Chapter 5, Solution 1. 1.5 MΩ 60 Ω
Chapter 5, Solution 1.(a)(b)(c)Rin 1.5 MΩRout 60 ΩA 8x104Therefore AdB 20 log 8x104 98.0 dBChapter 5, Solution 2.v0 Avd A(v2 - v1) 105 (20-10) x 10-6 0.1VChapter 5, Solution 3.v0 Avd A(v2 - v1) 2 x 105 (30 20) x 10-6 10VChapter 5, Solution 4.v0 Avd A(v2 - v1)v 4v2 - v1 0 20µVA 2 x10 5If v1 and v2 are in mV, thenv2 - v1 -20 mV 0.021 - v1 -0.02v1 1.02 mVChapter 5, Solution 5.IR0Rinvd vi - -Avd v0-
-vi Avd (Ri - R0) I 0But(1)vd RiI,-vi (Ri R0 RiA) I 0vd vi R iR 0 (1 A)R i(2)-Avd - R0I v0 0v0 Avd R0I (R0 RiA)I (R 0 R i A) v iR 0 (1 A)R iv0R 0 RiA100 10 4 x10 5 10 4 5v i R 0 (1 A)R i 100 (1 10 ) 100,00010 9 10 4 0.99999905100,0011 10()Chapter 5, Solution 6.vi -R0Ivd Rin -Avd vo-
(R0 Ri)R vi Avd 0Butvd RiI,vi (R0 Ri RiA)I 0I viR 0 (1 A)R i(1)-Avd - R0I vo 0vo Avd R0I (R0 RiA)ISubstituting for I in (1), R 0 R iA viv0 R(1A)Ri 0650 2 x10 x 2 x10 5 10 3 50 1 2x10 5 x 2 x10 6( ()) 200,000 x 2 x10 6mV200,001x 2 x10 6v0 -0.999995 mVChapter 5, Solution 7.100 kΩ10 kΩVS -Rout 100 Ω12 Vd-Rin -AVd Vout-
At node 1,(VS – V1)/10 k [V1/100 k] [(V1 – V0)/100 k]10 VS – 10 V1 V1 V1 – V0which leads to V1 (10VS V0)/12At node 2,(V1 – V0)/100 k (V0 – AVd)/100But Vd V1 and A 100,000,V1 – V0 1000 (V0 – 100,000V1)0 1001V0 – 100,000,001[(10VS V0)/12]0 -83,333,334.17 VS - 8,332,333.42 V0which gives us (V0/ VS) -10 (for all practical purposes)If VS 1 mV, then V0 -10 mVSince V0 A Vd 100,000 Vd, then Vd (V0/105) V -100 nVChapter 5, Solution 8.(a)If va and vb are the voltages at the inverting and noninverting terminals of the opamp.va v b 01mA 0 v02kv0 -2V(b)10 kΩ2V iava2V vb1V vo--(a)2 kΩ va-10 kΩ - ia(b)vo-
Since va vb 1V and ia 0, no current flows through the 10 kΩ resistor. From Fig. (b),-va 2 v0 0va va - 2 1 - 2 -1VChapter 5, Solution 9.(a)Let va and vb be respectively the voltages at the inverting and noninvertingterminals of the op ampva vb 4VAt the inverting terminal,1mA 4 v02kv0 2V(b)1V - vbvo--Since va vb 3V,-vb 1 vo 0vo vb - 1 2VChapter 5, Solution 10.Since no current enters the op amp, the voltage at the input of the op amp is vs.Hence 10 v ovs v o 10 10 2vo 2vs
Chapter 5, Solution 11.8 kΩ2 kΩ3Vvb 5 kΩ abio 10 kΩ4 kΩvo 10(3) 2V10 5At node a,3 va va vo 2812 5va – voBut va vb 2V,vo -2V12 10 – vo–io va vo 0 vo 2 2 2 1mA8484i o -1mAChapter 5, Solution 12.4 kΩ1 kΩ1.2V ab 4 kΩ2 kΩ vo
422vo vo vo4 233At node b,vb At node a,1 .2 v a v a v o2, but va vb v o 1434.8 - 4 x22vo vo vo33va vb 29.6vo 37is vo 3x 4.8 2.0570V71 .2 v a 1 .2 17 1.2 p vsis 1.2 -205.7 mW 7 Chapter 5, Solution 13.10 kΩab1V 90 kΩ io100 kΩ i2i14 kΩ50 kΩBy voltage division,90va (1) 0.9V100v50vb vo o3150v0But va vb 0 .9vo 2.7V3vvio i1 i2 o o 0.27mA 0.018mA 288 µA10k 150k vo
Chapter 5, Solution 14.Transform the current source as shown below. At node 1,10 v1 v1 v 2 v1 v o 5201010 kΩ5 kΩ10 kΩ20 kΩv110Vvov2 vo But v2 0. Hence 40 - 4v1 v1 2v1 - 2voAt node 2,v1 v 2 v 2 v o ,201040 7v1 - 2vov 2 0 or v1 -2voFrom (1) and (2), 40 -14vo - 2vo(1)(2)vo -2.5VChapter 5, Solution 15(a) Let v1 be the voltage at the node where the three resistors meet. ApplyingKCL at this node gives 1v vv1 vo (1)i s 1 1 o v1 R2R3RRR33 2At the inverting terminal,0 v1(2)is v1 i s R1R1Combining (1) and (2) leads to vvoRR RR R1 R3 1 3 i s 1 1 1 oR3isR2 R2 R3 (b) For this case,vo20 x 40 20 40 kΩ - 92 kΩis25
Chapter 5, Solution 1610k Ω5k Ωixvavbiy vo2k Ω 0.5V-8k ΩLet currents be in mA and resistances be in k Ω . At node a,0 .5 v a v a v o 1 3v a vo510(1)But810vo vo v a(2)8 28Substituting (2) into (1) gives1081 3v a v a v a 814Thus,0 .5 v aix 1 / 70 mA 14.28 µA5v vb v o v a100 .6 8iy o 0 .6 ( v o v a ) 0 .6 ( v a v a ) x mA 85.71 µA21084 14v a vb Chapter 5, Solution 17.(a)(b)(c)G voR12 2 -2.4viR15vo80 -16vi5vo2000 -400vi5
Chapter 5, Solution 18.Converting the voltage source to current source and back to a voltage source, we have thecircuit shown below:10 20 20kΩ31 MΩ(20/3) kΩ 50 kΩ 2vi/3 vo vo 1000 2v i 20 350 3vo200 -11.764v117Chapter 5, Solution 19.We convert the current source and back to a voltage source.24 (4/3) kΩ(2/3)V 4 kΩ0V4310 kΩ vo5 kΩ
10k 2 -1.25V 4 3 4x k 3 vv 0 -0.375mAio o o5k10kvo Chapter 5, Solution 20.8 kΩ4 kΩ9Va 4 kΩvs2 kΩb vo At node a,9 va va vo va vb 48418 5va – vo - 2vb(1)At node b,va vb vb vo 42va 3vb - 2voBut vb vs 0; (2) becomes va –2vo and (1) becomes-18 -10vo – vovo -18/(11) -1.6364V(2)
Chapter 5, Solution 21.Eqs. (1) and (2) remain the same. When vb vs 3V, eq. (2) becomesva 3 x 3 - 2v0 9 - 2voSubstituting this into (1), 18 5 (9-2vo) – vo – 6 leads tovo 21/(11) 1.909VChapter 5, Solution 22.Av -Rf/Ri -15.If Ri 10kΩ, then Rf 150 kΩ.Chapter 5, Solution 23At the inverting terminal, v 0 so that KCL givesvs 00 0 vo R1R2Rf vovs RfR1Chapter 5, Solution 24v1RfR2R1- vs R3R4vo-v2We notice that v1 v2. Applying KCL at node 1 givesv1 (v1 v s ) v1 vo 0R1R2Rf 1 1 1 v1 v s vo R RR f R2 R f2 1(1)
Applying KCL at node 2 givesR3v1 v1 v s 0 v1 vsR3 R4R3R4Substituting (2) into (1) yields(2) RRR R3 1 v svo R f 3 3 4 R1 R f R2 R3 R4 R2 i.e. RRR R3 1 k R f 3 3 4 R1 R f R2 R3 R4 R2 Chapter 5, Solution 25.vo 2 V vavo-va 3 vo 0 which leads to va vo 3 5 V.Chapter 5, Solution 26 vb 0.4V-8k Ω 2k Ωvo-vb 0.4 8vo 0.8vo8 2 Hence,io v o 0 .5 0.1 mA5k 5kvo 0.4 / 0.8 0.5 Vio5k Ω
Chapter 5, Solution 27.(a)Let va be the voltage at the noninverting terminal.va 2/(8 2) vi 0.2vi 1000 v 0 1 v a 10.2v i20 G v0/(vi) 10.2(b)vi v0/(G) 15/(10.2) cos 120πt 1.471 cos 120πt VChapter 5, Solution 28. At node 1,0 v1 v1 v o 10k50kBut v1 0.4V,-5v1 v1 – vo, leads tovo 6v1 2.4VAlternatively, viewed as a noninverting amplifier,vo (1 (50/10)) (0.4V) 2.4Vio vo/(20k) 2.4/(20k) 120 µA
Chapter 5, Solution 29R1vavb vi-va R2R2vi ,R1 R2But v a vbvb -R1R1voR1 R2R2R1vi voR1 R2R1 R2Orv o R2 viR1Chapter 5, Solution 30.The output of the voltage becomesvo vi 1230 20 12kΩBy voltage division,vx 12(1.2) 0.2V12 60ix p v 2x 0.04 2µWR20kvx0 .2 10µA20k 20k R2vo-
Chapter 5, Solution 31.After converting the current source to a voltage source, the circuit is as shown below:12 kΩ3 kΩ16 kΩ vov112 V 2vo6 kΩAt node 1,12 v1 v1 v o v1 v o 361248 7v1 - 3vo(1)At node 2,v1 v o v o 0 ix66v1 2vo(2)From (1) and (2),4811voix 0.7272mA6kvo Chapter 5, Solution 32.Let vx the voltage at the output of the op amp. The given circuit is a non-invertingamplifier. 50 v x 1 (4 mV) 24 mV 10 60 30 20kΩ
By voltage division,vo v20v o o 12mV20 202ix vx24mV 600nA(20 20)k 40kp v o2 144x10 6 204nWR60x10 3Chapter 5, Solution 33.After transforming the current source, the current is as shown below:1 kΩ4 kΩ4Vvi va 2 kΩvo3 kΩThis is a noninverting amplifier.3 1 v o 1 v i v i2 2 Since the current entering the op amp is 0, the source resistor has a OV potential drop.Hence vi 4V.vo 3(4) 6V2Power dissipated by the 3kΩ resistor isv o2 36 12mWR 3kix va vo 4 6 -2mAR1k
Chapter 5, Solution 34v1 vin v1 vin 0R1R2(1)R3voR3 R 4(2)butva Combining (1) and (2),v1 va R1Rv 2 1 va 0R2R2 R Rv a 1 1 v1 1 v 2R2 R2 RR R 3v o 1 1 v1 1 v 2R2R3 R 4 R 2 vo vO R3 R 4 R v1 1 v 2 R2 R R 3 1 1 R2 R3 R 4( v1R 2 v 2 )R 3 ( R1 R 2 )Chapter 5, Solution 35.Av voR 1 f 10RiviIf Ri 10kΩ, Rf 90kΩRf 9Ri
Chapter 5, Solution 36VTh VabButVThR1Vab . Thus,R1 R2RR R2 Vab 1v s (1 2 )v sR1R1vs To get RTh, apply a current source Io at terminals a-b as shown below.v1 -v2a R2R1voiobSince the noninverting terminal is connected to ground, v1 v2 0, i.e. no current passesthrough R1 and consequently R2 . Thus, vo 0 andvRTh o 0ioChapter 5, Solution 37. RRRv o f v1 f v 2 f v 3 R3 R2 R13030 30 (1) (2) ( 3) 2030 10 vo -3V
Chapter 5, Solution 38. RRRRv o f v1 f v 2 f v 3 f v 4 R4 R3R2 R1505050 50 (10) ( 20) (50) ( 100) 201050 25 -120mVChapter 5, Solution 39This is a summing amplifier.RfRf Rf5050 50 vo v1 v2 v3 (2) v 2 ( 1) 9 2.5v 2R2R3 2050 10 R1Thus,vo 16.5 9 2.5v 2 v 2 3 VChapter 5, Solution 40R1vaR2 v1- R3 v2-vb- Rf v3-Rvo-Applying KCL at node a,v1 v a v 2 v a v3 v a 0R1R2R3 v1 v 2 v3111 va ( ) (1)R1 R2 R3R1 R2 R3
Butv a vb RvoR Rf(2)Substituting (2) into (1)givesRvov1 v 2 v3111 ( )R1 R2 R3 R R f R1 R2 R3orvo R RfR(v1 v 2 v3111 ) /( )R1 R2 R3 R1 R2 R3Chapter 5, Solution 41.Rf/Ri 1/(4)Ri 4Rf 40kΩThe averaging amplifier is as shown below:v1v2v3v4Chapter 5, Solution 421R f R1 10 kΩ3R1 40 kΩ10 kΩR2 40 kΩR3 40 kΩR4 40 kΩ vo
Chapter 5, Solution 43.In order for RRRRv o f v1 f v 2 f v 3 f v 4 R2R3R4 R1to become1(v 1 v 2 v 3 v 4 )4RRf 112 Rf i 3kΩRi 444vo Chapter 5, Solution 44.R4R3v1v2aR1 bR2At node b,v b v1 v b v 2 0R1R2At node a,0 va va vo R3R4v1 v 2 R1 R 2vb 11 R1 R 2(1)vo1 R4 / R3(2)va But va vb. We set (1) and (2) equal.voR v R 1 v1 2 11 R4 / R3R1 R 2orvo vo(R 3 R 4 )(R 2 v1 R 1 v1 )R 3 (R 1 R 2 )
Chapter 5, Solution 45.This can be achieved as follows: R( v1 ) R v 2 v o R/2 R / 3 RR f ( v1 ) f v 2 R2 R1i.e. Rf R, R1 R/3, and R2 R/2Thus we need an inverter to invert v1, and a summer, as shown below (R 100kΩ).Rv1RR R/3-v1v2R/2 voChapter 5, Solution 46.v1 1RRR1 ( v 2 ) v 3 f v1 x ( v 2 ) f v 33 32R1R2R3i.e. R3 2Rf, R1 R2 3Rf. To get -v2, we need an inverter with Rf Ri. If Rf 10kΩ,a solution is given below. vo 10 kΩv2v110 kΩ 30 kΩ30 kΩ10 kΩ-v2v320 kΩ vo
Chapter 5, Solution 47.If a is the inverting terminal at the op amp and b is the noninverting terminal,then,vb 10 v a v a v o3(8) 6V, v a v b 6V and at node a, 243 1which leads to vo –2 V and io v o (v a v o ) –0.4 – 2 mA –2.4 mA5k4kChapter 5, Solution 48.Since the op amp draws no current from the bridge, the bridge may be treated separatelyv1as follows:i1 i2v2For loop 1, (10 30) i1 5i1 5/(40) 0.125µAFor loop 2, (40 60) i2 -5i2 -0.05µABut, 10i v1 - 5 060i v2 5 0v1 5 - 10i 3.75mVv2 -5 - 60i -2mVAs a difference amplifier,R80[3.75 ( 2)]mVv o 2 (v 2 v 1 ) 20R1 23mV
Chapter 5, Solution 49.R1 R3 10kΩ, R2/(R1) 2i.e.R2 2R1 20kΩ R4vo Verify: 2R 2 1 R1 / R 2Rv 2 2 v1R1 1 R 3 / R 4R1(1 0.5)v 2 2v1 2(v 2 v1 )1 0.5Thus, R1 R3 10kΩ, R2 R4 20kΩChapter 5, Solution 50.(a)We use a difference amplifier, as shown below:v1R1R2 v2vo (b)R1voR2R2(v 2 v1 ) 2(v 2 v1 ), i.e. R2/R1 2R1If R1 10 kΩ then R2 20kΩWe may apply the idea in Prob. 5.35.v 0 2 v1 2 v 2 R( v1 ) R v 2 R/2 R / 2 R R f ( v1 ) f v 2 R2 R1i.e. Rf R, R1 R/2 R2
We need an inverter to invert v1 and a summer, as shown below. We may let R 10kΩ.Rv1RR R/2-v1v2R/2 voChapter 5, Solution 51.We achieve this by cascading an inverting amplifier and two-input inverting summer asshown below:RRRRv2 va Rvo v1 Verify:Butvo -va - v1va -v2. Hencevo v2 - v1.
Chapter 5, Solution 52A summing amplifier shown below will achieve the objective. An inverter is inserted toinvert v2. Let R 10 k Ω .R/2Rv1R/5v3R v4voRRv2-R/4 Chapter 5, Solution 53.(a)v1R2R1vavbv2R1 voR2At node a, v At node b,vb R2v2R1 R 2But va vb. Setting (1) and (2) equal givesR v R 1v oR2v2 2 1R1 R 2R1 R 2 (1)(2)
v 2 v1 R1vo viR2vo R 2 viR1(b) v1vi v2R1/2 vAR1/2R2vaRgR1/2R1/2vbvB R2 vo At node A,v1 v A v B v A v A v a R1 / 2RgR1 / 2orv1 v A At node B,v2 vB vB vA vB vb R1 / 2R1 / 2Rgorv2 vB R1(v B v A ) v A v a2R gR1(v B v A ) v B v b2R gSubtracting (1) from (2),v 2 v1 v B v A 2R 1(v B v A ) v B v A v b v a2R gSince, va vb,v 2 v1 R v 1 1 (v B v A ) i 22 2R g (1)(2)
vB vA orvi 21R1 12R g(3)But for the difference amplifier,R2(v B v A )R1 / 2RvB vA 1 vo2R 2vo orEquating (3) and (4),R1vvo i 2R 22vo R 2 viR1(c)At node a,At node b,(4)1R1 12R g1R1 12R gv1 v a v a v A R1R2 /22R 12R 1vAva v1 v a R2R22R 12R 1vBvb v2 vb R2R2(1)(2)Since va vb, we subtract (1) from (2), 2R 1v(v B v A ) i2R2 R2vivB vA 2R 1v 2 v1 or(3)At node A,va vA vB vA vA vo R2 /2RgR/2va vA R2(v B v A ) v A v o2R g(4)
At node B,vb vB vB vA vB 0 R/2RgR/2vb vB R2(v B v A ) v B2R g(5)Subtracting (5) from (4),v B v A R2(v B v A ) v A v B v oRg R 2(v B v A ) 1 2 v o 2R g Combining (3) and (6), R 2 R v i 1 2 v o 2R R1g v o R 2 R 1 2 viR 1 2R g Chapter 5, Solution 54.(a)(b)ButA0 A1A2A3 (-30)(-12.5)(0.8) 300A A1A2A3A4 A0A4 300A420Log10 A 60dBLog10 A 3A 103 1000A4 A/(300) 3.333Chapter 5, Solution 55.Let A1 k, A2 k, and A3 k/(4)A A1A2A3 k3/(4)20Log10 A 42ThusLog10 A 2.1A 102 1 125.89k3 4A 503.57k 3 503.57 7.956A1 A2 7.956, A3 1.989(6)
Chapter 5, Solution 56.There is a cascading system of two inverting amplifiers. 12 12 v s 6v s4 6 vi o s 3v s mA2kvo (a)(b)When vs 12V, io 36mAWhen vs 10 cos 377t V, io 30 cos 377t mAChapter 5, Solution 57The first stage is a difference amplifier. Since R1/R2 R3/R4,v o′ R2100( v 2 v1 ) (1 4) 10 mAR150The second stage is a non-inverter.R R v o 1 v o ′ 1 10 mA 40 mV(given) 40 40 Which leads to,R 120 kΩChapter 5, Solution 58.By voltage division, the input to the voltage follower is:v1 3(0.6) 0.45V3 1vo 10 10v1 v1 7 v1 3.1525io 0 vo 0.7875mA4kThus
Chapter 5, Solution 59.Let a be the node between the two op amps.va voThe first stage is a summerva or 1010vs vo vo5201.5vs -2vsvo 2 -1.333v s 1 .5Chapter 5, Solution 60.Transform the current source as shown below:4 kΩ10 kΩ5 kΩ 5isv1 io3 kΩ 3Ω2 kΩAssume all currents are in mA. The first stage is a summerv1 By voltage division, 10(5i s ) 10 v o 10i s 2.5v o54(1)
v1 31vo vo3 32(2)Alternatively, we notice that the second stage is a non-inverter. 1 vo v1 2 v1 3 3 From (1) and (2),0.5v o 10i s 2.5v ov o 2i o 10i s33vo 10isio 5 1.667is 3Chapter 5, Solution 61.Let v01 be the voltage at the left end of R5. The first stage is an inverter, while thesecond stage is a summer.R2v1R1RRv 0 4 v 01 4 v 2R5R3v 01 v1 R 2R 4Rv1 4 v 2R 1R 5R3Chapter 5, Solution 62.Let v1 output of the first op ampv2 output of the second op ampThe first stage is a summerv1 R2Rvi – 2 voR1RfThe second stage is a follower. By voltage division(1)
vo v2 R4v1R3 R4v1 R3 R4voR4(2)From (1) and (2), R3 RR v o 2 v i 2 v o 1 RfR1 R4 R3 R2 R v o 2 v i 1 R1 R4 Rf voR1 2 RRviR11 3 2R4 R4 R 2R 4 R 1 (R 2 R 3 R 4 )Chapter 5, Solution 63.The two op amps are summer. Let v1 be the output of the first op amp. For the firststage,v1 R2Rvi 2 voR1R3(1)For the second stage,vo R4Rv1 4 v iR5Ro(2)Combining (1) and (2), R2 R R R v i 4 2 v o 4 v i R5 R3 R6 R1 R R R RR v o 1 2 4 2 4 4 v i R 3 R 5 R 1R 5 R 6 vo R4R5R 2R 4 R 4 voR 1R 3 R 6 R Rvi1 2 4R 3R 5
Chapter 5, Solution 64G4GG1 G31G 0Vvsv20V G2 vo--At node 1, v1 0 so that KCL givesG1v s G4 vo Gv(1)At node 2,G2 v s G3 v o GvFrom (1) and (2),G1v s G4 v o G2 v s G3 voorvo G1 G2 v s G3 G 4(2) (G1 G2 )v s (G3 G4 )voChapter 5, Solution 65The output of the first op amp (to the left) is 6 mV. The second op amp is aninverter so that its output is30(6mV) -18 mV10The third op amp is a noninverter so thatvo ' vo ' 40vo40 8 vo 48v o ' 21.6 mV40
Chapter 5, Solution 66.100 40 100 110(6) (2) (4) 2520 20 10 24 40 20 -4Vvo Chapter 5, Solution 67.80 80 80 (0.5) (0.2)40 20 20 3.2 0.8 2.4Vvo Chapter 5, Solution 68.If Rq , the first stage is an inverter.Va 15(10) 30mV5when Va is the output of the first op amp.The second stage is a noninverting amplifier. 6 v o 1 v a (1 3)( 30) -120mV 2 Chapter 5, Solution 69.In this case, the first stage is a summerva 1515(10) v o 30 1.5v o510For the second stage, 6 v o 1 v a 4v a 4( 30 1.5v o ) 2 120vo -17.143mV7 v o 1207
Chapter 5, Solution 70.The output of amplifier A isvA 3030(10) (2) 91010The output of amplifier B isvB 2020(3) (4) 14101040 kΩvAvB20 kΩa60 kΩ bvo10 kΩvb 60( 14) 2V60 10At node a,vA va va vo 2040But va vb -2V, 2(-9 2) -2-voTherefore,vo 12V
Chapter 5, Solution 7120k Ω5k Ω100k Ω40k Ω 2V-v210k Ω80k Ω 20k Ω vo- 3V-10k Ωv1 -30k Ωv350k Ω2050(2) 8, v3 (1 )v1 8530100 100vo v2 v3 ( 20 10) 10 V80 40v1 3,v2
Chapter 5, Solution 72.Since no current flows into the input terminals of ideal op amp, there is no voltagedrop across the 20 kΩ resistor. As a voltage summer, the output of the first opamp isv01 0.4The second stage is an inverter150v 01100 2.5(0.4) -1Vv2 Chapter 5, Solution 73.The first stage is an inverter. The output is50v 01 ( 1.8) 9V10The second stage isv 2 v 01 -9VChapter 5, Solution 74.Let v1 output of the first op ampv2 input of the second op amp.The two sub-circuits are inverting amplifiers100(0.6) 6V1032v2 (0.4) 8V1.6v v2 6 8io 1 100 µA20k20kv1 Chapter 5, Solution 75.The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE areinvolved as shown to measure vo and i respectively. Once the circuit is saved, we clickAnalysis Simulate. The values of v and i are displayed on the pseudo-components as:
i 200 µA(vo/vs) -4/2 -2The results are slightly different than those obtained in Example 5.11.Chapter 5, Solution 76.The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, thevalue of io is displayed on IPROBE asio -374.78 µA
Chapter 5, Solution 77.The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, thevalue of io is displayed on IPROBE asio -374.78 µAChapter 5, Solution 78.The circuit is constructed as shown below. We insert a VIEWPOINT to display vo.Upon simulating the circuit, we obtain,vo 667.75 mV
Chapter 5, Solution 79.The schematic is shown below. A pseudo-component VIEWPOINT is inserted to displayvo. After saving and simulating the circuit, we obtain,vo -14.61 VChapter 5, Solution 80.The schematic is shown below. VIEWPOINT is inserted to display vo. After simulation,we obtain,vo 12 V
Chapter 5, Solution 81.The schematic is shown below. We insert one VIEWPOINT and one IPROBE tomeasure vo and io respectively. Upon saving and simulating the circuit, we obtain,vo 343.37 mVio 24.51 µA
Chapter 5, Solution 82.The maximum voltage level corresponds to11111 25 – 1 31Hence, each bit is worth(7.75/31) 250 mVChapter 5, Solution 83.The result depends on your design. Hence, let RG 10 k ohms, R1 10 k ohms, R2 20 k ohms, R3 40 k ohms, R4 80 k ohms, R5 160 k ohms, R6 320 k ohms,then,-vo (Rf/R1)v1 --------- (Rf/R6)v6 v1 0.5v
Chapter 5, Solution 1. (a) Rin 1.5 MΩ (b) Rout 60 Ω (c) A 8x104 Therefore AdB 20 log 8x10 4 98.0 dB Chapter 5, Solution 2. v0 Avd A(v2 - v1) 105 (20-10) x 10-6 0.1V Chapter 5, Solution 3. v0 Avd A(v2 - v1) 2 x 105 (30 20) x 10-6 10V Chapter 5, Solution 4. v0 Avd A(v2 - v1) v2 - v1 20 V 2x10
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
About the husband’s secret. Dedication Epigraph Pandora Monday Chapter One Chapter Two Chapter Three Chapter Four Chapter Five Tuesday Chapter Six Chapter Seven. Chapter Eight Chapter Nine Chapter Ten Chapter Eleven Chapter Twelve Chapter Thirteen Chapter Fourteen Chapter Fifteen Chapter Sixteen Chapter Seventeen Chapter Eighteen
18.4 35 18.5 35 I Solutions to Applying the Concepts Questions II Answers to End-of-chapter Conceptual Questions Chapter 1 37 Chapter 2 38 Chapter 3 39 Chapter 4 40 Chapter 5 43 Chapter 6 45 Chapter 7 46 Chapter 8 47 Chapter 9 50 Chapter 10 52 Chapter 11 55 Chapter 12 56 Chapter 13 57 Chapter 14 61 Chapter 15 62 Chapter 16 63 Chapter 17 65 .
HUNTER. Special thanks to Kate Cary. Contents Cover Title Page Prologue Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter
Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 . Within was a room as familiar to her as her home back in Oparium. A large desk was situated i
The Hunger Games Book 2 Suzanne Collins Table of Contents PART 1 – THE SPARK Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8. Chapter 9 PART 2 – THE QUELL Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapt
