Chapter 2-1
4,5,6 shoψ that the gain is approxι
The missing entry for experiment
#3 can be predicted as foΙΙows:
2.1
exp€riments
The minimum number of pins re4uired by dualop-amp is 8. Εach op-amp haδ 2 iηput ιerιninals
(4 pins) and one output terminal (2 pins). Another
2 pins aτe required for poινeι
similaΙly the minimum number of pins required
by quad-op-amp is 14:
mately
ιω v/v.
: b: l'Φ : 0.oI V.
A lα)
(a)ι': ι._ Ua= 1.Φ_0.01 :0.99v
6\ τ'"
4x2+4xl+2:14
The missing entries for experiment #7:
t't
161
Refer to fig Ρ 2.2
All
I
_o-051 V
,, :
(c) v. = v, +
τJ' : ιJ'x lK
'1οοI
lκ+lM = ,r.
: 1u.L
o,:1,
'
' l0ο1
5.10
-
0.05i = 5.049 V
the resιuts seen to be reasonable.
2S
i: G.(ι _ ι')
'
A: lωlμ
: R.G.(ι _ v')
'
ι'
GainA:
=10οlx1
2
A =
,," : :ΣJ!
=
lα)
2Φ2ΥN
= R^G.
υ2_ vl
=2Χ1σX10Χl0r
: 20'ωο v/v
t1
Γ-*
+
ι|
υ":2.0ΦV
z : - 1.000 V
9r,..
Ι
==
+
τ'
For ideal op amp
ν.: ν :
_l.000V
voltage at positiνe input
Measur€d
Amplifiergain
:
_ 1.010
ν
2.6
A=
!+-aoo
: !(q + ν')
z
:
:
(2zιi000)ι
ι, 0.0l Sin
ι ,_ υ
?ι= Ucι _ z,/2: Sin(l20tr)l - 0.Φ5 Sin
z.":
- _ l.010_2.0ω
(' l.ω0)
:2ΦνN
-
l vSin (2π60)ι
1
1
2πnnt
ι'= ι-
2A
*
υ,l2:
Sin 120r, + 0.005 sin 200οr,ι
2.1
#
R (G"' ,, _ c.' , Refer to Fig. 2.4
')
':
υ': !':1ιι.: sιR (G-" υ, c.,ι')
υ
ο.(x)
o.0ο
0.α)
0.α)
ι.00
l.0ο
o.oο
ο.O0
(b)
3
(a)
4
l.α) l.l0
0.10
5
2-Ol
2'ω
_o.οl
6
t
-99
2.Φ
7
5.
to
1.OO
(c)
'. =
t.00
!
0.1
Ι01
_ο.99
99
ο.0l
1.00
ι0o
(d)
_5.lο
υ.:
ι2
νn(c. o2+
(oz
ιLR
G.+
G^ υ1*
- ιl\
'3c^
ιιd
(ν' + υ'\
+
'
iιRΔG..+
2u,cu
'
νve
haνe v, = Α. z,,
=> Ao
:
I d- ι",
: |Ι''RΔG^
1ιRG^, Agμ
\ιc.'')
CΙιapιeι 2_2
CMRR
-
20log,,, Α,
A"
:
80
+ A,ι =
dB
ΔG.
_
AcM
2.to
Α,
| = 26 1onΞ.
"|Aru|
" ΔG.
20 Iorl
' '
C'^- ' ^'^
10ι
: lοrxll:
lω
CMRR _ :ο IonΞ.
=
"
26
There are four possibiιiιies:
1ο
--J"ο.l
/ |00
2,4
Circuit
1ι
J
υ
- l00 = _lο
lo
b
-lo
lo
c
-t0
10
_
6ρ
o:
R'"(kΩ)
'\νtν'
lο kΩ
lo
1on
^G^
kΩ
: _2νΝ
1o
kΩ
10
_ιο
d
10
νinual ground no currenι in l0 kΩ
2.9
R2= l0 kΩ
lo:
'0.5 ν/v Rl" = l0 kΩ
lo:
_0.5 vΔr'
Rr= 10 kΩ
Closed loop gain is
τl
R'
vι
For
1
ιι
:
l0 kΩ
l0 kΩ
Rι
τr,: + l.Φ0V
'υ,,=
Aυ,:
+1
_1.ω0v
20 kΩ
x
1.α)ο
The two resisιors are 19ι, resistors
_ _l0(l_0.οt)
r!)
\ r, /.,"
l0( I + 0.01
= _0.98
V/V
)
_ _|0(l+0.οl ) _
11)
\ υll.,'
l0( I -o.ol)
Range is 0.98 ιo l.02 vΔy'
ιο
_ι-02 ν/ν
: _2νN
2.1 T
c.
e-
R,"
G: -l V/V
c : -0.1 v/v
:
5 kΩ
b. G = - lOv/v
d. G: lΦV/V
Chapter 2-3
2.12
G: -l V/V:
=Rz:lokΩ
*=*,
b' G : _2νN
-R, R. : lOkΩ, Ru:20kΩ
Ξ -------:
Rt'=,
a.
c.G= -0.5 V/V
: Rr=R, :20kΩ,
d.
Rι',
G: _1ωv/v
=
-R, +
Rr
___--Ξ
R.
:
'.
Ra:
ΞR2 =
19.95
Rt <
ι Mο
For laΙgest possibιe inpuι resistance choose
R, =
:
n,'
R'
1Mο
l MΩ -sο.t
19.95 -
= R,
:
50.l
ιο
kο
2.15
10kΩ
_,..'-Ι-l--Γ-
10kΩ, Rz = 1Mo
_Rz_'1Φ:
G:9=
lο
vi Rr
l0
v-: l0Y ,hΦ:0, ν-!: _5v
2.t3
2.16
Ι0k
'.:
z1 lk =
ι.: _1oιι
=
10
x
lο
v,/v
(0.5)
ln the figure
oo
i'' : 2kΩ
9: _ινlν
=
-5
2kΩ -
-Rz
Rr
25mA
Rr= l0 kΩ
=tR2:4Rr
Toιal resistance used is 1ω kC!
.'.Rl+R2: lωkΩ
R,+4R,=lωkΩ
R, = 20kΩ
and R, : 4ρ, : 36 1ρ
2.14
Gain is 26 dB
:
o-(u-)
l0 kΩ
:
η
dB
:
20
log lcl
: _ρ'gsνlν :
_3 mΑ
0.5
2-17
: &:
π,(r
+ft)
Rl
R1
r.?Ξ)
lω./
So 2X is the tolerance on the closed loop gain G
G:1o26'20=19.95
.''9
ιι
-2.s
This additionaι current is suppliΘd by the op amp.
16;r;11
26
tο kΩ
+0.5mA
i3: i"- i,=
+
o-(-5)
_R,
Rl
G:-tooYandx=1
Gain νaτiation - 102 <G <-98
Chapter 2-4
R,
2.18
Rz=5kΩ
:
l0o kΩ
-υn +
:
Rι
υo
lωl9___-]φ
_ιο
:
1.66
kο
]ω-
Rr=l5kΩ
2
kΩ resistor coηneοted to input be shυnted ιo
achieνe 1.66 kΩ is R., then
=
2
+R.
:
1.66
,-R,R,5
Rι
υi
,,
=
For
R|
/ R'
R,+2ko
R
kΩ i|
2 kΩ
9.76 kΩ
2.20
15
oY ι',: ι":5Υ
:t l 7o on R|, R,:
"" :
RΙ = 15 t ο.l5 kr}
R'?:5:!005k(l
'R, = ls&+
',--ξ,
R,
ts x
Rr= 1ΦkΩ
Rl=1kΩ
llΣ
t5.t5
vr
=υ<tsxΞΣ
"
t4 85
Ξ4.9v<r,<5.lv
For τl,
:
_ ι5:t 0.l5 v la.εs x
j€Σ
u'
t5-t5
Ξ,Ξl5.!5Χl!Σ
"
14.85
:
!":
-50
-
Rr
.'.R2=ilxlΦ:lΦkΩ
b. op. Amp has open loop gain of 20ω
o, : _! whereΑ:2α)ociνen
2.19
Α
ur
siηceRt"=lkΩ:Rl
94.85U<ι,=5.l5V
-7lrt
!" = _& = _tοονrv
lo
Αpply
KcL
v/v
at note y,
υι-ι| _ ι|_ υo
Rι
R2
300
v/v
Then subsιitute
-lo
5ο
η
and solνe foτ
9
_R2
!: :
Rz= 1Φ kΩ
= 1
'AυΙ
ι,,
Rt
r
*(I
* &')
.ι
\ R,/ A
_1ω/l
: _66.4ν/Υ
=
r+lr
+]Φ)^
\
1 .' 2ο(n
c. ι.et RΙ' .: R.llRl \rhete R, : l kf}
|
+
ιb
Need9:'1ωvΛr'
Αgain apply KcL at node V,, and R, is replaced
by Rr'
Αpply KcL at node yl
ιη_V1
Rr
υι
τ1_ ι,,
-
Ri
R2
R| = R2ιι
01
_ ι)|
-ao
substitute for
τ,ι
and
14
and
7'1
Substiιute
xι_0o
R1
yι :-oo=
λ
ιο
20ω
Chapter 2-5
and τ,'
'
:
solνe for
_
j9-
2.23
lοο
Rl'
(-%
- ".\
- uοοo/
R,': R,x\lω
f -% _,,)
\2ω0 ')
-t
l00
2ω0
1ω
κ
_ I _1
+
η
I
Rr' :
.'-
20ω
kΩ
0.9495
0.9495
"'Rl
Solve for
kΩ _ R' ll
R'R' _ R'
R - Rι+R,
l+R,
= 1kΩ
18.80
:t
output voltage .anges from
openloop gain is 200ο v/v
- ι0
to + 10v and
.'. voIιage aι inνenting input terminal Ψill νaΙy
ΞJ! 19 --!q.
2ω0 2ω0
ThUs ιhe νirtυal gτounι
oft5mV
i.e _5 mv
τ,
_4ι
=;n.+; 'l+Α
17o
2.21,
16η
1+Α
iRl +
Again 1,,:
ξ
& : 18.81 kο
Useξ = 13.31 19-
υ.: _Aυ : ι _i'R'
i,R,: v (1 +A)
-. _ i,R:
ιo l5 mν.
ψill depan by
a
!:
SoR.-:
'" i,
R.+ Rz
A+l
2.4
closedlooDρain
6:
,,:'(R2/R|'
.. LJ_!/&
υo
1
maΙimum
Gain
Errοr
ε: ljl&lj-i
Α
x trn
2-22
o.t%
too,
(' -
ft)
ιoo(t
+ft)
ro(r +ft)
Gain with resistance R,"
a)ForΑ:
ι. = 'i' R,
c,: Ul=0
R,=9:-RF
ιi
+
η
R,":Ψ:o
ιi
b)
ForA-finiιe|
",
= -Ψ' ι" =
υ,
i,Rp
+u._
- i,
" A!'_j,RFΞR.-Ψ__Ro
,**
n':!:
' ii
Rε
1+Α
R2
.-
t' R,R," \
ιR, *R"J
l+
1+&
Rr
chapιer 2_6
I
The effect of Rla in ιhe denominaιor is νery smaιl'
so ψe neglect ιhis
Rr.)
R2/Rr +
R:lr,
.: πι fuR. J: ηι'
l+--:
'*
R,
__Γ_
,
R'/
lrn
sο R|" =
Ι+--:
,*_;- R.
A
+
R,/RJl
ι+-j_ R"
}.
Rl
use
RI
l_,,
€,Rr
A
Rr
solving fo. Α we get
, : (,-frχ:
'00
R,
.------_.l
€
: ιI -
c.-,-,l
/I
ι-
l)
\
1)
lΦ : ,R2
Rr
R''=2kΩ=R'
'ξ
R,' = R.
Ιn order for
R
lt
G'
R2/ Rt'
G' -
:
1+ltR2/Rr'
Α
G:
R':2kΩ
R,: lωR, : 2ωkΩ
Again G.-", : - l0o and ε... :
..,4
R,' =--f,,
l+l+R2lRl', Rι
^
(,:
-R,/'
R,+R.
( r'f:-in,n.jt)
=R,'R'-11
R,R. *ι r-
1
^
l
R,
)Α '
ΑR.
+
t
'RιP,^,
R.
)
* n.,
R'Α=η+cR'+G&
R,. _ A-c
Rr
2.26
2.5
and Gn,,.,nu,
Gain error
.:lc
'
c
ttn
=
-
R2/
c*,'-,1
Rl
r*(r*&lze
\ Rι/
: - R2
Rι
I c"".'"
:l c -'l
lc-,,"",
I
-
l
l
'|
rl _ (_lοo)r(δl _ l)
:
909
2,27
- . Δlcl
ιal-
V/V
/lcl
I
l+ R,/Rt
Α
Α
ιιl 4]9] : o.5%. ^Δ :
ΔΑ
/
soq"
G
Α
0.ω5_ι+1ω
.5Α
A = 1ο1(.5) _ 10.1 k
(.ω5)
1+G
a. Equaιion
(l - G",,.'.,,)l:
\€
109o
\
:l0lx9
Α
R, R.
"'
:
/1
=
A
(R|
frlRr
|'- Α
R,
b.G,n.in.l =
''
L fr
1
To resιore ιhe gain to its norninal νaιυeof
R"
l+--i
RΙR,.
_
Α
π'
,'] =l
,-(t+lΔ)
R, \
I+
'LR|
=
100
I
I
l
Chapter 2-7
224
(nz ll n:)
(n, ll Rr) + R4
From example 2.2
!" = -!ι( ι*&n&'l
υ1 R,\ R2 R.-)
ιΙere R| : R:: R1 : 1MΩ
R,R.
" RrRr + RrR4 + RrRl
#,:1'+ +i)__β-lγ)
IMΩ
o'.
ιι
Rι
lM
f.:
' - (u'*z\
\v,
u.9
:
20ο
o'
I
Rn
R,
.,Rn.Ro
R,
_R|
= &('*&*&)
R,\ Rr Rr)
kΩ
231
A)R,
R,/R, = |QQQ,ρ,:
lωkΩ ΞRt :
a)R.:R,:l0oΩ
_Rr(t, Rη
*Rl) :
υl ". _
R, \
l'ι
R2 R.'l
R
l0ο0 Ω
_1666
η+ξ=
(R
||
R)+f :
R
Rη:
R)+r:
(R
||
R)+f :
R
(R: ll
: RΙ|Ξι|: Ι
Ι0 _ ι 1 ι - 2l+νι+2ιxξ:
2
RI + RΙ : RΙ2.,a Ι2 = 2ι
v:
Ιι=
RΧ2Ι
R/
Ι1+
+4!xξ:
Ιa= ΘΙ+4Ι)ι4:8Ι
R. οr -_
z, :
Rl
RΙ1
nι'-ι' = ιι,
Ιιz
i,:!."=-i,R,:-9R,
'RΙRl'
--:
Rι,
Ιn = 4Ι+ι2+4Ixξ:
230
ιι
r
ξ-ξ:
R: = (Rz ll
b)
-R._ l(nκ -π' = -!!9jlω0 - 2 _-
ζ: R| :1ωkfl
:
R, = (R ||R)+f =
:55.6kΩ
lMΩ _"-1_
Υ '
υl zYρ':_
t2+2)
υ'
Sο --.1 :
R,
R2
ιrV
|M
n'=_
'
(-20+2t
lωΩ
R2R3
.,R.
R,.
y
= '2o Υ
ΙfR.=R,
z,
ιo : 1)o/υ' :
ιι ιι / ι'
_ |M :5.o5
n,=
'
(_2ω+2\
ι'r'
v, _ RrR,*RrRn*RrRo
ι'
R!2
R.
___
1r:
t
PJ2
3
Ν'2
o
Rι
l,1l /,tl
1,1I Ι4
c)v,=dR=-/l?
ι z: _ι, R _ _2ΙR
1,
R : '4ΙR
': -Ιj
ιo_ _ι''R+Ιnξ: - 4/R
8/E
2
: -8/R
Chapter 2-8
2-32
Αssume Ιinear operation for op-amp
/. = l_Ι : ο.l mΑ
' lοK
ι:1l :0.1 mΑ
/2X l0kΩ = /lX1ΦΩ=V,
.'.y-
: y : οv
n,' : ]Σζ : Is ιο
a.
...
:
r.
= 0.1
:
Aρain1' iΙ
!!-!!
r. x
lff) Ω
X1Φ
l0 mΑ
.l mA
η:
:
/,
&
l0.l mA
X lokΩ:0.1 x 10kΩ: l V
lι' +
-0.1
Ι2andΙ2
v': v'- RLιl'= l _ Rι Χ 10.l
&: lωο'ζ: ο.01 v;Rl:lkΩ'
v.: 9.1ν
.. -9.1 V <4< 0.01 V
2_73
i, =
iι :
6.1
.o
3'1 mA
ι,[}n,
m R,
m_3.ι m
J .ι, : p'
ο.I
is fixed
7
i2= i1 i1
i1R1 : R2Q1 i)
- irR,
0, lu
Lεt
isneeded
+ov-ilRι irRr:0wherei:
ν.: lo'1' R' + v"
1-(-ll)
o
"ι_- -fi.l _ lω
Here _ l3 is loιvesι νalue of ζ
So R. * : 1.39 kΩ
c. lα)Ω<&<lkf)
1, stays fixed aι 1ο.l mA'.'. ΙL :
tι. V'
m
Thking a loop th.ough the input ιerrnina|s:
Now1,.:ζ+ζ:
and
_,,Α/Α
3'l m
ο.l
]
so R2
'
will b€ smaller than RI
lεt R, :
566
p,:
,I Ξ9Φ
ρ
=
R'-."* : i2 = (iΙ _
lsιο
,ι) : -3
mΑ
y,: ( 3m)(5ω): 1.5v
1.5 3.l m(Rι) + 12 : 0
R,. = (Ι2 l.5) = ].4kΩ
-1.1 m
iI
and
Chapιer 2_9
234
Ρotentiometer in the middΙe:
1o
ι|
.-t
lΓ_
ι|'
kΩ
5 *Σ*l)
".=_{
\5+R] lω l
ν|
!ι : _ι'tl νlν
Rι.
--)
236
Rr=50ιο
ι!
1gs'r: 1g;,
'l0kΩΧri:R(,; rJ
,,Ι |oΙΩ :
R=
ι:!ψ
Iι_|l
lυ ιι
a'
=
b. Ιnput Resistance
Rt =
outputResistance
ρ"
ο'!
ιι
ιι
=9 =
ιι
: ls :
:
l kΩ
V":
R,
i, + lOkΩ x l,. _ i,(I
,\
=
arrd
%
lno
&
c.
- !.lιkΩ
v. in the range of
oο
: t
10
v
r
ιο ._L
i, lo ιο'))
r,(ι kΩ X l0 + 10kο)
i'
:20Κx
!-ι : i'
20
..
2ο
a17
k()= ;,' -
--]2-
10
i,:
l5o 1), i _5ο?rr + _50rrrl\
: -t_
\lω'lω'|Φ')
.t \
Ξ_|/U| +:
\z) rr2|
Now v, = 2γ -6 υ"= _2ν
Γz + 1,
"
L 2
_zll
'J
237
',, = 112
d. From part a
i,:
/R, q+ R,
R, \
,, = _ιd
iΣυΣ+ lιυ\)
10
x
R1
,=l2=
20 kΩ
-
0.6 mA < i, < o.6 mΑ
0.2 m
235
Rr: lωkΩ_ ro<9< -r
Y
"iV
R,:R,
= 1φ1ρ
b:--&(R,*&*r')
v,
R,
\R. R'
fl,'r,Rr'
- 0+1:
/R. R.
1t : -lJ r,+Jr.
"
'
\R,
\
R, ')
I
and in problem
)
_R, _
_l=9f, _ lωkΙt
lta:10kΩ.+a =
ιi
The ιγeighted output of summe, circuit is
Io
kΩ' l0kΩ
_ "ι
'-(l0 & 'lookrt ''\
"/
+1ο: r!! + 1.!)+R' = ι.l2kΩ
\R,
I'
"" : -(2".*2\
\ ' 2)
i'=!!andι=!2
' Rr ' R)
Since i,, r, < Mor Mnput
.'.
Rl,
ιl"."
R2
&
R,
Σ
signals
10 kf,)
= 2.
if R, =
tOkΩ.R.- 20k{!
&
= 1=*' :2R'
R.2'
= 4o kΩ
Chapter 2-10
238
υ
'':
_(2ι
'
+
R,:&,=l0kΩ
4r,
+
R,' R" R. > ιο kf)
Rr:&.:5kο
8r,)
c) 1,.
\=z.\=η&:ε
Rl
R,
R,
R]
:
ιo
= (v,
+
5?,)
kΩΞ Rr : 80 k!l
R':20kΩ
R':40kΩ
239
a)r.:_(r,,+2τ'+3ιl,\
&:'=n,='οιο.
Rl
&
R1
=
R,: l0 kο
*, = 'oJn
z=n.: sιο
&:,=ρ.:]!ιο
',
R,
R":
R,, :
-1
lOkΩ
2
kf)
d) v,' = _6
f,ε= l0 kΩ
&,
:
ι,
R,=Ψ
l0kΩ
R":5kΩ
&,:3.3K
b) τl. =
_(v, + ι2+
20
j+ 2υ)
R,
:
1.67
kΩ
suggested confi guraιions:
ι'': _(2ι'+2v,+2ι.'
ι\
lb
R,: l=R': lg1ρ
Rr
Rr: l=R,
R2
&:
R1
= lρ 1ρ
2ΞR''2: ψ
kΩ
R, : 2ΞR.: l! kΩ
.,)
R4
υ":
(3
ιl,
+
3ι,)
Chapter 2-11
10
Ιn order to have coeficiΘnt
ιhe input resistors to
:
0.5. connect oπe
ι.. 9 :
ι'I
ko
0.5
For summer circυiι we should haνe
\:ι^a\=z
Rr
R2
Select R.
Thus R,
: 2&:
:
1ρ
1ρ
2Dkdt
andi/:R'=20kΩ
2.41
:
υ, Ι 2ι _ 3v, _ 4τ. : ConsideτFig. 2.1l.
"
Accordiπg to eq. 2.8 For a Ψeight€d sυmmer
υ
"
ciΓcuit:
υ- =
"
,R,R- R_i_ R, i
τ'==
'R,Rl,+ z'1Ξ1]J τ"-! υ"J
'Ro
'RrR, ',Rt
R" R._, R,R,_, R,_, R,_^
"π
''π-'
ππ_"ππ
2-4ιl
assυme:
R3 = 13.3
For v, and
u 1,
we assume
τ,l=3sinω,
v:: 1.5 V
Ra=10kO
The output signaι should be
τl.:'3sinωr-3
ι,' : ]sin ιot l
}r":
ιb = Ι.5 V l
Rι = l0kΩ+R.:40kf,}ΞR]:
= ll.3 kΩ
3sinωt_3
R,r40_r&xΦ:r
R, R,
R,:40kΩ,
R)
R'
Rb
: R,: R": l0kf,
40
3
Chapter 2-12
2.42
Refer ιo equaιion 2.8 and figure 2.l1 and make
appΓopriate change for ιhis problem.
υ : 2Si\ (2τ ωl) + 0.0l sin (2?, X l0ω t)
'
ι : 2S|n (2τ σ}t _ o.0ι sin (27, X l000 ι)
'
we wanι at ιhe outpuι
,"
:
lα) r|_
1o0
_12 V
_Ξ9
and
_
t6
r.
12 \,r'hen a" = aι = 42
Ι)
Ξa. Ξ |\νe
haνe ιhe peak νalue at ιr".
+ RF
: .'(ftχft)_ -(ft)
&
.&
R, R,
t6
(20x1+2x l +22x 1+2rX
From equaιion 2.8
..
<," <ο=&
:
12.8
kΩ
2.44
=,oo
&:
l0O
R1
Νoψ choose νaΙue οfR. and ιhen οbιain oιher
+
values.
Equation 2.9
:
'!
υι
v aR'
Rι
υι
Rι
".9:t:t*&
:
Ιf R2
100
setR,:0Ω'R,:10kf}
b. '"
" ,,R,:
kΩ
R!
R.=5ωR.=104kΩ
s;n""
& :
&. & = lω
R, Rt
Rl
c. b.
:.R":Rb:Rι:1ωkΩ
The output signal is
υ.: lΦ
100Ο
:
0
2 sin
' lω
v,
(2π X
and _2 < v <
alov
'
lωο
r)
2 which is less ιhan ιhe
limiι of
:ll=l+&
& =
U. = 1ω X 0.ο2 sin (2σ X
Rl
=R': l6kΩ
Rr
10, SetRl =
lOkΩ'&:
R2
99. seι
Rl
R, |oιΩ.R.:9qοkΩ
2.43
This is a weighted summer circuit:
(Rιιt' + Rrι,l * RΓl,,+π,ιJ
R. \
ιπ
η
&
,.
. ιo:5ιxon
o' : 5ιX aι
we maν \νnte:
'
ι2:5τΧa'
,l : 5rrx
"""
916 l2oaιl+2la|-
21a2Ι 2\α1
i:=
'
d.
Rl =
4r
az* or)
'\t6 a fl * 4 2)
'," = -R-(!!
lοOkΩ
1ω:1+&Rt
d. 'o
ι|
t. seιR
l0kΩ
shon_circυit R,:
υo:2
?!
10
kΩ
Chapter 2-13
short-circuiι R,:
1/RP,
_.
*'!/R-+,/l
''
I
Ro"'
',#'
: (ι
'ff)(*,# '
where:
R ρ : Rpι ll R", ll ... ll R,,
o"n
#""^)
\νhen all inputs aΙe present:
ι.=1
7lo: ιoΝ +
*
: -(o&*'*
fr"'
2.46
ι |: υ :
+R:
7]oP
1'o: R Χ
10 :
0.1 mΑ
i:
i'
(,--ξ)('*, *fr^,*
100μAwhen
tl)
R,,
υ,, * 2υ n
_ z. π.,,: ΙOkΩ+R.
'-zχ:
2okΙ}
ι=>n,":
=
ξ
!ι
2
RPt+ RPz
ν,ν2, ... ι,^/n is:
R,
RF l
ΓR.
*
,oN: -Jffirr,';""r'
***,
Then ιγe s€t !ινι : υιz = ... : o, ιhen:
Rπ : Rνι || Rlγ, || RΝ1 ... Ri/,
1
||
}
R
o. The outpuι νoltage that
to v/νΙ.
"'
)
(ι . &'Ι(&') _ z+ι!L: z+ρ-'
''
\ R]ν,,\RP2''
ρz
where R" = ΙiLΙ&
1;nno.;nn 4-1
'
"
2.47
Refer ιo ιhe circuiι in P 2.47:
a) Using sυperposiιion' we first set
resulιs in respons€
_2υ
('-ffχo) :
meter resistance does not affect i.
:
ι,':
&
l00 kΩ
Αs indicated, i only depends οn R and υand ιhe
ιP| = τP2''.
)*
Note that ifthe results from the last 2 constraints
differ. we 'ιvould use αt addiιional resistor coΙΙnecιed from the ρositiνe input to ground. (Rrc)
2ο
kΩ
|1
The circυit simplifies to:
",,μ: (rι * &),
R"l
I
li RPι
'
op,
llRP2
ι*,-J-*Ι*l-*,_ι R. *' ._tR""
' η'' η'' π
!pt
2,q
ι.: ιn+3υ''_2(ι"+3υ')
Refer to Ρ2.47
!ι
Rrt
= zir n',' _ t0kΩ>R. = 20kΩ
: ?9 :
&
"- 6
R"o= u=*.,.
Rπ
: Rιl || Rr. =
10
(t * R''lR" = l
\ Rλ./R, =(l\
:
l +9.06 Rρ
R, : iΠ ll
:
RPz |ι RP..
:.:
K
+
l| 3.3
k:
Ξ9'l&
2'48
Rρl
Ξ
ιο
RP
)RP'
2.48 kΩ
Chapter 2-14
l'l_ι
Rρl R"z Rro
(I_ Ι.)9 _ .ι+o.oο} .3+Rρ2::Rι
\
Rll
)
Rρι
R",
R,,: 'J,'f"
9+
R,, ll
RP
:
= 2.2sRp.
5
2'25RP || RPι'
: 2.25 R"o+ R"o :
Rr: I0kΩ=Rr6:18kΩ
R",:9X10k:9οkΩ
RE:3Xlok:30kΩ
+
Rp + Rpo
Rr = 20
1.8
R"
oo
kfι
υl
= ι +R,
Rι
:
Ι
+ I
-' :
χt
t
.υo _|
V1
0s.τ<l+->9=l
1η
1b
Αdd
a resistor as
shown
νο ' {l -τlΧ lOkO
τ - '_ ;-Ζτkfl + R
*,(f,).".:
2.49
R.
-
τ:o'19:1t
'R,+Ro
Un_U
υ
R'R:"-ι
R'ι
=":"(ι
R,)
F(orη the tΨo above equations:
u"
\' n,,/\r,
v, =(,'fr\/
Rl \:
lο =
10
R:lkΩ
"
,,
kf}
R
1+R2/Rl
+ no/
1+
R1lR4
tΟ
2.50
Refer ιo Fig. 2.50. setting ?,
output component due to
1]
.i = _2o ν
seιιing
duΘ to
U
τ, I
:
o, \μe obtain the
as:
! = o, we obtain ιhe ouιpuι component
v: as:
ou,:u,(t-Ψ)(#h):r*
The ιotal output νolιage is:
1,^:
ι,
'',
:
,: :
For v,
ι/., =
+
ι''':2Δ(ιl'_
loSin 2π x
losin 2rΙ X
ι')
ωι _ 0.ι
ωι + 0.1
4sin (2'Ι Χ l0ω
r)
l0 kΩ
R
sin (2rr X 1ω0 0
Sin (27 X 1α]0 r)
.l0 i'ι *10)
" ,,'l+lοι
l)
υ6: loυ,
,^ =
kΩ
+1-
t
chapter 2_ι5
2s3
o.999
ο.99ο
ο.9ο9
-Δ.1%
_1%
_g'lqο
?.(+)
Gain
eιτοr
2s.5
a) Source is ci_nnected
for an inverting amplifier
direcιly.
R.=R,. C:-&Rt
υ-:
" lgXl:10t o.οsq v
νn
i':
' tk(): 9!99 : 6.699.a
fol a non-inverting amplifieΙ:
c:'*&
R, =Φ
r
cυrrent supplied by the source is 0.099 mA.
case
b) inserting a buffer
RI
v/v
_ι0
Gain
a
kΩ
lοv
i.:1qJ:lοmA
" ιΚ
current supplied by the source is 0.
The load current iιcomes from ιhe power suρply
10kΩ
lωkΩ 1ωkΩ
10
kf)
c
_1
-2
d
+1
e
+2
f
+ll
c
_{.5 lokΩ 10kΩ
b
υo
Rt
Rtn
50
2.54
AccordiDg to
n_!o
6=
Rι
l +98/10 : ]9:
'I f I +9ο/lο |'2
R"1: 10 =
A+
I*(Ψ*99)
\t0 R.n/
..9ο.9ο
l0 R.h
50
_1
1ο
I
10οο
^($)
ε.::vrv
Compensated:
I
l"' lli-
,.R,
I + RJR,
A
error of Cain magnitude
|ψ
2.11:
50
ιn order to compensate the gain drop,
we can shυnt a resistor ψith Rl.
-
ut t+!
A
!(χ)
Ιo
5
Α:50V/Vl*&=tovnr
Rr
η.
X (510R,λ +
=
90Rsh + 9o0)
5ο X (l0R"h + 90Rsh + 9οo)
lωRsh
:
36ω
Ξ R,t :
kο
l00 kΩ
2.56
ifR,:1ρ1ρ=*l:ΦkΩ
oo:
100
50kο tωkΩ
lο kfl 10 kf,
10kΩ 10 krΙ
lοkΩ 1οο tΩ
kο
of the op-amp.
νo=υι_i
R2
36
kΩ
kf)
chapιeΙ 2'16
9ο
"τ=
kΩ
l0:
6,
ιl0
1γ7γ;
l0
Α(v/v)
162
2.Sa
for non_inverιing ampΙifier,
ΙfA
= l00 then:
G_
r+29
lο
=
;;@1ω
Uun.οπριπraleι!
Ι0:
1.t
s.ο9
ν/ν
ι
1.125
2.57
1.l
G^
:9-
Go
1+9ρ
cλ
G)A ι |Χl <Ι
Ιtlο : -!!t +99
A
l0οΞ Α > l']@ r]
n
G. = λ cn \ r - l
-_L
AV
.+A>GγF ιγhereF: l0ο-l - !ψ
..:Gr_σrl0o
co
Go
A
e%
(v/v)
l0
l0
l0
lω
lαx)
2
-0.83 t6
0.91 9
'ο.98 2
550
l0
9
-9.89 l.t
0.67 l3
is seι to
the bottom:
_ ιs
"
F
+
--ΞjjL
2ο+ ι0ο+ 20
_ Ι0.714 ν
\γhen set to the top:
_15+ 30Χ20
z-_
"
20 + t(n+20
Thus foι:
0.0l:
1 10 l02 lσ
Go (v/v)
1o4 1o5
Α(v/v)
|Φ
lοa
lo7
1o8
too high to
be pracιical
I
= 0.1: co (V/V)
Α(v/ν)
l:
2.l l:
2.59
Refer to fig. Ρ2.59. when potenιiom€ter
x'ι
τ=
Go
a-1
b
I
c_llω
d
l0
_l0
e
_lο
f
g+1
G^
r
,,1
(v/v)
1+--s
'τ =
lα5
case Go Α(v/v) c
v/v
co
103 1ο4
.. _G,_G*l0ο
Go
'*9ρλ
1ω
_t25:
η.
l04
for inνerting amplifier, Eq. 2.5:
G:
..90.90
l0 36
l+Ψ+Ψ
lο 16
c:
Go
1ο2 1o3
6,
1γρy1
Α(v/v)
l0
102 lO3 loa
|ο3 loa 1ο5 106
l
lο7
ι0 1ο2 lo3 lΦ
lο2 lο3 1oa 1ο5
1Φ
Ξ -ιο.714<
lo.7Ι4 ν
υr''= * 1o.7l4
poι has 20 tum, each tum:
2Χ
l,^:
"20
10'7|4
:
t.οτν
2.60
Rl : R: : lokΩ' R, : Ra : loοkΩ
ΕοUation2.|s&=&:lο
'RrRr
From equation 2.16
,ο:πηd
l
R1
&:
= !ρ =
ia Rι
From ηuaιion 2.20
'o
rlu
R,o:2R,:2xtOkΩ:20kΩ
chaρter 2-1'7
2.62
Refer to fig P2.62:
considering that
υ :
υ+i
u'+-_T=:;9υo
ot :
,. onlv: R, :
R
'ι
}Rrana }Rl
The tιvo resistance ,atios
diffef by 1%
...& = ο.sq&
it
Rr
Νow for the case from Α"- and Α7 fιom eqυalion 2.l9
^..:
(π+π)(,-ft,fr)
: lωlω
x(r
{ lo \
ν, only: R,
.ο.qq&.&,i
R.ι
R1l
i4ο.:0.0ο9
NegΙecting the effect of resisιance variation on
:2:zn
Ι
Α,
R'
tr':
"R,t0 : JlQ = tο vlv
CMRR : 20 log| Α,
I
lΑ"'l
|0
ιoρl
_i0.ο09l
:
20
=
60.9 dR
2.61
Ιfνr'e assume
R1
:
ea' 2'20: R'"
_
2R,
(Refer to Fig.
2.
1
|
,s
R1,
Ra:
ι''RΙ
= b: l V/V+R. _
Rr:Rz=Rr:Ra:lokt)
b)Αd_Rι
- P _ 2νινΞR2: ft.,:
ιlr
6)
uι
l?1
to
lOk(!
20k(}
_
Rι
16161
c| Ad = ξ'z = lΦ V/V+Rr _ l M!!
"RΙ
-
Ra
V/V+R: = 5kΩ =
Rη
RΙ:R1 :10kΩ
d)Αl' = * Rl
RΙ = R3
O.5
: lokt)
2 ιeminals:
R,:lr:2R
R2, then
Ξ R, = ?9 :
2
bet\γe€n
= υ2_ιl|
chapιer 2-l8
] connected to boιh
τr
&
l
U2:
R,:!ι:R
ι)'
υ+ = 1ι ::
^. iR, : R,
slnce
il-'
Rι
RlΞR: _
l*':s+l+RΙ
R,
R,,
R.
2.65
ηuation 2.l9
A"':!9=(rhχ'-ftft)
I (t R,. Rl)
,* &t Rr R4/
R1
,'*
"
R,
"ιπ ρ. + Rl
n., -*
n.
._lιmr",Ra_
'l
l?,
π
+
_
R,
I
RΙ
: ".. R:
Rι Rr + R,ι
R" + u"
i=i,+i,='"' R| R.+R4
Rl + R4
R'')
if we reΙ'lace & : ιγ;ιιι & : t& :
' Rl
Rr \Rr R,/
1= 1 *
R, Rι+R2 Rr+R4
|
Ξ
R5 = (Rι + Rr) |l (R1 + R4)
2,64
Ιn oιder ιo haνe an ideal differential amp:
R"+Rr _
R" +R1
R2
R,/RΙ +
]_
RrlRr
R4
R,/R3 +
1
Fοr
ιοιι & and &
RI
R,
Α l
haue ιo be al ιhe
minimum νalues
Αl1 resisιors are 1ο t.τ7o
...
l0
I0 +
!9lj
Ι - f! = l0 _.r
&
-τ
urα
l0 r=R:=10+,
10+τ R' 10*,
chα)se m'inimum νalues
ir _ 10_-r R: _ 10_.τ
ιo*ι
π'l0+,'π
.Α..."'
'' lο+,
_;#-,(,
R4/R1
'*-π'ι
-{
-5
|
.. ir(
t0 r r;'j -
i3,i,
110
ιο|,
-
r121
,
jt)
_
2x
lοι,
Chapter 2-19
foΓ worst case, minimum
CMRR φe haνe co max_
imize the denominator. ιηhich means:
:
R, :
R1
Neglecιing ιhe effeaι of resistance νariation
a."R= Δ =
π.n
ι
: ,*ι,
_
,obc(:)
+e) R] :
Rrn(l - €)
R, :
{2o : ξμ: κ
R1"(1
R'n
R't,
:
cι'rnn
ωωε
:
20 |oc
20 los lr +
-e)
R.n(l
+ ε)
-| + | l+ε + 11-ε
2Κ'_a 2Kl
-- +€
-1+€_ 1_€
Ι €
|ι
CMRR
Rr"(l
_ €'z)_l
(
Ι+€
l
'
|',t(l
-
€z)|
ll
l4cl
for e2 << l.
k: Aainn: 1ω' € : 0.ol
!9! = 6s 66
CMRR = 26 lnn-o()4
2.66
Refer to fig 2.16 and φ.2.i9:
^"-
=
πh(,_*fr
Ιn order ιo calculate Α,r,
principle:
υo
_
if
υn|
ι
",ι =
ιhen replace
",
:
\λ/e
use sup€Ιposition
R'
7ivr
+
2.67
R' ι _ R'\
ηπ
-πιl
π,
*ιo l
_
l' : \
,h"n
Α.. Α..
6.ρ1
RR =
Refer ιo ι, 2.t6: CM
'"4e
R' Γ R, /R" l ιrl,, R'Γ_l+ R./R'. + lr
+ --Ξl
lυ
'
2R,l ---l------.
RrlRr { tl " Rrl
Rr,/R4 + I I '"'
Ξ€
l.------a-
Ad
26 1on
CMRR:104
---:-| | +
= l0 ' Χ
0.25
&Γ, * :'1Ι' i :l weassυmedeat|ier *& : ft "*
+ l l
Α,
CMRR - 20 tonΙ
l = ,o ,o,' 2RlL R'/R'ιRrR.'.l
_lΑ..
| (, _
ιω=9ifRι _ lOklε
}:
R'
L'' 'ι' ππ/ ΞRa = lM(!
R?'lMΩi€
l!&Γz,&,*,,*.ll
_
R, ' 'll
π-'
CMRR _
20 tos
l2RrL
I
I
=
Α,' = l0o -R,
- &=R':
' ιMΩ
+ RzlR'
R.t
"
R,
=
R,Rr"R
+ Α", :
2 t*&
t+η
R.
ι'^
R,
\νe assume
ι
Ria:2R1 = 20kΩ+Rl : 10kΩ
CMRR _ 80db .20 bρ_ !!=9 !ι : ιo,
πv".
R'
l00
i: lΦ
""^_ξ
"r: o.-*ξ
+ r",
νo: _ R, + R.
?.rr,,
1+&
Αr:
I_&.η
l
R, Ro
l, , !R, , lR,,l
2Rr' 2Ro|
:
CMRR 20 r)ρ|'
-l Rr .R: L
I n, n. I
n,
wiιh ο
lo I
-
t
}€
e
o'25%
I
2.68
Refer to Fig.P2'68 and
""' :
Ro
Rr +
(,
R1\'
η.2.l9:
r: R,\
Rt
R4)
taJ
chΔPΙet 2_2o
_
r,
'tω+lωlmι'
=54,., = -:
R,
:
=
υt2
l00ΠΩ_llδτΩΞ-ll
('
lωkΩ \
ι' ' loοτπl ιo κt/
""2
&: &
RΙ R.'
Refer to 2.17:
b) Since Α.-
l0ο.|ω\
lοοlrmJ
Ξη]:
I
Now
\ir'e
ι/ot
1
v-
'-
V": ν"
i,'}
to vι, and v|,'
y" : ,_'
lΦ
υ.:
o.
-Ι(-M
ιff) kΩ Ιl lο kΩ
l00 kΩ || lo kΩ
-
l0ο kΩ
.
",4
lα) + l0ο
3c
2
ζand _2.5 <
vn < 2.5
Ξ-5=I/cΦ=5
c) we aply the
"a
}Ioιο
υ^ lωkΩΧ,;and
si.niru.l'l o" =
η:
I
o,
theo if we apply
we know
: ' υι * ιιzΞ Aa :
calculaιe Α".:
lω kΩ
Therefore_
υo2
sup€φosition
_ 1ΦkΩ X
UΛ'
}
l0 X τlo
,Ι.M + vn * lo X ιo
l''
principle ιo calculate
4..
1'.,,
is the output νoΙtage when
,
1,.'2
is ιhe output νoltage when
v,'
-R" Utt :
vnl = _;l
z^
|, = o
= o
\
:r,""(-r*rz (lω kΩtωkΩlo llkΩ)ιοkΩ
+ ιω kΩ,
||
_L : 1.,
=e
Now \re calculate 1,"u range:
_25 Ξ υD < 2.5 .,a
ι,lι
1ωkΩ
."
lοο kΩ ll l0 kΩ
kΩ || t0 kΩ) _ l0ο kΩ
_30V < ι"." < 30 V
(
10o
2.69
Refer to Fig. Ρ2.69; rve use sυpeφosition:
ul
βl
Chaptet 2-21
Calculate
τ,/
",:
_Ψ
_ Ψ: ι'+*-Ψ
2'22
"
- "o2
υo
_υ2
Uo|
:
I
l
β(v2_
,aol
1.oΥ
/Υ
2R
:
_ -Γ?!'! R' l 2,,.'L& n ηgκ'l
'," R, R" R, 'J
L2Rt '
u
: .4.' = -2&Γt + &l
R,L R"J
2.71
'-β
trι'!2,l-β
l "., _
β_1
^d_ υtq_
A':
Ri1 :
Fo"ι
""
a)
Refe.toη.2.17:
|-p
2
25
ο.9
_
MΩ=R :
R.+R.<Ξ=ρ.
1R. s
lω
6'8
-g:'
6.8 + 0.68
kΩ
25
kΩ
R5
Rr'Rn
Αd=t
MΩ
1
lο kΩ
&:680Ω
&:6.8kΩ
:ft=l.conn"","
and o together
a)
Ul)
Ξθ _
,4a
--ο.g
υ)
9:
_t vrv
:
+t vnr
i)
i,
Rt
,}: v so \νe can consider v', ,
short: il : υ,o/2R'=i, = !!
2Κ
ι'':i"':!μ
i0
A
ι
25
2R'
ιhen:
i7R2+υ^"+irR,
a νirtυaΙ
-
0
+ zr,
:
_!ia
,(l
kΩ
25
kΩ
r,
ι- : ι:ι 7Rz
"R"Rl
i' : i'+i_ : υia +υ,a Rι
" 2R, Rc R,
it : io+ i" : i,
..a ιo : _fi'3R1+ ιu^ Ι \Rr|
[2LR2 + υΒAι
The circuit on the left ideally has infiniιe inpuι
resistance
iiD
y9
:
+2νN
chΔpLer 2-22
2.73
Refer to Fig. 2.20. a.
&
ι Ι + R|/)
The ρain ο[ ιhe firξι staρe is: r
_ ,u'
,,
t 14 v :
'14 Υ τ'Ξ +14 vΞ _14 v < 10l ι.". =
=
_ο.14 v < rtc. < 0.14 v
=9
ιhe opamps of th€ firsι stage satuΓaιe at
ofcir-
ΑS explainΘd iη the texι, ιhe disadvantage
cuit in Fig. 2.20a is ιhaι
,*
gain equal to
*
1'
τ,Ι,.M
η&)
is amplified by a
in the first sιage and
ιherefore a very small U"- range is accepιable to
aνoid saturation.
b) Ιn Fig. 2.20b' rνhen
U".isapplied,ιl
for both
A' & Α, is
ιhe same and ιherefore no cuπenι
flows through 2R,. This means νoltage at the oυt-
put of Α, and Α, is the same as v,".'
iνr1:
τ,2 +1vrv
!=,,
2"
-
2.74
7ιi|
: ιcf,'
1'ιιl2
ι'2 : τca +
2
υ,1/2
Refer tο fig. 2.20a.
2.12
_ (2 _
2 + 0.04 sin Φ,
0.04 sin ιυι mΑ
:
2 * o.04s'nωl + R,ι
+ 50 X ο.04 sin ω,
: 2 + 2.04 sin ω, v.
ι, : 2
_
0.04 sin ιυl
:
_ R, i
50 X 0.(μ sin ω,
:2
v^
0.ο4 sin ωr)
outpυt of ιhe first stage: (1
*
2Rt
:
U.
l4+-14<Ui.m= l4
< υ,,<
This circuit alloΨ for bigger range of u...
:1
ιl,
14
.
2.ο4 sin ω,
ν^
Uι
_
_!
v^
:
ν^
:
2 + 0.o4sinωl
(,
:2
ι_ο2 sin ωΙ
4.08 siη ιυr
V
v
,+
CMRR
ft)ι,".
+
:
0.04 sin ω,
v.
_|
(, *
:
+
Φ
: f *&it
: (, ,
fr),..-a.^,,, '*
'ο
loc
|*
This ιlgure is for 2.72
+2ν
*,,l
,,,,l
ft)",=.axrl
No'ιν consider fig. 2.20.b
Rι=
4ο ιηv
_
*,,,,l
ft)ι,(ι
_
ft)ι,".
10
kΩ
Rr= l0 kΩ
R: = l0 kΩC
Rr= l0 kΩ
= o
(First stage)
t
+14 V
chapteτ 2_23
:
Uoι
/nt
:
CMRR
_ξ('-f,)
?c-
fi.sι
Ui2
' 2R'
:o-_a!ιaρ'!!L
'- 2 '2R,
:
ft)-
,r,,
:
:
1=Α,,Ξ
i(nΞl :
is zero b€aause
of ιhe
lΑ..Ι
ζ.
appear. Now one can eνaluate v" by analyzing
second sιage. The second stage analysis has been
done in problem 2.65. Αs shown in 2.65, ifeach
1Φ kΩ resistor have tx% tolerance, the Α- of
the differenιial amplifier is:
which is a|so the oνerallΑ..
l'^^=Χ:f=o-ο2
'5ο 5ο
lf2Rι =lkΩ
R' ι
R'ι
π ι'* π,,
: lω
= Φ.4 dB
kΩ
/
lω kΩ ι'
RΕ
= o.5
(1)
('
_
#;)=
('2)
al!!:
| *?9 =:νrν.
,r
lυ
uc: Ξ: _:vrv
τo
zolog |Αa| = o
CMRR: ω locl2ll
|99
- lω
-π;ft-)
2.77
Ιf all resistors are :t l7" differential gain Αd will
Ιemain almosι ιhe same but Α- \γill not be zero.
To calculate Α"., apply
ιo both inputs and at
both output termiηaIs offirst sιage y_ ψill
Χ:I
o.s(r
(1)'(2) ΞRtF = 0.505 kΩ:0.5kΩ
& : 50.25 kο : 50 kΩ
kο)
5 kΩ /
loo
differencing action ofthe second stage.
5ο
kΩΡot
+RΙε+ 1ω : 2R2
2&:
Refer to figure 2.20b.
B4νation 2.22,
Α". :
:
\2/
Malιimumgain
Α-
0.5
(,.ft)=.'|.,.-{hl
first stag€.
R:')
lω kΩ(,
l'" _ \(
R.\' ' R'] = lω kΩι l
= 2| νN
-R, & :
of ιhe second sιase is
Minimum gain
Ιn 2.2Φ, ιhe common mode νoltage is noι amplι
fiω and it is only propagated to the outputs ofthe
Α_:0
CMRR:
sο dB
stage_
:
: l+L
Rt
,"-l*l : zoωc(l *ft)
The commοn mode gain
:
1ω kΩ' i3 20ο kΩ
we use a series configuration ofRlΡ aηd
(Fixed)
R, (Ρot); Rι
1Φ
-.-'-_%..
CMRR
"lo.o2l
Α,.'.
Rι
1'ι
* (r *
20 loρ|20| |
2.16
R" \
Un2: Uιm*iι'/ ,*πJ
:
:
The cancιusion is that ιaΓge CMRR can be
achieνed by haνing laΙge differentiaI gain in the
ι,'Ι
υ-1 : ι'' _ R' Χ
!oz_ aοι
νN
ι^-:L:o'oz
'_ 5ο
Ucm oo,r+!ft,ι us
:
2o1
l0o kΩ\
5oo
,
b) υ.
lo
: υg_ ι" : ξ|l^a!9 : 6yμ
υι(V')
7tA
ch^ρler 2_24
c)
va(ι)
1.5
τ,h
and l,, can be 114 v or 28 v Ρ-P.
-28 < ι,,< 28
η'.-.
:
Io.ε
or 56 vP-Ρ-
v:
{J2
2,78
-1.5
Refer to Fig. P.2.78-a:
since ιhe inpuιs ofthe op_amp do noι draw any
,,
curτenι,
appeaιΙs across Rt
.
'. --01R
Fig. P2.78.8
ιρ: Z}o
'\λi
e use
υ
supeφosition:
v^
Zιιoι
on]|y:vB= --! :
2
zιi"ι
'2- ----
ur
:Ι!_i^|(Ζ,+R)
&
Rr
υ,(η
+τl,:i',R+i_,:1
.',
Now if only (_
ι,)
R
is applied:
υ' l
i,,' )/ (R +
- ----a-.zli.'' 'n = ----- z- zι)
υA : ιB+ _ι1+ zLio: : i,'2R + ioζL
lB
ι'=i'R=ι-'__-L
',.
R
The toιal current due ιo both sources is:
;:i
-
+i
1'r:y)
The circuit in Figure P2.78(a) has ideally infinite
input resisιance, aηd if requires that both termι
nals
be available' while the othercircuiι has
Ιlnite inpUι resisιance \rr'iιh one side otΖ
gΙounded.
ofζ
1.5
=υJ
fRR
Chapter 2-25
2.79
,,(s): _l _
sCR
';(S)
,ο(r)
I
jwCR
o-ι
'1
=
υi
:
jννx1xlo 9X lωX lo]
Ψ-, : loa rad/s and
J|ο
J9
1=
= 1591 μ,
"2τ
b. i = ι( j) : jΞ
I
phase
..
of9σ
b :
c)
u.(r) reaches
This indicaιes a
10V at,
:
2s
2.Eι
lη _ _]=
ιoRC
i.e. output Ιeads input by 90"
Ψ,
_ lo + 1flaι
= I0+Ι.ι
c
: 1ο+1
RC
v"(r) reaches o at , : l0RC:1ο
1
a'
:
if ιhe fιequency Φ is decreased
by a factor of lo, the ouιpυt \rill increase by a factor οf 1ο.
d. The phase does not chaηge, it wiιl be the same
relation as in part b.
Lao
then for |T|
ιω
if |T|
= lω V/V tor/- l
= l v/v,
kHz.
AΙso
/has to be l kHzX
Rc:4ιιT 2τΧlkΗzΧlΦ :
2.42
Rh
:
lπ:
<o
R, Thus R
ι
:
kΗz,
ιω
=
l.59μs
l0ο kΩ.
ιοRC=111
-=l. Rc
: l(χχ).167" : Ι*6
Rc
: -----.!-- : lgπF
lω0 X 1Φ2
ιγith a 2v-2ms pulse at the input' the output falls
iinearly until , = 2ms at ιγhich
υ-_ υ'-
R":R:1ωkΩ
where , in ms
Time constaηt Rc is
Rc
]
: 0.l5Ξc: 0'1 _
R
hen _
lv
ο'1
'ω
_
:
-]-L
lωK
1 = ---Ξ, = _2,
= --!
cRc
Thus v, = _4
kο
νo|ts
γ
υl(ι)
dc inpuι is applied the cuπent Ι is
/ = -Ι v =
R
υ^
Iο,..η
The voιtage r,. rises lineaily. Ιfat
l:
o, v.
:
l0
v
then
voQ)
(ms)
ι0v
chΔρtel226
η(v)
, (ms)
(ms)
with yΙ
:
τ,^(r)
2sin 1ω0, applied at the input'
2 Υ
___-J_
l0οο
Χ Io
,
sin(
ιωο, -
q0")
= 2sin (1ω0, + 90')
z/,,(r)
a) \γhen no
tt
2.83
?,_ιl) = ----:||-dι
R,"=R:20kc)
RCI
lrl = ωRc
-l:
lar
?.(ο.l)
ω:2nx l0kHz+C
:
C
=
2eτx1okΗzΧ2oΚ
0.796 nF
Refer ιo discussion in page l10.
ι-"
ιi
:
Rι
Rι / R
and ιhe finiιe dc _
ρain is -&_
R
I+SCR.
There fore for 40db gain or equiνalently 1ω
wehaνe':& :
b) ν/ith RΓ:
6.2s
v"(r)
62_8l
o< r<ο.ιms
v
:
-
7,,1οο;1i
e-ιlcRr1
(Refer to pg. l Ι2)
Ivx2M _ _lα,v
.la-)= lΧR.'
2οK
,,'(r): _100(l e ιll'5)
v/v
η(ι)
lω v/v
R
:
_
ΞfΙ, = l0οΧ20k:2Mf}
The comer frequency
_--J-
0.796m Χ 2M
vr
:
L=-l=/ IζF
i.,
628 ΙΙ7
, (ms)
(v)
)
"rlonenιial
2.U
Foro<Ι<0.5ms:
τ,.,(Ι)
, (ms)
-
:
_ I'
π;ro,
ι
ι
RC --_ !ms
_0.5 v
?l{rl=o_
ι"(0.5)
U.
Chapter 2-27
ιι(υ\
, (ms)
Then foι 0.5 < , < 1 ms' ιhe curΙent Ιlo\νs in the
opposite direction
:lms.
For
τ':
,.
τiser linearly reaching oo at
,
!2υ'.
we obιain ιhe to||oψing \λ/aveform: (assuming
ιime consιant is the same)
Ιf
is also doubled, then the waνefoιm becomes
ζ
ιhe same as the first case where
υ.(ι)
&.:lms.
τ,
,
=
:i:
1
V and
, (ms)
, (ms)
forο.5<t<1ms:
ι/' (,
)=
2,t5
ι,_{o.5)-Ι[_ra,
"
RCI
Each pulse lowers the output voltage by:
0.5
1''lι|:
'l l ms)
r,_(
o.5+
:
#f'"=
aι._ο.5)
RC
0.5 +
ρΞ :
ο
v
l0 μs
RC
_ 10μs _
1ms
'6.,,
Therefore a total of ιο0 puιses are required to
Another ψay of thinking about this circυit is as
cause a change of
Mn
,"(r).
folloιvs:
for o < , < o.5 ms a curτent
/ : lJ 1o*,
R
through R and c in the dire€tion
diagram. Al time t'qr'e \λτite:
ι'ι
= _cυ^ιΓl+υ^ιιι _
indicated on ιhe
_!ι - -!ι
cRc
ιγhich indicates that ιhe ouιput voltage is linearly
decreased, reaching -0.5 Vat ,
0.5 ms
:
m
l0
-20
-30
, (ms)
Chapter 2-28
,,,(2)
ι|
l-ΕΙ
,, l
22 = R2 ll
+
= Z':
Zι
ι,,,
ιι
and
Y|
Z' =
v
R,
|/R|
'
Y2 ' 1+
R7
" _Vo_l"
o'l.* ηlr
(RrlR,)
a
dc ρain of _
"Rl
"
&
sTc'
"(L)
Γ- 'Vo1
v^:
sc
I + sCR2
This function is of a
.:2
R
loιv pass circuiι haνing
and ]-dB freouencν
cR.
n,.:n,=iοιο
: 20 dB : 10
.'. lο _ &=f' = l0R, _ l0okΩ
R, '
,,o vo
η-
]t'
vt
-Rζ
vo
v
I l=
-ι.nc,
v^Γr*a*
"t AO AORC'J
vo
vι
:
vo
vι
:
RC'
I
RCs
('
dc gain
yr
.;('
.
#")
I
RC"
|
ιA^RC,
AoRC{ "
+
Rc. + !l
Yo:-#''o"t/':vΙ Rcs(iqo+ l)+ l
Rcs(Αo+ l)+
Ao
tt
lι
lι
l!
tt
tt
tl
STc)
(Loψ_pass
-
Dole freouencν
(kΗz)
.'.
ωo
C:
: 2r'X
l0 X l03
2r.x10x
_1
CRt
l03 Χ 1α) K
Unity gain frequency aι lα) kHz
0.5 nF
Rc(Αo + l)
The ideal integraιor only has a'poιe frequency at
the origin. The above poΙe frequency makes it a
more stabΙe circuit.
ΙfT=cR
3dB frequency at l0 kΗz
l
then
t.
and
f
vo
: _1_
vΙ
ω
=
2τr (Recall
that
T
ω
: 2d
T
scR
_1
'(*l)
:---:-159
I
I
jωCR
=
ilcR
,T
TCR
2π'CR
2?ΙcR
chapιer 2_29
For AO =
l,0Φ:
_l'οο0
2.90
ηn-i<''ω'l *
- ,. α)0
(J-zn1'rl,m)" +
ι'
l'-"
R=10 kΩ
'
: --159
2.8E
9(r): sRC:
: -lO
u-9o
ι
-sxo.o1
as
i.1 = _i.x lo '=
l'rl =
=
|
orl=
,,1
1.59
xlo-uxtoxi0'
t ιvhen ιν
:
|:e|
-
-w
x to'a
toarad/s
kΗz
for an input l0 times this ftpquency, the output
will b€ 10 times as large as the input: ι0v p€akιo_peεk. The Cj) indicates ahat ιhe output lags the
input by 90". Thυs
υ6Q\
:
_5
sin ( lo'5, + 9o') volts
dι
':rΦ:ο_lXlο-ux.lnn"
: O.lx10-6x 2
0.5
=
'
x lo-l
0.4 mA
The peak value of the oυtpυt sqυare waνe in = iR
:0.4mΑΧ10kΩ
:4V
Vι(v)
2f9
, (ms)
vo{
η
, (ms)
uo
= _cnΨ
For0<r<ο.5:
z^:_lmsxlV
= 2ν
"
0.5 ms
=0
l0V:4Vx2.5
ιhe valυe
therefo.e:
and Uo
The output ιvave has ahe same frequency as the
input signal
The aνeιage νalυe of the outpυt is zem
To increase the valυe of the output to
otherιι,ise
ofi?
has has to
: 10KX2.5
ω increasω
= 25kΩ
by 2.5 times
chapter 2_3ο
2.9r
οi
υi
φ = '90"
always
l'ol -'=r-
I
unity Rc
lηl
_(R2lRΙ)J
-
RC : lO']s ψhen
c = lomFΞ R : lΦkΩ
oo :
sRc, 19(jrr) : jνRC
r+-lRtC
If
|ιοal
= lkrad/s
cain is lο times
the unity gain, when
ilarlγ for w =
a
:
the frequency is 10 times the unity gain frequency. Sim-
'
ψ:
lο
lο krad/s,
krud/r. gain is o.l ν/v
.
ιlor
gain: lov/v)
for high frequency C is short circuiιed,
oo= R = lσ)=R': lkΩ
υ; Rι
ιlo _
RCs _ lO-]s
u, RtCs+ 1 l05s+l
Ξ
f
= lωkrad/s
ιrrdb
.',16
=
or
l5.9 kΗz
f (|Ηz)
This is ιhe transfer funcιion of a single time consιanι high pass fi|ιer haνing High frequency gain
:
(R2ιRJ
and 3dB lieqυency at
J_
.o = R,C'
.
Αt high frφuency the inpυt impedance
_1-
abbroaches R| as zo
select
Gaiη
Rι :
:
4ο dB
For a sain of
'Rl
"r* tl+,,, : 1.01 krad/s
if w = l0.lkrad/s: I'al : l0.l _ 10.
t.0t
φ=
s;
:
kf,}
:
Ιω.
1ω v/v
R:
: lω
Rι : l0 kcl
.'. R2 : l0kΩX lω
and
for unity gain:
ll0
l0
b€f,omes νery small
j'C
'
16
It
_95.7'1"
2.92
: tMο
For a 3dB freqυency of 5ω Ηz
_L : 2, 15φ
Rrc
C:
1
2πΧ5ΦXR, =
12nF
From the Bωe-ploι the gain
&
υi
reduces to
uniιy
Δt5Ηz
,*^Ιλ_J
2-93
Refef to the circuiι in ΓιgP 2'126:
oo
ιι
l .eL
Zr :
n,*!,2,
,sC
Cain : υo:
υi
Z-2
Zι
:Rz
=
_R2
π,nl
, sC
= -Zz = - I
Zι Z'Υ'
oo:-
R2/
(,
:,
(*, *
Rl
*;fi)ι'
*,*,.,l
*!)(or;. ".,)
chaρter 2-3
7]Λ
'
_R./ R'
.
"
(, -
_
where
l
,+a;},
+
ih,R2c2'
_ R2/ Rι
('.ft)(,.,;)
I --,:
= R,C,
'
w,'
a) for ιv =
u,
I
RrC,
<< ιr:,
υ^'' _R'/R, -R'R'
.R' ι,,
: 'Rt'ν,
-|'Jlυ)=-:--Ξ:#
η
|/J$l
(l + Ξ]
jwl
\
νν
υι'i.\ _-
-o.ι:
uor
b) for w, << w << ιv2
u-9(
,,,=.(t*fr)
&
:
"..(r
*$)
4 mV
2.95
R|
'"ιi,l=ffi;ffi
: i:2 mV
,o : 0.0ι sinω, X 2ω
:2sinωrao4v
voJ
c) for w >> w2 and ν2 >> w1:
: r(frχ*)
from the resuIts of a), b) and c) ψe can dΙaψ the
Bode-plot:
r+l
+ ,os X 2ω
2.96
Ιnput offset voltage = 5 mv
outpuι dc offset γoltage =
5 mv X closed ιoop gain
:5mVXl00Ο
- 2ωb/decade
The maximum ampιifier of an input Sinusoid thaι
results iιΙ an output p€ak amplifier of 13 _ 5 = 8v
is given by:
oesiεn:
,.:
' _!l0ο0
RzlRι
Rι]Rι
1+:r
l+:-.i_
w2
HP STC
LP STC
ft
= looo (ω
=
3-γ
Ιfamplifier is capacitively coυpled then
a.,,
=
1?
ffi:
13
mv
2-yΙ
dB gain in the mid-
fiEquency range)
|'4mυ
R;n for r.ι >> w,
:Rl:1kΩ+Rr:1μρ
/1 :lΦHz+w1 = 21τΧlω _ι
RtC
""
t
: 1.59 ,P
= C,
f1 : |0Ηz='w2 : 2τx Ιox l0r : I
i, c,
:
C,
15.9
pF
=+
2.94
Ιnverting confi gυration
1.4
100
2.98
10
MΩ
chΔpter 2-32
^)ΙB:
open input:
ιo:
: l t0.05 μΑ'
1r2:la0.05μΑ
1r,
(ΙB|+ ιB)l2
υ++ R2l31
9.31 = vos +
:
τlo5
*
R2l61
1ωωrrι
inpuι conn€cted
a)
υo=
=,,,"(ι*&+n,l,,)
X lο1 +
10ο00
:
Ι8t
Ξ 10oUο. : _o.22=> ιr," :
9 /rl : 93ο nΑ
Q)
-2.2 rοΥ
Ιg=Ιg=)!Qntr.
b)
ι". :
-2.2
Ιr'
For
l0 MΩ
=
0.95
μΑ.
0.95
X5
+
0.95
μΑ.
/r, :
:
:
Ξ
.γ
- u,fr)
Rr(/sr
+
Ιφ:
|ω(l.o5
1g2
ignore z;,r, compare to the νolιage drop across R.
2.46 holds:
therefore from Eq. 2.47:
Ιos
:
ΙosΧ R2+ Ios
80 nΑ
= l.05 μA
42.5 mV
42.5 mv
= υ6Ξ 51.5
πΥ
R,: sl ,*:
lοkf,} Ψhile
k(l
the pos-
soυrce.esisιance.
:
Therefore ιve should add 5 kΩ series resisιor to
the posiιive input ιerminal to maιe the effecιiνe
Resistance 5 kΩ + 5
kf} = l0kΩ Theresulι-
ing z6 can be found as follows:
υo
' _ΙιιX
lo I looιrs|
(1rr-1r2)X100
0.8
l0
Mο
: Ιosx lΦ :
rb: t10mv
7]o
a0.1 X
[f ιhe signal source ιesistance
,,,,#)
lΦ = 1l0
is
15
kΩ,
ιω kΩ
1ω kΩ
9
5 kfΣ
15
kΩ
-
mV
then the
rcsistances can be ηualized by adding a 5 kΩ
resistor in series wiιh ιhe negatiνe input load of
the op-amp.
2.φ
ο)
Therefoιe Ψe should add 5 kf,} source resistance.
c) Ιn this case, since R is too Ia.ge, we may
ιo :
- _ι,*,
l.o5X5+ lω(o.95 1.o5x*X9)
itive inpuι ιerminal sees 5
η,. << R/r, Also Eq
R, : l?, || π,
o.οs
From ιhe discussion in the text we kno\ir' ιhat to
minimize the dc outpuι voltage resultiog from the
inpuι bias cυπenι, \νe should make the toιal Dc
resisιance in the inpυι of ιhe op-amp equal. Currenιly. the negative input se€s a resistance of
R, 1l
R= l0 MΩ
0.95
57.5 mV
b)For
ι1ι'
0
2.45:
/rrRi
to gΙound:
υo: L'++r,(l,,+ff)
9.09 = aos
(1), (2)
η.
From
(1)
:
z,,.
chapιer 2_3]
: (.,_Ψ)
o.45 μA
5l
ι+ = -ιB7x (l0kο|l l MΩ) = '0.45 μAX9.9kf}
-4.46 mV
ιoz
:
Υ+*
=-
Rr:
fl.,
:
C,
0.16
2.102
sοz Ω
: 2o y lσ)= C, :
-!R,C'
: 3.l8 μF
_J- = 2l- x Iο+c. :
:
/
4.,t6 mΑ\
lMΩXι0.55μΑ* .πιn-/
99.5 mV
The Ψorst case dc offset voltage at the output is
(99.5 + 404)mv
503.5 mV
t*&:zω
Rι
R,,
ιlo"(l,' + ft)]
:
: 1φ1ρ
lΦk:
n,:
'
t99
4.46 |
[r
5ω X 2r' X
lω
I
l0oKΧ2π'XΙ0o
μF
2-101
Τh€ output component dυe to vos is
,, :
:
,rr(t
101
+
!$!) :
19,
o' : u' Ξ
,r,
8mV
x4mV
The outρut component due to input bias curιenι
/,
Vor(in μΑ)
lMΩ
= 4ο4 mv
V6
is:
:
:
!"r: I"+19!
R.= 1MΩ
: V1+ I uο x(;
2V." +
I
+
ft)
ιuοx(Ιρr+2yos')
\l M t k )
2ω3 vos
:2Φ3X4mV
Rl=l0kΩ
=8V
For capacitively couρled input
+
vo
=
:
β.,
0.55
-
Ψ)"^
μA
Ι"': Ι"-ΙΞ!2
chΦter 2 34
Vt:V:V65
Vo = Vo5
b) Largest output offset is:
'.'i =
V6 = l mVΧ 1ω+0.1
O
v'- _ y.. l
vΔ_ v"+ !MΩx-j-]ΖΙ
ι0ωyo\
IkΩ
= 1Φl Vos
=
Χ4mV
1001
l kΩ
A large capacitor placed in series \,r'ith
resisιοr
tMΩ
c) For bias current compensation Ψe connecι a
resisto. Rl in series with the positive inpuι terminal of the op-amp, wiιh;
r-"
= 4.Φ4V
V:2ΦmV:0.2V
:
]!9
:
Rl :
RΙ || R,
"" lο to n.η
Rl : ιokΩ|| lMΩ:10kf)
The offseι cuπent alone result in an outpuι offseι
νolιage of
lMΩ
: lox lO ex I x 106 : lOmV
d)Vr:1φrnu+10mv:llomv
losx
-'
R2
2-l0s
I
Rl=ι0kΩ
V*=V
=
Ve5
Νo dc cιrrrenι flows ιhroυgh Rt, c branch
..Vo:
= 2Vo51
=
See
VA+Vos
Vρ5
R' : lokοl| Ι MΩ : 9.9kΩ
V6: I6R7andY6:
= 3X4mV
12
R, ll
Fιom equaιion 2.Φ
3V os
:
ηυation 2.39
R, =
mV
o-2|
Rz=
Υ
lMΩ
2.103
Αt 0'c, \γe expecι
:ξ l0 Χ 25 Χ 1000
At 75'c,
μ=
Rl=l0ιΩ
:t250 ιnV
\νe exp€ct
:i10Χ50X1000μ::t5ΦmV
we expect ιhese quantities to haνe opposiιe po|ariιies.
2-lιA
lMΩ
V65:1mV
V- :
Ι BxR\ !
Ιf
vos
,82RrayOS , /srR] a yos
,
,",_
n,
&
100
a)
=
1+ft+nl =
= ,-*,(η1
lo.l kΩ
V, : l00Χ 1o'9X l Χ
106
:
O.1
V
+I :
Butf
Rr R2
-
il. .,θ - i)
1
R1
Chapter 2-35
ι _ ,3'
ι
'Dl
+vos
_
π
(Iιι'Ιιz): .3 : ,tlΞ! :
so the offset currenι
tο.l
is :t0.1 μΑ
LlΜ
a) To compensate for the effect of dc bias
Ιj'
μl
cυπent
we can consideι the lollowing model
Ao
fb(Ηz)
f,Glz)
lΦ
1Φ
107
1Φ
1
1Φ
lΦ
lΦ
1Φ
107
ΙαΙ
106
2x
lο
los
2x106
lMΩ
2.1αB
Εη.2'25:
t
λ:
ζ
Ao
Ι + jv'lwb
Αo:
80dB,
=l^l:
Α :40db@/: lωkΗz
_
[*($
-^i
- lAl - 2ololAo 2oloΕA
\.fnl- zοlog}l
:86 4Δ: 46 dΒ
_
:
l ' rlω}Hzγ
\ /" ι (|99.5)2+ ''f" 0.501 kHz
fo: Sot Hz
f, : Aofυ: 1.995 X lo'1 X 50ι
20 |og
simiιaΙ ιo the discussion leading ιo eqυaιion
(2.46) we have:
R1
: i|ι Rr:10kΩl| 1MΩ+R',:9.9
1ρ
86 dB
=
MHz
9.998
:
10
2.109
tdx
.
b) As discussed in section 2.8.2 the dc ouιput
voιιage of ιhe inlegraιor ιvhen ιhe inpuι is
gτounded
vo
:
=
Vo
is:
: mv(ι
0.303
:
V
0.313
:
vo
-
ibγ$)-
+ 0.01
V
V
lonΑ X l MΩ
Αo
= .8.3
Εq' 2.25|
X lo] v/v
A
:
Ao
e4.2.281
f' : λofι
-
t+iL
"
2.ι07
+
vos(l * &) * l,rn,
f
"
MΗz
Chapιer 2_36
5.t
x l0ι :
8.3
X tor
*
f-(!!q_ξηJ'
=
f6:
'
*
2.|ι1
('*7Ι")'
Eq.2.35:
kΗz
f,: Aofι
=
X l03 x 6o.',.kl'z
8.3
:
5o3 MHz
have:
J
Αo(db) =
:
2ιτ x 4'Ι.6 kΗz
y6
47.6 kΗz
_ls=rcA= l+ Ao
jflfυ=lι*;/l
| "ΙbΙ A
RI
_+:
",l n/t +
./ = |o/\.hΞ|Uo| = +
Jl + lω
l
106x6o.3 =
180.9
:
50
X
lο6
v/v
-4l
I /,,1- *A
l,
f
,:
Aof
υ:
_ lo-*J!]! _ .!Φ+!n _ lΗι
= l0 MΗz
r'''' _
50MΗz
υ| J|!
.]Φ
= [u
:
lokΗz
d)Αr:16rlω:1ω0vΔr'
o'l ) loa =
_
Γos = 7o
|' , n| |oΞ
Jι
l Jbl
/, : lοο0 X l0 MHz : l0 GHz
: 25 v/mv x lο :
l, ,4l _ Ι0.92'5!Hz
]ι
Ιι'l
25
1
=
=
Aof ι:
25
X loa Χ
X
104
v/v
.!Φ..> f o
2.51
X
Ι'
RI + --i
'
:
103
=
MΗr
lO
l
tω kHz
α)
t00
t+j/_
1-,
-5:'
For
6 : -6"
1dB
t-L
f tot
-f = "/:on x tan6" : 10.51 kHz
Φ : 84'' f : /3dBXtan84' : 95l kΗz
lυ MHz
2.\t3
al
& = _ lω ν/v. """
r,-. = lΦ
R'
η.235:
,, : ,.,n(l -
_
-
--------7J
/6= l50ΦX 10K:15oMHz
e)A
and cιosed loopgain
Rl
c)Α:ι5ωv/vΞΑo=15ΦVΛr'
f,
= l9.9vΛr'
: l *R,: lωv/ν
Rι
fo
lο"
|' * 4l = IοΞo't }
lhl
lι
l
v/v
2,1t2
MHz
lοv/v+Αo: l0X50Xl05
X
l9.9
1'
1ο. t
|'il
"/,
lο5
,Ι
_ o.l/',dhΞ|Uol
"r
6 X.
Α :5οX
:
_RzlRι _
_2o
υou,_ 1+ s
t *-llΣ2πΧ |06
,,.(r.&)
,\
: 20 log10 Ξ Ao : loA
a)Αo: 1ρx3x l05: 3X lo6Hzv/v
20 db
b)
ιρ,
2rr λ 106
l+Rrf' = l+20
F4.2.34ι
2ο dB + Α(db)
tO'] ,q,
: 60.3 Ηz
'+ 'f ι
f,: Aofu = 3x
:
ι''υυ
2.110
\ir'e
|oav/v
fι = |06ltz
2.65
60.1
_ != _roΑn,(Ι
G*".
2.5l kH,
62'7.5
ΜHz
ιl l l
r,
=
*&
fr)=
- lω v/v.
r,.(r *ft)
:
r,
:
/ι,jb
lω
kHz
k
- lΦ
,ovιr,
x l01 = l0.l MHz
kHz
Chapter 2-37
c)
: 2vlv,
1+fr
:
/366
lokΗz
:10MHzX2:2oΜΗz
Φ
_ft
:
-2VlV,
/, : loMHz(l +2) :
o
f
'
:
/.,oo
10kΗz
"
: -J- *
/,
:
+
(r
ι91on
:
ft)n,'
50
:
Χ
10
tO6
+
/laι =
7.96 NlΗz
:
MHz
Χ 7.96
79.6
2.117
a) Αssume two identical stages, each wiιh a gain
30MΗz
-fr : lωov/v, /3db : 2okΗz
: 20 kHz(| + lω) : 20.02 MHz
Go
Go
= 1+ jflΙ.
l+iy"φι
:
function:c
ηl+ ft:rvπ,/316:lMΗz
lMx1:
s)
-9,(r : -r,
lMHz
-rv"
:
ι , (Φl'
/,: l M(ι +|):2ΜΗz
1.oo
Gain:1*&: sε v/v
Rl
:
8 kHz
/10,
J':96x8=768k}Ιz
for f3dΒ : 24 kΙΙz
6r;n : E9 : 12 ΥtΥ
"/2.
r,^[rt _
2.115
f'
lcl
=
|Gl
: 0.99+/ :
:
Γ *(-'
*(-ιΥ
2.20
:
1
n;7
f,
"'"
rulP,r.
'
: --l -: ln.
2π Υ |o6 2τ'
to Apρendix F)
l +& = lov/v. R,
5τ
: lω: 1+*
Rι
l6ιH,
$
=
oι
and ιherefore a ]db frφuency
ιo'ιl"'
:
:
'
The overall froo
|o' {^Γ2
|
64.35 k|1z
which is 6 time greateι than the band\γidth
achieved using single op amp.
(case b above)
Ιι _ lω
: lιΩ. R, : 9kΙt
the time that
output νoltage to reach 99%
iι takes for the
ofits final νalυe,
then:sτ:100nsΞτ:2οns
y5
:
5Φ MHz ifsingleop-ampis
used.
ψith oμ_amp that has on|y
ιhe possible closed
l.ηl
2.1ι6
Ιf \νe consider
σo + Go
2.118
o.l42 ΜΙ1z
0.35 μs (Ref€r
RΙ
ι
should have 20db gain or
ιI} R! = lo
The folΙower behaves like a low-pass sTc circuit
wiιh a ιime constant
= 2Olog.E
RI
c) Each stag€
.r,:
= | ΜΗz
Note 3db
Ι' :lMHr:
lω
t *R,
f''.:
24
:
:
=
b) 40 db = 20 log
2.114
tr
will dΙop by 3db when:
Τhe gain
I MHz
:19
|ω
f, :
4o Ν|Hz
ρ gain aι 5
MΗz
'
is|
= 8v/v
a overall gain of 10ο, three such amplι
fier cascaded, ιrould be required. Noιγ, if each of
the 3 stages, has a lo\γ_frequency (d) closed loop
To obtεin
ρain K. ιhen iιs J_db freouencv
ιγ l Φ MΗz
ιa
.
Chapter 2-38
Thus for each stage ιhe closed loop gain is:
Set
v::
o,
RllRl
vο
_
vl -
l+Rt/R1
ωt
,
1,.-
which atf= 5 MHz becomes:
R,
f,
f.
=
= ----j-L = "-r- srnceJ
R,
2
..R.
Rl
. (*)'
setη:o'
,
f,
JrdR I
k= 5.7
Thus [οr each cascade sιaρe:
:
r,-. _ 1q
5
f16:1ΜΗz
The 3-db frequency of the overall
'''Ι
amplifierr, can
v.
:
zν.=\
.R,
I
= z
f,
R, - I +2
+ --r
R2
f_,
l
2.121
The peak νalυe ofthe largest possible sine wave
thaι can be applied at the input wiιhouι ouιpuι
clioninρ is: Ξ-.!2Υ
lα)
γ31ug
= !29 = 35
Jz
- ο.ιzv =
l20 mv rms
ηγ
2,122
2.119
R,
a)π = t. "/ldb:
J, :
R'
f,
l(nkΩ
I+ι
RI
GBP=GainX"fiυι
GΒP
1b
: k Ιb
I +t
L
b)l+ft=ι/',,1b:
cΒP : k+:
Ιl
k
a)le,:1Lρ
f,
for
τ.,
The non-inνerting amplifier realizes a higher
cBP
and
iι's independent of k.
when oυιρυι
2.1?,0
lο
-':::
_
16
.o
_ l0 + 0.l
_ lo
0- l mΑ ιherefore l'
Iω kΩ
: 10.1 mΑ is well υnder i.,.", : 20 mΑ.
b)Rι: lωd)
v1
lf oυιout is
v1
U.e sup"φorition ιo iaIuate
,"'tm
= J9
uP : 0'l V
is aι iιs ρeaλ. r. _
= 16γ.
at
iιs oeak' 1,
"
_ J!! _
0.t
l0omΑ
which exceeds i"* : 20 mA. Therefore
go as high as 1ο v. instead:
f,.,, for each case.
26.1
_
=
1-Ξ,:-=5o-', -9lrn Ω lωK
ξ
cannoι
20 =2v
ι0.0l
Chapter 2-39
u" =
'
3-:
l0ο
c)R, _
20 _
?.
0.l :
o.o2
v
:
t''-'.:20mΑ_ lοv-L l0v
R..,n
_JL 9 ρr.,"
Λι.ι^
2.123
op Αmp slew rate
:
:
0 ν/μs
For ιhe inpυt pυlse to rise 5v|, it
s
!0
5p =
2o mv
lω Κ
s62
ρ
The output
rate
0.1)Ρ
ι,
pυlse width
:
W
will be a triangular with
:
10
V/μs slew
o.2
l0 :
40 v/μs
_ 29J
T/2
=
sκ
Thur?9x2:
lgV/us
τ
4 ι,.s o.
For a sine ψave τ
Φl
0.5 μs
0.8-{
sloDe of the ιrianρle \νru"
Ξr _
will take
:
2,125
Ι
^_
.'. The minimum
-
(ο.9
f _ lT _
h:
υ
.Sin (2π
= 2π\25OX
l.,,
- =9 r"
"
dι
25okH,
x
250
|ol;, _
!οX
'" Iο6
'_
2τx l0r X 250 -
X
1o'r)
sR
t.-t/rv
2.126
υι
,"drdl
= lo"in-ι+Φ _ tο-"o.-ι=Φ
: l0 ιo
(V)
The highesι frequΘncy at which this outpuι is pos_
sible is ιhat for which:
99|
|-ξ* w
_Ξ
SR+ l0rυ,", ω/
dι l.,,
:6x10'1
Ξf
=
^ι'
2.127
lo+οΞωma!
45.5
kΗz
:
0.5, v. = 10 X 0.5 = 5V
ouιpuι distortion v/ill be due to slew_Rate limitation and will occur at the frequency for which
a) v,
d,"l :
dt 1.""
ιυ-"'Χ5 = 6' = 2,/ loξ
lο
-_] : 31.8 kHz
2-lu
Φl
dι
l,"^
Ξ
/mΔ"
ofY
that
= SR.
τl,=10u'sin2πΧ20X1σ
0.9
d""| : lo'ιl.Χ2τX2oΧ lor
dι l...
1/'lo 6
_
Thυs ,,. =
' 10 X 2π'Χ 20 X l0r
0.5
0.1
c) v, _
W=2μs
ο.2w:
ι,:ιF=0.2uS
rad,is
b) The output Ψill distort at the value
resutts in
,, + r,. =
SR
50mV
0.795
υ. = 500mV _ 0.5V
sleιry rate beΦns aι ιhe frequency for which
ο.4 μs
ο.5
:
sR
v
ωx
chapteι 2-40
_ !/l0 6 _
ιvhich siνes ω.
/:
ο5
2
X
106
3l8.3 kHz. Ηoιγeνer the small signal
radlsor
3
dB
frequency is
f''- _
f'
R,
I + --l
_2X106
!0
2ΦkHz
RI
Thus the useful frφuency range is limited aι
2ω
kHz.
d)
for/:
5
at the νalue
ω X
t0ι
ktlz, the sleψ Rate limiιaιion occurs
of η given by
=
l/|o 6
SR+υ' _
2τ / 5 \ !0ι\
: 3,r8 V
l0
such an inpυt voltage' however would ideally
resulι in an outpuι of 3l.8v \νhich exceeds ι4'.",.
ThUs
y.'-' :
ν
j-9_EΙ
lο
:
I
v
Deak-
chapteΙ 3_
3.1 Using the expression in (3.2) using
B:7.3x
lorsCm rK 3':
* = 8.62X
l0
seV/K:
T : -70"C :
203 K:
Ε": 1.l2Υ
'we
haνe:
n, =
'Ν
x ιο
5.33
ιι, : ρr:
'8
Thaι is, οne oυt of every 5.33 x 10|8 silicon
atoms is ionized at this temp€ratuΙe.
T : 0'C: 213 Κ:
n',N_ l.52 X loecm ι: 1 =
T = 2o'C :
Τ = l00'C :
1:
_
'N l.43
7 : 125'C:
'Ν-
n,
x
412
1.72x
lOrr
2.87x
lo
t0r2
:
:
2.2
x
1
=
x l0
9.45
Using
'ι ι"ι(l ' a"l. ln 5".*)
loΙoι30o.1!']e
!!t :2.25 x
lo2 cm-r.
:
s,
ni:BT3neΕ8ΙΚτ
Χ lot5(3ω)32
l.5
x
"
ι'ιzιιzxε'οzxιo
\rιω\
l0lο holeycm3
beloιγ intrinsic level by a factor of l 07
Hole concentration in Ρ doped s, is
":
p_
1.5
x-loro
lο'
Ρhosphorus doped
:
1.5
Si
x lorcm
so
nn = Νo : Pnno: nι2
nh :
:,
n' ιp'
( I._5
x
I5
ιo-lcm'
μ, =
1350 cm2lvs and
p, : 480 cm2lVs, we have:
ρ : 2.28 Χ 1o5 Ω_cm.
!oΙ0l2
x !O'
r
Α:
R:
7.59
(b) ,ι,
-
3
x
1O-8 cm2' ιγe haνe
Χ ιο9 Ω.
ΝD
2
cm and
:
1016
cm
3
:
p" =
'ιi
Using
μ, =
1110 cm2Λy's aπd
μ. :
l l lO cm2lVs and
= 2.25x loacm
1
,.i
Ιn phosphorus doped sr', hole concentration drops
.'.
\
"AR = o.4 wiιhl=0-0l
3.
7.3
2.2]
Usinρ
3.4 Ηole conceηtraιion in intrinsic
:
:
n,2
Ν,
5xloε;
P:n_ni=1.5xtQlοcm3
,,
Usiπg (3.3) we have
=
7.3
rl
3.3 since 1γr>>ni, we can write
n
x 1015 x 13931]r" ιlliι:,ι.ιlxlο
432 x l}tzlcm3
:
:
P^:2.ΔXιoglcm3
cm ':
lOΙ8cm
lOa/cm3
3.6 (a) The resisιivity of silicon is given by (3.t7)
For inιΙinsic silicon'
106 carriers/cml
Po-Νι =
lo16
At 398 K' ho|e coηceηtration
e'E8l(.2Κτ'
x
3.56
(t.5x loro)i
= 2.25x roaxcm.
"
Substiπting the νalυes given in the problem
a,
19I67"a]
: p": 2,25 x
T : |25"c :273 + |25 :398Κ
At398K, ni : BT3ne ΕEι2Κτ)
P.=
398 K:
3.2 Use equation 3.2 to find
ιι' : BT3n
|o
373 K:
X l0l2cm ιl 1 _
n.
x l0
].05
o-=ll
ι'n_
ΝD -
si
Hole concentration
293 Κ:
8.60X lOocm r:
a.'Ν
No: ι" : 1.5 x 1Ol7 P atoms/cm3
3.5T : 21"c : 273 +27 : 3ΦΚ
At 3Φ Κ, π, : 1.5 x lgiο/cmr
Phosphofoυs dop€d
x l0'cm ll t =
2.67
1
: 4ω cm2lv's, ιγe haνe:
ρ:0.56ο-cm;R: 18.8kΩ.
(c) a, - lv, : lο'8cm 3;
o'2 : 2.25 Χ lo2 cm'3
po :
μρ
Using
ψ : 400 cmz/Vs, we have:
rο-cm;R = 188Ω.
ρ:5.63 Xlo
Αs expecιed, since ND is increased by lω, the
resisιiνity decreases by ιhe same facιoι
chapιer 3_2
pρ-No:
(d)
:
ρ=
2.25
X
1.56
Ω-cm;
since ρ
(e)
lOa
:
52.1
be 2.8
kΩ
X ι0'6
Ω-cm, we
R = 9.33Χ lo-'?f}.
5V
10 μm
5V
μm l0. l0
l0
l
Need 'Ι6,in =
R:
iS giνen to
3.7 Electric Field
:
E = lv
2
cm-l
direcιly calculate
-.
1;n, :lι
1016cm
'9r'o '
l0'cm"
=Ξ Ν
3.Ι
5V
lOx l0 6m
D:
_
:qNDμoΕ
m^]μfi1
l.6 Σ
4'63
X
ν/cm
acm = l01
|o
|o
ΝD
Y
l.15o
/ l0ι
:
2.25
x
1o|7/cm3
0
o-,
-
n,t (1.5x lor")2
D
tο|6
N
l.a/cm,
From Figure p3.10
10x lO acm
'|o"
Ρ,.
Φ : _10"p-_ p','-_
_
d-τ
lV
6.lxloo
a
:
since o.1 μm
0.l x l0 cm
Φ _ loxX2.25Xlo1
dr
ο.1 X104
: 2.25 x l0t1
5000 V/cm
Ηence
+5
v2 a.ir, = μ,,Ε
:
V-
480 X 500ο
= 2.4 Χ 1Φ cm/s
υn ι,ih : [ιnΕ: 1350 x 5οω
:6.75 X lο6 cm/s
_ 6.75 X 1ο6 :
υP 2.4 x 106
ι': 2.8125 υn
τ,,
oι
2.8125 or
aIterηaιiνeιy,
υ':!"Ε:μ,_1350
uo ΙlnΕ L\ρ 480
2.8Ι25
3.E Cross section area of Si ba(
:5Χ4:20μm2
since
:
20
lμm: l0
x
acm,
we get
l0'8 cm2
:20x10 8x1.6x10re
lv ,
(t0|ξ x 5ω | lot x 12ω) x
l0 \ l0 '
: 16 μA
Here n :
:
q (nμi + ρμP)
x (_2.25 Χ
D" : η
lL.
ll' ρ
DP: μPvT
3.11ηuaιion 3.2l
Dn = μnvTaΙ\d
E
N, and since it is η - Si, one can assume
ρ << n and ignοre the ιerm ρμl,
Αlso
1o|1)
: ν,
D,l
Do
Ιntrinsic
l
35ο
48ο
35
t2.4
1Ο'"
l 10ο
400
28.5
10.4
l0''
7οο
260
18.1
6.7
l0'-
360
15ο
9.3
3_9
ζ :
l0'" cm' and
3.12 Using (3.22) 1v" :
rli
1.5 Χ 1ρl0 grn ] ψ9 hnνg ζ, = 695 mY
:
Using (3.26) and €s
currenι Ι = Aq (ρμ,, + rμ,1)ε
3.9 Jdirι
12
Doping
μ,
lιP
Concentration
cm,/V.s cm,/V.s
Caffiers/cm.
it can be shown as
:
Φ
Jι' : _οD
' μdx
: _ 1.6 X ιo '9 x
: o.432 Νcm2
w:
42A
:
x
l1.7 Χ 8.85
x
10
'' F/cm,
|σ' cm = 0.424 μm. The
exιension ofthe depιetion width into thel, andΡ
regions is giνen in (3.27) and (3.28) respectiνely:
Ψe have
t' =w No =o2r2um
Nr+ND
No
:
' /' w'λr + Ν,, = o2|2ιm
Since both regions are doped equally, the depletion region is symmeric.
Chapter 3-3
:
Using (3.29) andΑ
106 cm1the charge
ιude on each side ofthejunction is:
:
Qr
3'39
x
magnι
x= 0.1 μm
,{: l0μm X 10μm:
K
at 3α)"
:
Using equation 3.22, built in voltage yo
v" = v'h(rya!4!:
,n[ lo'o
l( l.s
25.8
/
\ l0
Ιι
l
u"
:
Ηeιe
W:
Depletion \rith
o_f;!.",
ι.α*
equation 3.26
ιο'1 l'
{ l.oγ lo '" \106
X 10-'cm = O.95l μm
Use ηuation 3.27 and 3.28 ιo find
l
1rn.o',.
loι5,'
η
and
x/
ΝΛ _ ο_95ι X 10'u
X:w
''
Νι| Νo
lο|6+ Ιο|5
: 0.8642 μm
Νo = 0.95l y toιs
Χ':w Nι+Νo
''
!0|6 lιo|5
use
0.8642 μm
ηuation 3.29 ιo calculate chaΙge storω on
eiιher side
0,
:
__
on(#:#")
4oO μm2
Χ ι0 'cm
1.6 Χ ιo-ι9 X 1016 x (o.3 X 10
:
= 4ω x ιο.8.
4ω
x
Io_8cm2
)
,.u,
'" ^ ,o
'" ,"[ι;τlο'".lo''
, ,d.]
= 0.951 X lO-a
Ηence'
oj:5.53Χlol4c
0.3
2XΙ_MX10
V
: A ι (ψ"
*
11
N,^
o)
w
-
4)2
Ι2
ιq1ι'ι o1 w
3.t6vo: v, lnf&&)
1
ιli.
Ιf lVΑ or lVD is iηcτeased by a factor of 1 0, then
neιγ value of yo wiιι be
Ν1N")
v, : v, ln(l0
tr,')
The change in ιhe value of Vρ is
:
ln( ι0)
ln(N,Υ,'l
t n,t
'
3.17 Using (3.22)
with e=
Νr,: 1016cm_3,aπdni=
Vo
:
635
mY
Using (3.31) and
W wherejuncιionarea
:
0.3 μm
Since iy'1 >>
<-
=
0.69
Q,
xn
ΧP
= 0.951
:
wΞ
:
/z,,
1o_8 Χ o.l
since ΝΑ >> l{D
ο'o)'.Ι
SoVρ:
,=
lV
3.15 Equation 3.26
V
= 0.633
cm
fC
=16
x lo't I
,
a
Χ ιο acm
Xl0Xl0'cm
25.8 mV
1 n,.
10
:1ΦXl0ocm2
Sq or: 1.6 x 1o_l9 x lω x
xlo . x 1016
3.13 From table 3.1
yr
aΑxΝ
: 0.l X l0
Here
c.
'ισ\1
:
Q,
3.14 charge stored
1.5
10Ι6cm-3,
x
10t0, we have
yi: 5 Y ιγe haνe
: 2.83 μm.
: 4 X 1σ6 crn1, Ψe haνe
W _ 2.83 Χ 10'cm
Using (3.32) with Α
Qr: 4'l2 Χ |σ" c'
3.18
ηυation 3.3ι
:
]?(;;+fi)ιv,+
v,l
:lTG;ilFπh
Chapter 3-4
3.2| n,
| + a.lv^
l\(
s ιN, N"l "
.,' tz.,υ.
y |o''x (lω)''' z" "'(
: 1.4939 x l0lo/cm2
,,'1uι 30ο K1 : 2.232 x 1o2ο
E4uation 3.32
f
o
(!u!n,
*
)
"
.
*. t u,
_^FΛm;τq
_
'n
^
η':7.3 ι tοl)x
"
2u r"u'u'
L'Ν
Ι'' =
Aοn2
Ιu
%.ι'
I )
,',
o
ιν^
:
o''i;(+η-
x lo-o x
l8
+
1o''
10
:1.44X lO 6A
= 1 .44
='79
x 1o
μλ
16
Χ
-Ι''= tron''!-r"u'u,
' 'L,No
ll
_11
x lo
't
x
lo'o
_ lon. lο_8.,n2' ,.u,,0
- oon''!ι_
:3.6 X
1ο
1olο)2
lo
15
10X10 4Χ1ο16
Α
Ι: Ι'(e '_l):o.5xlο'
,.1 vil)r9.
3.οxi0 |'|e ι
ι/
:ft;)
3.23
=
V, =
0.6645
12
Ιο 'l/.
\
l|_o.sxlo'
v
Y
Raιe power dissipaιion of diode = 0.25 W
Ιfcontinuous current '7' raises ιhe poΨer dissipa-
lο"'1'
tion to halfthe raιed νalue. ιhen
12νxΙ
Ι=
.2=
1xo.zsw
Ιo.42 mA
For raιed value of0.25 w. the currenι will
e1lx|
l
25'9
'o
ν1ν_
Ξy
2Φμm2:2Φxl0 3 cm2
2ωX10 3ΧΧ 1.6X l0 ''x11.5 x
5
'..Ι
X (l.5 X
Νow / : /ρ+ Ι. : 1Φ Ι.+/,= l mΑ
LmΑ: O.Φ99 mA
1_:
" lοl
Ιo : |'Ι" = 0.990l mA
l0
vlν'ι
' l,
: Aι'nL-Ξ-L!ρ
'L"No
1'
Ιu:ιo
3.41''
soι
l)
For ιhis case using equaιion 3.41
Ι" = D' Lι'ND
l0.- lο.- l0I8
20 5 l ο'u
3.20 Equation
oΙ
"
]" :1aη'z_Ρι1"ν'uτ
I)
''L.No
Fot p* n junction y'vΑ >> ND
''
: lon'
{nu'ut, , !,u
LrNo.
1,
lO20
&)1"'"
L"N
ι _ ιqn,'(!+,
' ' \ι,,ΝD
ξ;)"''"'
-ι lr'Jr'rnz' ,"
3ω Κ)
3.22 ηuation 3.39'
= Qtn
/e
l0ro
K) : 4.63l x
,,'(ut 305 K) :
.n
2-ls2
,r,2(at
Ι : Aqni2(ξ;*
(3o5)'''
,ι,'?(at 3o5
3.19 ηuation 3.39
Here
x
=2.152
*!ι
vο
'.,0(#!u"")u,
ln .. .")
Αt 305 K.
^β*1141γ.μψ1
=Qn l
ΒT3Ι2e-EolΙ2Κτ'
n, = 7'3
η]
ο, _
=
Αι 30ο K,
:
x
0'25
W
1
20.8 mΑ
:
Ι
|2
ω
''
""J
chaρter 3_5
Ιf in 20 ms, breakdown occers only for 1 0 ms'
then during l0 s the cuπeπt can be
,
?9
10
substifute for α from equation 3.43
o
,,'l
zο.ε
E*.-!o!o
"'ΝΑ + /νD
_ \Ι
z^]n. +
:41.6 mA
3.24 ηuaιion 3.48,
η
X J€s
Γs'
',':oFffiψ
:
v,
=
ιr, ln(ΝoΥ'')
t n,'
X t0 .
2.5.9
,n[ tο'" x lo'' )
,
[1t.s
x
:
Io'oy'/
:0.635 V
: 4ΦX
C;.
\/10Ι"
.rι
f
F
q :
:
C'
'
Γ
14.65
fF
:
-y"
X loΙ5\
I
---:-
ο.635
ro
pr =
r , ro '
[----':-),
Χ lο -.'
\
25_9
τΙ: 10X 1o-''
: 259 pS
x25'9
ForΙ=0.lmΑ
c,: ΙjΞlxl
ForYi=lV ar:
=
c': =Ξ':'
(,-ξΓ
OU'O,
('.#)'
0.45 pF
Forη=loYc;: -qqξ
('.#)'
:0.25 pF
3.26 ηυation 3.43
o Γ*^-!'! "
Λ/ "'NΑ + λD
Equaιion 3.45
'
c.: lΙ)l
l0t" + l0I110.635
43.7 f F
|+
3.25 Equaιion 3.49'
.:
w
3.27 Eqυation 3.57,
10'8
1.o4X1o_t2Xl.6X1ο
:43.7
€sΑ
2Jvo+
(259
\
25.9
:1pF
Χ ιo '.'z') X
x to '''
X ιo
r
3.28 Εquaιion 3.51,
-'ρ LΙ
(lo x ιo n)'
--πιcm
6'ι,_
note Iμm:10
: lω ns
Qp :
τpΙ ρ
=lωX1o'xo.2xιo'
l2c
= 20Χ 10
c'
=
1
[!!|ι
lφ x lο-' 1 r
r lο '/
\zs.q
vR
ο.1
772 pP
o.2
x I0
,
chapιer 3_6
3.29
a,
Pn.np
ln-regron
Lλ
fut--
no({
nο {x)
',^
2',"
{_ _ _
ll
-1πp"
I
b. The current / = 1, +
ι,enειh of p and n
:
λ^
AqD,,fi
_ n;(e '- l)
u
= tron,,
,--!4 X"ΙΝ
_ Ι) x
' ' \w"
q""t'
t)
_ l11w
Χ.11eu',
'
_
X'1
_l1
,
D
-i
x1"v/ν-'
l"Aqn!
'lWo
"
Χ,,']Ν ι
= Aοn''|
''n''' ------------ι:-
^dodΙ
" dv 'dv
But 1 : Ιr ("u'. _'|)
dΙ
Similarly
(e
+ 1eν/v'
o..
x')
"
Φ
'ι μ = AoDvdι
'
|ιw;
oo
ρ'.lψ''
''l
(w"_ x^'2 ι
_'
'
- 1
'"
η
.!4.,ror lry->> Χ^
2 D"
w,- x,
x
L
L,
("ν/u _ 1Ι
'Γ Γ)
lιοtι'."u"' -
: ont*,r.
dι - P"(.X") ρ""_P..e '_P""
W"-X"
W"-X"
dx
π"ιw-
fegion
:
2
Ι\n
'*
I
Qo: Aqxξιn'ιx'l_ n'.lι',
12
p'1x'1 : ρ."ev/u' τ'd l,,'', : )
Io: AJo:
Wn
l
/,
Fiηd curτent component
Xn
f T:Ι:;"--l
l
o^,*,
'":ψλ^
Xp o
-Wp
p"(x")
l)
D-I
+ ---'''''''''''!!
lη' lw',
xρ'ΝA]
- l\
The excess change, Qr,, can be obtained by mulιiplying ιhe aΙea of the shedω triangle ofthe
p"(r) distibution graph by Aq.
dvη
_ντ
-Ι
Ιse
'
soCo=1t'L
vτ
d.-," = !w: tr to'' _ 8., lo,/F
2 t0
solνe for
W"
:
η
63'25
25_9a
μτι
1ρ'
chaριer 4_l
4A
4.1
(a)
v
t-5
The diode can be reverse-blased and thus no curιent would ffow. or forward_blased ιvhere currenι
would flow.
(a) Reνers€ biased /
oΑ V, 1.5
:
:
(b) Foward biased /
:
γ
yD : 0v
1.5Α
ve+=|0ν vp_:0ν
/:
1
kH3
(b)
4,2
ov
(a) Diode is condυcιing and thus has a
across it.
consφu€nιly
1ο
(b) Diode is cυt
V: 5V
: 0.l
kΩ
drop
mΑ
ofι
/=OA
(c) Diode is conducting
1= 5_(_5) :0.t
10
kΩ
yp+
/=
mΑ
=0v
1
yρ-
=
_1ov
kH:
(c)
(d) Diode is cuι off.
V=_5V /=oΑ
43
(a)
o,
y'"u'oo
+1V
vo=0V
Neither Dt nor D2 conducts so the.e is πo oυtput.
+2ν
(d)
Ι
:
2
_(_5)
2kΩ
-5V
= 3.5 mΑ
Vρ,=
10v
yp_
=
0v
/=1kH3
Both Dt and D2 conduct when Y, > 0
+,='-l*"
:2τοA
D,\
' 'cιltοlT
(e)
chapι€r ,ι-2
Vr+ =
10V yρ =_10v /= l kH3
Dr conducts when
< 0.
τ,/> 0 and D2 conducts
ο)
when
1,/
Thυs ιhe oυιput follows ιhe input.
(η
Vo+=
10V Yr=-5V /=
when ,/
Vo+=
conducιing
loV vP-=oν f= 1 kΗr
_Dl is cutoff when
ι/Ι <
when
0
.
0,
va <
I kH1
outpuι follo\λ,s the inpuι as DI is
> 0, the
Dt is cut offand ιhe circuit b€comes
divideι
a volιage
(k)
G)
-10v
yρ+=οv yρ_=_lov /=1kΗ1
D| shorts ιo gτound \ιhen ,/
when
U, <
> 0 and is cυt
οff
0 \r,/hereby ιhe output follows τrr
(h)
Vr. = lV V,-=-9V /= I kH3
when ι/ > o, Dl is cutoff and D2 is
conducting.The output becomes 1 V
\Ιy'hen ,/< 0' DI is conducting and D2 is cuιoff.
The output becomes:-
Uo =
as
0v - The
οutput is aΙways shorted to grcund
Dι conducts.rhen z/>
0 and D2 conducιs when
?lζ0.
4.5
Forζ<3V.D1
is on and D2 is off
For η>
isoff
3
(η
ιDl
V'(ι)
Vr+=loV yρ-=_5v /=
i"( 0i
ri
1kH3
When υ1> 0, D1 is cutoff and vρ follorvs
z1
When τ1< 0, D1 is conducting and the circuit
60
mA
0
becomes a voltage divider \γhere the negative
φak is
IkΩ
l kΩ+ lkΩ
. tοv:
5v
l'6,n"o1,
= Φ mΑ
and
η
is
J ir=0
on +ιε
=
60 mA.
Chapter 4-3
'"o'",
=
For μak
ζ
mΑ
Ψ.60
:
24.2
mA
(b)
+.1
reduced by 107o
v
ir,n""r, = 60 mA
:
'"o""'
mΑ
Ψ.60
:
23.5
o.+
15
i\.
4.6
1:Omn,f,
Ι=AB
ο
0
0
0
0
I
0
1
I
0
0
I
I
1
1
I
y=A+B
-τ
and,
,
and y aΙe opposite
A=B
3-(-3):
kΩ
1ο
mA
aΓe ιhe same
ml
o.+
.r
Dr
D2
Cutoff
Conducting
_3+ ο.4 (5)
-3V
4.10
(a)
fo.
(l0
if A * B
||
20)
kΩ
4i
The case for the highest current in a single diode
is when only one input is high:
V,=5v
ν
J<ο2mAΞRΣ25ιΩ
R
(ι0||
4.8
The maximum inpυt current occurs when one
input is low and the other two aΙe high.
v
Σ_j s
(b)
R
o.t mn
=
(
,,2o
1011 20) + 20
:
O.15 mA
x4=3v
',y'= i5
R>50kΩ
ι/
V+
l0||ιo 5kΩ cutoff
4.9
(a)
5kΩ
20) + 20
*
ξ:o.u.o
1:0.3m4
_ ''= _ο's v
Ι=0A
l0 x 6
10+ 10
Ι
10|lι0
5
kΩ
5(l0)
10+ t0
:2.5Υ
Ι
4,11
n=Ψ-&'=..4γn
Conducting
ψff=o.:,"l
-3V
50
The laΙgest reνerse voltage appearing across the
diode is φual to ιhe peak input voltage
l2o^Γ2
:
169.'Ι
ν
chapιer
μ
4.12
4.14
Vι(ι)
,,:|,,
'2'
3ν
-l v
voG)
2ν
D staΙts to conduct when
ι,a >
0
v;k"8):
.oφ"rt1
i6,,ur,
:
:
ν
l.5
30
15
mΑ
mΑ
4.13
For ξ > 0V: D is on, Vρ = |'' io=Vi / R
Forη<οv: Disoff' vo=o' iD=0
viG)
2ν
2ν
voΘ
aν
:
Ιo
""oo':
1r,o,*,
Ιeverse
diωe voltage: Vρ
Vr
: V9:
ι^^"l:
2
2o
10
V1
Υ
mλ
mA
condυcιion occurs
z1
For
:
Αsinθ = ι2-conducιion across D occurs
a condυcιion aηgle Qτ _ 2Θ) that iS 20% ofa
cycιe
ιτ_2Θ _
2τ5
θ:0.3zr
|
chapter
Α:
12 /
:
sinθ
ι5
v
14.83
;. Ρeak_ιo-peak siηe waνe voιtage
xt"cl
Vr[mV]
-40
20
0
23.5
40
21
ι50
36.5
=2^=29.67ν
Given the aνerage diode currenι to be
l IAsinΦ _
2nJ
R
12
d6 = !Φ mA
l Γ_ |4.83cos6 _
2τL
R
R:3.75Ω
|2Φ1Φ o'1τ
].=o.."
Peak diode current =
r:
-
12 : o.75 A
R
A + \2 : 26.83Υ
A
:
Ρeak rcνeΙse voιtage
for yΖ = 25 mv
16.8"C
4.18
10ω
:
15
Ιs
For resistοrs specified to only one sigηigicant
.'.
digiι and peak_to_peak voltage to the nearest voΙt
l5 so lhe peaΙ{o-peaΙ sine
then choose A
ι = 0.173 V
atυ : o.7Υ
i: Ιr"o.,'o'.': l.45x
:
wave
R=
3 Ω
Asinθ =
15sinθ : l2
θ = 0.93 rad
7_θ
voltage: 30V and
conduction staτts aι U, =
coηduction stops at
.'.
12
'=
ιo_'
l>: ..''-Iι : e sΞ
:.r
10-'
:
iz 0.335 μA
=2O%
Αverage diode current =
+Γ-u-*g]|=,,,,:
Peak diode currenι
"u,"o
_ 15_12 _ ιΑ
3
Peak ιeνeιse voltage :
Given ID and
yD
find /, Usiηg:
Ι,
!
in
a
_-
s=
forψard_biased diode, ιγe can
Ιo'
e " '
3.46
Vo:0''71 vΞ/D:
RED GREEN
oΝ oFF
OFF OFF
OFF ON
v,
:
25
A , which ιγe
-
Dι condυcts
No cυπent flows
- D2
conducts
ι-€t
a decrease
whereι:1.38X1o-'3J/K
T = 273+x'C
by a factor of 10 in /D result in a
of vD by
decreas€
b|o
:Ψ
q
ο.746 ιrιη
Vr: 0.80V=1ρ :27.32ΙnA
Vo: O.69 V+1, : 6.335ιntr
yD = ο.ωvΞID :9.l1 μA
Ιo:
l9c
x 10
16
using
can use to calcυlate the following diode cυπents:
4,16
4 = l.ωΧ lo
4.?,O
mv, we have
A+12:27Υ
V
3ν
O
-3V
qi
υ.o.25
2ττ
v,
o1Ιν-
il=l,e
10l2/s
Fracιion ofcycle ιhaι cuπenι flows iS
n_2Θxlω:2o'5
4.11
4.19
"'"0025
Ι
,eu
Δv
:
o'n'
= Ιr''uo
^v)/vτ
:
vDl
ιse
vτ Δv/vτ
Taking the Ιatio of the aboνe two φuations' we
have:
lo: e
'=ΔV:57.56mV
Thus the result is a decrease in the diode νoltage
by 57.56
mV
chapter,ι-6
4.21
,s can be found by using
I
s
:
JD'
e
u9
ι
4.23
The νoιιage across three diodes in series is 2.4 v:
thus the volιage across each diωe must b€ ο.8
Vτ
v
Let a dec.ease by a facιor of l0 in /D resυlι in a
decrease of V2 by
. lνι'
ιn
-:ι-":ι."
lο "
ΔV
Using
:
. vo'ντ
6vl lτ
ιν vτ
0.65
424
vD:
0.65
Vο=o.64Υ
Υ'Ιo:
of Ιο Biνes
10Ψο
__
7
mΑgrs:
Vo=o'59 Υ
v'/D:
l0% oflo gives Vo
vD
o3o V,
lo μΑ = /,
0.59
V
=
/, =
l0% oΙ Ι ρ giνes Vp=
16
7.69
:
x
76.9
mΑ+/s :
lo_7
l0_I5Α;
X l0
Ι8Α;
using Ι s : ]D'e
Let an increas€ by a factor of 10 in /D resuιι in an
)
increase of yD by
lo= Ιse"
Δv
'
:
1r"vo'vτ
'"ιv/vτ
l0 mA=9rs
:
lo.7 Χ lo_l5A:
/DX10givesvD:0.76ν
vD = o3oV, 1r: l mΑ+/5: l-07X
X l0 giνes Vo
a
mY
decrease in the diode voltage
4,25
we can write the following
diοde currents:
Ιρ2: l1_ Ι2:
Ι91
: Ι2:2
8
KcL ηυations
for the
mΑ
rnA
Vρ2_ Vρ1
lf ,2 has saιuration cuπent Is. then Dι, which is
l0 times laξe, has saturation cυrrent 10/s. Thus
-
|
οι'Vτ
toι = lol<e "' '
Taking the raιio of ιhe tψo equations aboνe' we
haνe:
lo_'5Α;
: 0''76 Υ
Vr: 0.8OV, Ie: 10A+Is: 2lοΧ l0 15Α;
/D X 10 giνes Vo : 0.86 v
Vo: 010 V,/, = 1ρρAΞ/s : 1ο.7X lo-|8Α;
/D X ι0 gives Vo : o.'Ι6 Υ
ID
17.6
-
ω,nv
Thus th€ resυlt is an increase in ιhe diode νoltage
by 60 mV
o.1o V' /, :
'ΞΔy = l7_6mν
2:e
we can write:
have:
vD :
have:
V=
Ι3e(vD+Δv)/vτ
"o'"'=Δy:
2'
-Δv vτ
we can write the folιowing KvLequations for the
diode voltaφs:
:
Taking ιhe ratio ofthe above two φuations, ιγe
lο:
VD'Yτ
Taking the raιio of the above tιra φuations' Ψe
X l0 l5A: by
νDlvτ
/s can be found by
lνD 5|l vτ
ι^
:=ι"ρ:ι"P
Thus the resυlt is
0.ΦΥ
4.22
loΙo:
ιhe required currenι is
V,/, = 1Α+/r:1.ρ7X lοl2A:
|UΙο of Ιo giνes
yD:
.
Connecιing an identical diode in paΙallel φould
reduce ιhe current in each diode by a factor of 2.
wriιing expressions for the cuπenιs' \νe haνe:
Thus the result is a decrease in the diode voliage
o1o
'
is
1ο:ro''"=Δv:6ο.v
vD:
"
1D will be
yD=
6.9 mΑ. ινhich rγouId giνe
ο.794v giving
an οUιpuι volιage of 2.]9 v Thus ιhe change in
outpuι νoΙtage
l0.Ι5 mV
have:
ιnV
/.e
found to be 7.9 mΑ.
lf l mΑ is dra\νn away from the circuit'
Taking the ratio ofthe aboνe two equations, Ψe
by 60
Ι, =
l |Vnι |ol'!ι
Ι^'
:=4=-P .
Ιo,
lo
:92.2mν
+v
To instead achieve y
ΙD2
Ιo,
_ ΙI_ 12:
Ι2
:
50
mι
l
v'vτ
l0
we neω:
-|onn,ronr,
= 6.739
'|o
Solνiηg the above equation wiιh r/ still at 10 mA,
we find
12
:
5.75 mΑ.
Chapter 4-'7
4.26
we can write the folloψing
diode currents:
Ιoz =
KcL
520 rnv
520 kf)
equations for the
Aι ω'c,
|0mA_V/R
1
4μA
=
Ιo' = V/R
vy'e can \Ι/Ιite the
folloιηing
diode volιages:
v : vo,
KvL ηuations
for ιhe
vo,
we caΙΙ write the folloι'iηg diode equations:
.
- Voιl|τ
.
. Voι'Vτ
480
Taking the ratio of the two φuations aboνe,we
Vz =
have:
Ι'oz
Io,
_ |oπ|,
'
--V/
V /R
To achieνe y = 80
/D2
Ιo,
R=
mY
"'vD2-νDι|/vτ
_
νr
"ν
we ne€d'
_ lοmΑ_o.o8/R _
o.o8/f
-o.oq/0.o254 _
23.1
solνiηg the above equaιion 're haνe
R: t94 Ω.
427
For a diode condυcting a constaΙt cuπent, the
diωe γoltage decreases by appΙoximately 2 mv
μr increase of l ' C.
T : _2o"c corresponds to a temperanΙ.e
decrease of Φ' C ,ιγhich results in an increase
the diode voltage by 80
T:
+1o"
C
mV Thus yD
mV
50' c , ιγhich resυlts in a decrease of
diode volιage by I Φ mV Thus Vo = 590 mV.
+Ιov
2
kdι
+
Dr
lol4
4.29
FoΙ a diode conducting a constant curΙ€nt, the
diode νolιage decreases by approximaιeIy 2 mv
of 1" C.
Α decrease in yD by 100 mv corresponds to a
junction ιempera|ure increase of 5οo
The power dissiparion is given by:
c.
:
(ts A)(0.6 v) = 9 w.
The thermal ιesistance is given by:
5ο"
= 116" (: / \ν
9 w
PD
c
430
,
'[,
Rr
25
: 514.6 mV
Vρ7:4μA X520kf}:2.08v
et ο"c.l : l ,'l
4'
Vz: 5Φ-2.3xlx25|oΕ4
: 525.4 mV
v",
^4: !x 520 : o.ll v
^T
PΙ'
4.8
xIx
mV
per increase
of
corresponds to a t€mperature
incΙease of
the
= 770
48O + 2.3
520
Given two different voltag€y'cυπent
foτ a diode- we can write:
:
Ιo,
Ι reno''
Ιo, = Ι
reu
measu.ements
''
o"'u'
Taking ihe ratio of the aι'ove two equations, we
have:
D2
i:
_
νo)/v'
_
= V ol V oz
""
0.5 mΑ, we have:
:
Foτ /p =
Αt
20"c
ΔV
:
Vnt:Vz=520mV
Rl : 520 kΩ
: Vτ,,(oΞ#a):
ΞyD=552mv
For ID = 1.5 mA, we haνe:
_248mV
r,
^(E)
chapιer 4-8
\y _
y"
l.5 X
|n(
ι
lo
_ _z-zο.ν
Α.)
' \ lοΑ
yD
: 580 mv
Ξ
)
ΑssumingνD changes by _2 mν per 1'
For ID
ForlD
:
:
c
:t25"
ιγe have, for
iη t€mperanrre,
increase
chaflges:
0.5 mA, 5ο2 mv < v D
1.5
c
Ξ
602 π|ν
mA, 530 mV = YD < 630 mV
Thus the oveιaIl range of yD is between 502 mv
and 630
:
τ,
mY
:
,:
τl
o.6v.
ι-2!
νoι\'vτ
- ,' "'uo'
For ιhe firsι diode, witb
o : '7σ) mv,
1, : 0.5 mΑ:
V
:
ΔV
Vτ
:
1D
'-u^, v^
:
,D
0.65 V,
0.7
voo- vo
,
ιD
R
',(3j)
:
,10
mv
Ξ
yD = 74o mv
67.7
mV+ v^ _'761'1 mν
0.6
l.ο
0.8
,:
0.65
v
to
Fori:0.3mΑ,,:u,'(i)
mV = Vo = 5l0 mV
luz.u mν
Ξ
vD
:
Fori=0.4mA,
/1\
0'025 Χ Ιn|
\
: 0.661V
r: ο.668 v
---::
lο ''l
I
Now ψe can refine ιhe diagτam to haνe a better
.4 mΥ
0.5 mA and 1D
= l.5 mA,
the
estimate
ιhe two diωe !olιages is
fixed diode current, ιhe diode
νolιage changes with temp€lature at a constant
mv since'
.aιe (_2 mv
for
(v)
From this sketch one can see thaι the operating
difference b€ιween
23ο
υ
wehaνe:
ΔV : Vτ,"(ψ,ψ) = _ l90
/, : l'5 mA:
ln(0'Ψl5) =
ιv _ v,'\tl
0.4
point must lie beιwe€η
0.5 mA:
:
ο.8
ο.6
0.4
o.2
0
Ιo = Ι A andvD = 7(n mv,
Fοr ωth ID
i(mA)
0.1 mA and
l.5 mΑ:
537
mA
Make a skeιch showing these points and load line
and deιermine ιhe operaιing point. The poiπts for
the load line are obtained using equation 4.7
u,t,(P)
_
For the second diode, rvith
=
0.2 mΑ
1.0
\0.1./
:
i:
V, i:1.45
we haνe:
Δv = v'' rn(!Ξ) _
/D
?
= lO t2 eo6t0n2s
: 0.03 mΑ
4J1
Giνeη t\i,io different voltage/currenι measurements
for a diode. we haνe:
i : ι.
a
per'c
temp. inc.ease). ιhis voltage
difference ψill be independent of temperaιure
!
4i2
R=lkΩ
voo
1V
0.660
/s= l0 lsA: l0
calculaιe some poinιs
12
mA
ο.664
0.67
From this graph we get the operatiηg point
i:
0.338
m'
ι
=
0.6635
V
chapter
Nou, one compare graphical resυlts with the exponential model.
Αti:0.338mA
,,'(i)=
"=
:
0.6637
The differeηce between the exponential modeι
:
=
-07 :1.5mA
1^:1
" ο.2 k{)
b)
Diωe
1
0.6637 _ 0.6635
0.0002
has 0.7
ν
drop at
v
0.2 mV
v:0.7V
i: 1-0.7
=
o'2 Κ
2υ=
o'7
1.5
R=lkΩ
3 υ:0.'7
+
*z.:rο.οzsl"ε(f)
+
23
0.2 kΩ
/s:l0"A:l0''mA
4
Use Ιteraιive AnaΙysis procedure
v": v,h(?):
=
0.6607
o.orsh(-g+)
V
:
o.oε:ε
v
Ι0
'loiff"
'"
=
434
2οο
lv
v
i_l-0.7ο93:|4517mΑ
=
0.7093
ο-2 kΩ
sιop
aS we are getting the same resυIt.
Ω
by:Ιs: ιDe " ' wiιh/r= lmAand
Vo = O.7
V This gives /s
:
6.914X
lO_16
v
Α.
acfoss the 4 series-condiωe droρ musι be 0.75
V
Αpplying this νolιage to the diode gives cυrrent
0.6635
v
0.3365 mΑ
sιop here as we are geιting almost same value of
Ιpaτ,d Va
+
23 X o.o2sloc(l 4i37)
Ιn order ιo have 3.ο
nected diodes, each
t-ffut':0.3362m4
4 vD _ 0'o25kl(9!Ξe] =
\
/
=
0''1+
we first find the νalue of Ιs for the diode, given
0.3393mΑ
3 vD = o.o2sh(qΨ)
η
=
1.4517 τnλ
435
,, ='_ofΨ':
,":
0.3 mA
υ
mλ
Xο'o25loc(ι!#)
: 0.7093 V
_
;= l 0.7093 :
t vo:11.7v,r, : 'i*' :
current. Use
mA
: 0.7101 V
' _ t _ 0.7101 = l.4494
0.2
433
2
l mΑ
υ7: ι|+ι:v'ωε(?)
V
is :
x1
φuatiοn 4.5
o.oδXl"(η+)
and graphicaι resυIts
ζ9
/p = 7.39 mΑ. 'ιlr'e can then find ιhe resistor νalue
using:
ρ: l0V_3v
7.39 mA
ra.r
r'
436
constant νoltage drop model:
ir, :
v
using
,D.:
υsins
υ^: 0-6v+i^' _v
0.7 V
+
-
0.7
0'6
For the difference in currents to vaη/ by only
1%+
io, = Ι.o|io'
v-0.6: 1.01(v-0.7)
V = 10V
ForV=2V&R = lkΩ
Chapter 4-10
Aι y^
_
y^
-
0_7
ν ι^' = 2
o'1
_
I._t
43E
mA
0.6
ν i^'
2
0'6
_
coNsTΑNT VOLTAGE
Refer ιo example 3.2 -
I
DRoΡ MoDEL
l.4 mA
(a)
I
+lοv
2=lj:l.οε
l-3
ioι
Thus ιhe percentage difference is
@t.εο-I:
8%
0.86
mΑ
@
o.;
1ο
kΩ
i '0 =,o.o
v:ovΘ
*Θ'o
437
Ανailabl€ diωes haνe ο.7v drop aι 2 mΑ currenι
Since 2YD = 1.4 V is close to 1.25 V
Use /v paΓallel pairs of
diωes ιo spliι ιhe l mΑ
v
currenι eνenly.
'lο v
)i
lmΑ
7
\
7
\
+
(b)
v
+lοv
The νolιage drop across each pain ofdioder is
v .'. volιage drop across each
61o6" _ !Ξ _ 0.625 v Eοuaιion
1.25
)'
Vz_Vι
=
4.5 is
2ryrloc(i)
ιν^-v'\/ν^
lz: lιe '
^
:
(ο.625 υ7l/υυ]5
-!0v
0.Φ957 mΑ
so current thΙough each branch is 0.09957 mA
|
' in = 0.09957 =
lmA\νill sr,ιiι
l0.M
branches.
choose N = 1ο
There a.e 10 pain ofdiodes in parallel
.'.
,^'
"'
τ,D
- l0 ι t5l0)
=
- l0+
1o
kΩ
-
0.7
-
l.29(10)+o.7
4i9
(a)
Need 20 diodes
curreoι ιhrough each pair of
diodes
..
:1ΞΔ:
t0
o.l .,n
Voltage across each pair
: z*Γu
t
:
+
z..lv.loρΓ9J)l
-\ 2./l
l'25M Υ
5V
-5+0-7:
5+4.3
l0
0.93 mΑ
t.29mΑ
-4.3V
:
3.6V
chapιer 4_l l
(b)
,
Cutoff
kο
10
v
v
I
Ι
-5V
-5V
s-/(10): sv
οΑ
V = 2-0.7
:1.3v
, _ r.3-(-5)
(c)
:
3.15
mA
(b)
.
-5V
4.1 +5
t0
:
9.3 mΑ
v:5-0.7
D2
(d)
Ι:5_l.'l:1.65mA
{r
v
V: |+o.7:1.1
4.41
(a)
*
-5V
Ι:oΑ
Υ
tT'' :oou*9
Θo.7 v
-3V
Chapter
4-12
4.43
(b.)
l0 kΩ
-"*:-9J:o.rsrmA
o0A+
lt.2.ι5
+
0.'τ:
0.5.15
vΘ
v:-3+0.3s3(s)
v=
l.235
v Θ
R
-
l2o
{2
o.'|
50
Reνe.se
νoltage
:
:
3.]8 kΩ
\2o
rt :
rcgJ ν.
The design is essentially the same since the supply volιage >> 0.7
v
3V
4.44
use exponential diode model to lind the percentage chaπge in ιhe currenι.
4.42
(a)
(l0
;_ _ Ui.,",L_0'7=sο--,ncικ
R
||
20)
kΩ
(ν'
i^'
Ioι
=:e
t20\
6" \τl
ν
Δ"lvτ
1'lvτ
For +5 mV change
Ψ:"5ι15:l.22l
Ioι
_-!_ !]- :
ι = (|οl|
20) + 20
qο
chΔnge
o.l24 mA
V:2oΙ:2.48Υ
:
ior_ io'
:
22.1%
ioι
x 1ω
: 1Ξ?LJ
l
1
1U0
For -5 mV change
iΡ_"5n5=o'zιz
(b)
Ι:0Α
<--
qο chΔn1e
: -
5kο
0.818_ l X ια)
I
la.\%
Maximum alιo\γable volιage signal change when
V+
3V
: 'oz._'ol y 1φ :
loι
ιhe cuπent change is
limitω to
t
1ο%.
so ιhe cuπent varies from 0.9 ιo l.1
2
Ι
.'Ξ<]
"rtnrf '2 2
../ = oA
v:' 22 ] = -ο-s
:2.5Υ
Ι
iot
Δy =
Forl.1 Δy :
For0.9
:
25 ln(l.l) :
25 lπ(o.9)
_2.63 mV
+2.38mv
For :!:|oΨo cuπent change ιhe νolιage signal
change is from _2.63 mv to +2.38 mν
chapter Ζ'_l3
25mV,
Nowυr:5mvx
:
ο.α)5
x
0'025
ο_o25
+ lο],
1mΑ
Ten diode connected in parallel current operates
at a total current of 0. ι A. so the current through
each diode : QJ :
1ο
l
o.oι
diode =
L:25.V:
iD 0.οl Α
ηuivalent resistance,
Req,
'q
4:
=
|ο
o.zs
z.s Ω
of 0 diodes connected
1
a
Ιf theιe is one diode condιΙcting 0.1 Α cuπeηt,
then ιhe small Signal resistance of this
: 4ΞY :
61o6"
0.1
A
2.5
ρ
This value is the sam€ as of
Ω
10
diodes connecιed
is the resistance for makiηg connectioη,
:
ra + O.2
=
2.5 + O.2
= 2.7
For a parallel combinaιion of l0 diodes. equiνa-
Ιf ιhere is singιe diode conducting alI the 0.l A
curτent. ιhe connectioη resistence needed for ιhe
be
5.0
mV
=
vτ
Rsι
ιο
υι
ντ +
o.ο25
o(ns + lo'Ι
0.025
+ tο4,
ο_o25
Herelts:10kΩ
-!!:
P
iot
t
limiιω
l0%. using
Δι
=
25
:09ιοl_l
x lο-rIn(b-2)
\in'l
mA ΔV :
Foι l.1mΑ ΔV :
o.21
o.25 = o.02 Ω
The va.iation is _2.63
4.6
The dc cuπenι / flows ιhrough ιhe diode giνing
L)
"\ ι]
diωe resis,-"" .,(=
ano ,n"
small_signal equiνalent circuit is repfesented by
ana ιιe.e
_2.63 τnΥ
2.38 mV
mv
to 2.38
mv foΙ
t
10%
νariation Το obtain the ιimit on ιhe i!Ιpua signal,
divide this by the gain given in the problem.
Δoo _
" (ι6lι1)
^,,"
rise ιo ιhe
.-
4.47
As shoιγn in Ρrob 4.46
For 0.9
o,zta
singιe diode \γill
1μΑ
dl,
lent resistance, Req, is
nq :2!:
lο
mV
exponential model Ψe get
the resistance in each
branch
1.0
The cu.τenι changes aΓe
in parallel.
Ιf 0.2
0.1 ιηΑ
.
+Ι:25ι'A
in parallel
n
o.|22 roν
l..
roaυo=;vs:rJx-;
small signal Ιesistance of each
10'l
25 mV +
2.63
mv ιo 2.38 ι2l
υ1lq
Νow for ιhe given νalues of (v6lvr)
and Δrs using φuation (l) aΙd (2)
υ6Ιυ,
caΙculate
ιo
/in mΑ
0.5
0.ω25
-5.26 to 4.76
0.1
o.0225
-26.3 to 23.8
0.01
0.2475
-263 to 238
ο.0o1
2.4975
-2630 to 2380
v,
in mV
&
r,
Lx
"^: ' ,r*R,
vτ
"yΙ+/Rs
V,ll
"Vτ'o
Ι"
Ι
Chapter 4-14
4.49
4.48
when both DΙ and D2 aΙe conducting ιhe small_
signal model
r
rιl
a.
since
cl
and c2 are large capaciιors, ιhey are
replaced by a short circυit foΙ
f,ι2 _
7li= tιι|+ rιι2
Ac.
vτ
7]o
lm_Ι _ Ι
lm
* v'
Ι ιrn_Ι
\k
The currcnι ιhrouρh each diode is
"ιlι
v
τ
2v,
"2
0.ο5
t
From the equivalent circuit
ιo:
1,i
R +
R
_
R
(2ro||2ro\ R+r,ι
Noψ 1is expressed in mA
:
1?
/
where 1is in
υo
mΑ
NoψI= oμΑ, 19:
υi
0μA
o
/: lO μA, ls: lor10 3:0.0lV
1
:
l0o
μΑ, ls:
ιι tωΧt0
1= 5Φ μA,
1
:
=
900
μΑ,
1σ
μA
k:
υι
t
r:0'lV
sωx ΙO ] = 0.5V
0
lμΑ
5ο kΩ
0.167
l0 μΑ
5kΩ
o.66-Ι
lΦ μΑ
5ωΩ
o.9524
lmA
5οΩ
0.995
mΑ
5Ω
0.9995
10
19:sοoxl0r:0.9V
1s =
l0οo
r
l0_3 =
lV
ηυivalent circuit
Chapter .1- I 5
ιo
b. For signal current ιo be limited
!lo%
ol Ι
(1is the biasing currenι) the change in diode νoltage can be obtained from the eqυation
+2.5-
+2.5-
b:
Ι
Δυo
"';)
;Φ'
l0 mV
"ouo'u'
:
:
:
ο.9 to l.1
_2.u' mv to
:t
+2.3
mv
2.5 mV
so ιhe signal νoltage across each diode is
limited
to 2.5 mv \νhen the diωe cuπent .emains vr'ithin
l0% of the dc bias cuπent.
4.50
.'.3o: l0 _2.5 _2.5:5πΥ
: ΞJI! : 0.5 uA
^n6;
t0 K
Th€ cυrrent through each diode
: 9Σ *l :
2
0.25
uΑ
The signal cuΓrent i is 0.5
of the dc biasing currenι
.'. Dcbiasingcurrent
c.Now/:1mΑ
.'.
1,
:
ρ.5
μΑ'
/:
0.5
and this is 10%
Χ lo
:
5 μΑ
,o
Maximum cυrrenι derivation l0%
.'_
i,' :
0'5
i
lο
:
0-o5
mΑ
= 2io = 0.l mA
.'. Maximum ,o : i X
aιd
:
0.1
10
kο
x10
Each diωe exhibits 0.7 v drop aι 1 mA cuπent.
υsing diωe exponential model ινe haνe
Αlso Q of each diode
: L:
ΙD
soο
Αccording to current νariations (see the figure)
ιhe ac νolιages across diodes are shown and
ι1 :2.63mΥ
= l.Φ5 V
+ 2.38 mv +
- r, : V-' fn(!)
\i, /
:
and Ut
0.7v, il :
,,
iv
1mΑ
+z:o.7+y-ι"rj)
' ιl'/
: 7ω + 25ln(,)
calculaιion for different values of vo
υo
: o'
io = 0,
the
cuπent
/ : 1mΑ,
diνides eqυaliy in D3, D4 side and Dt, D2 side.
Chapter 4-16
i,: l': i' = i':1:2 ο.s.l
ι/ : 7ω + 25 ln(0.5) :683 mv
0: !1 : ι::683mV
,.'"(v)
From circυiι
+τ/o: 683+683+0: οv
- lν_i- - lο|κ - o-lmΑ
ι!:
Uι+tr1
Fοrr^
Because of symmetry ofιhe circuit
i' _ i' _ Ι+!
22
and ia =
i,
= 6.5+0.ο5 = o.55mΑ
= ρ.45
7ω . 2.5ln(!] _
\l'l
7,Δ
=
7ι|
= 7ω
+
685 mv
:
680 mv
(mΑ)
(rηv)
25ln(i!)
ι,'(υ'
(mA)
(mΑ)
(mv)
v,
} τιlnV
ο
o
0.s
0.5
683
683
0
+1
0.1
0.55
o.45
685
680
1.005
+2
o.2
0.6
0.4
-647
677
2.010
+5
0.5
0.75
0.2s
-693
665
5.028
+9
0.9
0.95
ο.05
-699
425
9.O74
+9.9
0.99
0.995
0.ω5
-7ω
568
ι0.Φ
9.99
0.999
0.999s
0.α]05
-7ω
51ο
10.18
t0
I
I
ο
7οο
0
lo.7
= _ι1
U2+τrο: -0.680
+ 0.685+ I = 1.0ο5 v
ι]t
+
4.51
similarly othe. values are caIculaιed in the table
foι boιh posiιive and negative νalues of τ,,,
The largest valυes of r., on posiιiνe and negaιiνe
side are +l0 V and -10 V resμcιiveΙy. This
1 mΑ
restrictioπ is imposed by the current 1
A similar ιable can be generated for the negaιiνe
νalues. Ιt is symmetricaI.
Representing diode by the small signa! resistances, the circuit is
:
For
V, > 10, τb wi1l be
because/:1m4.
Fοr Ι
5V
=
saιuraιed at l0
ο.5 mΑ_ will satυraιe at 0_5
v
and it
+
ιi
C
+
is
,, _-vt
mA Χ lo K =
"
ιb=
υ.
Phase
I
-
|
_
ιo
1
1+scr,
shift = -,rn - '(sC'a)
\1./
chapter 4_ ι 7
= tan
:
ιan
Ι(ωcrd)
b. Ιf m
,r..!]
\ ,./
For phase shift of 45"
45 . ιan '(zτ xιαlx tοJX toΧ
\
+I =
Noψ
Ι
ΔVo
Δy+
151
Range is 15.7
f ι,ι
μA to
Range ofphase shift
1570
to r57
x
μA
is 84.3" to-5.71'
10
aτe in series
mra :
mrι+ R
mV
rο
ιιΑ
νaries from
diωes
μA
'x
mvτ+ ΙR
o'-'
_
o'ofs)
mVτ
'{"!Ι'":'ruΔVo _
vτ
^v*
Form:3
:
vτ+v+ -
ΔVo _
2.68 mV/V
o.'7
mVτ
Δy* mvτ+lo
mxo.7
ror4l?=smv
/ι
mΑ
4s2
v* - 0.7
5mv
-Lx
ΙD vτ+v''o.7= mΑ
i.".
t0.025.- lο _o_7 l'-mv
l Ι" 0.025 + !0_ 0.7l -mΑ
Ι D:--
/,3
^
5
4.987 mA
mA
_vt-o.1
ΙD
ro
0.7
5 mΑ
R = 1.86 kΩ
^ ΔVo
Δy*
-
_
FoΙ no load
rι! _
R + rd
vτ/ Ι
vR+J
Ι
vτ
,
v'-o1m
"R
ΔΙ1.d
ΙR+ vτ
1
Diode should be 5 mΑ diode
c. For m diodes coηnect€d in series
: V*
vo
R
vr' : mx -!
"
ΙD
so no,v
4& : ΙL
I
1.f
R mrγ
Io
*
-o'7m
: _.vt
v*
mVτ
ΙD
v+
: -av,
ΙD
Small-signal model
Ιo
mVτ
v*
-
o.'7 m
y* -0.7m
-o.7m+nv,
Chapter,tr-18
4.53
10
ΙD
5
-
0_7
mA
R = l.86 kΩ
Diode should be 5 mΑ diode
c. For m diodes connecι€d in series
v' 0.7 m
,
'o_-____FvΙD
so now
Δf
__
1
t,1
R mr,1
Io
v*
- 03 rn
^V,
lo
Small-sign-al model
a.
:
-
: _ι,'l| nl
ΔVo
b.Atdc 1ρ
v* _o.'Ιrr,
o.7
m+mvτ
4s4
v+ - o.j
=
y*-0.7m
Ιοv'
Ι, (r2|| R)
Ιo
mVτ
mV7
mVτ
From small-signal model
Δv o
+
__
+5V
lι
rι
I
R
l^
Ι^
-__-__,!_ + ,,:!
''
D
v+ -
vτ
I
vτ
v+
ντ
-
o.1
vτ
l
0.7
v+ - o.j
Diode has 0.7 V drop at 10 mA cuπent
'D vτ+v'-o'7
ΔV'=
mA
/. 'V
v, v*-oi
.ι'e' _ζXy,-v-
F.,.
v.,
,5mv
oJ= rΑ
l0_0.7 l_.
Γο.025 .A
0.025 + l0 ο.7l
L l,
/ρ:
5
mA
v
\γhen
i, :
l50 Cl
ν1v-
s
Ιo-_ 4.987 mA
= 1.5
10x 10 ] :
Ξ/s:
/s.ο7'n025
6.9l X lο_ι5Α
volιage drop acfoss each diode =
:.ιo:1r"v|vτ =
6.9i Χ 10
: 73.9 mA
/ι=1.5/l50:10mΑ
jjlζ
=
0.75
V
"'" "n'""'n"
Chapter ,r-19
Ι:
I':
Ιo+
:
Ioι _ ' o"' " n'
w"h^""
73.9 mΑ + 10 mΑ
π_'
83.9 mΑ
.'.ρ:5-ι's
83.9
mΑ
:
diodes
4l.7 Ω
Δv /2
Use small signal model to find voltage u, when
load resisιor, Rι, has loΨeι valυes
:0.025 _
,'" : V,
Ιo 7].9
0._]4
Ω
Fo.
.'./r:7:
""
:
,, :
diαle.
rO
33.9
ur"(:oη):
0.753
So No load,
V
,,
=
o.o25
2V,:
.(S3#)
2 Y 0.753 = 1.506 V.
:
Ιncrease in voΙtage
1.506
Now load is changed
lΞ:
R.: tgoΩ: l':
Χ
_ l.5 : 0.ω6 v
Iο0
The diode curΙent reduced by = l5 _ 10
.'-^vo: _5 mΑxrr: _l.7
:
5
=75Ω:
^v
8,
.'.
ιoad Reρuιaιion
'
:
-60.2 mv
ιl.2 mν
:
ΔV
o
Δl'
:
39
5
: -7.8 mV/mA
Using l5 mΑ source, i.e / : 15 mΑ
Νow the diode cυrrent changes from 8 (: l5_7)
πΑ
(=
l5_2)
: 2yτh(l3!) :
Loaα reρυlaιlοn
"
7, =
,v,h(f6)
Δv: zy-ιn(!]
' \1ο., =
mΑ
mV
:
Δy" γaΙiation is _ω.2 mv to _ l l.2 mv
Peak ιo peak ripple : -ω.2 ( ||.2' = _39 mV
Δyo
lΞ:26rna
75
Diode current reduced by 20 t0 : l0 mΑ
.'. Δvo : - 1q ra = _ 10 Χ 0.34 : _3.4
R,:50Ω: 1,: lΞ:36.a
5ο
Diode cu.rent reduced by 30- l0 : 20 mA
Δvo = _2o r7: 6.8mV
R'
a,
10
to 13
ls,,,l,
_3 ,o9
to
10
lο
when load is disconnecιed all the cuπent / flo\rs
through ιhe
vτ
2v'tacause there aΙe 2
_24'3 mΥ
ΔI'"
^v
24.3
\1
_2l
_4.9 mv/mΑ
=
clearly ιhe l5 mΑ cuπeηt source provides lower
mV
load regulation but it is at the cost of higheΓ current and higher power dissipation.
The l0 mA current source ιγill be a better choice.
For l0 mΑ soυrce when ιhe load cυπenl νaries
oνer full.range' i.e from 2 to 7 mΑ; the output
νoltage νaΙiation is Δ% = -39 mV.
4-56
4J5
+5V
τg: l.5 V
+
Δι5
ΙoI
l50 Ω
I.varies from 2 to 7 mΑ
To supply
a load cuΙΓedt of 2 to 7
Ι must be greater than 7
mΑ or
15
mΑ, the curτent
mΑ. so ι can b€ only
lο
mΑ.
No\ν leι us eνaluate load regulaιion for both 10
mA and 15 mΑ sources, l0 mΑ source:
Since 1. varies fmm 2 to 7 mΑ, the current ID ψilΙ
varry from 8 to 3 mΑ
'
.'.
I = Ιι+In
Both diodes are 0.7 v. l0 mΑ diodes
First find % wiιh no ιoad, i.e 1ι : 0 and 1
Use iteration to fiΙld v" and /,
r^=5-:-9Jjl:zome
"
18ο ll
v2- vt = 2.3v,bE(?)
:
1..
chapιer 4 2ο
Vz
:
0.7 + 2.3
ν
:0.7l'7
=
'"
V,
'
:
o|r{
'
:
x 25 Χlo-r Χ loc(?8)
"' :
23.19
'ι ιο
' tnn(23
ν
0''72|6
ιo = -2'_x0.lzιε:
5
V':
'
:
πA
./
19.76 mΑ
:.
:
x
2
0.722
180
24.2
πA
:
o
V
Υ' Ιo :
n'722
mΑΞ,
t9.8l
v aηd ,D : 19.8l
Sovo=zΧo.717=1.434Υ
o.'ll1
mA
Ι,' 2 \0.'122
= 96 mA
t50
So ι.: A)-9'6 : 14.6 π|^
.'.vo: o1 + 23 x v, bc(ff)
= 0.7095
Vo:
2
V
V
o='|.42 Υ
.'. Loaded output νoltage
of 150 Ω is connected
a. Load
b.
:
:
2vo
zlo:
+
2.3v.
:
-
1.401
0.1
:
ln = lO
Δy _
New
lO
<
:
a Δνl vτ
'e
'
- l.30l _
l.401
2
!..10|
I:
.,
ι.301
4.2
X |m
l_50
(a)
v:
Ω
:
v
l.3ol/2
l.434
v
:
0.6505
v
v
_
l0.4 X 100 Χ
V.
:
V"=
to'4.2
ofthe νolιage supply = 5 +
5.8
/^:5.8_2Χ0.7
=
"
18ο
P:
/.7
ν
v. No\λ, do ιhe problem again as
done in the begining and in paπs a and b.
4.2)
V
:9.6+8Xl0oΧlor
: lO.4 V
:
V.ρ +
(c)6.8:6-6
16.1
supply can be Ιowered
d. New value
V' : V-o! r-x2Ι',
24'4mA
10-3: 1.o4W
r_1.1
9'l=v,,,+30Xl0xι0_3
η"=8.8v
η: 8.8 + 30 x 20 x 10 3:9.4V
P:9.4x20Χ l0 3:188mW
8.7mΑ
+/. =
Po\ir'er aaιing:
(b)
,4
value of 5V supply :
ID
v-o + r.Ι-τ
l0 = 9.6+Γ"X50Xlo'1
where
ο.05
v. :
r.:8Ω
l80 Ω X 16.l ΙnA' + vo= 4.2ν
so the 5
(5
Υ
='Ι.4%
aηd % is lowered
1.301
volιage across each diode
:.
lo.25 mΑ
v
1.,ω1
c. with 150 Ω load connecιed
by 0.l v ofiιs nominal value.
,," =
- 1.42
5.8
r"c(l!#)l
Αs found earlier, ινith no load y"
vo
l.42
4.57
so noψ
:
:
e. Ρercentage change in output νoltage
1':0'717xz:9.56mΑ
'
15ο
:. Iτ' = Ι| 9.56 : 19.81 9.56 =
vo
24'2 mA
Now \rr'hen l50 Ω load is presenι
ιion here
o
5.8
:
Ιt is almost similar to earlier result' \νe stop iιeraV
Ι0./
ν
0.722
Doing one more iteraιion, almost same value is
obtained
.,
: o.'7l7 Υ
,^ _ 5_2Χ0'717 "
18ο
-' bρ(24.4\
'\
o'1 + 2'3v
:
z'ιv. πn(|9'76\
-\ 10
o:] +
v|+ 2.3vτbB(?)
79)
Z.lY-
O.? +
:
v2
+ 2 Χ 1Ξr
1Φ mΑ
6.6+2Χ2uJx 10 3:7V
P:'1 x2ΦΧ 10-3:1.4W
(d)18:17.6+r.X5Χl0_3
Γ.:80Ω
ι,:17.6+ 80X 10X 10 ]:18.4V
Ρ : 18.8 x 10 x l0_3 = 184 mW
chaρlet 4_2l
(e)1.5: V.,,+
v-
1.5
x
200
x 10
= 1.2Υ
Vz:'7.2 + l.5 X 4ω X lo_3
P:7.8X4Φx
458
3
10
:
7.8
x
V
= 3.l2Vr'
:
V zeners provide 3 X 6.8 20.4 V
10 = 30 Ω resistance. Neg|ecting R, we
(a) Three 6.8
wiιh 3
r.:30f'
/r* : 0.5 mΑ
Vz: '7-5 ν
Ιz: 12 mA
3
haνe
Load Regulation = _3ο mν/mΑ.
5.l v zeners we use 4 diodes to proνide
20.4νwiιh4x.10_ t20 Ω resisιance.
(b) For
=
Load.e8uIatioΠ
l20 mv/mA
4.59
82Ω
small-signal mod€l
5
Δr, = 5+82
Now Δτι : 1.0 V
.'.
Δvo
:
:
4.60
V,
:
9.1
:vzo+
For 1'
For/z
ν
a :
1-:
' t.2 6.25 -ρ,
sELEcT / : 10 mΑ so that
/. = 3.7mΑ \Ιr'ΗΙcH Ιs
>
,zκ
R:10-7.5:25oΩ
. 10
For ΔV+ : :tl V
I.2 || 0.03
: ll , X
ΔV
9'o^ _
^
δ25o+ (|a-l οJ-'
: 10.1 v
THUS yo : +7 .4 ν 'Γo +7 .6 ν
wΙTΗ y* :llv
ΑND ,.:0
1l/,
v-: v.^+
x o.o-l
vo
:
ο.25
7.55
ν
5
:
εΣ
x
t.o
mV
V'o+ r'Ι"'
x
5
x
28
vzo + 5/.
:
8.96 Χ 51"
V"o: a.96Υ
νz:
= 7.|4
87
;Δvs
57.5
Ξ vzo
Ξ-'
From the smaΙl-signal model
Δ,o
7'5:V2ρ+12Χ30X1o_3
10-3
: 10mΑ η:9.0l v
: lω mA V.:9.ι6ν
9-7.155
ο25 t
4.6t
_
lo_
lο
7.5
!
= 7.38 mA
250Ω
9V
25o
Ω
= 7.ι4 + 0.03 X 0.5
JR,
{
^
"MΙN
:
7.ι55
7.38 _
1.o4
ο._5
kο
and
V, :
7.155
γ
chapteτ 4_23
13.5
_
< l5 (9.825 + 7(0.ω5))
25
V
S 205.6 Ω.
.'. use i :
205 Ω
_
7 _
205+7
(c) Line
reρυιaιion
'
!25%
chΦge in vs = :t5
V6
chaπges
by :t5 X 33 :
*165
X
coπesDondinρ to
t0
:
or
Loψest output voltage = 8.83
:
-
Line Reρulaιion
tz
R * rz
Δv o
V
=
30
4.64
(a) Vzτ
:
R)
:
-29.1
gf
load
2ο +
=
1ι = ο
!ιzοl
4'
2αl
20 mΑ and
_ 0.25)
=
l5 v.
ιω -
* |.65%
:
:
20
mA':
_ 135.4 mV
_0'l354 Χ
t0
lω :
_1..l5%
:
aηd the sυpply aι
zs ν.
25Υ
(b) The minimum zener cuΙrent of 5
,ι :
:t165 mV
Vr: VroΙrrΙ,
Vzo * r'Ιzτ
l0 : V ro + 1(O.O25)
+ Vro = 9.325 Y
when
x
6.17
ys
mA occurs
is aι its minimum of
see the circuit below:
t5v
)ξ_v:9825+7x--:-----:L
2O5
r
-15
_ (Vzo
+ rzΙz)
20+5
V
(e) The maximum zener cuιτent occurs at no
v
: -(rrll
=
corresDondins ιo
3ω+30
:96mV
Load Regulation
_(7|l 205)
4.77 νl^
mv
V
(rzιl R)
_6.77 {ι
Regulation
(d) Load
_]1
V, :
205
26519.925\ +',t(25)
9vz = |0.326 ν
_
P-:
lo.326 Χ r25
'\205
: 739.4 mW
10.326)
Αltemate circuit to calcυlate
1
yz
- 1V,
chaρtel 4-24
V2x:
9.825
Υ
(b)
:1dι
,
ιz
25_ νzo
-0.7
250+7
_
25
-
v
9.82.5
205+1
_
71.6 mΑ
12ν
O
9.825+0.0716Χ7
10.326
(c) The diode conducιs aι an angle
V
o
as aιΦve!
: .in-,l9J) :
aιιτ_θ:
4.65
\12 )
]-}4o & sιo's
|'76.65"
Thus ιhe conducιion angle is
:
π
2θ
I73.111" or 3.025 rad.
π0
ιt
Uo.,,": :-.l | (t2sinΦ
'
zτJ
_ 0.7)dΦ
θ
: lt-
2τ'
using the constant voltage drop model:
:
vDΔ
ideal :0.7V rD:0
l2"o,δ _ 0_76lΙ
_ ltιz ,
2τ'
_4.572
-
0-7
t.5 x Iol
(e)
_
ι,
+ 0.7
V,
0, ioι ι.
=
For vs = '0.7
0.7 V
Υ
v
2.5j mA
PΙν occurs when ,s is
ιo:o'
ΡΙV : l2V
=
2cosΘ _ o.?(7 _ 2Θ)
(d) Peak cuΓrenι in diode is:
t2
(a) uo
o
at
iιs the peak and
|
Chapter ,1-25
,r: ξ
',
: !_ο'mτ
1
4.66
t2
:
T
O.43
ιo(aνe') =
1Χ
area of shaded triangιe
!"es
=
0.7)r(;,,,)
, l.ε, rl!
r\4)
=Ι
υD
_
υI'
=
o'1'A
ι,b ID(,' lmΑll/vΙ
i"
iD( ι mΑ)
ν
ο.οz)
4.6A
I'D(at
lmA) = v.ln
υt'taΙ
= νs_
ιγhere R is in
ι.l1
υD(al
* v,
ideal
i2λ
ο.7
v
tn["/{l
-'.]
L lο
lmA)_vrln(f)
kΩ
4.67
to = l2J1 -
:
o.1
16.27
ν
conduction begins aι
υr:12{7i16=6.1
,: tt '(#)
:
0.M12 rad
τ- θ
angle = π _
conduction ends at
.'. conduction
2Θ
:
3.06 rad
The diode conducts for
lΦ x
tοο =
48_7% of ιhe cvcle
n_0
ι(
V" *" _ !- | (l2J2sinΦ
' 2τJ
0
=
;r.*.
2.5
First find rr and ,2
2.5
_
!ι'
4
+
O.'7
rl =
0.07 'Γ
5.06
v
: '?"" :
5.0,6
mA
_ 0.7)dΦ
chaPteΓ 4_26
4-69
R : Drt
Ρeak νoltage acfoss
: l2^Γ2
: 15.57 V
Dl
J'
12ν
1
J.
lkΩ
^
2V o
l.4
'. ιzν ^,D'
Cont.
:
6
.;n ,-14 =
t2J2
0.0826 rad
Fraction ofcycle that Dt & D2 condυct is
n_2Θx1Νl =
4l '4Eo
Note D3 & D4 condυα in the other half cycle so that
;o
_ 2rt
_ Voo =
cοnduction starts at
6
16.2'7
: ";n '-9a
12J2
= 0.M12
and ends at
angle
:
π _ θ- conduction
τ _ 2Θ
: 94.8% conduction inte.vaΙ.
τ0
^r
(lzJzsinΦ 2vD)dΦ
ιo^''=
' ?
ιτJ |
is 2(47.4)
ther€
Υ
τad
:
t\,r'o
diodes
:
conau"t
:
2(3'Φ)
2ιr
x
tα.l
_
2(|2{2cosΘ)
:
9.44
Noιe ιhat during 97.490 of the cycle ιhere will b€
conduction. Holreνer each of the two diodes conducts for only ha|fιhe ιime, ie. for 48.770 of ιhe
cycle.
: 24 :
1".'ε R
l
'
_ l.4(τr 2θ)
g.44
-a
-r
J..
_ ο.7) dΦ
Itt
i^_-:lο'|2=l0_l2mΑ
rko
l.4Φ];
4-1t
τ0
= Ι0.l2 V
'
Υ
vo.arε
9'Ι .49ο
tt
ν^ ".^ l l ιιzJzsinΦ
τJ
nrtcosΦ
7τ
3.06 rad in each half
cycle. Thus the fraction of a cycΙe for which one
of the
Θ
: 1t
+
R
tI a+
1J
Refer to Fig 4.23.
Foτ2V9g<<V''
4.70
uo
: ?V,
*"'ππ
(a)
For τt,
u
D\
'
:
Tums
"'.
Ξ'l
2v
oo = ?γ, _
=
10
V
1.4
:
17.91
'.o
v
Ratio : Ψ-Φ = g 'ιll ,o
t'7.91
ι
Chapter
(b)
:
100
101.4
:
For vρ',".
v, :
;t
+27
V
1s9.3
v
iι
Raιio _ Aort :1.g65 tol
Tums
Jο
159.3
4.12
Refer to Fig. 4.23
For 2Vpn<< V,
?v, -
v..",
=
(a) For
ξ,n ,
D4
+
3Y,
: 2v
R
D2
t.4
l0 V
=
rcν=?-v"
'..v':
o"
2v
v.-
l'r
Ja,
l.4
11.4|;) = l7.9ν
Tιrms ratio =
!4-&'
Ι7.9
(b) For yonνs = l οl0
:
9.5 ,o
1
v
lωv=2.y"_1.a
+y":tοι.4rΞ):t59v
"
\2)
Turns ratio = Ψ!2 : ι 'ol ,n ι
o,,',- _ ||1v.sinΦ
ιτJ
=\_o:l:
159
4.73
nort !. rcsο. 24rt !
Ξ turηs Ratio = 5:l
assumω ys
2V 5|^^'
= 2vs
X
1.1
ιroking
_
PΙΥ
o.7
v
:z-'66
Υ
:
:
49.32
ν
at Da
Vs_ Vδ
= vr + (vs - 0.7)
:2Vs-O.1
using a factor of l.5 for safeιy we select a diode
having a PIV rating of55 V
:
48.6
V
Ιf οhoosing a diode, allow a safety maΙgin
l.5ΡIV = 73v
4:74
The circuit is a full ιγaνe rectifier with centre
tapped secondary winding. The circuit can be ana-
lyzed by Ιooking aι
15
ThuS νoltage acrosδ secondafy winding
_ Voo
: 2xΨ
: 36.6 V
:
ιs
v."2= 15+0'77τ:
nqo
n.=24rt-ιoq"
"2
PlΥ =
>> 0.7
- o.τ)dΦ
ι$
and
ιo
seperately.
v.+
4-15
of
chapιer 4_28
(c)(i) use eq (4.31)
ιD'""c:
Ι,'(1-"ffr)
Vo.*J,,
l2(Vp
-τ-ι'-"^]01ψ"-r"",
:
_
(1) ι],
= (v P
v
DδL
F4.
φ.28)
(ii) ip
vDδ: ψ"_ voδtn
o.|(vP_
:
0_lλ60\I0-
: o.οl(V, ,"δ :
ff('."8)
233 mA
: !$1l
"',
=
l66:t ρ'F
(ii) for
v,
:
Voo)
+
"'."D-ω)
735 mΑ
: vp
NB next user /, = yΡlR
voo
R
vP vDο
ψ#
but heΓe aΓe used iD' a"e =
\ir'hich is more accurate.
(a)
: vP_vDo_'vΓ
(i) vo'.,ε
= 12{2
=
(2rt_
:
15.5
ii)
=
o.llo.ι
Ψ)
16.19
_
t5.42(t +
:
.ι49 mΑ
ι'ιz{z oιl(ι Ψ)
V
4,76
(i) v,
(b)
i) using eq (4.30) ιve have ιhe conduction angle =
.Μ=,!2v.lψo_ν-1
:
.'.
:
Joj
=
η)
:
1455
r"ffi;
mΑ
v"a: !#Ρ
Tn : L
l,
2(60)
2t
loι X 0.l
8.1..]
(ii)c:____ι-:-=833μF
2(ω) ι0'0.0l
conduction 0'447 X 1ω
(a) i1
:7.1%
: 2"6.61(Ve
'
E\
trΨ(, -
0.1(Vp_
^l
u
'i 0.1,,
:
.=-l^(2FR)0.l
Fracιion ofcycle for
Fracιion of cvcle
2,
zτ Fιv
+
οnlν half of ιhe οeriοd
0.447 rud
ιii) ωΔ/
l,(t
'v,
,.l\
The facιor o[ 2 accounιs'for dischrge occurin8
c
=
l0' \
(ii) i2',""1
V
=
vo""_q
o.7 \rtzJ1_
o.,z)(1 _
=
(d) (i) iD'p"^k
v
o.'7)
y n0
= 9!!1
2ιτ
:0.141
:
V6 = Vt
: (Vρ_
rad
:
2.25'Ιo
(ii)
(
Vo
t6.2?)l
I
-
voo
I
- 1ιl
,r(,- Ψ)
_ 0.t)
τ)
yo = (t6.27)(l _
$)
:
lο.lo
v
μF
chapιer ,1_29
4ia
(b)
(i) Fraction
_
ofcγcle
"
2ιυΔι
2zt
:@",*
Nιτ
: :J2(ο.l)
:
r*
= Vr_Vp6:
Vp : 16"1 Υ
υo'p"uι
45%
v.-.: ξ :
(c) use eq (4.34)
ι'(t.ffi|
(i)io.."s=
=
(b\
D44 rιA
^/
=:ro.ι
Ψ('- "#)
(d) use φ(4.35)
Ιiliρ
ιιυio
-
2",___L-) - 23.ι mΑ
Jz1ο'I1,,
'
_ 735 mΑ
= ι'(ι *z,Ξ')
\
/o'02 '
ι, :
g.71γ"_ Voox2)
discharρe occυrs onlν oνer
(Vp-2VDd
^
(iil
c
=
E4(4.28\
t6
ωXCX150
C:889μF
1,'(1 ,
v
vo-voo
4.71
1)
^
τ
(Vp- v Dot;i
:
V.
16
rI.ε
J2
E(l *" Ε.1 :
I\
2(0.1)-/
ie',".:
(ii)
Vrms
60Hz
|4.2%
:'ΨΡ'
r.ltonrocy"t"
(ii)
v
0.7
'φ
120
lΦ =
>
,
!7
2
v P2v
-vP
po
2fcR
_Ι
PΙV
2f
:
I
----!-:8]3
2(0.01)/R
Φ) (i) Fraction
:
83.3
μF
ιr.F
:
Vp_ vDo_v'ι2+vP
= vo'Δ'ε+vP
:
:
15
31.7
Iω :
:
(ii) i2'"". =
V
PΙV
ofcyc1" : 19&11go
(ii) Fracιion ofcycle
(c)i) iD'a"e
16.7
FoΓ a 50% safety ma.gin
(d)io'^ε:
: ]@Χ
Nπ
+
using
4.5%
Ψ(' *,fi) : ,'o,e
Ε4μ
(d)(D ?o
:
,'',; :
ι5i9(r -
Ψ('
+ nt^!-o.oz1
-
z"
fi)
r",ffi):
=
=
356 mΑ
zzz
-ι
7o4mA
:
/.
(e) io,ρ""ι
x 31.7
41 .6 Υ
1.5
',('-'ff|
ξιε
;".-= *(,
:
:
:
I.36
=
f;
*e ιaue
."P)
Α
: ι r(ι + zo
|5!ι
rso\+z. ^,t@\
2t
2.61 A
chapιer ,ι-30
4.80
4.79
vr=
Dl
J
.v"
t-
-l
ο7v
cΞ Ξ R.
J+
v
'ι
ξ=
rοv
i,
16 + 2y oo
= 17.4y
v
,tilfE
v
.'.i,s: 16 +Υoo:16;Υ
16
RMS Voltage across secondary
_ 2x
=
RMs seconday volιage = .j-:j: = |2.3
J2
Ι)
(a)λ,:
:l l v, R= l50Ω
15
t6.7
J'
v,,:
x
15
I
v R:
l50f)
= 23.6ν
vP
(b)η=
2fcR
l6
2 X ω X c X
c = 444.4 μF
15ο
note: we got ιhe same valυe for c as ιhe fu|Ι waνe
rectifieΓ as dischaΙge is over ιhe same amι of ιime
T/2 and the peak is ιhe same _ι6
V
c)
(b) Using Eq.(4.28)
.. v16
' )fcR 2l6οΧcXt5ο
c= 444.4 μF
PΙV : (Vρ _ vDo'
= 2(16.7)
:
32.7 ν
.'. Using
ΡΙv=
-
iρ..,r=
vP (For D|)
:16+0.7
:
η
v
Α|lowing a 50
I + τ ^]i nl zν
1
*. fJ-)2)
^l2λ
= 0.734
=
voltage aι anode
of D,=0 Vω
Y
PIV=16_(_Vοo) :_vDυ
η.(4.35)
Ιll
V
16.?
7o
margin
(4.34)
= .11(r
l50\
iρ'^^'=
ν
for D3
a 50 % maτgin
Ι1-(
(e) Using
_ 16"Ι
O.7
1.5(32.7)= 49
(d) Using
+
+
zn^[_vrlzv'1
]Σ(l + 2. E\,2Ι
l50\
12
= l.36Α
=
=
(d) Using 4.34
i2."'r=
l1-(
|+
= -jiΙ
l5ο'
l
+
τβ
"tzv
16.7
25ν
11
n1l6/2x)\
= 0.734
(e) Using (4.35)
ir.."*= /.(
l
+
21τ
Jτ;-/2v Γ')
Χ l.5
Chapter 4-31
.
*zoΓwz'z)
|ftιl
l.36Α
ι0
T/4
^ι
4.E1
ο.9}
Δ,
= o.o2325T
= 0.02325 ms
(c).'.
Charge gained during condυction
:
Chaηge lost duιing discharge
C
cη
i".",ξ Δr=
,
'..aνι
lωXlo
Cur_
_
_ -i]--
6
------------_-_-
ο.02.]25
Xo.93
X l0
= 4.ο Α
aD'ιιp
R
io,",g=i",u,.+i".,.
t
toY
η
ι
M=
a
= 4.6
4.rnr
1ω
cψl
il
(d)r...ar=
l",nn*, nr.noοu.,ιon
I
c
t<_T
e-"Rc
v"= 9.3
j.η=
-
9.3
= 9.3(ι
_ T/cR)
_ 9.3(l _ T/cR)
9.3Τ
ΝB ιhi(
= -jj1
fcR
ic F,ο{4'l8l
= ο.g3v
υn'^""=Vo_Voo lΙ2ιη
93 !
2
ο.q;
= 8.84 V
(b)
l0.v
9.3
:
=
4+9.3/lα)
= 4.O9 A.
Note that in this case iD-!
(ι
:
is constant and
iD-. during to the lin-
iι is approximately
constant).
o1
=
iρ''",= i6 .,* * iη.,,
=4+υo.^u'/ΙΦ
ear input
CR
aι
=lΦXt0-6X40X1or
_ lms +t
During the diode's ofiintervaΙ, the capacιor diachaΙges throυgh the resisιor R according ιo :
aot
-
4-42
Refer to Fig P 4.73 aηd let capacitor
8.4
be conR. The
tlvo
supplies, η* and τlg are identical. Each is a full_
ιγave r€ctifier similar to ιhat baδed on the centertapp€d-transformer
circuit for each suρpιy' ιhe dc
outpuι is 15 v and the ripple is l v peak{o-peak.
Thus τ6: 15 :t l/2 V Ιt folloιrys that the peak
νalue of ζmusι be l5.5 + ο.7 _ l6.2V
.'.
Voltage across secofiary =2(16.2)
.
9.3
v,
c
nected across each of the load r€sistors
RMs
= 32.4 Υ
'rt
across secondafν
Tιrms Ratio =
Ψ=
22.9
=
s'zι,ι
Use η.(4.35) to find
ia.^^'=
Ι/|
= ο.2(1 + zτ
= 3.7O A
+
32Δ= 22'9γ
2τ]vJff')
"Γss
tz'
,-"
chapιeΙ 4_32
η
4.U
v2fcR
'
,Ι > 0 D| coΠducts D2 cutoff
vr<oD|cuιoff
(4.28)
DISCHARCE OCCURS OVER 7,' 2=
^
1
2f
D.conducts
'
15.5
- k=
-l
ltl
x60x75
2
where 20o mΑ =
]l
R
R= .!Ξ= 75 Ω
vt
0.2
C=
1722
rJ
Consider
μF
when looking at
a)υι: |v
ι6:oV
υA: _o''ιv
PIV
Κeeps D, off so no current flows through
ΡΙv oνer Dl
R
virtυaι gnd as fdbk is closed ιhrough D,
(b)
τ' : 2y
υ6:0V
υ
J,
PΙv=
iη1
=
+ 16.2= 3ι.7 v.
15.5
+
=
(c) υι
0V
:
Allowing for 50 % safety margin
ΡΙV=
1.5
x
31.7= 4'1'6
v
use Eq(4.34) to find
ip''""= Ι γ(| + τL_v"lzν11
= o'2(l +
=
τΓl5'5/2')
A
1.95
υ1
ι
=
1.7Υ
oV
- vinuai gnd
as negaιive feedback is closed
throυgh R.
(d)
4-83
υ"=υι(l+R/R)
= 2η when ιhe diode is conducting
(a) z,:11 γ υcι=2Υ υq: l.7Vι : η:lV
(b) v,=2 γ νo:4Υv^:4.'lΥ ι :2Υ
(c) rl:
tv τ,. = ' l2v - diode is cυι off
:
lv
ιι: _2ν+υo: 2ν
ι1 : 2.7 Υ
v :0V
4.85
a.
+2ν
uo=0V
ι=oV
(d)
υ':
2y
ιι:_l2Υ
l0v
lkΩ
5V
Diode staΙts condυcting when v, :2.5
Diode conducιs fully ιvhen η :2.7 v
5V
ιlρ."',
:
vo:ov U :ov
5
V
V
Chapter 4-33
Diode δtarts conducting ιγhen ,, : _ 1.5
conducts fully Ψheη ,, : _ 1.3 v
b.
+2ν
v
and
voΝ'
lkΩ
Diode stops condυcιing when
ιhΔι νι
:
U,
:
l.5
v
and after
υn
v"N)
d.
tkΩ
2.7
2.5
2
_2ν
Diode conducts for vΙ <
ylσ)
ducting for
_2.7
Diode stops con-
q> -2.5 V
V, (V)
v'(Υ)
Vr
(V)
2.7
2.5
2
t.5
ylσ)
4.U
For
tkΩ
2.5.<
η < 2.5
For υ' > 2.5 v
'
-2ν
andη
the diοde D| is
rates for ,Ι > 2.7
mιionaι ι,,'2_
both D,
2ι
v
+
oN
are off arΙd
aηd it satu_
at a νalue of 2.7
!ΚΧ
V
1 m= 3.2
The satυ_
V
chapter ,ι_-34
+2ν
oFF D| & 12
D2ON OFF
D|
DΙ
oN
D2OFF
4-88
lkΩ
-2ν
y"
ιv)
Each diode has 0.7 v drop \Ι/hen conducιing
The zener has 8.2 v drop \νhen conducting. so
the limiteι ιhreshoιds are
!(2 x O.'l + 8.2)
x9.6 V
:
4.E7
+2ν
v,
**n
4-89
I
kfl
Diodes have 0.7 V drop at
..
l mΑ
For diode D,
i^
lmA
ι|o-ι\1\/ντ
io: t\ |ο '"
ι|^U - |l'7't ν-
currenι
(v)
Chapter 4-35
ιo=o..1
+r.'(#)
10
iDΧ t kΩ
KΩ
Using ιhese equations calculate ?, for ιhe different
νaIues of
,,"r
τr.,
FoιD2υ,: ,o_iDx l kο
υ" (|η
ι' (η
y
0.s003
ο.5
1
ο.6
o.62
o.7
1.7
0.8
55.4
t0
KΩ
I
D2 on
0
ο
*0.5
_ο.50ο3
-0.6
-o.62
-o.7
1.7
-0.8
- 55.4
ι/"
(v)
Ιn the lirniting region
,o:
,/
R
Ξ
lΦ0 -n.r'
10οο + R
52.6 d'
vι(v)
lt is
a hard
L:
limiter wιh
0.7V
a
gain K
:
1
and
L, :
0.7 V,
4.92
4.η
a.
10
KΩ
ο.l mA diode
va
Chapter 4-36
Diodes exhibit 0.7 V drop at 0.1 mA
.
.
y" the
is
voltage drop across a diode when conducting
v^ = o? +
"
io
:
O.1
iD
Y-ln(
' \o.l mA/l
x e(v
o''])/vτ
rna
in mA For Vo > 0
o.7
o.t
1.4
1.9
o.73
o.332
1.46
3.t2
o.74
o.495
1.48
-3.955
0.75
0.718
t.5
-5.19
-
D, and D, arc oFF and Ι, = ο
I
:
'
1,
v'
= r K()
Vo: Vg* 5/, ιisι
V οι'Vou
/ (mΑ)
(v)
ofρoints for yΑ > 0
η(v)
ye(ν)
Vι= V,
+vD3+
Voι
0.4
-o
ο
ο.8
ο.8
0.5
o-0οoο3
-0
t.0
l.o
0.6
o.(n2
0.ω2
t.202
t.212
o.1
ο.l
ο-l
l.4l
l.9l
0.73
o.332
0.332
'!.792
3.452
0.715
ο.4ο6
ο.406
1.876
3.91
o.14
0.495
0.495
1.975
4.45
45
ο.6ο5
ο.605
2.O95
5.12
o.7
V,η
3kΩ
v, (v)
-2
_2ν
4
-5
For Currents
Ι'
v"(\η
v^
(mA)
0.8
0.5
-o
-ο
_ο_8
l.Ο
- t.o
0.6
o.(n2
t.2
-
o.4
Υ
= l-?l +
1
: 0.71 v, sο that v1 > 5.7 V
υ6:1.1l+lKxiDi
List of points for η< 0
(v)
υoι = o"Ι
l'βl. νD|
ir, > I mA
1
.2t
"ι'_
l'7t.
4
ι^=!+l.2a25
slooed'o
=.1
'
dιιι
4
(V)
Chapter
Table l for
υ1
4-37
> 5.? V
1,
}Ιere
ι1(V)
ι"(η
5.8
2,1325
6
2.1825
7
1.0325
8
3.2825
9
3.5325
10
3;1825
ι" = |o _ tX Io-] :6.9l
' vn/vr o,7 /0.025
Table 2
Foι z, > 0 diode D2 not conducting
'
X lο
|6Α
, so iD2
= 0
id l
+ τη",
ιo-6
o.521
1.524
I
Ιο_5
ο.584
O-4
o.642
1 .'7
lo-3
o.'l
o-2xto
:
_2.ω1
7.2xt0-
-2.O1
ιoz(Υ)
l0-6
o.527
2,524
-2.531
lo-1s
1ο'5
0.585
2.595
-2.625
-2.10
3.8x l0-r4
tο-4
o.642
-2.7U
-1.O24
-2.2
2-l
3
0.659
-2.459
3.459
- 2.5
33.5X l ο_8
0.5Χ lo_3
0.682
-3.r28
-4.624
-2.6
l8X tο_6
0.6x 1ο_3
0.687
-1.247
-5.087
-2.7
l x 1ο-3
l.7X lο-3
0.713
-3.413
-8.5 r6
1.5x 1o-3
2.2x tO-3
o.720
-3.43
-to
x10
t6
o.2x
12
through 1
kΩ
B_ D3 starts conducting when ther€
t.625
C- V6 = 2.7 V , the 0.7
42
2.M2
t7
ν : vΛ+2
4
3
20.14
2
ΙΙσ€
(1)
ior: ι,
(2)
"''o''u'
ιD!+. vA+2
l kσ
"',: v'h(T)
.^.
ιJ,
φ)
φuaιions
drop
across D3 clamps the
ιΙcrc Dι λ!s almsι consιa'ι
νoltlg€ dfup ard i|' i:
I
=4Κ.ioι*,ιoι
v
vo{9
8,717
caΙculation for ,,, ,, etc. \Ρhen D, and D] are condυcιing. Αlso ,, < 2.5y
:
v
is 0.5
volta8e across R3 and D3 controls ιhe cuΠent iD2
5.'Ι
v.|"(ξ)
i2 flows
resistor
voιn8e _ _
3-4
Y
Υo:νoι+lKXrDl
ιl,:3Κio'+vD|+1Κ.iD|
ιD2
lo
A- For small ip2,D3 is OFF, D2 is ON,
1.595
s.14
FoΓ table 3, use ιhe following
v,(v)
iρ2(A)
across D3
2.7
o.74
o-5Χ 1o-2
,odΑ)
_5313
2.7
o.717
2
16Α
v1(V)=4K
-1.291
I
Ye(v)
2.7 |
1.297
10
X 10
Table 3
o.291
1ο-
,,,
u,(v)
'or(V)
6.91
(5)
- loz
ιρ_3kΩΧio'
Points for _ l0 = ιrΙ = 6 are caιculated as shown
in tables 2 and 3
Diodes exhibiι a 0.7 v drop at a cυπent of l mA
ior(A)
:
'_?
4 6 810
-2
-3
-4
4.94
η \-/Λ
lοv.-.
ιo^/Σν'
o-ι
"_ΞF-Γ--
+
+
Ανerage (dc) νalue of output = 1O'Γ2
= 14.14 V
YrN)
η(v)
c
Chapter 4-38
4,95
lool
(a)
+1ον
:
v; =
=
(b)
1vr1
lv'(t -
+20ν
Noιγ
0v
Vtt + lV 21
and
]v]| +
(c)
" "c"''
lvrl
aι lhe end of T,
\rhere
v,
t:
T' υo=v,1
" ''""''
fi7)
: ιu,ιt'' r"l
: 20-Vt
:
+
2oΞv|+
|V1|_αv'-20(l)
|v2|_ 2α|v7| = 20
(2)
from (l ) & (2) we find that
(d)
(e)
Vt:2Vt
-ΠΙ_,"
Then usiηg (1) aηd noglecting α yι yieΙds
-ΓLΓ_.'
3|v2ι: 20Ξ|vr|:
Yr : 13.33 v
6'67
Υ
The result is:
+ 13.33
V
+10v
l0v
(c)
+18V
(f) Here there are ιwo differenι time constaηts
inνolνed. To calculat€ ιhe output leνels Ψe shalΙ
ageΓate the discharge and charge waνe forms.
During7l,
τ;o
: Vι e'τlRc
2ν
(h) Using a meιhοd
similar ιo that employed for
case (f) above we obtain
A t=Tι:T=ιa:v|
+13.33
_6'67
<-T1*'Ι'2'
Where for f<<
CR
yl = y,(l - T/cR)
\νhere α << 1
During the period T:
: V'(l -
α)
V
ν