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# Linear Circuits MAE 140

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This 40 page Class Notes was uploaded by Tevin Aufderhar on Thursday October 22, 2015. The Class Notes belongs to MAE 140 at University of California - San Diego taught by Mauricio De Oliveira in Fall. Since its upload, it has received 34 views. For similar materials see /class/226815/mae-140-university-of-california-san-diego in Mechanical and Aerospace Engineering at University of California - San Diego.

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Date Created: 10/22/15

Active Circuits Life gets interesting Active cct elements operational amplifiers OP AMPS and transistors Devices which can inject power into the cct External power supply normally comes from connection to the voltage supply rails Capable of linear operation amplifiers and nonlinear operation typically switches Triodes pentodes transistors MAE140 Linear Circuits Active Cct Elements 1v Ampli ers linear amp active mm Signal processors Stymied until 1927 and Harold Black Negative Feedback Amplifier Control rescues communications Telephone relay stations manageable against manufacturing variability Linearity Output signal is proportional to the input signal Note distinction between signals and systems which transform them Yes Just like your stereo amplifier Idea controlled current and voltage sources MAEMD Linear Circuits Linear Dependent Sources Active device models in linear mode Transistor takes an input voltage v and produces an output current i0gv where g is the gain This is a linear voltage controlled current source VCCS O O O 139 1 Ii quot1 1i 311 O CCVS r transresistance CCCS 3 current gain o O V V VI 1 W O o VCVS M voltage gain VCCS g transconductance MAE140 Linear Circuits 69 Linear dependent source contd Linear dependent sources are parts of active cct models they are not separate components But they allow us to extend our cct analysis techniques to really useful applications This will become more critical as we get into dynamic ccts Dependent elements change properties according to the values of other cct variables n 3r i i i r11rlSv0 IS 0 1i quot17quot0 0 O Source on Source off MAE140 Linear Circuits 70 Cct Analysis with Dependent Sources Golden rule do not lose track of control variables Find i0 v0 and PO for the 5009 load 5 Current divider on LHS ixgiS 3 Current divider on RHS i0 48ix 12iS Ohm s law v0i0500 6000iS Power p 1 v 72 00012 0 0 0 a S MAE140 Linear Circuits 71 Analysis with dependent sources I Power provided by 1C5 PS5025i 5i Power delivered to load 720mg Power gain 67 254320 Where did the energy come from MAE140 Linear Circuits 72 Nodal Analysis with Dependent Source at nOde C GIVC VSlGzVC VSzGBvcGPVc VDO KCL at node D GPVDVCGEVD3i30 CCCS element description iBGpvC VD Substitute and solve G1G2 GB GPVC GPVD G1VSIG2vSZ 3 1GPVC b 1Gp GE vD 0 MAE140 Linear Circuits 73 TampR Example 4 3 p 148 R 3 Find v0 in terms of VS What happens as p900 Node A G1G2 G3 VA G3VB G1VS Node B VB A Vx MVA Solution V0 VB A VA MGI VS G1GzlMG3 For large gains p 1MG3gtgtG1G2 G R VO M 1 V 3V5 1MG3 R1 39 This is a model of an inverting opamp MAE140 Linear Circuits 74 Mesh Current Analysis with Dependent Sources Dual of Nodal Analysis with dependent sources Treat the dependent sources as independent and sort out during the solution Rm R1 R2 R3 gR2R3l39A R313 VS 1 R3 gR2R3l A R3 R4I B 0 75 MAE140 Linear Circuits Example 4 5 BJTransistor MAE140 Linear Circuits Needs a supermesh Current source in two loops without R in parallel Supermesh entire outer loop Supermesh equation izRE Vy i1RB VCC 0 Current source constraint H b Solution VCC Vy 13 l1 RB 3 1RE 76 Example 4 6 Field Effect Transistor vx VS 8 R1 4 Since cct is linear v0 K1v51 K2vS2 Solve via superposition First 1231 and vS20 then vSI0 and 1252 This gives K1 and K2 77 MAE140 Linear Circuits A Brief Aside Transistors Bipolar Junction Transistors E C Semiconductors doped silicon D F B n doping mobile electrons Si doped with Sb P or As p doping mobile holes Si doped with B Ga In E F C B Two types npn and pnp B Heavily doped Collector and Emitter Lightly doped Base and very thin Collector and Emitter thick and dopey Need to bias the two junctions properly Then the base current modulates a strong CeE current Amplification iC3iB MAE140 Linear Circuits Tra nsistors chc Common Emitter Amplifier Stage Biasing resistors R1 and R2 Keep transistor junctions biased in 1 amplifying range Blocking capacitors CBl and CE52 V Keep dc currents out Feedback capacitor CE Grounds emitter at high frequencies MAE140 Linear Circuits 79 Operational Amplifiers OpAmps Basic building block of linear analog cicruits Package of transistors capacitors resistors diodes in a chip Five terminals Positive power supply VCC Negative power supply VCC Inverting input Vp Non inverting input vn Linear region of operation v0 Avp vn Ideal behavior 105 lt A lt108 VPquot Saturation at Vccj VCC limits range 39 cc MAE140 Linear Circuits 80 5 E E I I A I I I Nuninverting 39 Ii Invenina inpuI l I input i lt 39 m I i i i iquot r quotquotquotquot m i I i QIU i 0395 I 39 I I i i i xquot 019 39 I Offset gt5Ukn iTseu 011 i g Cy null I u i 39 Z i I L0 4 I i U n 50 km 5UQlt I kn I 5 I I I lt i I i V OFFSETNULL E1 8 NC INVERTING INPUT E 2 7 V NONINVERTING INPUT I 3 6 I OUTPUT v I 4 5 I OFFSET NULL MAE140 Linear Circuits 81 IdealOpAnua Vo Vcc EqUIvalent linear CIrCUIt Dependent source model 106 lt R1lt1012S2 0092 s VpVn 10 lt R0 lt100S2 09 105 lt A lt108 Need to stay in linear range VCC S V0 S VCC V e V V p Rfi 3 vp V 3 if Vii V0 Ideal conditions V 1 1 vp v 0 82 MAE140 Linear Circuits Non inverting OpAmp Feedback What happens now Voltage divider feedback R2 V V n R1R2 O Operating condition v vS p R R V0 1 2V R25 Linear non inverting amplifier R1 1132 2 Gain K MAE140 Linear Circuits 83 Example 4 13 Analyze this 139 0 Vp R2 S VS R1 1 32 OpAmp has zero output resistance RL does not affect v0 AMP R VP 4 V R2 R3R4 KTt1KSKAMP O 021 V5 R1R2 R4 84 MAE140 Linear Circuits Voltage Follower Buffer Feedback path Vl V0 lp0 vpvS Ideal OpAmp VP vquot V0 VS i0 Loop gain is 1 Power is supplied from the Vcc Vcc rails MAE140 Linear Circuits 85 OpAmp Ccts inverting amplifier Input and feedback applied at R1 R2 same terminal of OpAmp R2 is the feedback resistor So how does it work KCL at node A T Invemng amp VN VS VN V0 V0 iN 0 v0KvS hence the name Noninverting amp MAE140 Linear Circuits 86 Inverting Amplifier contd Current flows in the inverting amp VS 11 Rm R1 1 87 MAE140 Linear Circuits OpAmp Analysis Example 4 14 Compute the input output relationship of this cct Convert the cct left of the nodeVs A to its Th venin equivalent v v R2 v T 0C R1R2 S R1R2 R1R2 R1R3 R2R3 R1R2 R1R2 Note that this is not the inverting amp gain times the voltage divider gain There is interaction between V0 RivT the two parts of the cct R3 RTRmR3 This is a feature of the R4R1R2 R2 V inverting amplifier R1R2R1R3R2R3 R1R2 configuration R2R4 MW 88 MAE140 LmearCnrcunts 1 2 1 3 2 3 Summing Amplifier Adder 39 So what happens Node A is effectively grounded vnvp0 Also l39N0 because of Rm SO l1 l2 l O v v v A 72 70 0 R1 R2 RF This is an inverting summing amplifier Ever wondered about audio mixers How do they work MAE140 Linear Circuits 89 Mixing desk Linear ccts R Q RF RF RF RF Virtual ground at vn V0 17 V1 IT V2 1 Vm 1 2 m Currents add Summing junction Permits adding signals to create a composite Stringsbrasswoodwindpercussion Guitarsbassdrumsvocakeyboards K1111 K2v2 Kmvm MAE140 Linear Circuits 90 Design Example 4 15 Design an inverting summer to realize v0 5v113v2 R R Inverting summer with l 5 l 13 R1 R2 13KQ 65K 2 1 1K9 56KQ VIO IVVV V V0 VIHVVV I V0 5K9 431m V2 V2 W Nominal valueis Standard values If v1400mV and VCC15V what is max of v2 for linear op Need to keep vOgt 15V 15 lt 5vl 13v2 15 gt 5v113v2 v2 lt15 5gtlt04 MAE140 Linear Circuits IV 91 OpAmp Circuits Differential Amplifier Use superposition to analyze O v20 inverting amplifier v R2 v 01 R1 1 1210 noninverting amplifier plus voltage divider T R4 R1R2 V02 V2 R3 R4 R1 V0 V01 V02 R R R R 2 v1 4 v2 K1 inverting gain R1 R3 R4 R1 K2 nonInvertlng gain K1V1K2V2 92 MAE140 Linear Circuits Exercise 4 13 What is v0 This is a differential amp v is 10V v2 is 10V R11KQ1KQSOOQ R2R3R41KW V0 K1V1K2V2 R2 R1R2 V1 R1 R1 R4 R3 R4 203XX10 5V MAE140 Linear Circuits 93 Lego Circuits V1 V KR1R2 Noninverting ampli er R R R1 Inverting ampli er MAE140 Linear Circuits 94 Lego Circuits contd R1 RF V1 Differential ampli er MAE140 Linear Circuits 95 Example 4 16 OpAmp Lego VF 101m 97V IOKQ 3 3KQ IOKQ IOKQ 9K9 5K9 VC W VCCi15V So what does this circuit do It converts tens of 0F to tens of 0C Max current drawn by each stage is 15mA MAE140 Linear Circuits 96 OpAmp Cct Analysis OpAmp Nodal Analysis Use dependent voltage source model Identify node voltages Formulate input node equations Solve using ideal characteristic vpvn Rest of Vp 139 Clrcult Rest Of Rest of olrcult circuit MAE140 Linear Circuits 97 OpAmp Analysis Example 4 18 Seemingly six non reference nodes AE v0 Nodes A B connect to reference voltages V and v2 Node C E connected to OpAmp outputs forget for the moment Node D G1G2VD G1VC G2VE 0 Node F G3 G4VF G3VE O OpAmp constraints VA V1VD VB V2 VF G1VC G2VE G1 G2V1 G3VE G3 G4V2 G3 G4 G3 G1 G2 G1 G V1Ei VC V2 98 MAE140 Linear Circuits OpAmp Analysis Exercise 4 14 R2 Q R Node A VAZVS Node B Node C Constraints vBvpvn0 Solve MAE140 Linear Circuits V0 G2G3G4XG1v 4 VS G4 G2 5 R2123 R2124 R3R4 v S R1R3 99 Comparators A Nonlinear OpAmp Circuit We have used the ideal OpAmp conditions for the analysis of OpAmps in the linear regime vnvp inip0 if Avp vnsVCC What about if we operate with vp v That is we operate outside the linear regime We saturatell V0 VCC Vp gt Vl V0 VCC Vp ltVn Without feedback OpAmp acts as a comparator There is one of these in every FM radio MAE140 Linear Circuits 100 Analog to digital converter comparators Current laws still work 8V 0 3R Parallel comparison V03 Flash converter 2R 3 bit output v Not really how it is done l 02 Voltage divider switched 2R vs l V01 Input V01 V02 V03 R V 5V 1gtVS 0 0 0 CC VCCOV 3gtVSgt1 5 0 0 5gtVSgt3 5 5 0 VSgt5 5 5 5 MAE140 Linear Circuits 101 OpAmp Circuit Design the whole point Given an input output relationship design a cct to implement it Build a cct to implement v05v110v220v3 Inverting summer followed by an inverter Ezng 39 lOOKQ lOOKQ V0 4 Summer gt lt Inverter gt r MAE140 Linear Circuits 102 V0 R eqvlRqu2Reqv R R1 R2 m MAE140 Linear Circuits Example 4 21 How about this one V OKV Non inverting amp vpevo 100gtlt103294gtlt103 3 v 35v 294X10 P P P KCL atp node with ip0 V2V V1Vp p V3Vp 0 4 2gtlt104 10 05gtlt104 35vp 05v1 v2 2V3 Non inverting summer Fewer elements than inverting summer Req R1R2R3 39 39 39 Rm 103 Digital to analog converter Voltage Conversion of digital data to analog voltage value Bit inputs O or 5V Analog output varies between vmm and vmax in 16 steps MAE140 Linear Circuits 104 Signal Conditioning Your most likely brush with OpAmps in practice Signal typically a voltage representing a physical variable Temperature strain speed pressure Digital analysis done on a computer after Anti aliasing filtering data interpretation Addingsubtracting an offset zeroing Normally zero of ADC is 0V Scaling for full scale variation quantization Normally full scale of ADC is 5V Analog to digital conversion ADC Maybe after a few more tricks like track and hold Offset correction use a summing OpAmp Scaling use an OpAmp amplifier Anti aliasing filter use a dynamic OpAmp cct MAE140 Linear Circuits 105 Th venin and Norton for dependent sources Cannot turn off the ICSs and IVSs to do the analysis This would turn off the DCSs and DVSs Connect an independent CS or VS to the terminal and compute the resulting voltage or current and its dependence on the source T Compute V3 in response to is v5 VT z39SRT ls 106 MAE140 Linear Circuits

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