Microelectronic Circuits
Adel S. Sedra
University of Waterloo
Kenneth C. Smith
University of Toronto
Tonny Chan Carusone
University of Toronto
Vincent Gaudet
University of Waterloo
New York Oxford
OXFORD UNIVERSITY PRESS
Chapter1
SolutionstoExerciseswithintheChapter
Ex:1.1 Whenoutputterminalsare open-circuited,asinFig.1.1a:
Forcircuita. v oc = v s (t )
Forcircuitb. v oc = is (t ) × Rs
Whenoutputterminalsareshort-circuited,asin Fig.1.1b:
Forcircuita. isc = v s (t ) Rs
Forcircuitb. isc = is (t )
Forequivalency
Rs is (t ) = v s (t )
vs (t) Figure1.1a is (t)
Figure1.1b
Ex:1.2
voc vs Rs isc
v oc = 10mV
isc = 10 µA
Rs = v oc isc = 10mV 10 µA = 1k
Ex:1.3 Usingvoltagedivider:
v o (t ) = v s (t ) × RL Rs + RL
vs (t) vo Rs RL
Given v s (t ) = 10mVand Rs = 1k
If RL = 100k
v o = 10mV × 100 100 + 1 = 9.9mV
If RL = 10k
v o = 10mV × 10 10 + 1 9.1mV
If RL = 1k
v o = 10mV × 1 1 + 1 = 5mV
If RL = 100
v o = 10mV × 100 100 + 1K 0.91mV
For v o = 0.8v s , RL RL + Rs = 0.8
Since Rs = 1k , RL = 4k
Ex:1.4 Usingcurrentdivider: Rs is 10 A RL io
io = is × Rs Rs + RL
Given is = 10 µA, Rs = 100k For RL = 1k , io = 10 µA × 100 100 + 1 = 9.9 µA For RL = 10k , io = 10 µA × 100
+ 10 9.1 µA For RL = 100k , io = 10 µA × 100 100 + 100 = 5 µA
For RL = 1M , io = 10 µA × 100K 100K + 1M
0.9 µA
For io = 0.8is , 100 100 + RL = 0.8
⇒ RL = 25k
Ex:1.5 f = 1 T = 1 10 3 = 1000Hz
ω = 2π f = 2π × 103 rad/s
Ex:1.6 (a) T = 1 f = 1 60 s = 16.7ms
(b) T = 1 f = 1 10 3 = 1000s
(c) T = 1 f = 1 106 s = 1 µs
Ex:1.7 If6MHzisallocatedforeachchannel, then470MHzto608MHzwillaccommodate
806 470 6 = 23channels
Sincethebroadcastbandstartswithchannel14,it willgofromchannel14tochannel36.
Ex:1.8 P = 1 T T 0 v 2 R dt
= 1 T × V 2 R × T = V 2 R
Alternatively, P = P1 + P3 + P5 +··· = 4V
Itcanbeshownbydirectcalculationthatthe infiniteseriesintheparentheseshasasumthat approaches π 2 /8;thus P becomes V 2 /R asfound fromdirectcalculation.
Fractionofenergyinfundamental
= 8/π 2 = 0.81
Fractionofenergyinfirstfiveharmonics
= 8 π 2 1 + 1 9 + 1 25 = 0.93
Fractionofenergyinfirstsevenharmonics
= 8 π 2 1 + 1 9 + 1 25 + 1 49 = 0.95
Fractionofenergyinfirstnineharmonics
= 8 π 2 1 + 1 9 + 1 25 + 1 49 + 1 81 = 0.96
Notethat90%oftheenergyofthesquarewaveis inthefirstthreeharmonics,thatis,inthe fundamentalandthethirdharmonic.
Ex:1.9 (a) D canrepresent15equally-spaced valuesbetween0and3.75V.Thus,thevaluesare spaced0.25Vapart.
v A = 0V ⇒ D = 0000
v A = 0.25V ⇒ D = 0000
v A = 1V ⇒ D = 0000
v A = 3.75V ⇒ D = 0000
(b)(i)1levelspacing:20 ×+0.25 =+0.25V
(ii)2levelspacings:21 ×+0.25 =+0.5V
(iii)4levelspacings:22 ×+0.25 =+1.0V
(iv)8levelspacings:23 ×+0.25 =+2.0V
(c)Theclosestdiscretevaluerepresentedby D is +1.25V;thus D = 0101.Theerroris-0.05V,or 0.05/1.3 × 100 =−4%.
Ex:1.10 Voltagegain = 20log100 = 40dB
Currentgain = 20log1000 = 60dB
Powergain = 10log Ap = 10log (Av Ai ) = 10log105 = 50dB
Ex:1.11 Pdc = 15 × 8 = 120mW
PL = (6/√2)2 1 = 18mW
Pdissipated = 120 18 = 102mW
η = PL Pdc × 100 = 18 120 × 100 = 15%
Ex:1.12 v o = 1 × 10 106 + 10 10 5 V = 10 µV
PL = v 2 o /RL = (10 × 10 6 )2 10 = 10 11 W
Withthebufferamplifier:
v o = 1 × Ri Ri + Rs × Av o × RL RL + Ro = 1 × 1 1 + 1 × 1 × 10 10 + 10 = 0.25V
PL = v 2 o RL = 0.252 10 = 6.25mW
Voltagegain= v o v s = 0.25V 1V = 0.25V/V
=−12dB
Powergain (Ap ) ≡ PL Pi where PL =6.25mWand Pi = v i i1 ,
v i = 0.5Vand
ii = 1V 1M + 1M = 0.5 µA
Thus, Pi = 0.5 × 0.5 = 0.25 µW and
Ap = 6.25 × 10 3 0.25 × 10 6 = 25 × 103
10log Ap = 44dB
Ex:1.13 Open-circuit(noload)outputvoltage=
Av o v i
Outputvoltagewithloadconnected
= Av o v i RL RL + Ro
0.8 = 1 Ro + 1 ⇒ Ro = 0.25k = 250
Ex:1.14 Av o = 40dB = 100V/V
PL = v 2 o RL = Av o v i RL RL + Ro 2 RL
= v 2 i × 100 × 1 1 + 1 2 1000 = 2.5 v 2 i
Pi = v 2 i Ri = v 2 i 10,000
Ap ≡ PL Pi = 2.5v 2 i 10 4 v 2 i = 2.5 × 104 W/W
10log Ap = 44dB
Ex:1.15 Withoutstage3(seefigure)
v L
v s = 1M 100k + 1M (10) 100k 100k + 1k
×(100) 100 100 + 1k
v L v s = (0.909)(10)(0.9901)(100)(0.0909) = 81.8V/V
ThisfigurebelongstoExercise1.15.
Ex:1.16 ReferthesolutiontoExample1.3inthe text.
v i1 v s = 0.909V/V
v i1 = 0.909 v s = 0.909 × 1 = 0.909mV
v i2 v s = v i2 v i1 × v i1 v s = 9.9 × 0.909 = 9V/V
v i2 = 9 × v S = 9 × 1 = 9mV v
× 9.9 × 0.909
= 818V/V
v i3 = 818 v s = 818 × 1 = 818mV
v L v s = v L v i3 × v i3 v i2 × v i2 v i1 × v i1 v s = 0.909 × 90.9 × 9.9 × 0.909 744V/V
v L = 744 × 1mV = 744mV
Ex:1.17 Usingvoltageamplifiermodel,the three-stageamplifiercanberepresentedas
vi Ri Ro Avovi
Ri = 1M
Ro = 10
Av o = Av 1 × Av 2 × Av 3 = 9.9 × 90.9 × 1 = 900V/V Theoverallvoltagegain
v o v s = Ri Ri + Rs × Av o × RL RL + Ro
For RL = 10 :
Overallvoltagegain = 1M 1M + 100K × 900 × 10 10 + 10 = 409V/V
For RL = 1000 :
Overallvoltagegain = 1M 1M + 100K × 900 × 1000 1000 + 10 = 810V/V
∴ Rangeofvoltagegainisfrom409V/Vto 810V/V.
Ex:1.18 ii io Ais ii RL Ro Rs Ri is ii = is Rs Rs + Ri
io = Ais ii Ro Ro + RL = Ais is Rs Rs + Ri Ro Ro + RL
Thus, io is = Ais Rs Rs + Ri Ro Ro + RL
Ex:1.19 Ri Ro Gmvi RL Ri vi vo vs v i = v s Ri Ri + Rs
v o = Gm v i (Ro RL )
= Gm v s Ri Ri + Rs (Ro RL ) Thus,
v o v s = Gm Ri Ri + Rs (Ro RL )
Ex:1.20 Usingthetransresistancecircuitmodel, thecircuitwillbe Ri Rs is ii Ro RL vo Rmii
ii is = Rs Ri + Rs
v o = Rm ii × RL RL + Ro
v o ii = Rm RL RL + Ro Now v o is = v o ii × ii is = Rm RL RL + Ro × Rs Ri + Rs = Rm Rs Rs + Ri × RL RL + Ro
Ex:1.21
v b = ib rπ + (β + 1)ib Re = ib [rπ + (β + 1)Re ]
But v b = v x and ib = ix ,thus Rin ≡ v x ix = v b ib = rπ + (β + 1)Re
Ex:1.22 f Gain 10Hz60dB 10kHz40dB 100kHz20dB 1MHz0dB
110 10 10 10 10 10 10 f (Hz)
Ex:1.23
Gain (dB) 20 dB/decade 3 dB frequency
RL Ro Ri Vi Vi Gm Vo CL
V2 Vs = Ri Rs + 1 sC + Ri = Ri Rs + Ri s s + 1 C (Rs + Ri ) whichisanHPSTCfunction.
f3dB = 1 2π C (Rs + Ri ) ≤ 100Hz
C ≥ 1 2π(1 + 9)103 × 100 = 0.16 µF
Ex:1.25 T = 50K
ni = BT 3/2 e Eg/(2kT )
= 7.3 × 1015 (50)3/2 e 1.12/(2×8.62×10 5 ×50) 9.6 × 10 39 /cm3
T = 350K
ni = BT 3/2 e Eg/(2kT )
= 7.3 × 1015 (350)3/2 e 1.12/(2×8.62×10 5 ×350)
= 4.15 × 1011 /cm3
Vo = Gm Vi [Ro RL CL ] = Gm Vi 1 Ro + 1 RL + sCL
Thus, Vo Vi = Gm 1 Ro + 1 RL × 1 1 + sCL 1 Ro + 1 RL
Vo Vi = Gm (RL Ro ) 1 + sCL (RL Ro ) whichisoftheSTCLPtype.
ω0 = 1 CL (RL Ro ) = 1 4.5 × 10 9 (103 Ro )
For ω0 tobeatleast wπ × 40 × 103 ,thehighest valueallowedfor Ro is
Ro = 103 2π × 40 × 103 × 103 × 4.5 × 10 9 1 = 103 1.131 1 = 7.64k
Thedcgainis
Gm (RL Ro )
Toensureadcgainofatleast40dB(i.e.,100), theminimumvalueof Gm is
⇒ RL ≥ 100/(103 7.64 × 103 ) = 113.1mA/V
Ex:1.24 RefertoFig.E1.24
Ex:1.26 ND = 1017 /cm3
FromExercise1.26, ni at T = 350K = 4.15 × 1011 /cm3
nn = ND = 1017 /cm3
pn ∼ = ni2 ND
= (4.15 × 1011 )2 1017 = 1.72 × 106 /cm3
Ex:1.27 At300K, ni = 1.5 × 1010 /cm3
pp = NA
Wantelectronconcentration
= np = 1.5 × 1010 106 = 1.5 × 104 /cm3
∴ NA = pp = ni2 np
= (1.5 × 1010 )2 1.5 × 104
= 1.5 × 1016 /cm3
Ex:1.28 (a) νn drift =−μn E
Herenegativesignindicatesthatelectronsmove inadirectionoppositeto E Weuse Sedra,Smith,Carusone,GaudetMicroelectronicCircuits,8thInternationalEdition©OxfordUniversityPress2021
νn-drift = 1350 × 1 2 × 10 4 ∵ 1 µm = 10 4 cm
= 6.75 × 106 cm/s = 6.75 × 104 m/s
(b)Timetakentocross2-µm
length = 2 × 10 6 6.75 × 104 30ps
(c)In n-typesilicon,driftcurrentdensity Jn is
Jn = qnμn E
= 1.6 × 10 19 × 1016 × 1350 × 1V 2 × 10 4
= 1.08 × 104 A/cm 2
(d)Driftcurrent In = AJn
= 0.25 × 10 8 × 1.08 × 104
= 27 µA
Theresistanceofthebaris
R = ρ × L A
= qnμn × L A
= 1.6 × 10 19 × 1016 × 1350 × 2 × 10 4 0.25 × 10 8
= 37.0k
Alternatively,wemaysimplyusethepreceding resultforcurrentandwrite
R = V /In = 1V/27 µA = 37.0k
Notethat0.25 µm2 = 0.25 × 10 8 cm2
Ex:1.29 Jn = qDn dn(x) dx
FromFig.E1.29,
n0 = 1017 /cm3 = 105 /(µm)3
Dn = 35cm2 /s = 35 × (104 )2 (µm)2 /s
= 35 × 108 (µm)2 /s
dn dx = 105 0 0.5 = 2 × 105 µm 4
Jn = qDn dn(x) dx
= 1.6 × 10 19 × 35 × 108 × 2 × 105
= 112 × 10 6 A/µm2
= 112 µA/µm2
For In = 1mA = Jn × A
⇒ A = 1mA Jn = 103 µA 112 µA/(µm)2 9 µm2
Ex:1.30 UsingEq.(1.44), Dn μn = Dp μp = VT
Dn = μn VT = 1350 × 25.9 × 10 3
∼ = 35cm2 /s
Dp = μp VT = 480 × 25.9 × 10 3
∼ = 12.4cm2 /s
Ex:1.31 Equation(1.49)
W = 2 s q 1 NA + 1 ND V0
= 2 s q NA + ND NA ND V0
W 2 = 2 s q NA + ND NA ND V0
V0 = 1 2 q s NA ND NA + ND W 2
Ex:1.32 Ina p+ n diode NA ND
Equation(1.49) W = 2 s q 1 NA + 1 ND V0
Wecanneglecttheterm 1 NA ascomparedto 1 ND , thus
W 2 s qND · V0
Equation(1.50) xn = W NA NA + ND W NA NA
= W
Equation(1.51), xp = W ND NA + ND since NA ND W ND NA = W NA ND
Equation(1.52), QJ = Aq NA ND NA + ND W Aq NA ND NA W = AqND W
Equation(1.53), QJ = A 2 s q NA ND NA + ND V0
A 2 s q NA ND NA V0 since NA ND
= A 2 s qND V0
Ex:1.33 InExample1.10, NA = 1018 /cm3 and ND = 1016 /cm3
Inthe n-regionofthis pn junction
nn = ND = 1016 /cm3
pn = n2 i nn = (1.5 × 1010 )2 1016 = 2.25 × 104 /cm3
Asonecanseefromaboveequation,toincrease minority-carrierconcentration (pn ) byafactorof 2,onemustlower ND (= nn )byafactorof2.
Ex:1.34
Equation(1.64) IS = Aqn2 i Dp Lp ND + Dn Ln NA since Dp Lp and Dn Ln haveapproximately
similarvalues,if NA ND ,thentheterm Dn Ln NA
canbeneglectedascomparedto Dp Lp ND
∴ IS ∼ = Aqn2 i Dp Lp ND
Ex:1.35 IS = Aqn2 i Dp Lp ND + Dn Ln NA
= 10 4 × 1.6 × 10 19 × (1.5 × 1010 )2 × ⎛ ⎜ ⎜ ⎝ 10 5 × 10 4 × 1016 2 + 18 10 × 10 4 × 1018
= 1.46 × 10 14 A
I = IS (e V /V T 1) IS e V /V T = 1.45 × 10 14 e 0.605/(25.9×10 3 )
= 0.2mA
Ex:1.36 W = 2 s q 1 NA + 1 ND (V0 VF )
= 2 × 1.04 × 10 12 1.6 × 10 19 1 1018 + 1 1016 (0.814 0.605)
= 1.66 × 10 5 cm = 0.166 µm
Ex:1.37 W = 2 s q 1 NA + 1 ND (V0 + VR ) = 2 × 1.04 × 10 12 1.6 × 10 19 1 1018 + 1 1016 (0.814 + 2)
= 6.08 × 10 5 cm = 0.608 µm
UsingEq.(1.52),
QJ = Aq NA ND NA + ND W
= 10 4 × 1.6 × 10 19 1018 × 1016 1018 + 1016 × 6.08 × 10 5 cm
= 9.63pC
Reversecurrent I = IS = Aqn2 i Dp Lp ND + Dn Ln NA
= 10 14 × 1.6 × 10 19 × (1.5 × 1010 )2 × 10 5 × 10 4 × 1016 + 18 10 × 10 4 × 1018
= 7.3 × 10 15 A
Ex:1.38 Equation(1.69),
Cj0 = A s q 2 NA ND NA + ND 1 V0
= 10 4 1.04 × 10 12 × 1.6 × 10 19 2 1018 × 1016 1018 + 1016 1 0.814
= 3.2pF
Equation(1.70),
Cj = Cj0 1 + VR V0 = 3.2 × 10 12 1 + 2 0.814
= 1.72pF
Ex:1.39 Cd = dQ dV = d dV (τ T I )
= d dV [τ T × IS (e V /V T 1)]
= τ T IS d dV (e V /V T 1)
= τ T IS 1 VT e V /V T
= τ T VT × IS e V /V T
∼ = τ T VT I
Ex:1.40 Equation(1.73),
τ p = L2 p Dp = (5 × 10 4 )2 10 = 25ns
Equation(3.57),
Cd = τ T VT I
InExample1.6, NA = 1018 /cm3 , ND = 1016 /cm3
Assuming NA ND ,
τ T τ p = 25ns
∴ Cd = 25 × 10 9 25.9 × 10 3 0.1 × 10 3
= 96.5pF
SolutionstoEnd-of-ChapterProblems
1.1 (a) V = IR = 5mA × 1k = 5V
P = I 2 R = (5mA)2 × 1k = 25mW
(b) R = V /I = 5V/1mA = 5k
P = VI = 5V × 1mA = 5mW
(c) I = P/V = 100mW/10V = 10mA
R = V /I = 10V/10mA = 1k
(d) V = P/I = 1mW/0.1mA = 10V
R = V /I = 10V/0.1mA = 100k
(e) P = I 2 R ⇒ I = P/R
I = 1000mW/1k = 31.6mA
V = IR = 31.6mA × 1k = 31.6V
Note: V,mA,k ,andmWconstitutea consistentsetofunits.
1.2 (a) I = V R = 5V 1k = 5mA
(b) R = V I = 5V 1mA = 5k
(c) V = IR = 0.1mA × 10k = 1V
(d) I = V R = 1V 100 = 0.01A = 10mA
Note: Volts,milliamps,andkilohmsconstitutea consistentsetofunits.
1.3 (a) P = I 2 R = (20 × 10 3 )2 × 1 × 103 = 0.4W
Thus, R shouldhavea 1 2 -Wrating.
(b) P = I 2 R = (40 × 10 3 )2 × 1 × 103 = 1.6W
Thus,theresistorshouldhavea2-Wrating.
(c) P = I 2 R = (1 × 10 3 )2 × 100 × 103 = 0.1W
Thus,theresistorshouldhavea 1 8 -Wrating.
(d) P = I 2 R = (4 × 10 3 )2 × 10 × 103 = 0.16W
Thus,theresistorshouldhavea 1 4 -Wrating.
(e) P = V 2 /R = 202 /(1 × 103 ) = 0.4W
Thus,theresistorshouldhavea 1 2 -Wrating.
(f) P = V 2 /R = 112 /(1 × 103 ) = 0.121W
Thus,aratingof 1 8 Wshouldtheoretically suffice,though 1 4 Wwouldbeprudenttoallow forinevitabletolerancesandmeasurementerrors.
1.4 Seefigureonnextpage,whichshowshowto realizetherequiredresistancevalues.
1.5 Shuntingthe10k byaresistorofvalueof R resultsinthecombinationhavingaresistance
Req ,
Req = 10R R + 10
Thus,fora1%reduction,
R R + 10 = 0.99 ⇒ R = 990k
Fora5%reduction,
R R + 10 = 0.95 ⇒ R = 190k
Fora10%reduction,
R R + 10 = 0.90 ⇒ R = 90k
Fora50%reduction,
R
R + 10 = 0.50 ⇒ R = 10k
Shuntingthe10k by (a)1M resultsin
Req = 10 × 1000 1000 + 10 = 10 1.01 = 9.9k a1%reduction;
(b)100k resultsin
Req = 10 × 100 100 + 10 = 10 1.1 = 9.09k
a9.1%reduction;
(c)10k resultsin
Req = 10 10 + 10 = 5k
a50%reduction.
1.6 Usevoltagedividertofind VO VO = 5 2 2 + 3 = 2V
Equivalentoutputresistance RO is RO = (2k 3k ) = 1.2k
ThisfigurebelongstoProblem 1.4 Allresistorsare5k
Theextremevaluesof VO for ±5%tolerance resistorare
VOmin = 5 2(1 0.05) 2(1 0.05) + 3(1 + 0.05) = 1.88V
VOmax = 5 2(1 + 0.05) 2(1 + 0.05) + 3(1 0.05) = 2.12V
Theextremevaluesof RO for ±5%tolerance resistorsare1.2 × 1.05 = 1.26k and 1.2 × 0.95 = 1.14k
1.7 VO = VDD R2 R1 + R2 Tofind RO ,weshort-circuit VDD andlookback intonodeX,
10 // 10 // 10 3.33 k
Voltagegenerated:
+3V[twoways:(a)and(c)with(c)havinglower outputresistance]
+4.5V(b)
+6V[twoways:(a)and(d)with(d)havinga loweroutputresistance]
Tomake RO = 3.33,weaddaseriesresistanceof approximately200 ,asshownbelow,
Toincrease VO to10.00V,weshuntthe10-k resistorbyaresistor R whosevalueissuchthat 10 R = 2 × 4.7.
1.11 Connectaresistor R inparallelwith RL
Tomake IL = I /4(andthusthecurrentthrough R,3 I /4), R shouldbesuchthat
6 I /4 = 3 IR/4
⇒ R = 2k
4
1.12 Theparallelcombinationoftheresistorsis
R where 1 R = N i=1 1/Ri
Thevoltageacrossthemis
V = I × R = I N i=1 1/Ri
Thus,thecurrentinresistor Rk is
Ik = V /Rk = I /Rk N i=1 1/Ri
Tomakethecurrentthrough R equalto0.2 I ,we shunt R byaresistance R1 havingavaluesuch thatthecurrentthroughitwillbe0.8 I ;thus
0.2 IR = 0.8 IR1 ⇒ R1 = R 4
Theinputresistanceofthedivider, Rin ,is
Rin = R R1 = R R 4 = 1 5 R
Nowif R1 is10%toohigh,thatis,if
R1 = 1.1 R 4
theproblemcanbesolvedintwoways:
(a)Connectaresistor R2 across R1 ofvaluesuch that R2 R1 = R/4,thus
R2 (1.1R/4)
R2 + (1.1R/4) = R 4
+ 1.1R 4 ⇒ R2 = 11R 4 = 2.75R
R
=
(b)Connectaresistorinserieswiththeload resistor R soastoraisetheresistanceoftheload branchby10%,therebyrestoringthecurrent divisionratiotoitsdesiredvalue.Theadded seriesresistancemustbe10%of R (i.e.,0.1R). 0.1R 1.1R 4 Rin R 0.8 I 0.2 I I
Rin = 1.1R 1.1R 4 = 1.1R 5 thatis,10%higherthanincase(a).
1.14 For RL = 10k ,whensignalsource generates0 0.5mA,avoltageof0 2Vmay appearacrossthesource
Tolimit v s ≤ 1V,thenetresistancehastobe ≤ 2k .Toachievethiswehavetoshunt RL with aresistorRsothat (R RL ) ≤ 2k
R RL ≤ 2k
RRL R + RL ≤ 2k
For RL = 10k
R ≤ 2.5k
Theresultingcircuitneedsonlyoneadditional resistanceof2k inparallelwith RL sothat
v s ≤ 1V.Thecircuitisacurrentdivider,andthe currentthrough RL isnow0–0.1mA.
0 0.5 mA is vs RL
1.15 (a)Betweenterminals1and2:
(b)Sameprocedureisusedfor(b)toobtain
(c)Betweenterminals1and3,theopen-circuit voltageis1.5V.Whenweshortcircuitthe voltagesource,weseethattheThévenin resistancewillbezero.Theequivalentcircuitis then
Now,whenaresistanceof3k isconnected betweennode4andground, Sedra,Smith,Carusone,GaudetMicroelectronicCircuits,8thInternationalEdition©OxfordUniversityPress2021
I = 0.77 12.31 + 3 = 0.05mA 1.17
V
1
3 I2 I1 I3 V
(a)Nodeequationatthecommonmodeyields
I3 = I1 + I2
Usingthefactthatthesumofthevoltagedrops across R1 and R3 equals10V,wewrite
10 = I1 R1 + I3 R3
= 10 I1 + (I1 + I2 ) × 2
= 12 I1 + 2 I2
Thatis,
12 I1 + 2 I2 = 10 (1)
Similarly,thevoltagedropsacross R2 and R3 add upto5V,thus
5 = I2 R2 + I3 R3
= 5 I2 + (I1 + I2 ) × 2
whichyields
2 I1 + 7 I2 = 5 (2)
Equations(1)and(2)canbesolvedtogetherby multiplyingEq.(2)by6:
12 I1 + 42 I2 = 30 (3)
Now,subtractingEq.(1)fromEq.(3)yields
40 I2 = 20
⇒ I2 = 0.5mA
SubstitutinginEq.(2)gives
2 I1 = 5 7 × 0.5mA
⇒ I1 = 0.75mA
I3 = I1 + I2
= 0.75 + 0.5 = 1.25mA
V = I3 R3
= 1.25 × 2 = 2.5V
Tosummarize:
I1 = 0.75mA I2 = 0.5mA
I3 = 1.25mA V = 2.5V
(b)Anodeequationatthecommonnodecanbe writtenintermsof V as 10 V R1 + 5 V R2 = V R3
Thus, 10 V 10 + 5 V 5 = V 2
⇒ 0.8V = 2
⇒ V = 2.5V
Now, I1 , I2 ,and I3 canbeeasilyfoundas I1 = 10 V 10 = 10 2.5 10 = 0.75mA I2 = 5 V 5 = 5 2.5 5 = 0.5mA I3 = V R3 = 2.5 2 = 1.25mA
Method(b)ismuchpreferred,beingfaster,more insightful,andlesspronetoerrors.Ingeneral, oneattemptstoidentifythelowestpossible numberofvariablesandwritethecorresponding minimumnumberofequations.
1.18 FindtheThéveninequivalentofthecircuit totheleftofnode1.
Betweennode1andground, RTh = (1k 1.2k ) = 0.545k
VTh = 10 × 1.2 1 + 1.2 = 5.45V
FindtheThéveninequivalentofthecircuittothe rightofnode2.
Betweennode2andground, RTh = 9.1k 11k = 4.98k
VTh = 10 × 11 11 + 9.1 = 5.47V
Theresultingsimplifiedcircuitis
5 0.545 k 4.98 k
V 5.47 V
I5 = 5.47 5.45 4.98 + 2 + 0.545 = 2.66 µA
V5 = 2.66 µA × 2k = 5.32mV
1.19 WefirstfindtheThéveninequivalentofthe sourcetotherightof v O .
V = 4 × 1 = 4V
Then,wemayredrawthecircuitinFig.P1.19as shownbelow 5 V 4 V 3 k1 k vo
Then,thevoltageat v O isfoundfromasimple voltagedivision.
v O = 4 + (5 4) × 1 3 + 1 = 4.25V
1.20 RefertoFig.P1.20.Usingthevoltage dividerruleattheinputside,weobtain
v π v s = rπ rπ + Rs (1)
Attheoutputside,wefind v o bymultiplyingthe current gm v π bytheparallelequivalentof ro and RL ,
v o =−gm v π (ro RL )(2)
Finally, v o /v s canbeobtainedbycombiningEqs. (1)and(2)as
v o v s =− rπ rπ + Rs gm (ro RL )
1.21 (a) T = 10 4 ms = 10 7 s
f = 1 T = 107 Hz
ω = 2π f = 6.28 × 107 rad/s
(b) f = 1GHz = 109 Hz
T = 1 f = 10 9 s
ω = 2π f = 6.28 × 109 rad/s
(c) ω = 6.28 × 102 rad/s
f = ω 2π = 102 Hz
T = 1 f = 10 2 s
(d) T = 10s
f = 1 T = 10 1 Hz
ω = 2π f = 6.28 × 10 1 rad/s
(e) f = 60Hz
T = 1 f = 1.67 × 10 2 s
ω = 2π f = 3.77 × 102 rad/s
(f) ω = 1krad/s = 103 rad/s
f = ω 2π = 1.59 × 102 Hz
T = 1 f = 6.28 × 10 3 s
(g) f = 1900MHz = 1.9 × 109 Hz
T = 1 f = 5.26 × 10 10 s
ω = 2π f = 1.194 × 1010 rad/s
1.22 (a) Z = R + 1 jω C = 103 + 1 j2π × 10 × 103 × 10 × 10 9
= (1 j1.59) k
(b) Y = 1 R + jω C
= 1 104 + j2π × 10 × 103 × 0.01 × 10 6
= 10 4 (1 + j6.28)
Z = 1 Y = 104 1 + j6.28
= 104 (1 j6.28) 1 + 6.282
= (247.3 j1553)
(c) Y = 1 R + jω C = 1 100 × 103 + j2π × 10 × 103 × 100 × 10 12
= 10 5 (1 + j0.628)
Z = 105 1 + j0.628
= (71.72 j45.04) k
(d) Z = R + jω L
= 100 + j2π × 10 × 103 × 10 × 10 3
= 100 + j6.28 × 100
= (100 + j628), 1.23 (a) Z = 1k atallfrequencies
(b) Z = 1 /jω C =− j 1 2π f × 10 × 10 9
At f = 60Hz, Z =− j265k
At f = 100kHz, Z =− j159
At f = 1GHz, Z =− j0.016
(c) Z = 1 /jω C =− j 1 2π f × 10 × 10 12
At f = 60Hz, Z =− j0.265G
At f = 100kHz, Z =− j0.16M
At f = 1GHz, Z =− j15.9
(d) Z = jω L = j2π fL = j2π f × 10 × 10 3
At f = 60Hz, Z = j3.77
At f = 100kHz, Z = j6.28k
At f = 1GHz, Z = j62.8M
(e) Z = jω L = j2π fL = j2π f (1 × 10 6 )
f = 60Hz, Z = j0.377m
f = 100kHz, Z = j0.628
f = 1GHz, Z = j6.28k
1.24 Y = 1 jω L + jω C = 1 ω 2 LC jω L
⇒ Z = 1 Y = jω L 1 ω 2 LC
Thefrequencyatwhich |Z | =∞ isfoundletting thedenominatorequalzero:
1 ω 2 LC = 0 ⇒ ω = 1 √LC
Atfrequenciesjustbelowthis, ∠Z =+90◦ . Atfrequenciesjustabovethis, ∠Z =−90◦
Sincetheimpedanceisinfiniteatthisfrequency, thecurrentdrawnfromanidealvoltagesourceis zero.
1.25 is Rs Rs vs Thévenin equivalent Norton equivalent
v oc = v s isc = is v s = is Rs
Thus, Rs = v oc isc (a) v s = v oc = 1V is = isc = 0.1mA
Rs = v oc isc = 1V 0.1mA = 10k (b) v s = v oc = 0.1V is = isc = 1 µA
Rs = v oc isc = 0.1V 1 µA = 0.1M = 100k
v o
vs vo
v s = RL RL + Rs
v o = v s 1 + Rs RL
Thus, v s 1 + Rs 100 = 40 (1) and v s 1 + Rs 10 = 10 (2)
DividingEq.(1)byEq.(2)gives
1 + (Rs /10)
1 + (Rs /100) = 4
⇒ RS = 50k
SubstitutinginEq.(2)gives
v s = 60mV
TheNortoncurrent is canbefoundas
is = v s Rs = 60mV 50k = 1.2 µA
1.27 Thenominalvaluesof VL and IL are givenby
VL = RL RS + RL VS IL = VS RS + RL
Aftera10%increasein RL ,thenewvalueswillbe
VL = 1.1RL RS + 1.1RL VS
IL = VS RS + 1.1RL
(a)Thenominalvaluesare
VL = 200 5 + 200 × 1 = 0.976V
IL = 1 5 + 200 = 4.88µA
Aftera10%increasein RL ,thenewvalueswillbe
VL = 1.1 × 200 5 + 1.1 × 200 = 0.978V
IL = 1 5 + 1.1 × 200 = 4.44µA
Thesevaluesrepresenta0.2%and9%change, respectively.Sincetheloadvoltageremains relativelymoreconstantthantheloadcurrent,a Théveninsourceismoreappropriatehere.
(b)Thenominalvaluesare
VL = 50 5 + 50 × 1 = 0.909V
IL = 1 5 + 50 = 18.18mA
Aftera10%increasein RL ,thenewvalueswillbe
VL = 1.1 × 50 5 + 1.1 × 50 = 0.917V
IL = 1 5 + 1.1 × 50 = 16.67mA
Thesevaluesrepresenta1%and8%change, respectively.Sincetheloadvoltageremains relativelymoreconstantthantheloadcurrent,a Théveninsourceismoreappropriatehere.
(c)Thenominalvaluesare
VL = 0.1 2 + 0.1 × 1 = 47.6mV
IL = 1 2 + 0.1 = 0.476mA
Aftera10%increasein RL ,thenewvalueswillbe
VL = 1.1 × 0.1 2 + 1.1 × 0.1 = 52.1mV
IL = 1 2 + 1.1 × 0.1 = 0.474mA
Thesevaluesrepresenta9%and0.4%change, respectively.Sincetheloadcurrentremains relativelymoreconstantthantheloadvoltage,a Nortonsourceismoreappropriatehere.The Nortonequivalentcurrentsourceis
IS = VS RS = 1 2 = 0.5mA
(d)Thenominalvaluesare
VL = 16 150 + 16 × 1 = 96.4mV
IL = 1
150 + 16 = 6.02mA
Aftera10%increasein RL ,thenewvalueswillbe
VL = 1.1 × 16
150 + 1.1 × 16 = 105mV
IL = 1
150 + 1.1 × 16 = 5.97mA
Thesevaluesrepresenta9%and1%change, respectively.Sincetheloadcurrentremains relativelymoreconstantthantheloadvoltage,a Nortonsourceismoreappropriatehere.The Nortonequivalentcurrentsourceis
= VS RS = 1 150 = 6.67mA
v 2 S R2 L (RL + RS )2 × 1 RL
v 2 S RL (RL + RS )2
Sincewearetoldthatthepowerdeliveredtoa 16 speakerloadis75%ofthepowerdelivered toa32 speakerload, PL (RL = 16 ) = 0.75 × PL (RL = 32 ) 16 (RS + 32)2 = 0.75 × 32 (RS + 32)2 √16 RS + 32 = √24 RS + 32
(√24 = √16)RS = √16 × 32 √24 × 16 0.9RS = 49.6
Open-circuit (io 0) voltage 0 Rs vs vs vo is io Slope Rs Short-circuit (vo 0) current 1.31 Rs RL RL is vs vo vo Rs io io
RL representstheinputresistanceoftheprocessor
For v o = 0.95v s 0.95 = RL RL + Rs ⇒ RL = 19Rs For io = 0.95is 0.95 = Rs Rs + RL ⇒ RL = RS /19 1.32
Case ω (rad/s) f (Hz) T (s)
a 3.14 × 1010 5 × 109 0.2 × 10 9
b 2 × 109 3.18 × 108 3.14 × 10 9
c 6.28 × 1010 1 × 1010 1 × 10 10
d 3.77 × 102 60 1.67 × 10 2
e 6.28 × 104 1 × 104 1 × 10 4
f 6.28 × 105 1 × 105 1 × 10 5
1.29 Theobservedoutputvoltageis1mV/ ◦ C, whichisonehalfthevoltagespecifiedbythe sensor,presumablyunderopen-circuitconditions: thatis,withoutaloadconnected.Itfollowsthat thatsensorinternalresistancemustbeequalto RL ,thatis,5k 1.30 vs vo Rs Rs io is vo io v o = v s io Rs
1.33 (a) v = 10sin(2π × 103 t ),V
(b) v = 120√2sin(2π × 60),V
(c) v = 0.1sin(2000t ),V
(d) v = 0.1sin(2π × 103 t ),V
1.34 Comparingthegivenwaveformtothat describedbyEq.(1.2),weobservethatthegiven waveformhasanamplitudeof0.5V(1V peak-to-peak)anditslevelisshiftedupby0.5V (thefirsttermintheequation).Thusthe waveformlooksasfollows:
1.37 Thermsvalueofasymmetricalsquare wavewithpeakamplitude ˆ V issimply ˆ V .Taking theroot-mean-squareofthefirst5sinusoidal termsinEq.(1.2)givesanrmsvalueof, 4
= 0.980 ˆ V whichis2%lowerthanthermsvalueofthe squarewave.
1.38 Iftheamplitudeofthesquarewaveis Vsq , thenthepowerdeliveredbythesquarewavetoa resistance R willbe V 2 sq /R .Ifthispoweristobe equaltothatdeliveredbyasinewaveofpeak amplitude V ,then
Averagevalue = 0.5V
Peak-to-peakvalue = 1V
Lowestvalue = 0V
Highestvalue = 1V
Period T = 1 f0 = 2π ω0 = 10 3 s
Frequency f = 1 I = 1kHz
1.35 (a) Vpeak = 117 × √2 = 165V
(b) Vrms = 33.9/√2 = 24V
(c) Vpeak = 220 × √2 = 311V
(d) Vpeak = 220 × √2 = 311kV
1.36 Thetwoharmonicshavetheratio 126/98 = 9/7.Thus,thesearethe7thand9th harmonics.FromEq.(1.2),wenotethatthe amplitudesofthesetwoharmonicswillhavethe ratio7to9,whichisconfirmedbythe measurementreported.Thusthefundamentalwill haveafrequencyof98/7,or14kHz,andpeak amplitudeof63 × 7 = 441mV.Thermsvalueof thefundamentalwillbe441/√2 = 312mV.To findthepeak-to-peakamplitudeofthesquare wave,wenotethat4V/π = 441mV.Thus,
Peak-to-peakamplitude
= 2V = 441 × π 2 = 693mV
Period T = 1 f = 1 14 × 103 = 71.4 µs
Thus, Vsq = ˆ V /√2.Thisresultisindependentof frequency. 1.39 Decimal Binary 0 0
1.40 (a)For N bitstherewillbe2N possible levels,from0to VFS .Thustherewillbe(2N 1) discretestepsfrom0to VFS withthestepsize givenby
Stepsize = VFS 2N 1
Thisistheanalogchangecorrespondingtoa changeintheLSB.Itisthevalueofthe resolutionoftheADC.
(b)Themaximumerrorinconversionoccurs whentheanalogsignalvalueisatthemiddleofa step.Thusthemaximumerroris
1 2 × stepsize = 1 2 VFS 2N 1
Thisisknownasthequantizationerror.
(c) 5V 2N 1 ≤ 2mV
2N 1 ≥ 2500
2N ≥ 2501 ⇒ N = 12, For N = 12,
Resolution = 5 212 1 = 1.2mV
=
=
Notethattherearetwopossiblerepresentations ofzero:0000and1000.Fora0.5-Vstepsize, analogsignalsintherange ±3.5Vcanbe represented.
Input Steps Code
+5 0101
0101
6 1110
1.42 (a)When bi = 1,the ithswitchisin position1andacurrent(Vref /2i R)flowstothe output.Thus iO willbethesumofallthecurrents correspondingto“1”bits,thatis,
iO = Vref R b1 21 + b2 22 +···+ bN 2N
(b) bN istheLSB
b1 istheMSB
(c) iOmax = 10V 10k 1 21 + 1 22 + 1 23 + 1 24 + 1 25 + 1 26 + 1 27 + 1 28
= 0.99609375mA
CorrespondingtotheLSBchangingfrom0to1 theoutputchangesby (10/10) × 1/28 = 3.91 µA.
1.43 Therewillbe44,100samplespersecond witheachsamplerepresentedby16bits.Thusthe throughputorspeedwillbe44,100 × 16 = 7.056 × 105 bitspersecond.
1.44 Eachpixelrequires8 + 8 + 8 = 24bitsto representit.Wewillapproximateamegapixelas 106 pixels,andaGbitas109 bits.Thus,each imagerequires24 × 10 × 106 = 2.4 × 108 bits. Thenumberofsuchimagesthatfitin16Gbitsof memoryis
2.4 × 108 16 × 109 = 66.7 = 66
1.45 (a) Av = v O v I = 10V 100mV = 100V/V
or20log100 = 40dB
Ai = iO iI = v O /RL iI = 10V/100 100 µA = 0.1A 100 µA
= 1000A/A
or20log1000 = 60dB
Ap = v O iO v I iI = v O v I × iO iI = 100 × 1000
= 105 W/W
or10log105 = 50dB
(b) Av = v O v I = 1V 10 µV = 1 × 105 V/V
or20log1 × 105 = 100dB
Ai = iO iI = v O /RL iI = 1V/10k 100nA
= 0.1mA 100nA = 0.1 × 10 3 100 × 10 9 = 1000A/A
or20log Ai = 60dB
Ap = v O iO v I iI = v O v I × iO iI
= 1 × 105 × 1000
= 1 × 108 W/W
or10log AP = 80dB
(c) Av = v O v i = 5V 1V = 5V/V
or20log5 = 14dB
Ai = iO iI = v O /RL iI = 5V/10 1mA = 0.5A 1mA = 500A/A
or20log500 = 54dB
Ap = v O iO v I iI = v O v I × iO iI
= 5 × 500 = 2500W/W
or10log Ap = 34dB
1.46 For ±5Vsupplies:
Thelargestundistortedsine-waveoutputisof 4-Vpeakamplitudeor4/√2 = 2.8Vrms .Input neededis14mVrms .
For ±10-Vsupplies,thelargestundistorted sine-waveoutputisof9-Vpeakamplitudeor 6.4Vrms .Inputneededis32mVrms
For ±15-Vsupplies,thelargestundistorted sine-waveoutputisof14-Vpeakamplitudeor
9.9Vrms .Theinputneededis9.9V/200 = 49.5mVrms
Av = v o v i = 2.2 0.2 = 11V/V
or20log11=20.8dB
Ai = io ii = 2.2V/100 1mA = 22mA 1mA = 22A/A
or20log Ai = 26.8dB
Ap = po pi = (2.2/√2)2 /100 0.2 √2 × 10 3 √2
= 242W/W
or10log AP = 23.8dB
Supplypower = 2 × 3V ×20mA = 120mW
Outputpower = v 2 orms RL = (2.2/√2)2 100 = 24.2mW
Inputpower = 24.2 242 = 0.1mW(negligible)
Amplifierdissipation Supplypower Output power
= 120 24.2 = 95.8mW
Amplifierefficiency = Outputpower supplypower × 100 = 24.2 120 × 100 = 20.2%
1.48 v o = Av o v i RL RL + Ro = Av o v s Ri Ri + Rs RL RL + Ro vi vi vs Ri Rs RL Av ovi Ro
Thus,
v o v s = Av o Ri Ri + Rs RL RL + Ro
(a) Av o = 100, Ri = 10Rs , RL = 10Ro :
v o v s = 100 × 10Rs 10Rs + Rs × 10Ro 10Ro + Ro = 82.6V/Vor20log82.6 = 38.3dB
(b) Av o = 100, Ri = Rs , RL = Ro :
v o v s = 100 × 1 2 × 1 2 = 25V/Vor20log25 = 28dB
(c) Av o = 100V/V, Ri = Rs /10, RL = Ro /10:
v o v s = 100 Rs /10 (Rs /10) + Rs Ro /10 (Ro /10) + Ro = 0.826V/Vor20log0.826 =−1.7dB
1.49 (a)
v o v s = v i v s × v o v i
ThisfigurebelongstoProblem 1.49.
= 1 5 + 1 × 100 × 100 200 + 100 = 5.56V/V
Muchoftheamplifier’s100V/Vgainislostin thesourceresistanceandamplifier’soutput resistance.Ifthesourcewereconnecteddirectly totheload,thegainwouldbe
v o v s = 0.1 5 + 0.1 = 0.0196V/V
Thisisafactorof284× smallerthanthegain withtheamplifierinplace!
(b)
Theequivalentcurrentamplifierhasadependent currentsourcewithavalueof 100V/V 200 × ii = 100V/V 200 × 1000 × v i = 500 × ii
Thus, io is = ii is × io ii = 5 5 + 1 × 500 × 200 200 + 100 = 277.8A/A
Usingthevoltageamplifiermodel,thecurrent gaincanbefoundasfollows, io is = ii is × v i ii × io v i = 5 5 + 1 × 1000 × 100V/V 200 + 100 = 277.8A/A
1.50 InExample1.3,whenthefirstandthe secondstagesareinterchanged,thecircuitlooks likethefigureabove,and
v i1 v s = 100k 100k + 100k = 0.5V/V
io
Figure1
Figure2
ThisfigurebelongstoProblem 1.50
Av 1 = v i2 v i1 = 100 × 1M 1M + 1k = 99.9V/V
Av 2 = v i3 v i2 = 10 × 10k 10k + 1k = 9.09V/V
Av 3 = v L v i3 = 1 × 100 100 + 10 = 0.909V/V
Totalgain = Av = v L v i1 = Av 1 × Av 2 × Av 3
= 99.9 × 9.09 × 0.909 = 825.5V/V
Thevoltagegainfromsourcetoloadis
v L v s = v L v i1 × v i1 v S = Av v i1 v S
= 825.5 × 0.5
= 412.7V/V
Theoverallvoltagehasreducedappreciably.This isbecausetheinputresistanceofthefirststage, Rin ,iscomparabletothesourceresistance Rs .In Example1.3theinputresistanceofthefirststage ismuchlargerthanthesourceresistance.
1.51 Theequivalentcircuitattheoutputsideofa currentamplifierloadedwitharesistance RL is shown.Since
io = (Ais ii ) Ro Ro + RL wecanwrite
1 = (Ais ii ) Ro Ro + 1 (1) and
0.5 = (Ais ii ) Ro Ro + 12 (2)
DividingEq.(1)byEq.(2),wehave RL Ro Aisii io
2 = Ro + 12 Ro + 1 ⇒ Ro = 10k
Ais ii = 1 × 10 + 1 10 = 1.1mA
1.52
Thecurrentgainis
io ii = Rm Ro + RL
= 5000 10 + 1000
= 4.95A/A = 13.9dB
Thevoltagegainis
v o v s = ii v s × io ii × v o io
= 1 Rs + Ri × io ii × RL
= 1 1000 + 100 × 4.95 × 1000
= 4.90V/V = 13.8dB
Thepowergainis
v o io
v s ii = 4.95 × 4.90
= 24.3W/W = 27.7dB
1.53
Gm = 60mA/V
Ro = 20k
RL = 1k
v i = v s Ri Rs + Ri
= v s 2 1 + 2 = 2 3 v s
v o = Gm v i (RL Ro )
= 60 20 × 1 20 + 1 v i
= 60 20 21 × 2 3 v s
Overallvoltagegain ≡ v o v s = 38.1V/V
ThisfigurebelongstoProblem 1.52
10 k vs vi
o ii Ri 5-mV peak
Av ovi 1 k
1.54 vi vo 100 500 1 M 100vi
20log Av o = 40dB ⇒ Av o = 100V/V
Av = v o v i
= 100 × 500 500 + 100 = 83.3V/V
or20log83.3 = 38.4dB
Ap = v 2 o /500 v 2 i /1M = A2 v × 104 = 1.39 × 107 W/W
or10log (1.39 × 107 ) = 71.4dB.
Forapeakoutputsine-wavecurrentof20mA, thepeakoutputvoltagewillbe20mA × 500 = 10V.Correspondingly v i willbeasinewave withapeakvalueof10V/Av = 10/83.3,oran rmsvalueof10/(83.3 × √2) = 0.085V. Correspondingoutputpower = (10/√2)2 /500 = 0.1W
1.55 vi 200 k 1 M 1 V vo 20 100 1 vi
v o = 1V ×
× 1 ×
= 1 1.2 × 100 120 = 0.69V
Voltagegain = v o v s = 0.69V/Vor 3.2dB
Currentgain = v o /100 v s /1.2M = 0.69 × 1.2 × 104 = 8280A/Aor78.4dB
Powergain = v 2 o /100 v 2 s /1.2M = 5713W/W
or10log5713 = 37.6dB (Thistakesintoaccountthepowerdissipatedin theinternalresistanceofthesource.)
1.56 (a)CaseS-A-B-L(seefigureonnextpage):
v o v s = v o v ib × v ib v ia × v ia v s = 10 × 100 100 + 1000 × 100 × 10 10 + 10 × 100 100 + 100
v o v s = 22.7V/VandgainindB20log22.7 = 27.1dB
(b)CaseS-B-A-L(seefigureonnextpage):
v o v s = v o v ia v ia v ib v ib v s = 100 × 100 100 + 10K × 10 × 100K 100K + 1K × 10K
10K + 100K
v o v s = 0.89V/VandgainindBis20log0.89 = 1dB.Obviously,caseaispreferredbecauseit provideshighervoltagegain.
1.57 Eachofstages#1,2,..., (n 1) canbe representedbytheequivalentcircuit:
v o v s = v i1 v s × v i2 v i1 × v i3 v i2 ×···× v in v i(n 1) × v o v in where
Thisfigurebelongsto 1.56,part(a).
Thisfigurebelongsto 1.56,part(b).
v i1
v s = 10k 10k + 10k = 0.5V/V
v o v in = 10 × 200 1k + 200 = 1.67V/V
v i2 v i1 = v i3 v i2 =···= v in v i(n 1) = 10 × 10k 10k + 10k = 9.09V/V
Thus,
v o v s = 0.5 × (9.09)n 1 × 1.67 = 0.833 × (9.09)n 1
For v s = 5mVand v o = 3V,thegain v o v s must be ≥ 600,thus
0.833 × (9.09)n 1 ≥ 600
⇒ n = 4
Thusfouramplifierstagesareneeded,resultingin
v o v s = 0.833 × (9.09)3 = 625.7V/V andcorrespondingly
v o = 625.7 × 5mV = 3.13V
Thisfigurebelongsto 1.57.
1.58 Deliver0.5Wtoa100- load. Sourceis30mVrmswith0.5-M source resistance.Choosefromthesethreeamplifier types:
Chooseordertoeliminateloadingoninputand output:
A,first,tominimizeloadingon0.5-M source
B,second,toboostgain
C ,third,tominimizeloadingat100- output.
Wefirstattemptacascadeofthethreestagesin theorder A, B, C (seefigureabove),andobtain
v i1 v s = 1M 1M + 0.5M = 1 1.5
