1EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu1Midterm 1 Announcements Review session: 5-8pm TONIGHT 277 Cory Midterm 1: 11:30-1pm on Tuesday, July 12 in Dwinelle145. Material covered in HW1-3 Attend only your second lab slot this weekEE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu2Review: Second-Order Filter CircuitsC+–VSRBand PassLow PassLHigh PassBand RejectZ = R + 1/jωC + jωLHBP= R / ZHLP= (1/jωC) / ZHHP= jωL/ ZHBR= HLP+ HHP2EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu3Lecture #9OUTLINE The operational amplifier (“op amp”) Ideal op amp Feedback Unity-gain voltage follower circuit Summing, difference, integrator, differentiator, active filterReadingCh. 14EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu4The Operational Amplifier The operational amplifier (“op amp”) is a basic building block used in analog circuits. Its behavior is modeled using a dependent source. When combined with resistors, capacitors, and inductors, it can perform various useful functions: amplification/scaling of an input signal sign changing (inversion) of an input signal addition of multiple input signals subtraction of one input signal from another integration (over time) of an input signal differentiation (with respect to time) of an input signal analog filtering nonlinear functions like exponential, log, sqrt, etc3EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu5Op Amp Terminals 3 signal terminals: 2 inputs and 1 output IC op amps have 2 additional terminals for DC power supplies Common-mode signal= (v1+v2)/2 Differential signal = v1-v2-+V +V –Inverting input v2Non-inverting input v1positive power supplynegative power supplyv0outputEE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu6Op Amp Terminal Voltages and Currents• All voltages are referenced to a common node.• Current reference directions are intothe op amp.+–V +V ––Vcc++Vcc–+v1–+v2–common node(external to the op amp)i2i1io+vo–ic+ic-4EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu7Model A is differential gain or open loop gain Ideal op amp Common mode gain = 00ioARR→∞→∞=v1v2Ri+–RovoA(v1–v2)i1i2io+_ Circuit Model()121212,2(),0cm docmcmddocmvvvvvvvAv AvSince v A v v A+==−=+=− =EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu8Model and Feedback Negative feedback connecting the output port to the negative input (port 2) Positive feedback connecting the output port to the positive input (port 1)v1v2Ri+–RovoA(v1–v2)i1i2io+_ Circuit Model5EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu9Summing-Point Constraint Check if under negative feedback Small viresult in large vo Output vois connected to the inverting input to reduce vi Resulting in vi=0 Summing-point constraint v1= v2 i1 = i2 =0 Virtual short circuit Not only voltage drop is 0 (which is short circuit), input current is 0 This is different from short circuit, hence called “virtual” short circuit.EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu10Inverting Amplifier Negative feedback Æchecked Use summing-point constraint12 1222122110, 02.()( )ovinin outooinvClosed loop gain Avvv iiUse KCL At Nodevv v viRRRvvRvInput impedance Ri==== ==−−===−==Ideal voltage source – independent of load resistor_+vin+-R2R1RLv1v0v2i26EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu11Non-Inverting Amplifier Ideal voltage amplifier_+vin+-R2R1RLv2v012 1202221121,02.()(0)()ovininoininvC losed loop gain AvvvviiUse KCL At NodevvviRRvRRAvRvInput impedancei==== ==−−==+===→∞2EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu12Voltage Follower_+vin+-RLv2v0210222112 2110()(0)()11oinRRvvviRRvRR RAvR R=→∞−−==+== =+=7EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu13Example 1 Switch is open Switch is closedv1v2vin+-RRLv02R_+i1i2R121 313124502502,0 0()00()1,inininoininvvi ivvivvviiRvvivvvRvARv==→=−=→== →=→=−=→==== →∞i5i3i412 1 32024500, 0 0() ()1,2ininoininvv i ivv vviiRRvvvRARv===→=−−===−=−==− =EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu14Example 2 Design an analog front end circuit to an instrument system Requires to work with 3 full-scale of input signals (by manual switch): V For each input range, the output needs to be V The input resistance is 1MΩ01,010,0100±± ±:: :010±:_+vinR2R1RLv2v02bacvbvav1211111(1 )oinabinabcainabcRvvRv v Switch at cRRv v Switch at bRRRRv v Switch at aRRR=+=+=++=++8EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu15Example 2 (cont’d)21212121110 (1 )1(1) 0.10.1 (1 ) 0.0110 , 90 , 9009in a b cvab abvabc abcaavabc abcabcRRRRMRM ax A Sw itch at cRRR RRRA Sw itch at bRRR R RRRRRRA Sw itch at aRRR R RRRRkRkR kRR=++=Ω==+++== + ∴ =++ ++== + ∴ =++ ++∴=Ω=Ω= Ω=EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu16Summing Amplifier_+v3+-v0R1R2+-+-R3R0v2v19EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu17Difference Amplifier_+v2+-v0R4R1+-R3R2v1EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu18Integrator Want What is the difference between:oinvKvdt=∫R+-CV0+-vin10EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu19Differentiator Want _+vin+-Rv0CEE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu20A DAC can be used to convert the digital representation of an audio signal into an analog voltage that is then used to drive speakers -- so that you can hear it!0 0 1 0 10 0 1 10 1 0 01.520 1 0 10 1 1 00 1 1 11 0 0 02.533.541 0 0 11 0 1 01 0 1 11 1 0 01 1 0 11 1 1 01 1 1 14.555.566.577.5BinarynumberAnalogoutput(volts)0 0 0 00 0 0 10.5MSBLSBS1 closed if LSB =1S2 " if next bit = 1S3 " if " " = 1S4 " if MSB = 14-Bit D/A8V−+V05K80K40K20K10KS1+-S3S2S4−+Application: Digital-to-Analog Conversion“Weighted-adder D/A converter”(Transistors are used as electronic switches)11EE40 Summer 2005: Lecture 9 Instructor: Octavian Florescu21Digital Input 0123456780246810121416Analog Output (V)11111000010000000001Characteristic of 4-Bit DACEE40 Summer 2005: Lecture 9 Instructor: Octavian
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