EE 230 Lecture 23 Spring 2010

EE 230 Lecture 23 Spring 2010 - EE 230 Lecture 23 Nonlinear...

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EE 230 Lecture 23 Nonlinear Op Amp Applications – Waveform Generators
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Quiz 17 An oscillator based upon a comparator with hysteresis is shown. If V STAH =12V and V SATL =-12V, determine the peak value of
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And the number is ? 1 3 8 4 6 7 5 2 9 ?
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Quiz 17 An oscillator based upon a comparator with hysteresis is shown. If V STAH =12V and V SATL =-12V, determine the peak value of Solution: The peak value of the V OUT1 waveform is determined by the boundaries of the Hysteresis window V OUT1MAX = 1 SATH 12 R 2K V 12V = 2V R +R 12K =
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Modifications of Comparator with Hysteresis 1 12 R θ = R+R V OUT R 2 R 1 1 R θ = V IN V REF V SATL V IN V SATH Hysteresis Region V OUT SATH - θ V SATL - θ V V OUT R 2 R 1 1 2 R θ = R V IN V SATL V IN V SATH Hysteresis Region V OUT ( ) REF SATH 1- θ V+ θ V ( ) REF SATL 1- θ θ V V SATL V IN V SATH Hysteresis Region V OUT SATH θ V SATL θ V Many other ways to control position and size of hysteresis window Note this is the basic inverting amplifier with op amp terminals interchanged Correction from Last Lecture
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• Serves as an amplifier directly • Stable • No hysteresis loop • Not useful as an amplifier directly • Unstable • Serves as comparator with hysteresis If ideal op amps both have gain 2 FB 1 R A= 1 + R SATH θ V SATH V SATL V SATL θ V V OUT v IN SATH θ V SATH V SATL V Region 1 SATL θ V Region 3 Region 3 Comparison of basic noninverting amplifier structures Review from Last Lecture
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1 12 R θ = R+R Waveform Generator ( ) ln SATL 1 SATH SATL V θ -1 tR C θ V- V ⎛⎞ =− ⎜⎟ ⎝⎠ V SATH V SATL V OUT t θ V SATH θ V SATL this process repeats itself the rise time and the fall times are identical the period of the nearly triangular waveform is thus 2t 1 ( ) 2l n SATL 1 SATH SATL V θ -1 T2 t R C θ V == () 1 ln SATH SATL SATL 11 f R C θ V V θ -1 If V SATL =-V SATH , this simplifies to 1
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EE 230 Lecture 23 Spring 2010 - EE 230 Lecture 23 Nonlinear...

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