EE 230 Lecture 23. Nonlinear Op Amp Applications. Waveform Generators

Transcription

1 EE 230 Lecture 23 Nonlinear Op Amp Applications Waveform Generators

2 Quiz 7 An oscillator based upon a comparator with hysteresis is shown. If STAH =2 and SATL =-2, determine the peak value of

3 And the number is? ? 6 9 7

4 Quiz 7 An oscillator based upon a comparator with hysteresis is shown. If STAH =2 and SATL =-2, determine the peak value of Solution: The peak value of the OUT waveform is determined by the boundaries of the Hysteresis window OUTMAX = R 2K SATH = 2 = 2 R +R 2K 2

5 Correction from Last Lecture Modifications of Comparator with Hysteresis OUT SATH Hysteresis Region θ SATH IN R θ = R+R 2 θ SATL SATL IN OUT OUT SATH ( -θ) +θ REF SATH R 2 R R REF θ = R+R 2 OUT ( -θ) +θ REF SATL SATL OUT SATH IN Hysteresis Region R R 2 R IN θ = R 2 Note this is the basic inverting amplifier with op amp terminals interchanged Hysteresis Region -θ SATH SATL -θ SATL IN Many other ways to control position and size of hysteresis window

6 Review from Last Lecture Comparison of basic noninverting amplifier structures If ideal op amps both have gain A FB R =+ R 2 SATH Region OUT SATH Region 3 v IN θ SATL θ SATH θ SATL θ SATH SATL SATL Region 3 Serves as an amplifier directly Stable No hysteresis loop Not useful as an amplifier directly Unstable Serves as comparator with hysteresis

7 Review from Last Lecture Waveform Generator OUT SATH R θ= R+R 2 θ SATH θ SATL SATL t this process repeats itself t ( ) SATL θ- = RCln θsath-satl the rise time and the fall times are identical the period of the nearly triangular waveform is thus 2t ( ) SATL θ- T = 2t = 2RCln θsath-satl f = = T 2RC θ - SATL ( θ-) ln SATH SATL If SATL =- SATH, this simplifies to f = 2RC +θ ln -θ

8 Review from Last Lecture for SATL =- SATH R θ= R+R f= 2RC +θ ln -θ 2 Square and distorted triangular output waveforms Slope of square wave is determined by SR of Op Amp

9 Waveform Generator with Linear Triangle Waveform Goal: Determine how this circuit operates, the output waveforms, and the frequency of the output

10 Waveform Generator with Linear Triangle Waveform Lets first check stability Since stability is determined by the poles of a linear network, must first assume devices are operating linearly

11 Waveform Generator with Linear Triangle Waveform Lets first check stability Since stability is determined by the poles of a linear network, must first assume devices are operating linearly OUT OUT IN IN Noninverting Comparator wth Hysteresis What is the linear model of this comparator? Noninverting Comparator wth Hysteresis Linear region is area where slope is negative Recall, in this region, OUT= -K0IN -K 0 Linear Comparator Model

12 Waveform Generator with Linear Triangle Waveform Lets first check stability Since stability is determined by the poles of a linear network, must first assume devices are operating linearly Linear circuit model C OUT R -K 0 IN Linear Comparator Model Inverting Integrator Linear Circuit Model with Excitation (Recall do not need to provide excitation to find poles but details will be discussed later)

13 Waveform Generator with Linear Triangle Waveform Lets first check stability C OUT R -K 0 IN Linear Comparator Model Inverting Integrator Linear Circuit Model with Excitation IN ( sc+g ) = sc+outg OUT = -K0 T( s ) = OUT IN = -K0 s+ RC K 0 s- RC Single pole at s = K 0 RC The system is unstable!

14 Waveform Generator with Linear Triangle Waveform Since the comparator will be in one of two states, the current in the resistor will be constant when OUT2 = SATH and will be constant when OUT2 = SATL Analysis strategy: Guess state of the OUT2, solve circuit, and show where valid when OUT2 = SATH, I R will be positive and OUT will be decreasing linearly when OUT2 = SATH, I R will be positive and OUT will be increasing linearly

15 Waveform Generator with Linear Triangle Waveform SATH SATL DD SS Observe T = t 3 -t = (t 2 -t ) + (t 3 -t 2 )

16 Waveform Generator with Linear Triangle Waveform SATH SATL DD SS

17 Waveform Generator with Linear Triangle Waveform SATH SATL DD SS Guess OUT2 = SATH will obtain t 2 -t t = - dτ+ t RC ( ) OUT SATH OUT t ( t ) = OUT HYH valid for t < t < t 2

18 Waveform Generator with Linear Triangle Waveform SATH SATL DD SS Guess OUT2 = SATH t = - dτ+ t ( ) RC ( ) OUT SATH OUT t t = OUT HYH at t=t 2, OUT will become SATL Substituting into integral expression for OUT we obtain t RC valid for t < t < t 2 2 = - dτ+ HYL SATH HYH t

19 Waveform Generator with Linear Triangle Waveform SATH SATL DD SS Guess OUT2 = SATH valid for t < t < t 2 t2 = - d τ + HYL SATH HYH RC t t2 = - d τ + HYL SATH HYH RC t t2 = - HYL SATH ( τ ) + t HYH RC = - ( t t ) + 2 RC HYL SATH HYH

20 Waveform Generator with Linear Triangle Waveform SATH SATL DD SS Guess OUT2 = SATH valid for t < t < t 2 = - ( t t ) + RC HYL SATH 2 HYH t-t=rc 2 - ( ) HYH SATH HYL

21 Waveform Generator with Linear Triangle Waveform SATH DD SATL SS Guess OUT2 = SATL will obtain t 3 -t 2 Following the same approach observe t = - dτ+ t RC 2 = - ( t -t ) RC ( ) OUT SATL OUT 2 t t = ( ) OUT 2 HYL It thus follows that HYH SATL 3 2 HYL (valid for t 2 < t < t 3 ) t-t=rc ( ) HYL HYH SATL

22 Waveform Generator with Linear Triangle Waveform SATH DD SATL SS T = (t 2 -t ) + (t 3 -t 2 ) t-t=rc 2 t-t=rc 3 2 ( - ) HYH SATH ( - ) HYL HYL SATL HYH T=RC( - ) - HYH HYL SATH SATL SATH f= = t RC - - SATL ( )( ) HYH HYL SATL SATH

23 Waveform Generator with Linear Triangle Waveform SATL SATH f= RC HYH - HYL SATL - SATH ( )( ) If we use the noninverting comparator with hysteresis circuit developed previously and if R If SATH = DD, STAL = SS =- DD θ= R+R then θ = HYH -θ DD - θ = HYL -θ -θ f= 2RC θ DD 2

24 Example: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume R =2K, R 2 =8K, R=0K, DD +5, SS =-5

25 Example: Solution: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume R =2K, R 2 =8K, R=0K, DD +5, SS =-5 =θ + ( -θ) HYH SATH R =θ + ( -θ) HYL SATL R R θ= R+R = OUT 2

26 Example: Solution: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume R =2K, R 2 =8K, R=0K, DD +5, SS = = OUT 2 R θ= 02. R+R = 2 HYH HYL = θ = θ Upper Circuit SATH SATL ( ) ( ) +-θ =3+4=7 R R +-θ =-3+4= HYH HYL = θ = θ Lower Circuit SATH SATL ( ) ( ) +-θ =3-4=- R +-θ =-3-4= -7 R

27 Example: Solution: HYH HYL = θ = θ SATH SATL ( ) ( ) +-θ =3+4=7 R R +-θ =-3+4= HYL CENT HYH HYH HYL = θ = θ SATH SATL ( ) ( ) +-θ =3-4=- R R +-θ =-3-4= -7 HYL W HYH CENT

28 Example: Solution: = OUT 2 4 Assuming SATL =- SATH OUT SATH -7 W - 7 IN SATL

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30 Poles are inherent and unique characteristics of any linear network.

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49 End of Lecture 23