1 / 56

Stability and Frequency Compensation

Stability and Frequency Compensation. Chapter 10. General Consideration. Unstable if. Alternatively,. (to grow in magnitude). (to add in phase). Body Plots. ( GX,Gain cross over frequency). (PX, phase crossover point). Worst Case Scenario ( β =1). β increases. ( β increases).

seoras
Download Presentation

Stability and Frequency Compensation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stability and Frequency Compensation Chapter 10

  2. General Consideration Unstable if Alternatively, (to grow in magnitude) (to add in phase)

  3. Body Plots (GX,Gain cross over frequency) (PX, phase crossover point)

  4. Worst Case Scenario (β=1) β increases (β increases) Assumptions: 1. β does not depend on frequency. 2. β ≤1 The magnitude plots are shifted down. The system becomes more stable as β is reduced. βH(ω) with β=1 represents the worst case stability. H(ω) is often used to analyze stability.

  5. Review Slides

  6. Laplace Transform/Fourier Transform for RC LPF (Laplace Transform) Complex s plane |+p| (Fourier Transform) -p p=1/(RC) Location of the polein the left complex plane )|=1/|+p| Phase=-tan-1(/p)

  7. Rules of thumb: (applicable to a pole) Magnitude: 20 dB drop after the cut-off frequency 3dB drop at the cut-off frequency Phase: -45 deg at the cut-off frequency. Phase is more significantly affected by the pole than magnitude. 0 degree at one decade prior to the cut-frequency 90 degrees one decade after the cut-off frequency

  8. Laplace Transform/Fourier Transform for RC HPF (Laplace Transform) Complex s plane |+p| (Fourier Transform) -p p=1/(RC) Zero at DC. Location of the polein the left complex plane )|=||/|+p| Phase=90-tan-1(/p)

  9. Zero at the origin. Thus phase(f=0)=90 degrees. The high pass filter has a cut-off frequency of 100.

  10. Time-Domain Response of a System Versus Position of Poles (constant magnitude Oscillation) (unstable) (exponential decay) The location of the poles of a closed Loop system is shown.

  11. One-Pole System (one-pole feedforward amplifier) Ione pole system is Unconditionally Stable.

  12. Root Locus Plot for a One Pole System As the loop gain increase (e.g. β), the pole moves away from the origin.

  13. Two-Pole System When β is reduced, the system becomes more stable. Assumption: β does not depend on frequency. The system is stable since the loop gain is less than 1 at a frequency For which the angle(βH(ω))=-180.

  14. Root Locus Plot for a Two-Pole System (1)

  15. Root Locus Plot for a Two-Pole System (2)

  16. Three-Pole System

  17. Relative Location of GX and PX • Case 1: <βH(jω1)=-175o • Case 2: <βH(jω1) such that GX<<PX • Case 3: <βH(jω1)=-135

  18. Case 1: <βH(jω1)=-175o The system is technically stable, but it suffers from ringing.

  19. Case 2: <βH(jω1) such that GX<<PX

  20. Case 3: <βH(jω1)=-135 (30% peaking)

  21. Phase Margin Phase Margin=Phase at GX +180o

  22. Transient Response Versus PM • Peaking is usually correlated with ringing in the time domain! PM=60o, usually the optimum value.

  23. Caution • PM is useful for small signal analysis. • For large signal step response of a feedback system, the nonlinear behavior is usually such that a system with satisfactory PM may still exhibit excessive ringing. • Transient analysis should be used to analyze large signal response.

  24. Example: A system with PM =60 but still exhibit ringing when driven by a large signal

  25. Frequency Compensation • Reason for frequency compensation: • |βH(ω)| does not drop to unity when <βH(ω) reaches -180o. • Possible Solution: (Push PX Out) (push GX in)

  26. Option 1: Push PX OUT (minimize the # of poles) What’s the problem? Each stage contributes a pole. Reduce # of stages implies difficult trade-off of gain versus output swings.

  27. Option 2: Push GX In Problem: Bandwidth is sacrificed for stability.

  28. Typical Approach • Minimize the number of poles first. • Use compensation to move the GX towards the origin next.

  29. Review:Small Signal Equivalent Model (Transmission Zero)

  30. Review:Differential Pair with Current Mirror (Slow Path) (Fast Path) Caution: Not a good Idea to cancel the non-dominant pole with The zero. You may have settling problem due to a step response.

  31. Telescopic Op-Amp CGS5+CGS6+CDB5+2CGD6+CDB3+CGD3 (Non-dominant Pole, and a zero) (Dominant pole)

  32. Bode Plot of the Telescopic Op-Amp (Typically Limits the Phase Margin) (Zero at 2ωp is neglected)

  33. Compensation Strategy (1) Key points: The phase contribution of the dominant pole near the PX is approximatelyl -90 degrees. Moving the dominant pole towards the origin affects the gain, but not the phase near PX.

  34. Compensation Strategy (2) Assumptions:1. The second non-dominant pole occupies a much higher frequency. Design for a phase margin of 45 deg. The desired phase at ωp,A is -135 deg. Steps: Determine the desired location of the the dominant pole. (ω'p,A). Start from ωp,A, Assume -20 dB/dec roll-off. The load capacitance must increase by ωp,A/ω'p,A. 3. The unity-gain bandwidth is ωp,A

  35. Increasing the Output Resistance does not Compensate the Op-Amp

  36. Fully Differential Telescopic Cascode If M5 and M3 have the same gmoverid, gm5=gm3. Since μp<μn, W5,6>W3,4and N,K are more capacitive than X,Y. X, Y are the non-dominant pole. (reason: 1/gmof M3,4) (Dominant Pole)

  37. Effect of the Capacitance at N,K Mirror pole is avoided! N, K lowers the dominant pole, but does not introduce an additional pole.

  38. Two Stage Op-Amp Cascode, High impedance node, dominant pole High impedance node, CL may be large, dominant pole (low impedance node, high freq pole)

  39. Bode Plots for the Two-Stage Op-Amp Stability Problem! Assumption: ωp,E< ωp,A Compensation strategy: Move GX below PX. Move the unity gain bandwidth below or equal to the second pole. Problem: A large compensation capacitor may be required!

  40. Miller Compensation Create a larger capacitor at E using Miller effect. Net effect: a moderate value capacitor (Cc) can be used to lower ωp,E,thus saving chip area.

  41. High Frequency Model of CS Stage (Review)

  42. Dominant Pole Approximation (Review) A zero in the right-half-plane.

  43. Pole Splitting as a Result of Miller Compensation (Simplified Model) (Cc provides a low impedance path, thus reducing the resistance at node E to 1/gm.

  44. Observations • By using Cc, • Only a moderate value capacitor is required. • The output pole is shifted to a higher frequency, allowing a greater unity bandwidth.

  45. Zero In the Right-Half Plane Zero Frequency: Zero is produced by Cc and CGD9 which introduced a low impedance path from input to output. Effect of a zero: A RHP zero slows down the drop in the magnitude by contributing 20dB/dec. A RHP zero decreases phase by contributing a phase of –tan-1(ω/ωz) Net effect on phase: GX is pushed to a higher frequency. PX is pushed to a lower frequency

  46. A RHP Zero’s effect on PM GX is pushed out PX is pushed in

  47. Moving the RHP Zero (Derive this!) (Cancel the non-dom. pole with the left plane zero)

  48. Downside of Miller Compensation Dominant pole Increase CL will lower ωp2 See next slide.

  49. Susceptibility of Miller Compensation Due to an Increased CL (Revised) As CL increases, the stability of the circuit can degrade.

  50. Other Compensation Techniques (1) Feedback path : CC Feedfoward path Feedfowardcurrent flows through M2 to the power supply.

More Related