EE 230 Lecture 25. Waveform Generators. - Sinusoidal Oscillators The Wein-Bridge Structure

Transcription

1 EE 230 Lecture 25 Waveform Generators - Sinusoidal Oscillators The Wein-Bridge Structure

2 Quiz 9 The circuit shown has been proposed as a sinusoidal oscillator. Determine the oscillation criteria and the frequency of oscillation. Assume the op amps are ideal. X C C A A

3 And the number is? ? 6 9 7

4 Quiz 9 The circuit shown has been proposed as a sinusoidal oscillator. Determine the oscillation criteria and the frequency of oscillation. Assume the op amps are ideal. X C C Solution: X A A X X sc G X sc+g - G X+ = 0 sc ( X ) ( ) ( ) Ds=sC+sCG - G +G ( ) G - G C 2 X Ds=s+s + Oscillation criteria: G=G X ( C) 2 ω OSC = C

5 eview from Last Time: Sinusoidal Oscillation A circuit with a single complex conjugate pair of poles on the imaginary axis at +/- jβ will have a sinusoidal output given by OUT ( ) ˆ β θ ) X t =2a sin( t+ k The frequency of oscillation will be β rad/sec but the amplitude and phase are indeterminate

6 eview from Last Time: Sinusoidal Oscillation Criteria A network that has a single complex conjugate pair on the imaginary axis at ±jω and no HP poles will have a sinusoidal output of the form X 0 (t)=asin(ωt+θ) A and θ can not be determined by properties of the linear network

7 eview from Last Time: Characteristic Equation equirements for Sinusoidal Oscillation OUT Characteristic Equation Oscillation Criteria: If the characteristic equation D(s) has exactly one pair of roots on the imaginary axis and no roots in the HP, the network will have a sinusoidal signal on every nongrounded node.

8 eview from Last Time: elationship between Barkhausen Criteria and Characteristic Equation Criteria for Sinusoidal Oscillation Characteristic Equation Oscillation Criteria (CEOC) If the characteristic equation D(s) has exactly one pair of roots on the imaginary axis and no roots in the HP, the network will have a sinusoidal signal on every nongrounded node Barkhausen Oscillation Criteria A feedback amplifier will have sustained oscillation if Aβ = - Differences: Challenge:.Barkhausen requires a specific feedback amplifier architecture 2.Sustained oscillation says nothing about wave shape It is impossible to place the poles of any network exactly on the imaginary axis Sinusoidal Oscillator Design Approach: Place on pair of cc poles slightly in HP and have no other HP poles With this approach, will observe minor distortion of output waveforms

9 eview from Last Time: Sinusoidal Oscillator Design Strategy Build networks with exactly one pair of complex conjugate roots slightly in the HP and use nonlinearities in the amplifier part of the network to limit the amplitude of the output ( i.e p =α ±jβ α is very small but positive) Nonlinearity will cause a small amount of distortion Frequency of oscillation will be very close to but deviate slightly from β Must be far enough in the HP so the process and temperature variations do not cause movement back into LHP because if that happened, oscillation would cease!

10 eview from Last Time: Sinusoidal Oscillator Design Consider the following circuit: K Ds ( ) = s+s C C C CC Oscillation Condition: K C C C = This is achieved by having K satisfy the equation C K = + + C 2 2 ω = OSC CC 2 2

11 eview from Last Time: Sinusoidal Oscillator Design Consider the special practical case where = 2 = and C =C 2 =C: C 2 K = C = 2 ω OSC = C This is termed the Wein Bridge Oscillator One of the most popular sinusoidal oscillator structures Practically make K slightly larger than 3 and judiciously manage the nonlinearities to obtain low distortion

12 The Wein-Bridge Oscillator Another Perspective: C 2 K = C =

13 The Wein-Bridge Oscillator Another Perspective: C 2 K = C = 2 Note this is a feedback amplifier with gain K and Z β = Z+Z 2 Lets check Barkhausen Criteria for this circuit

14 The Wein-Bridge Oscillator Lets check Barkhausen Criteria for this circuit Z β = Z+Z 2 C 2 K = C = 2 Z K Cs Kβ = = Z +Z s C C +s C + C + C + Setting Kβ= -, obtain K C s ( ) 2 scc+sc+c+c ( ) + = ( ) 2 s 2CC 2+s C 2+C +2C2 KC +=0

15 The Wein-Bridge Oscillator Lets check Barkhausen Criteria for this circuit Z β = Z+Z 2 C 2 K = C = 2 Putting in s= jω, obtain the Barkhausen criteria ( ) 2 -ω 2CC 2 +j ω C 2+C+C 2 2 KC =0 Solving, must have (from the imaginary part) C K = + + C 2 2 And this will occur at the oscillation frequency of (from real part) ω = OSC CC 2 2

16 The Wein-Bridge Oscillator Basic implementation for the equal, equal C case + K X 0 Z 2 Z ω = OSC C

17 The Wein-Bridge Oscillator Amplifier Transfer Characteristics OUT 3 SATH ω = OSC C IN SATL Slope slightly larger than 3 Amplitude of oscillation will be approximately SATH (assuming SATH =- STATL ) Distortion introduced by the abrupt nonlinearities when clipping occurs

18 The Wein-Bridge Oscillator Amplifier Transfer Characteristics OUT 3 SATH ω = OSC C IN OUT SATH 3 SATL IN SATL Abrupt nonlinearities cause distortion Better performance (reduced nonlinearity) can be obtained by introducing less abrupt nonlinearities to limit amplitude

19 Amplitude Limiting in Noninverting Amplifier Structure ( 2+ε) Observe: Amplifier gain changes from 3 + ε for < IN < 2 to 0 for IN < or IN > 2 SATH and SATL strongly dependent upon op amp bias voltages DD and SS This nonlinearity in the real amplifier will limit the output signal amplitude Can cause rather significant distortion

20 Obtain slope >3 for < IN < and slope <3 for < IN < 2 and for 2 < IN < - Limit IN to interval 2 < IN < 2 Can we do this? If possible, hard nonlinearity associated with amplifier saturation will not be excited Dramatic reduction in distortion is anticipated

21 Consider: NLD Transfer Characteristics D = XX for I D > 0 I D = 0 for D < XX (will assume XX > 0 )

22 Consider: Analysis of Amplifier Circuit: Case : I D = 0 and I D2 = 0 D = XX for I D > 0 I D = 0 for D < XX Solution: ( + ) 2 3 OUT = + IN Must determine where this part of the solution is valid

23 alid for D < XX and D2 < XX but D = - 3 and D2 = 3 thus, valid for 3 > - XX and 3 < XX but = ( ) 3 3 o in = in = 2 3 in 3 = 3 in valid for < and > 3 3 IN XX IN XX < < XX IN XX 3 3

24 Graph of solution for Case XX 3 (2 + 3) + XX 3

25 Case 2: NLD 2 is in the conducting state ( D2 = XX ) NLD is in the nonconducting state (I D = 0)

26 Solution for Case 2 Continued Applying superposition we obtain 2 + = OUT IN xx XX 3 This solution is valid for D < 0 and I D2 > 0 But: D = - XX and - - I = - D2 OUT IN XX XX Substituting the validity conditions, we obtain IN XX < 0 and - - OUT IN XX - XX >0 2 3 The first of these inequalities is valid provided XX >0 and substituting the expression for OUT into the second, we obtain after simplification > 3 XX

27 Solution for Case 2 Continued XX >0 IN > 3 XX Assuming XX >0, the region where Case 2 is valid is thus determined by the second inequality + 2 XX 3

28 Case 3: NLD 2 is nonconducting (I D2 = 0) NLD is conducting ( D = XX ) = + 2 OUT IN XX XX 3 + 2

29 Solution for Case 3 continued: This solution is valid for D2 < 0 and I D > 0 But: IN < D2 = - XX and 3 XX - - I = - D IN OUT XX XX Substituting the validity conditions, we obtain XX < 0 and - - IN OUT XX - XX >0 2 3 The first of these inequalities is valid provided XX >0 and substituting the expression for OUT into the second, we obtain after simplification XX 3 + 2

30 Case 4: NLD and NLD 2 both conducting (this case never happens and need not be considered since we already have a solution for all inputs) Thus, if we neglect the saturation of the op amp, we can write an expression for the output as 2 + IN + xx IN > XX "2" OUT = + IN XX < IN < XX "" IN xx IN < XX "3" 3 This is shown graphically, along with the saturation of the op amp, on the following slide

31 Overall Transfer Characteristics + 2 XX 3 (2 + 3) + XX 3 + 2

32 Overall Transfer Characteristics D = XX for I D > 0 I D = 0 for D < XX If XX =0.6, this represents a good approximation to the transfer characteristics of a silicon diode. We thus can replace the NLD with a diode and obtain the amplitude stabilized Wien-Bridge oscillator

33 Wein Bridge Oscillator with Amplitude Stabilization 2 < > 2 ωosc C

34 Wein Bridge Oscillator an alternative view of same circuit using feedback concepts Note double inversion ( ) β s +Cs = +Cs + +Cs Cs [ ] A =+ FB 2 < 2 [ 3-A 2 FB ] > 2 Ds=s+s ( ) + C C A =3+ε FB ωosc ( ) 2 C

35 End of Lecture 25