EE 230 Lecture 23 Nonlinear Op Amp Applications - waveform generators
Quiz 6 Obtain an expression for and plot the transfer characteristics of the following circuit. Assume =2K, 2 =8K, =0K, DD +5, SS =-5 5 2 IN OUT -5 2
And the number is? 3 8 5 4 2 6 9 7
And the number is? 3 8 5 4 2 6 9 7
Quiz 6 Solution: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume =2K, 2 =8K, =0K, DD +5, SS =-5 5 2 IN 2 OUT -5 =θ + ( -θ) HYH =θ + ( -θ) HYL 2 θ= + 2 + 2 = OUT 2
Quiz 6 Solution: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume =2K, 2 =8K, =0K, DD +5, SS =-5 5 2 IN 2 OUT -5 2 θ= 02. + = HYH HYL 2 = θ = θ +-θ =3+4=7 +-θ =-3+4= HYH HYL = θ = θ +-θ =3-4=- +-θ =-3-4= -7
Quiz 6 Solution: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume =2K, 2 =8K, =0K, DD +5, SS =-5 5 2 IN 2 OUT -5 2 HYH HYL = θ = θ +-θ =3-4=- +-θ =-3-4= -7 HYL OUT W HYH IN OUT HYL HYH IN HYH HYL = θ = θ +-θ =3+4=7 +-θ =-3+4= CENT CENT
Quiz 6 Solution: Obtain an expression for and plot the transfer characteristics of the following circuit. Assume =2K, 2 =8K, =0K, DD +5, SS =-5 HYH HYL = θ = θ +-θ =3-4=- +-θ =-3-4= -7 HYH HYL = θ = θ +-θ =3+4=7 +-θ =-3+4= OUT OUT HYL W HYH IN HYL HYH IN CENT CENT
eview from Last Time: θ= + 2 Shifted Inverting Comparator with Hysteresis HYL CENT W HYH =θ( - ) W θ + 2 = + -θ CENT If = DD, STAL = SS =- DD =2θ W CENT DD = -θ Shift can be to left or right depending upon sign of
eview from Last Time: Inversion of Hysteresis Loop θ= + 2 DD SS Noninverting Comparator with Hysteresis HYH θ = HYH θ- θ = HYL θ- If = DD, STAL = SS =- DD HYL θ -θ = = -θ -θ HYH DD HYL DD
eview from Last Time: Shifted Inverted Hysteresis Loop θ= + DD 2 SS Shifted Noninverting Comparator with Hysteresis HYL HYH CENT θ = + HYH -θ θ- θ = + HYL -θ θ- θ = + CENT -θ θ- 2 If = DD, STAL = SS =- DD θ -θ = + = + -θ -θ -θ -θ HYH DD HYL DD
Waveform Generator with Linear Triangle Waveform Goal: Determine how this circuit operates, the output waveforms, and the frequency of the output
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 = and will be constant when OUT2 = Analysis strategy: Guess state of the OUT2, solve circuit, and show where valid when OUT2 =, I will be positive and OUT will be decreasing linearly when OUT2 =, I will be positive and OUT will be increasing linearly
Waveform Generator with Linear Triangle Waveform DD SS Observe T = t 3 -t = (t 2 -t ) + (t 3 -t 2 )
Waveform Generator with Linear Triangle Waveform DD SS
Waveform Generator with Linear Triangle Waveform DD SS Guess OUT2 = will obtain t 2 -t t = - dτ+ t C OUT OUT t ( t ) = OUT HYH valid for t < t < t 2
Waveform Generator with Linear Triangle Waveform DD SS Guess OUT2 = t = - dτ+ t C OUT OUT t valid for t < t < t 2 t = OUT HYH at t=t 2, OUT will become Substituting into integral expression for OUT we obtain t = - dτ+ C HYL HYH t
Waveform Generator with Linear Triangle Waveform DD SS Guess OUT2 = valid for t < t < t 2 t = - d τ + HYL HYH C t t2 = - d τ + HYL HYH C t t2 = - HYL ( τ ) + t HYH C = - ( t t ) + 2 C HYL HYH
Waveform Generator with Linear Triangle Waveform DD SS Guess OUT2 = valid for t < t < t 2 = - ( t t ) + C HYL 2 HYH t-t=c 2 - HYH HYL
Waveform Generator with Linear Triangle Waveform DD SS Guess OUT2 = t = - dτ+ t C 2 = - ( t -t ) C HYH 3 2 HYL OUT OUT 2 t t = will obtain t 3 -t 2 Following the same approach observe It thus follows that OUT 2 HYL t-t=c HYL HYH + 3 2 valid for t 2 < t < t 3 -
Waveform Generator with Linear Triangle Waveform DD SS T = (t 2 -t ) + (t 3 -t 2 ) t-t=c 2 t-t=c 3 2 ( - ) HYH ( - ) HYL HYL HYH T=C( - ) - HYH HYL f= = t C - - HYH HYL
Waveform Generator with Linear Triangle Waveform f= C HYH - HYL - If we use the noninverting comparator with hysteresis circuit developed previously and if If = DD, STAL = SS =- DD θ= + then θ = HYH -θ DD - θ = HYL -θ -θ f= 2C θ DD 2
Stability and Waveform Generation Waveform generators provide an output with no excitation Waveform circuits are circuits that, when operated in quiescent linear condition, have one or more poles in the right half-plane Will now investigate the pole locations of waveform generators Conditions for oscillation Triangle/Square/Sinusoidal Oscillations
Poles of a Network T(s) can be expressed as X T( s ) = X OUT IN ( s) ( s ) Ns T( s ) = Ds where N(s) and D(s) are polynomials in s D(s) is termed the characteristic equation or the characteristic polynomial of the network oots of D(s) are the poles of the network
Poles of a Network Ns T( s ) = Ds Theorem: The poles of any transfer function of a linear system are independent of where the excitation is applied and where the response is taken provided the dead networks are the same Equivalently, the characteristic equation, D(s), is characteristic of a network (or the corresponding dead network) and is independent of where the excitation is applied and where the response is taken.
Poles of a Network Ns T( s ) = Ds Ds Dead Network
Poles of Networks some examples OUT T( s ) = = +Cs IN D s = + Cs Dead Network
Poles of Networks some examples OUT IN Cs OUT T( s ) = = +Cs IN D s = + Cs Dead Network
Poles of Networks some examples OUT IN OUT T( s ) = = I +Cs IN D s = + Cs Dead Network
Poles of Networks some examples I OUT T( s ) = = I +Cs IN D s = + Cs Dead Network
Poles of Networks some examples OUT T( s ) = = 2 +Cs IN D s = + Cs Dead Network
Poles of Networks some examples 2 OUT T( s ) = = I +2Cs IN = + 2 Ds Cs Dead Network Note dead network has changed as has D(s) and thus the pole