Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 196.002CIRCUITS ANDELECTRONICSThe Operational Amplifier AbstractionCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19 MOSFET amplifier — 3 portspowerport inputportoutputport +–Iv+–Ov+–SV Amplifier abstraction+–Iv+–SV+–OvIvOvFunction of vIReviewCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19 Can use as an abstract building block for more complex circuits (of course, needto be careful about input and output). TodayIntroduce a more powerful amplifierabstraction and use it to build morecomplex circuits.Reading: Chapter 15 from A & L.IvOvFunction of vIReviewCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Operational AmplifierOp AmpOUTv+–+–INvMore abstract representation:inputportSVoutputportpowerportSV−+–+–+–Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Circuit model (ideal):i.e. ∞ input resistance 0 output resistance “A” virtually ∞ No saturationOvAv∞→A+–+–vv+v–0=i+0=i–Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19(Note: possible confusion with MOSFET saturation!)Using it…+–VVS12−=−LROv+–12V+–12VVVS12=DemoINvμV10μV10−OvV12V12−610~Abut unreliable,temp. dependentsaturationactive regionINvCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Let us build a circuit…Circuit: noninverting amplifierEquivalent circuit model1ROv+–2RINv+v−v()−+−vvA+–0=i+0=i–op amp1ROv+–2RINv+–+v−vCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Let us analyze the circuit:Find vOin terms of vIN, etc.What happens when “A” is very large?()−+−= vvAvO⎟⎠⎞⎜⎝⎛+−=212RRRvvAOININ212OAvRRAR1v =⎟⎠⎞⎜⎝⎛++212INORRAR1Avv++=Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Let’s see… When A is largeGain: determined by resistor ratio insensitive to A, temperature, fab variations212INORRAR1Avv++=()221INRRRv+≈gainDemoSuppose610A=R9R1=RR=2RR9R101v10v6IN6O++⋅=10vvINO⋅≈101101v106IN6⋅+⋅=212INRRARAv+≈Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19e.g. vIN= 5VSuppose I perturb the circuit…(e.g., force vOmomentarily to 12V somehow).Stable point is when v+ ≈ v-.Key: negative feedback Æ portion of output fed to –ve input.e.g. Car antilock brakesÆ small corrections.Why did this happen?Insight:+–RINOv2v =+–RINv+v−vnegativefeedback2vO5V5V10V0i=–A12V6V6VCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Question: How to control a high-strung device?Antilock brakesMichelinnoyesfeedbackyes/nois itturning?it’sall aboutcontroldiscv. v. powerful brakesapplyreleaseCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19More op amp insights:Observe, under negative feedback,0AvRRRAvvvIN121O→⎟⎠⎞⎜⎝⎛+==−−+−+≈ vvWe also know i+ ≈ 0i -≈ 0Æyields an easier analysis method (under negative feedback).Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Insightful analysis methodunder negative feedback+–1ROv+–2RINvINvc221INORRRvv+=gINvb0=ie2INRvd2INRvf0i0ivv≈≈≈−+−+INvaCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Question:+–Ov+–INv+v−v?01=R∞=2R221RRRvvINO+=orwithINOvv≈INvcINvbINvaCite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].6.002 Fall 2000 Lecture 19Buffervoltage gain = 1input impedance = ∞output impedance = 0current gain =∞power gain =∞+–Ov+–INvINOvv≈Why is this circuit
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