ECEN 457 (ESS)

Op-Amps Stability and Frequency Compensation Techniques

Analog & Mixed-Signal Center Texas A&M University

Reference: Sergio Franco, “Design with Operational Amplifiers and Analog Integrated Circuits” 4th Edition , Chapter 8, 2015 Stability of Linear Systems

 Harold S. Black, 1927  Negative feedback concept  Negative feedback provides: • Gain stabilization • Reduction of nonlinearity • Impedance transformation  But also brings: • Potential stability problems • Causes accuracy errors for low dc gain  Here we will discuss frequency – compensation techniques

2 Stability Problem

LOAD INPUT xd xo A(s) xi

xf β

A(s) 1/  A(s) H (s)    CL 1 1 As 1 1T (s) A(s)

 Feedback forces xd to become smaller

 It takes time to detect xo and feedback to the input

 xd could be overcorrected (diverge and create instability)

 How to find the optimal (practical) xd will be based on frequency compensation techniques

3 Gain Margin & Phase Margin

|T| (dB)  G is the number of dBs by T0 M which | T ( j   180  ) | can increase f x f-180º f (dec) until it becomes 0 dB GM 1 GM  20log |T j180 |

0 fx f-180º f (dec)  Phase margin ϕm is the number Tj  -90º of degrees by which x

ϕ can be reduced until it reaches -180º m –π (-180º) -270º m 180  T jx 

Conceptual Gain Margin GM or and Phase Margin ϕ m T jx  m 180

* ωx is the crossover frequency 4 Gain Margin & Phase Margin   At the crossover point, 

jm T j x 1T j x 1m 180 e

 The non-ideal closed loop transfer function becomes

Ajx  Ajx  Aideal H j     CL x 1 Aj 1Tj 11/T j x x  x A A  ideal  ideal  jx 1 e 1 cos m  j sinm 1  1 H CL j x  AIdeal , AIdeal  2   1 cos m  sin m

5 Gain Margin & Phase Margin

 Observe that different ϕm yield different errors. i.e.

H j  • In practical systems, ϕm CL x ϕm = 60º is required 90º 0.707

60º 1.00 • A worst case ϕm = 45º for a typical lower limit 45º 1.31 • For ϕ < 60º, we have 30º 1.93 m Ajx   Aideal 15º 3.83 indicating a peaked closed- ∞ (oscillatory 0º loop response. behavior)

6 Why a dominant pole is required for a stable amplifier?

 A good Op amp design implies:  (i) A ( j  )  1 over as wide a band of frequencies as possible (ii) The zeroes of  A ( j  )  1  0 must be all in the left-hand plane Note that A( j)  0 These two conditions often conflict with each other. These trade-offs should be carefully considered. Let’s consider a practical amplifier characterized with these poles,  A   A    A(s)   0  0 1 2 3 1 s / 1 1 s / 2 1 s /3 s  1 s  2 s 3 The characteristic equation becomes     A   A(s) 1  0 1 2 3 1  0  s  1 s  2 s  3   s  1 s  2 s  3  A0 1 23  0

7        Critical Value of βA0      3 2 s  s 1  2  1  s 1 2  1 3  2 3  1 2 3 1 A0  0

Note that βA0 is the critical parameter that determines the pole locations for a given α1, α2 and α3 (0≤ β ≤1). Furthermore, when βA0 = 0, the roots are at -α1, -α2 and -α3. Therefore, for small βA0, the roots should be in the left-hand plane (LHP). However, for

βA0>>1, two of the roots might be forced to move to the right- hand plane (RHP). This can be verified by applying Routh’s stability criterion. Let us write the polynomial as 3 2 b3s  b2s  b1s  b0  0 In order to have, in the above equation, left half plane roots, all the coefficients must be positive and satisfy

b2b1  b3b0  0

8 Critical Value of βA0  The condition for imaginary-axis roots become  b j 3 b j 2  b j   b  0 3 2  1 0 2 2 b0  b2  j b1  b3  0

Now, for s=jω being a root, both real and imaginary parts must be zero. That is, 

2 2 b0  b2  0, b1  b3  0 or b3b0  b1b2

Then the two roots are placed at b b    j 0   j 1 p2,3 b2 b3

9 Critical Value of βA0

b2b1 b0  b3 Then

1 231 A0 C  1  2 3 1 2 13  23 

Thus, the critical value of βA0 becomes       A  2  1  1  2  2  3  3 0 C  2 3 1 3 1  2

Thus when βA0 becomes (βA0)C, the amplifier will oscillate at   b1  OSC   1 2  1 3  23 b3

Also when βA0 > (βA0)C, the amplifier has RHP poles, therefore is unstable.

10 Critical Value of βA0

5  Let us consider some numerical examples. Let A0 = 10 , (i) Three equal poles α =α =α =107rad/s. 1 2  3 The amplifier oscillates at 7  OSC  1 3 10 3rad/s

5  A0 C  8, C  8/ A0  810   (ii)   2  3 , then the critical loop gains yields 1 104 104

4  2 4 210 A0 C  2  210 , thus the amplifier is stable if C   0.2 1 A0

Since   R 1 /  R 1  R 2 ,   0 . 2 causes  R 2 / R 1   4 , which means that for an inverting (non-inverting) configuration, the gain must be greater than -4 (5) to keep the amplifier stable.

11 Critical Value of βA0

(iii) Let us determine A0C under the most stringent condition β=1. Then from previous equation,

1 1  2  2 3 3 A0  2        2 3 1 3 1  2

In order to have a large A0, the poles must be widely separated. i.e. α1<<α2<<α3, then the A0 inequality can be approximated as

 2 3 A0   1 1

To obtain a conservative A0, let α2=α3, which yields

 2 A0  2 1 This inequality bounds the DC gain to provide a stable closed loop configuration. 12 Peaking and Ringing

 Peaking in the frequency domain usually implies ringing in the time domain    Normalized second-order all-pole (low pass) system 2  1 H (s)  0  s j 2 2   s 2  0 s  1 0 1 2  j  Q s j 0 Q 0

13 Peaking and Ringing   GP: Peak gain – We have the error function,  1 1 E( j )     2 (1) 2 2 D   1  1     2   2  0  Q 0 To find out the maximum value of |E(jω)|, calculate the derivative  of D(ω) and make it equal to zero   3 d  1  4 2  1  2 D( )  2  2    0  1  OR   0  2  2 4 *  2Q2  0 d  Q  0 0  

For Q  1 / 2 , use ω* in Eq. (1) and we get the peak gain 2Q2 GP   Ej0 1 (2) 4Q2 1

14 Peaking and Ringing

 OS: Overshoot – Inverse Laplace transform b Inverse s - domain : Laplace time - doamin : eat sinbt u t (s  a)2  b2 For a 2nd order all pole error function, the impulse response is   2 2 b v(t)  (t) H (s)  0  0  in s j  2 2 2 2 0 b s  a  b 0 s  s  0 a  Q  2Q 2 Inverse  1 Laplace h(t)  0 eat sin(bt)u(t) b  1  b 4Q2 0 Consider the normalized step response of this system,

t  b 0 at  1 for v (t)  u(t)   (t)dt y(t)  h(t)dt 1 e sinbt  ,  tan in  (3) 0 b a 

 2 Use the damping factor  to represent, a  0 ,b  1 0

15 Peaking and Ringing

Usually, the damping factor ξ=1/2Q is used to characterize a physical 2nd order system. Thus, we rewrite the normalized time-domain equation (3)  1 y(t) 1 0 eat sinbt  1 e0t sin1 2 t    2 0 b 1   (4)  2 b 1 2  1  tan1  tan1  , for small  a     For under damped case ξ < 1 (Q > 0.5), the overshoot is the peak value of y(t). To find the peak value, we first calculate the derivative of y(t), and make it equal to 0.  d  y(t)   e 0t sin1 2 t   2   0 d0t 1    0t 2  e cos1 0t  

16  Peaking and Ringing

 d y(t)  0 dt   0   e 0t sin1 2 t   e0t cos1 2 t   2  0 0 1   2  2 1   tan1 0t    tan

To satisfy the equality above, we have n 0t  ,n  0,1,2,... 1 2 In other words, the normalized step response y(t) in Eq.(3) achieves extreme values at time steps of n=0, 1, 2, …

17 Peaking and Ringing

n=0 n=1 n=2 n=3 n=4 n=5 n=6

Peak Q=5 Value

Time t

18 

 

 Peaking and Ringing    n  0, t  0, y(t)  0  0          1  1 2 1 2 n 1, 0t   , y(t) 1 e sin 1 e 1 2 1 2   2 2    2   2 2 1 1  1 n  2, 0t  , y(t) 1 e sin2  1 e 1 2 1 2  3 3   3 1 1 2  1 2 n  3, 0t  , y(t) 1 e sin3  1 e 1 2 1 2    Therefore, the global peak value of y(t) is achieved when n=1.

Thus, the overshoot is defined as

  Peak Value  Final Value 2 OS(%) 100 100e 1 (5) Final Value

19 Peaking and Ringing

 ϕm : Phase Margin For a 2nd order all-pole error function T (s) 1 1 E(s)    2 1T (s) 11 T (s) s 1 s 2  1 0 Q 0 Therefore, the loop gain is given by 1 1 The cross-over frequency thus can be obtained  1  2 2    2     T ( j )   x    x  1 x   2      0  0Q       20  Peaking and Ringing  Solve the equation and get the crossover frequency,   1/ 2   1 1  1/ 2   1     4 4 1  2 2  x 0  4 2  0  4Q 2Q  And thus the phase margin is   1 1  180  T j  cos1 1    cos1  4 4 1  2 2  m x  4 2   4Q 2Q   Study the relationship between phase margin and gain peaking (Eq.2) or overshoot (Eq.5), we have GP(60)  0.3dB OS(60)  8.8% Q  0.82 GP(45)  2.4dB OS(45)  23% Q 1.18

21 Peaking and Ringing

30 30 100

90

24 24 80

70

18 18 60

50 Q Factor 12 12 40 Gain Peaking (dB) Overshoot (%) Overshoot

30

6 6 20

10

0 0 0 0 10 20 30 40 50 60 0 8 16 24 32 40 48 56 64 Phase Margin (degree) Phase Margin (Degrees)

22 The Rate of Closure (ROC)

 One effective method of assessing stability for minimum phase systems from the magnitude Bode plots is by determining the ROC.

 The Rate of Closure (ROC) Determining the ROC is done by observing the slopes of and 1⁄ at their intersection point (cross-over frequency ) and deciding the magnitude of their difference. ROC  SlopeA Slope/  f  fx

The ROC is used to estimate the phase margin and therefore the stability (How ?)

23 The Rate of Closure (ROC)

 Observing a single-root transfer function 1 H ( jf )  1 jf / f0

f  f0 /10 SlopeH  0dB/ dec H  0 f 10 f0 SlopeH  20dB/ dec H  90

f  f0 SlopeH  10dB/ dec H  45  Empirical Equation

H  4.5 SlopeH  This correlation holds also if H(s) has more than one root, provided the roots are real negative, and well separated, say, at least a decade apart.

24 The Rate of Closure (ROC)

 In a feedback system, suppose both |A| and |1/β| have been graphed. ∠ ∠ ∠ ≅4.5 Thus, the ROC can be used to estimate the phase margin (Page 4)

m  180  T ( jf x )  Cases  ROC  20dB/ dec  m  90  ROC  30dB/ dec  m  45  ROC  40dB/ dec  m  0

ROC  40dB/ dec  m  0

25 The Rate of Closure (ROC)

Aol 1/β1 A(s)  100

Aol **  s  s  80 1 1       p1  p2  * 1/β2 60 Gain (dB) 40 ** 1/β3

* 20dB/dec ROC 20 *  “Stability” 1/β4 0 ** 40dB/dec ROC 1 10 100 1K 10K 100K 1M  “Marginal Stability” Frequency(Hz)

26 Stability in Constant-GBP OpAmp

 A(s)  t  Constant-GBP OpAmp (i.e. s  j j  ) • Unconditionally stable with frequency-independent feedback, or    0 . (e.g. in a non-inverting or inverting amplifier, the feedback network contains only resistors) • Stable for any β≤1. • In feedback systems, since now we have T  A  A , A  90 these circuits enjoy

m 180 Ajfx 18090  90 • Typically, due to additional high-order poles in OpAmps,

60 m  90

27 Feedback Pole

 Feedback Pole Feedback network includes reactive elements  Stability may no longer be unconditional  jf  0 f(dB) |A| |1/β| 1 jf / f p |A(jfx)|

1/β fz 0

 A pole fp (or a zero) of β ft f (dec) f β f becomes a zero fz (or a x 0 t pole) for 1/β. β0 fp

 For the case fz  0 ft

The effect of a pole within the feedback loop

28 Feedback Pole

 Examine the error function

, , ⁄

Using OpAmp high-frequency approximation: 1  1 E(s)   1 f f 2 1 1 j   A ft 0 ft 0 f z Refer to page 14, and we have s=j2πf. The peak value of E(s) can be obtained. For Q >> 1, the approximate result is 

fx  fz 0 ft ,Q  0 ft / fz

29 Feedback Pole

 The lower fz compared to β0ft, the higher the Q and, hence, the more pronounced the peaking and ringing.

 Derive the phase margin 1 T( jfx )  A( jfx ) 1/( jfx )  90 tan fx/fz  f  f x  0 t fz fz

f   f  As z 0 t ,  T ( jf x )   180  and ROC = 40dB/dec The circuit is on the verge of oscillation!

30 Differentiator

CR  Feedback pole example: differentiator

Vi a Vo  Assume constant-GBP OpAmp

Ajf  GB/ j  ft / jf Uncompensated Differentiator ZC 1 1   ,ZC  ZC  R 1 jf / fz jC RS CR

Vi  To stabilize the differentiator, a V add a series resistance Rs. o Compensated Differentiator

31 Differentiator

 At low frequency, RS has little effect because RS<<|1/jωC|

 At high frequency, C acts as a short compared to RS, The feedback network becomes |1/β| = 1+R/RS.

H(dB) H(dB) A0 UncompensatedA0 Compensated

|A| |A| |1/β| |1/β|

f (dec) f (dec)

f0(fz) fx ft f0 (fz) fx ft

32 Differentiator

 Assume RS << RC, the series resistor RS introduces an extra pole frequency fe  1 ZC  R 1 1 jf / fz 1   ZC   RS , 0 1, f x  ft f z ZC 0 1 jf / fe j C

 Choose R S  R / f t / f z , we have f e  f t f z

T( jfx )  A( jfx ) 1/( jfx ) 1 1  90 tan fx/fe  tan fx/fz   135

Therefore, ROC = 30 dB/dec, ϕm ≈ 45º

33 Stray Input Capacitance Compensation

 All practical OpAmps exhibit stray input capacitance. The net

capacitance Cn of the inverting input toward ground is

Cn  Cd CC / 2Cext

Cd is the diffrential Cap between input pins, Cc/2 is the common-mode cap of each input to ground, and Cext is the external parasitic cap. Cf

R R  In the absence of Cf , 1 2  there’s a pole in feedback

1 Vi  1 R2 / R1 1 jf 2 R1 // R2 Cn a Cn Vo ROC ≈ 40 dB/dec (See page 29) 34 Stray Input Capacitance Compensation

 Solution: Introduce a feedback capacitance Cf to create feedback phase lead.

 In the presence of we have 1 1⁄ 1 1⁄ ,

 To have 45°(i.e. ROC=30 dB/dec): Make the cross-over frequency exactly at

H(dB) ≅ → ≅1 A0 |A| Since ϕm ≈ 0º

→ 2 1 ϕm ≈ 45º

Solve the equation to get |1/β| ϕm ≈ 90º 1+R2/R1

fz fp ft f (dec) 35 Stray Input Capacitance Compensation

 To have 90°(i.e. ROC=20 dB/dec): Place exactly on the top of to cause a pole-zero cancellation Thus using simple algebra R (Neutral Compensation) H(dB)

A0 |A| ϕm ≈ 0º

ϕm ≈ 45º

|1/β| ϕm ≈ 90º 1+R2/R1

fz fp ft f (dec)

36 Capacitive-Load Isolation

 There’re applications in which the external load is heavily capacitive.

 Load capacitance CL • A new pole is formed with

output resistance ro and CL • Ignore loading by the feedback network 1 1 • The loaded gain is Aloaded  A1 jf / f p  , f p  2roCL  • ROC is increased and thus invite instability

 Solution: Add a small series resistance RS to decouple the output from CL

37 Capacitive-Load Isolation

R 2 1  R1   ro  RS  rO C  1   C R f     L 2  R2   R2 

38 Uncompensated OpAmp

 The poles of the uncompensated OpAmp are located close together thus accumulating about 180° of phase shift before the 0-dB crossover frequency .  Unstable device, thus efforts must be done to stabilize it.  Example of uncompensated OpAmps is 748 , which is the uncompensated version of 741.  They can be approximated as a three-pole system 1⁄ 1⁄ 1⁄

1 1 1 2 2 2 Three-pole OpAmp model 39 Stability of Uncompensated OpAmp

 With a frequency-independent feedback (i.e. 1⁄ curve is flat) around an uncompensated OpAmp, we have curve can be visualized as the || curve with the1⁄ line is the new 0-dB axis

• For 1⁄ ° ROC ≤ 30 dB/dec

45° • For °1⁄ ° 30 dB/dec ≤ ROC ≤ 40 dB/dec

0° 45° • For 1⁄ ° ROC < 40 dB/dec

0° Three-pole open-loop response 40 Stability of Uncompensated OpAmp

 The uncompensated OpAmp provides only adequate phase margin only in high-gain applications (i.e. high 1⁄ )  To provide adequate phase margin in low-gain application, frequency compensation is needed. • Internal compensation  Achieved by changing • External compensation  Achieved by changing

41 Internal Frequency Compensation

 How to stabilize the circuit by modifying the open loop response ? • Dominant-Pole Compensation • Shunt-Capacitance Compensation • Miller Compensation • Pole-Zero Compensation • Feedforward Compensation

42 Dominant-Pole Compensation

 An additional pole at sufficiently low frequency is created to insure a roll-off rate of -20 dB/dec all the way up to the crossover frequency.

 Cases:

• :

ROC = 30 dB/dec 45° • :

ROC = 20 dB/dec ≅ 90°  This technique causes a drastic gain reduction above 43 Dominant-Pole Compensation

 Numerical example:

• rd=∞,ro=0

• g1=2 mA/V, R1=100 kΩ,g2=10 mA/V, R2=50 kΩ • 100 , 1 , 10 • Find the required value of for 45° with 1

For ϕm=45º, we have Draw a straight line of slope -20 dB/dec until it intercepts with the DC gain asymptote at point D and get . Thus:

1 Requires EXTREMELY LARGE passive components

44 Shunt-Capacitance Compensation

 The dominant-pole technique adds a fourth pole  Extra cost and less bandwidth.  This technique rearranges the existing rather than creating a new pole.  It decreases the first (dominant) pole to sufficiently low frequency to insure a roll-off rate of -20 dB/dec all the way up to the crossover frequency.

1 1 1 2 2 2

45 Shunt-Capacitance Compensation

 Cases:

• :

ROC = 30 dB/dec 45° • :

ROC = 20 dB/dec ≅ 90°

 Since is chosen to insure roll-off rate of -20 dB/dec all the way up to the crossover frequency. Thus:

46 Shunt-Capacitance Compensation

 The first pole is decreased by adding an extra capacitance to the internal node causing it.

 Given the value of from the desired value of ,we can find 1 → ≅ 2 2

47 Shunt-Capacitance Compensation

 Numerical example:

• rd=∞,ro=0

• g1=2 mA/V, R1=100 kΩ

• g2=10 mA/V, R2=50 kΩ • 100 , 1 , 10 • Find the required value of for 45°with 1

For 45° , 1 Then, 10 → 159 EXTREMELY LARGE Unsuitable for monolithic fabrication

48 Miller Compensation

 Miller’s Theorem

If Av is the voltage gain from node 1 to 2, then a floating impedance ZF can be converted to two grounded impedances Z1 and Z2: V1 V2 V1 V1 1   Z1  Z F  Z F Z F Z1 V1 V2 1 Av

[Liu] V V V V 1 1 2   2  Z  Z 2  Z 2 F F 1 Z F Z2 V1 V2 1 Av 49 Miller Compensation

 Applying Miller’s theorem to a floating capacitance connected between the input and output nodes of an amplifier.   1 1 Z j C 1 Z F j CF 1 Z  F  F  Z1    2 1 1 1 A 1 A j1 A C 1 1 j1 1 C v v v F Av Av   F  Av 

[Liu]  The floating capacitance is converted to two grounded capacitances at the input and output of the amplifier.

 The capacitance at the input node is larger than the original floating capacitance (Miller multiplication effect) 50 Miller Compensation

 This technique places a capacitor in the feedback path of one of the internal stages to take advantage of Miller multiplication of capacitors.

 The reflected capacitances due to and the DC voltage gain between and ( ) yields , 1 and , 1 , ≅ and , ≅  A low-frequency dominant pole can be created with a moderate capacitor value.

51 Miller Compensation

 Accurate transfer function

⁄ ≅ ⁄ ⁄

 Pole/zero locations

RHP zero ≅ Dominant Pole

≅ Second Pole

52 Miller Compensation

 Right-half plane zero:

• The RHP zero is a result of the feedforward path through

• The circuit is no longer a minimum-phase system. • It introduces excessive phase shift, thus reduces the phase margin. • In bipolar OpAmps, it is usually at much higher frequency than the poles → 1⁄ ≅1

53 Miller Compensation

 Pole Splitting

• Increasing lowers and raises

• The shift in eases the amount of shift required by → Higher bandwidth

• Increasing above a certain limit makes stops to increase. 2 ≅ ≫, 54 Miller Compensation

 Numerical example:

• rd=∞,ro=0 , g1=2 mA/V, R1=100 kΩ,g2=10 mA/V, R2=50 kΩ • 100 , 1 , 10 • Find the required value of for 45°with 1

From and , we can calculate 15.9 and 3.18 Assume is large → 83.3 Since → is the first non-dominant pole

For 45°, 10 Then, 100 1 Much smaller than that of shunt- capacitance compensation 31.8 2 Suitable for monolithic fabrication 55 Pole-Zero Compensation

 This technique uses a large compensation

capacitor ( ≫) to lower the first pole .  It also uses a small resistor ( ≪) to create a zero that cancels the second pole .  The compensated response is then dominated by the lowered first pole up to  Transfer function 1⁄ 1⁄ 1⁄  Pole/zero locations 1 1 1 ≅ , , ≅ 2 2 2

56 Pole-Zero Compensation

 and lowers the dominant pole ≪, creates a zero , and creates an additional pole ≫  Choose such that cancels  The open loop gain now becomes 1⁄ 1⁄ 1⁄

- To have 45°, the cross-over frequency should be - Since the compensated response is dominated by pole up to 1⁄

- Thus, ⁄

57 Pole-Zero Compensation

 Numerical example:

• rd=∞,ro=0 , g1=2 mA/V, R1=100 kΩ,g2=10 mA/V, R2=50 kΩ • 100 , 1 , 10 • Find the required value of for 45°with 1

+ Cc + R1 C1 V1 R2 C2 V2 - R - g1Vd c g2V1

From and , we can calculate 15.9 and 3.18 For 45° , 10 Then, 100 → 15.9 Relaxed compared to shunt- is chosen such that → 10 Ω capacitance compensation, but still LARGE Since 1≫ → It will not affect the phase margin

58 Feedforward Compensation

 In multistage amplifiers, usually there is one stage that acts as a bandwidth bottleneck by contributing a substantial amount of phase shift in the vicinity of the cross-over frequency  This technique creates a high-frequency bypass around the bottleneck stage to suppress its phase at , thus improving

59 Feedforward Compensation

 The bypass around the bottleneck stage is a high-pass function ⁄ ⁄  The compensated open-loop gain is

 At low frequency: ≪ ≅ The high low-frequency gain advantage of the uncompensated amplifier still hold.

 At high frequency: ≫ ≅

The dynamics are controlled only by → Wider bandwidth & Lower phase shift 60 Summary of Internal Frequency Compensation

 Dominant-pole compensation:

• It creates an additional pole at sufficiently low frequency. • It doesn’t take advantage of the existing poles. • It suffers from extremely low bandwidth.

 Shunt-capacitance compensation:

• It rearranges the existing poles rather than creating an additional pole. • It moves the first pole to sufficiently low frequency. • The value of the shunt capacitance is extremely large → Extra cost

61 Summary of Internal Frequency Compensation

 Miller compensation:

• It takes advantage of Miller multiplicative effect of capacitors, thus requires moderate capacitance to move the first pole to sufficiently low frequency. • It causes pole splitting, where the dominant pole is reduced and the first non-dominant pole is raised in frequency.

62 Summary of Internal Frequency Compensation

 Pole-zero compensation:

• Similar to shunt-capacitance technique, a large capacitor is used to shift the first pole to sufficiently low frequency. • A small resistance is used to create a zero that cancels the first non- dominant pole  Feedforward Compensation

• It places a high frequency bypass around the bottleneck stage that contributes the most phase shift in the vicinity of

63 External Frequency Compensation

 How to stabilize the circuit by modifying its feedback factor ? • Reducing the Loop Gain • Input-Lag Compensation • Feedback-Lead Compensation

64 Reducing the Loop Gain

 This method shifts |1/β| curve upwards until it intercepts the |a| curve at , where is the desired phase margin. º  The shift is obtained by connecting resistance across the inputs.

[Franco]

Uncompensated 1 1 1 Shifts the curve upwards

to improve ϕm

65 Reducing the Loop Gain

 Rc is chosen to achieve the desired phase margin : 1 ° Then,

° ⁄ [Franco]

66 Reducing the Loop Gain

 Prices that we are paying for stability: • Gain Error: 1 1 11⁄

The presence of reduces , thus resulting in a larger gain error.

• DC Noise Gain: 1 0 ≅ 1 The presence of causes an increased DC-noise gain which may result in an intolerable DC output error.

THERE’S NO FREE LUNCH ! 67 Input-Lag Compensation

 The high DC-noise gain of the previous method can be

overcome by placing a capacitance in series with . • High frequencies:

 is short.  curve is unchanged compared to the previous case. [Franco] • Low frequencies:

 is open  1 , we now have much higher DC loop gain & much lower DC output error.

68 Input-Lag Compensation

 Rc is chosen to achieve the desired phase margin : 1 ° Then,

[Franco] ° ⁄  To avoid degrading , it is good practice to position the second breakpoint of 1⁄ curve a decade below °. °. Then, °

69 Input-Lag Compensation

 Advantage(s):

 Lower DC-noise gain due to the presence of .  It allows for higher slew rate compared with internal compensation techniques: Op-amp is spared from having to charge/discharge internal compensation capacitance.

 Disadvantage(s):  Long settling tail because of the presence of pole-zero doublet of the feedback network (||).  Increased high-frequency noise in the vicinity of the cross-over frequency.

 Low closed-loop differential input impedance which may cause high- frequency input loading

, 1⁄ ≪

** is the open loop input impedance of the Op-Amp.

70 Feedback-Lead Compensation

 This technique uses a feedback capacitance to create phase lead in the feedback path.

 The phase lead is designed to be in the vicinity of the crossover frequency which is were is boosted.

[Franco]

71 Feedback-Lead Compensation

 Analysis: ⁄ 1 1 ⁄

where, [Franco]

, 1 • The phase-lag provided by is maximum at .

• The optimum value of , that maximizes the phase margin, is the one that makes this point at the crossover frequency.

1 • The cross-over frequency can be obtained from

1 1

• Having , the optimum can be found

1⁄ 2

72 Feedback-Lead Compensation

 How much phase margin can we get ?

• At the geometric mean of and , we have ∠ 90°2tan 1 [Franco] • The larger the value of 1⁄, the greater the contribution of 1/ to the phase margin.

• E.g. 1⁄ 10 →∠ 1⁄ 55° Thus, ∠ ∠∠ ∠ 55° - The phase margin is improved by 55° due to feedback-lead compensation.

73 Feedback-Lead Compensation

 Advantage(s):

 helps to counteract the effect of the input stray capacitance as we discussed beforehand.  It provides better filtering for internally generated noise.

 Disadvantage(s):  It doesn’t have the slew-rate advantage of the input-lag compensation.

74 Decompensated OpAmps

 These OpAmps are compensated for unconditional stability only when used with 1⁄ above a specified value 1 1  They provide a constant GBP only for 1⁄  They offer higher GBP and slew rate.  Example

• The fully compensated LF356 OpAmp uses ≅10 to provide 5 and 12 ⁄ for any 1V/V.

• The decompensated version of the same OpAmp, LF357, uses ≅ 3 and provides 20 and 50 ⁄ but only for any 5V/V.

75