Stability and Frequency Compensation

Chapter 10

General Consideration

Unstable if

Alternatively,

(to add in phase)

(to grow in magnitude)

Body Plots

(GX,Gain cross over frequency)

(PX, phase crossoverpoint)

Worst Case Scenario (β=1)

(β increases)

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

β increases

Review Slides

Laplace Transform/Fourier Transform for RC LPF

𝑉 𝑜𝑢𝑡𝑉 𝑖𝑛

(𝑠)=1/(1+𝑠𝑅𝐶)

𝑉 𝑜𝑢𝑡𝑉 𝑖𝑛

( 𝑗ω)=1 /(1+ 𝑗ω 𝑅𝐶)

𝑠= 𝑗ω

)|=1/| +p|

Phase=-tan-1(/p)

p=1/(RC)

(Fourier Transform)

(Laplace Transform)

𝑗ω

-p

𝑗ω| +p|

Location of the pole in the left complexplane

𝜎

Complex s plane

Rules of thumb: (applicable to a pole)Magnitude:1. 20 dB drop after the cut-off frequency2. 3dB drop at the cut-off frequencyPhase:3. -45 deg at the cut-off frequency. Phase is more significantly affected by the pole than

magnitude.4. 0 degree at one decade prior to the cut-frequency5. 90 degrees one decade after the cut-off frequency

Laplace Transform/Fourier Transform for RC HPF

𝑉 𝑜𝑢𝑡𝑉 𝑖𝑛

(𝑠)=𝑠𝑅𝐶 / (1+𝑠𝑅𝐶)

𝑉 𝑜𝑢𝑡𝑉 𝑖𝑛

( 𝑗ω)= 𝑗ω𝑅𝐶 /(1+ 𝑗ω𝑅𝐶)

𝑠= 𝑗ω

)|=| |/| +p|

Phase=90-tan-1(/p)

p=1/(RC)Zero at DC.

(Fourier Transform)

(Laplace Transform)

𝑗ω

-p

𝑗ω| +p|

Location of the pole in the left complexplane

𝜎

Complex s plane

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

Time-Domain Response of a System Versus Position of Poles

(unstable)(constant magnitudeOscillation)

(exponential decay)

The location of the poles of a closedLoop system is shown.

One-Pole System(one-pole feedforward amplifier)

Ione pole system isUnconditionally Stable.

Root Locus Plot for a One Pole System

As the loop gain increase (e.g. β), the pole moves away from the origin.

Two-Pole System

The system is stable since theloop gain is less than 1 at a frequencyFor which the angle(βH(ω))=-180.

When β is reduced,the system becomesmore stable.

Assumption:β does not dependon frequency.

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

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

Three-Pole System

Relative Location of GX and PX

• Case 1: <βH(jω1)=-175o

• Case 2: <βH(jω1) such that GX<<PX• Case 3: <βH(jω1)=-135

Case 1: <βH(jω1)=-175o

The system is technically stable, butit suffers from ringing.

Case 2: <βH(jω1) such that GX<<PX

Case 3: <βH(jω1)=-135

(30% peaking)

Phase Margin

Phase Margin=Phase at GX +180o

Transient Response Versus PM

• Peaking is usually correlated with ringing in the time domain!

PM=60o, usually the optimum value.

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.

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

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)

Option 1: Push PX OUT

(minimize the #of poles)

What’s the problem?Each stage contributes a pole.

Reduce # of stages impliesdifficult trade-off of gain versus output swings.

Option 2: Push GX In

Problem:

Bandwidth is sacrificed for stability.

Typical Approach

• Minimize the number of poles first.• Use compensation to move the GX

towards the origin next.

Review:Small Signal Equivalent Model

(Transmission Zero)

Review:Differential Pair with Current Mirror

(Slow Path)

(Fast Path)

Caution:Not a good Idea to cancel the non-dominant pole withThe zero. You may have settling problem due to a step response.

Telescopic Op-Amp

(Dominant pole)

(Non-dominant Pole, and a zero)

CGS5+CGS6+CDB5+2CGD6+CDB3+CGD3

Bode Plot of the Telescopic Op-Amp

(Zero at 2ωp is neglected)

(Typically Limits the Phase Margin)

Compensation Strategy (1)

Key points:1. The phase contribution of the dominant pole

near the PX is approximatelyl -90 degrees. 2. Moving the dominant pole towards the origin

affects the gain, but not the phase near PX.

Compensation Strategy (2)Assumptions:1. The second non-dominantpole occupies a much higher frequency.2. Design for a phase margin of 45 deg.The desired phase at ωp,A is -135 deg.

Steps:3. Determine the desired location of the the dominant pole. (ω'p,A). Start from ωp,A,

Assume -20 dB/dec roll-off.

2. The load capacitance must increase byωp,A/ω'p,A.

3. The unity-gain bandwidth is ωp,A

Increasing the Output Resistance does not Compensate the Op-Amp

Fully Differential Telescopic Cascode

If M5 and M3 have the same gmoverid, gm5=gm3.

Since μp< μn, W5,6>W3,4 andN,K are more capacitive than X,Y.

X, Y are the non-dominant pole. (reason: 1/gm of M3,4)

(Dominant Pole)

Effect of the Capacitance at N,K

N, K lowers the dominant pole, butdoes not introduce an additional pole.

Mirror pole is avoided!

Two Stage Op-Amp

(low impedance node,high freq pole)

Cascode, High impedance node, dominant pole

High impedance node, CL may be large, dominant pole

Bode Plots for the Two-Stage Op-Amp

Assumption: ωp,E< ωp,A

Stability Problem!

Compensation strategy:1. Move GX below PX.2. Move the unity gain bandwidth below or equal to the second pole.

Problem: A large compensation capacitor may be required!

Miller Compensation

Create a larger capacitor at E using Miller effect.

Net effect: a moderate value capacitor (Cc) can beused to lower ωp,E, thus saving chip area.

High Frequency Model of CS Stage (Review)

Dominant Pole Approximation (Review)

A zero in the right-half-plane.

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.

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.

Zero In the Right-Half Plane

Zero Frequency:

Zero is produced by Cc and CGD9 which introduced a lowimpedance path from input to output.

Effect of a zero:1. A RHP zero slows down the drop in the magnitude by contributing 20dB/dec.2. A RHP zero decreases phase by contributing a phase of –tan-1(ω/ωz)

Net effect on phase:3. GX is pushed to a higher frequency.4. PX is pushed to a lower frequency

A RHP Zero’s effect on PM

GX is pushed out

PX is pushed in

Moving the RHP Zero

(Derive this!)

(Cancel the non-dom. polewith the left plane zero)

Downside of Miller Compensation

Increase CL will lower ωp2

Dominant pole

See next slide.

Susceptibility of Miller Compensation Due to an Increased CL (Revised)

As CL increases, the stability of the circuitcan degrade.

Other Compensation Techniques (1)

Feedback path : CCFeedfoward path

Feedfoward current flows through M2 to the power supply.

Other Compensation Techniques (2)

The DC of output is not tied to the DC of X.

Note the current of the feedback current.

The feedforward path is broken sinceVX is not sensitive to small signal at Vd2.

Misc. Slides

Slewing in Two-Stage Op-Amps

The negative feedbackof the amplifier makes itsuch that V+ ~V-

Small Signal Intuition

Virtual ground

Positive Slew Rate

Slew Rate:

If M5 is not wide enough to supportISS+I1, VX drops to drive M1 into triodeRegion.

Negative Slew Rate

If ISS=I1, M5 will be turned off.If ISS>I1, V3 will enter the triode region.