region:
so, W6 : 20

To find lVrrl, we use 1, :

)"c"^(l)v2,,

Qt o, Qz Qq Qs Qd Qt Qt
1ο(μΑ) l|2.5 112.5 ||2.5 ι12.5 225 225 225 225

lvovl(v) 0.2s o-2s o-2s 0.25 0.25 o.25 0.25 0.25

|yorl(v) 1.ο 1.0 1.ο l.ο Ι.Ο 1.0 1.0 1.0

/mΑ\ Ο.9 0.9 0.9 0.9

8'lvJ
1.8 1.8 1.8 1.8

rρ(kΩ) 8ο 8Ο 8ο 40 40 40 40

This table is for 8.107

 Chapter 8-38

Vυl : -755*V65j : 1'5+1 : 0'5v (b) For the common-mode input range:
The lower Ιimit is when Q5 is Ιeaving saturation,
with lVrrl : \V oul, this would be rvhen
Vοs: : -tV+o.2ν _-ο.8v
yss+|yos|
Vsr : -0.5+0.25 - -0.25V
yin1nιn, : Vcsι*Vos: vΙΛ+vov+νD5
Vinrin : Vsl Vsc : - 0.25 - 1 : 1.25ν
: 0.4 + Ο.2 - ο.8 - -0.2 V
The upper limit is when Q5 leaνes saturation: The upper input Ιimit is when Q1 and Q2 leave the
yrrno, : Voo-\Vou|: 1.5 - o.25 : 1.25ν saturation regiοn:

V1n,no.,_ Vsru*- V56 : 1.25 - 1.Ο : +0.25V Vοι: Voυ-Vrog: Ι -(Ο'4+0'2): g'4γ
Vοst : |vov| : 0.2Υ, so
so, range is (-1.25 V to +0.25 V)
For ιhe οuιpuι range. Vρn,o* is Vin(ma*) : Voι vov+vGS1

: : : _ Vοι#V,n:0.4 v: ο.8v
V gπax V oo - |V ou\ l.5 - 0'25 1.25 Υ
Sο. ιhe fange of input volιage is
Vo^in : - Vss * lVoul : - 1.5 + 0.25 C0.2 V tο +0.8 V)
(c) The maximum οutput voltage is
: -1.25 V
(-l.25ν
Vo(^u*): Voo-lvoul: 1-0.2: +0.8V
so the output range is +1.2-5 V.)
Vo6rn) - - Vss * lvoul - - 1 + 0.2
8.108
Referring to Fig. Ρ8. 108
_ -Ο.8 v
Ιυg: Ιοι'ι: Ιnεr: so range is (_0.8 V tο +0.8 V)
20Ο μΑ
Ιo, : 2Ιj1 : : o1 5V :25kΩ
400 μA (d\ 162 - ,o^ \Y
lnz
- 0.2 mΑ
Nο requirements are given for Qu anιl Q7, so
lv,l 5V
choose
* Ιοs _ : :
ι06 - ιo1 - lnε
- 0.4 mΑ
12.5 kΩ
Ιo1 2Ιo5p 400 μΑ

8n1 -8n2:t*"
|Ι,|
- Ο.2 mΑ : 2 mAlΥ
o.2ν/2
Qι Qz o, Qa Qs Qο Or Oo
o^1 _ Ο.4mA _ 4mΑ/V
n. -
δmh |Ι
V οvl2 o.2l2
25 25 100 lΟo 5Ο 2ΟΟ 50 25
(Υ A, gn,1('r6'll ,o) : (2 mΑ/v)(25 k l| 25 k)
_
: 25 ΥlΥ
A, : -g,,,2(rooΙ| :
ιo: \t'ιwlι)V}u so, 'o) -4 mΑ/V(12.5 k1| 12.5 k)
: -25 ν/ν
(Υ),',', =
2Ι r',
- 2(20o μ^) Aο : Aι' A2 : 25(-25) : 625ν/ν
k'n(V o)t 4oO μΑ/v2(Ο.2 v)2
:25 8.109
1r
ι : ;k
(y),. : r*
2Ι 2(200 μΑ) v'ov
k',,(V ou)2 40o μΑ/v2(O.2 V)2
: 100 |\aΙ vov - jτm

(v),, : V,
2(2Ι RΕF) 2(400 μA) Ιf K increases by 4 -+ decreases
k',(v o')' 4O0 μΑ/v2(ο.2 v)2 by ll2
-50 g',' _ 2ΙlVou: k'Vov
4Ιnεp 4(200 μΑ) --> ifk increases by 4
(Υ),: k'n(V oi2
=
gm increases by x
1ΟO μA/V2(ο.2 v)2 2
(b) Α1 = gπR61
_ 20Ο
: --> Α' increases Χ 2 as doesΑρ
ΙdeaΙly, Vo@c) o
(c) offsets due to ζ mismatch are unaffected.

others reduced xI sinceΑρ increases X 2

2
 Chapter 8-39

8.110
, -Wr, ,-.: !+ - O.5
'e) 25mV :50{}
ιo':illnεr:-x90μΑ
- 50.. l/r'l mΑ
R,ι : (β Ι 1)r"5 : (101X50) : 5.Ο5 kΩ
: μA A1 : G^1Ro, : (4 mΑ/v)(5.Ο5 k) : 2o.2Υ /Υ
112.5
output offset οurrent : Ιοz - Ιοο
For the common-emitter stage (p5),
: 112.5 - 9ο _ 22.5 pA
ΞVo : 22.5 ψ (r"6|| r"r) o -:::Ι.,
δr) O.5 mΑ
- 20mΑ.zV
VT 25mA
,., - lo : 88.9 ko Τhe load is essentially -
112.5 μ
(Rr+ r") (β + 1) - R/(β + 1)
lVo : 22.5 p (111 k ll 88.9 k)
Αs: -8.s(RιXβ+1)
: 1.11 V
: -2o mΑ/V (10 kX101) : '20,200ν /Υ
v^":Vo-l.Ιlv:lmV 16-
A" 1109 I so,
Ao : Aι' As. Aο : (20.2)(-20' 20ο)(1)
8.111 - 408,040 v/v
offset current : Ιr2 - Ιrq or
: Ιρι - Ιοη
-A"(dB) : 2Οlog,n(408,04Ο) : 112 dΒ
,^: Εz(Vcs_V,)'
8.113

Ιo. : (vt+ Ι3 : 225 μA

ξσor- ^Vt))2 :
Ψu Cοx 180 μA/V2
Ιo : Ιor- Ιoo
Ψo Co" : 6o μA/Υ2
_
} ιιvo, -Vt-vGS+vt+Δy,)X For Q3 & Qq:WlL: 60/0.5

(vcs- v,*vcs-vt- : I,t'o

^v,)l +17,,1
ke(w L)
- ΔV, .\{ru r, - 2V, /
'2
- K(Vc.s- V,).^V,
^v,)
ly l :-,V60μΧΙ20
|,o,l,(.9 @:0.25ν
Ι,, : g', LV, ' 2ι^ 'lx225μ
then 8πs.q :
Recall 1o : G^ι. V u" vλ
and G,, -- g-, : 1.8 mΑ/V
Sinοe g. of Qιo' Qι & Qη are identical to g, of
+vo, : ζtt.^v, Qε & QgtΙιenVonrr: 0.25 V
8 nι1'
Thus for pt,
FοrΔV,:2mV
'12 _
(Ο.2_5 2x225ψ
,^.-0'3mx2m_ ΖmΥ l80 μx (W / L)ιι
Ο._] m
--> (W / L)r. : 40 i.e, (20 10.5)
Since Q12 is 4 times as wide as ρ/j, then
8.11.2
(a) Referring to Fig. P8.1 12, (w / L\," - 4xο.520 : 8o/0.5
Ι.': Ι"': l22- Ο.2 mΑ _ O.l mΑ
zL)ιz
Rs: ( '/L114- l))
L'-l.":!_ 2 O.1 mΑ
1r. : 0.5 mΑ r@_,.)
since the output voltage is held at 0
116:lmΑ
(b) considering the first stage,
!

+ RB :
180μx Ψx225ψ ι'HΓl
-
Umι:8ηl:τ:zs.v ιΙ"J aJ_!Δ : 4 mA/Υ The voltage drop on,Ι?g is
555.6 Ω
:

X 225ψ :0.125ν.
since alΙ f n' : Φ, the load of the differential
555.6
Tο obtain the gate voltages:(assum e |V,,| : |Vφl
:
stage is just the inpuι of pr.
0.7 v)
 Chapter 8-41

_R'll (a) Taking the Thevenin equivalent ofthe input

a, : f" : Riι=3O4 kΩ bias network with respect to the - 10V supply ter-
r " i Ι{"
minal, we have
_ -3.09 v/v
ν'BB- (VCC_vΕΕ)RB2 - [ι0 -(_l0)|lωk
andΑ1

andΑ : 8513.3'09 _ 4rc4ν /Υ RB|+RΒ2 82k{ l00k

6.42
Τhe gain has been reduced by a factor of 2.07 and
: 10.98 V
οan be restored by reducing Ra by this same factoτ Ras _ Rεrll Rrr: 82kι| 1ωk : 45'1 kΩ
to increasel3. Thus Ra : 1.1 l kΩ , v ,, - v ,, 10.99 0.7
(Νote that this is a first order approximation). 'Ll
't'ι _ __-_i- Rr, 4?Ι; k
,.εt+
R-,
i
9.5 k +
ρ1 |0l
8.1 17 : 1.03 mA
-R., - 303.5 {1' '_' lο0( 1.Ο_] !!ι,r, : l.02 mΑ
(a) Α.l :
,-ζ'Ω: 2,325 Ι,-, - αΙ'':
jl'rε' -- lOlι ι.v.l mΑ)
:
β
- 13ο.5 vΔ/ Sinceβ >> 1,

l'e 13 tS lncreaseo oy 130.5 Vcι' Ξ

οιz
:20.33 v cC - Ι : 10 - 1'02 mΑ(5'1 k) :
C|Rc1 4'8 V

+Α:8513X20.33 , Vrc-Vε"z-Vc'
.t'z__-______Ξ--
: 1'73.1X 1ο3 v/v I( rz
(b) Let the output rΘSistancο ofthe οurrent source 1ο-Ο.7-4.8V : 1mΑ
4.5 k
be R-+ * Ro" :3 kll" {.= { , + r"'t ::ιο
\β+l ') Ι,, : (#?)(',,l : (j#),' mΑ) : Ο99 mΑ
The amplifier οan be modelled as shown:
vcz -
vo Ι c2Rc2 + V :BB - 10 V : 0.5 v
(o'99 mΑ)( 10.6 K)
V o(dc) V sεz : 0.5 - 0'7 : -0'2 ν
: V aι
,ιε3-Vo-VεΕ: -0.2-(-lο) - O.98mA """_'
R- Iοk
Thus,
1., : : (i#)(0.e8 mA)
: o'e7 mA
Arolυ:
A' Rt uft1'".
R1. + Ro .
(b)8.r : Ι", 1Ο2 mΑ : 40.8 mΑ/V
: 113.1x 10'l00 109 i:
+ 30ω
: .
'πl : β' : ιοο :2'45k!''
-rr_o
5583 V/V 40.8 mΑ/V
8m|
For the original amplifier:
R,n : Rs"ll .., : 45.1 klι 2'45k: 2'23kd'
Aω,qο-85l3Xπ#52 = 337s
- 25mV:25.5Ω
+
' '':V,
?J Ι r' Ο.98 mA
8.118
Rou, _ Rε3ll
' Γ + ηc:-l
+1οv [r", 1β.- 1)]

'" k" lιrr Γο.ozss

= 10 L"'"-""
κ" + !f-\l
lol .l
_ νg Ω

(c) icr : 8^ιVi)'f : ε^r: 4Ο.8 mA/V

By cιrrrent division,
i!2 : Rcι
l0.6 kr,3 \ι,
irι l rn' Rrr

r'z : (β Ι |)r"2: (101)(25_mv) : 2.53 kΩ

 Chapter 8-42

sο.' 1Ξ : 5'1 k _ 0.67 AlA

io, 5.1 k + 2.53 k
;^^
Υ: β': Ι00Α/Α ΙD Ι12 Ι
lτz

w w w
Αgain, by current diν'ision, 8m
Ι Ι Ι 2Ι 2Ι
vλ
Ι_

lυι R,, o
icz - Rc, + (β + 1)(r", * Rr.)

ryλ rλ ψλ
ΙΟ.6 k 2lv zlv 2lv 2lv
ol o1 Ιv o|
10.6 k + (101X25.5Ω + 10 k) Ι Ι Ι Ι Ι
Ι
1η : o.οtοso A/a
ιcz
(b)To find the differential gain' apply -Ψ ro n,
'Ξ:Br+Ι:1Ol
ιιl
andV,,,/2to Q2
ιln : io1 Rr1 : j", (10 k)
v,, - ε,,,'(,", Ιl ,., il u,r,,
: *)
? e,)*(#)(flHe,) ..
v ,ir." -1- ,n, ll .,,.
τ,
: (101 )(10 kXΟ'01036)(Ι00)(ο.67)(40.8 mΑ/V) ,r, - t,rr(*)'?
: 28'603 V/V or 2.86 x 104νlν
i,15 - -V rs B.s : 8. (*)
8.1 19 "'")?
,s : v rt :
Referring to flg. P.8. I 19, assume the system is
"il *): "'(*)
v i,1. (r.,r ll .
balanοed.

|ιal 'Ι
Ιοι_r'r: since grrg: gr,1,
2
V,ε _ -'ml(*)Ψ
Ιοe.a: ,(ξ)='
sinceg,,. :28.t
, : l1'l , so thut
: l1'ι(J-)ιzε,'rl : Ι1''lV
iιt
lv oul
2
Ψ ιa

δml )'/
Ι/2 Ι with Ιl aοοlied to o,.
Vοv/2 lvou\
+
2"
8π6'8:6:w Ι2Ι v
14
: -8,n2(r"rl1r,,oil
*)
2
V,o
ι ι
: V,ι _ -+nl\π)1.V'o
: l,Yo,l so that
i
''
"^ |Ιol since9,,,6:8nqΧ2,

foι s.l _ 2lv ol ion: ( 8.ι'(*^)Ψ

-g.a(2)
'
rοο'x : _τ
|vλ iaι : 8-ι Vιa
io : 8n,ι Viι l 8-, Viι : 2 8,oι Via
Ιn Summary,
Aλ io Rn
τ:, - #: - 2 8'' (r"'' |l r'*)

ε.':
Ι roo-
-. rοg-_ _Τ
lvΛ
1v-|
A,ι ^ l r|\lyol vA
ζ,-'|v*|\1) Ι - v*
 Chapter 8-43

(ο) Ιf each input transistor (Q1and, Q2) is repΙaced

6mt
o_:- Ι2Ι
V"' vov/2 vou
with a current ,ora""
------ -'2Rrr'
of
vA/Vov
IMRR: |Aλ_:
lAr"l rooll r6s.
-
I
From e2,with a transfer ratio of (1
*), η" 8-rη-
vA/vov
ι D6 - 2 Rrrι' g*roo, VA/2Ι. 1

From vA/Ι 2Ι .Vι

Vou Ι
: - #*) o'
Qι'i υε
:ffi(' - #,)(' :MRR : t uL(,u ?u : 4(v 1/ v 6)2
i6 - ios* iou t(V οv\

ι^:Υ:*Γ_
" Rss L 1 t |roq.,1.r-
\ g.'
l ) '\ηl
8n,ι ro') (e) Τo find the input common-mode rangΘ, con_
l ')l sider both upper and lower limits: Lower limit is
U-
Γl
g-,,nr)J when the current source begins to leave the satura-

g... : g.oandro., : foι, tion region at V o, : V o,

Since
So,
: :
":Ψ[(l-n}χl #)J Vr(.i,) Vcs, * Vov + Vss
The maximum limit ocοurs when p1 or Q, begins
Vt + 2vov - VDD

Vo: io(roull rrr) andsince -_1- >> I to leave the saturated region: For example, when
Vsl : Voο - Vsc: - Vuu,
: Vt+Vov+.Vsr
'" =-;#(#){,.uil .,,)
Yr1.a*1
yr1πa,.1 _ vt + vov * Vοo
- vι + 2V ov

ltA-,.t :lvol:\r"ull
ι 'y'|
roo\
- I Vr(max):Vυο-Vov
|v,r"l Rss 8ntrol So the range is
(d) Ιfthe current sοurce is fabricated as a simple (- V oo + v t + 2V o, = V lcnι Ξ V oo - V or)
current mirroη Rr, : L

fοs : fo8: .", r.οll roε

vA 8.120
? =
2Ι (a)

+5V

Qc 20a μA
t\ +4.3 V
o. Qz
+3.6 V l oο ra tοοra
'|, '|,

l ο0 ρAt
u"

rnA
'f,l

-5V
 Chapter 8-44

DC Analysis t't' : I 8-r xlx l.osx lor = 3.1ooJ

R_ 3.6 (-4.3)
- : ko
100 μΑ
79 '',"2ν
υo
Node voltages: -7
Vl : -43Υ Vυ:
11c3
-0.'7ν
VC: +0.'7ν Vp _ oΥ Thus, I : 3300 V/V (Polarity correct)
υi
VΕ _ + 3.6ν Vr : + 43Υ (d)R,,:2r.1
Vc : + 4jΥ
=2x]Ψ:50kΩ
4
(b)
:
Transistοr 16{mΑ) 8,,(mΑ/v) η,(mΩ) R,,, rρp|| rρa|l
[".
-
=iLΨ]
Qι ο.Ι 4 2 2l| lο0 X 2l
Qz 0.1 4 2
= ο.2lΙ0.2ll Γzs.tο
L-"'" + ^
ror I

: 16.4 kΩ
Or 0.1 4 z
(e),,ic,ιll,ιnl _ - 4.3 - 0.4 + 0'7
Qq 1.0 40 0.2
= -4V
Qs 0 0 Φ' υιcπ1max1 _ vG + 0'4 : + 4'7 ν
(f) The voltage at the base of Qa can rise to Ι/r3.
Qt ο.l
(Vε)+ 0.4: +4V
Qο o.2 before p3 saturating. Τhus υ9 can go up to + 3.3 v
2 Τhe voltage at the output can go down to V6rr. of
Qc 0.1
QD+0.4- vA 0.4: - 4.3-0.4: -4.7Υ
Qυ 1.0 0.2
Thus the linear range at the output is - 4.7 V to
Qε 0.1 +3.3 V
(g) Αt the posit'ive 1im'it of υρ
Qr 0.1

Qc 0.2

G
(c) Total rοsistance at co1'lector Q3 is
:9ιroι|l (β+ + 1)(rool1 ,oo) ,|, zoοre
'oΔ1
|| 2|l
= 1O0 X 2 1Ο1(0.2 |l 0.2)
'|,
zοora
= 1.65 MΩ +3-6 ν 10c) 10.1mΑ

o^ι
Large
(ignore.) 100ρA ψ
1mA +3.6 V
Qo RL
t - gmtVt/ 2

,u'
2

i.e.υo : + 3.3 v and Q2 just cut off

^ v
^ι - ql rrΑ
-1.3

=363Ω
(this is the minimum allowed R7- for * 3.3 V out-
puη
 Chapter 8-45

At thο negative limit of υo i.e. υ, : * 3.3 V and

:
For Q6:ro j x 40 x
frvoro -
v,r'
Q1 has cut-off. Q3 will a1so be cut-oξ and'Qawil1
cut-οff.
=Vcsε:1.50V
Thus,
Vι: -33Υ vΒ: -|'7Υ
Vc:+1.5V Vo:0Υ
ι0l X ο.1 : mΑ
-€
1Ο.1
Vε: +lΥ V.: +3V
Vc:+3.3Υ Vπ:+2.7Υ
{r-n
(c)
Transistor ΙD Vcs gm rs
(μΑ) (v) (mA/V) (MΩ)
R,_ - , 4'7 - 42.1 (} This is ιhe minimum
t.7 28.3
l1.1 mΑ Qι 10 s
aΙlowed R'fota - 4.7Voutput. 1.7 28.3
Qz 1Ο 5

8.121 et lo 1.7 28.3 5

DC analysis
o, 20 1.7 56.6 2.5
Ιnεr: :!xqox}{voro-v,)'
es
(a) 10 μA
ro 1.7 28.3 5
_ V o'o: 1.71 v : |.7 Υ
Qο 5ο 1.5 200 I

,o_jX2oXΞ,rorrr-l)' o'Ο _1.5* 0 ω

ΞVcs^-_2ν o" 10 1.7 28.3 5

R _ 3-(*3.3):660k{) Qι 20 1.7 56.6 2.s

10 μΑ
(b) See figure above ec 10 1.7 28.3 5

Vcsι : Vcsz : Vo'o- |.7 Υ eo 5o 1.7 141.4 1

,csr _ r-----------:=,,
r) Exlo -, - l.7l
_
V= 1.7 V Qε 10 2205
/zο x 19
γ_r er 10 2205
V css - V"rj : 1.7 V * Cut-off.

This figure belongs to 8.121

+2.7Υ
'}zora
-_> 10ρΑ

2ν
+ 1V- {
ro,,a rο rA t {
ro"a
{sο"a
1ΟρΑ
i vo

OV
-3.3 V
,|, so ra
{zo"ι

--5V
 Chapter 8-46

(g) Q6 cuts off

thus,

1Ι - .sο,a
RL

R,-
_ |v:2OkΩ
vi 50μA
2

Jr
'VΙ
Γoο

(h)Maximum possible voltage at drain of Q5 is

+2v Αt theis value.rνe have:

Total resistance at the drain of Q5, R is: ι)

/^ : 50ιl,Α + j--! mΑ
R : (8,,, ro)0oιll ,o)|| ,o, 2

= [(2s.3 X 5)(2.5lΙ 2)] ll 5

= 4.9 ΜΩ
Τhus, ai :
1),
g^lR +2v σJ
= 28.3 Χ 4.9 : 138.'7 Υ /Υ
σool| 'oo) 50ρΑ
!9 -
and
υιs (roo|| ro) + J_

(I ll 1) Ι o : Lμ'C o"lr{, - V n I V,)'

rll
---l
rl+-.!-
' 200 .9 v,,:b.17V
Ψ: nt.'lνlν For the lowest posible output, the circuit becomes
υi

Rin:-
: rr, 4.3 v o-J
R,,,, || 1 / g'ο
|| rro

= 1ll 1ll 1/2Ο0MΩ

=5kΩ
(e)ιlcmΙ^u" _ V6*V,
= + 4.3V Where:
Q6 cνts off and Q1 conducts
l,/cMIrinl - Vοs, |- Vrmin
: V"
= Vcst + vA- vτ Ι^
"2 - O.O5 mA
=l"7- 3.3-1_-2.6ν
(0 Vo ru* : Vc .u* V cru : t,, + -
)*,r""(Ψ)ι- 4-3 1)2

=Vε*|V,|-Vo'u +Vu : r.4-5 v

-+l +1-1.5:+0.5V Τhat is, the range of υo is
Uo^Ιn : Vo- V,: - 3.3 - | : -4.3v + 0.Ι7V
-1.45Vtο