Download

CHAPTER 10 

Frequency Compensation 

Analog IC Analysis and Design 10-1 

Chih-Cheng Hsieh

Outline 

1. General Consideration 

2. Phase Margin 

3. Frequency Compensation 

4. Compensation of Two-Stage Op Amps 

5. Other Compensation Techniques 


Chih-Cheng Hsieh

General Consideration 

Y 

X 

( s) 

= 

H ( s) 

1+ βH 

( s) 

• Consider the negative feedback system, where β is assumed constant 

• If βH 

( s = jω1 ) = −1 

, the gain goes to infinity, and the circuit can amplify its own 

noise until it eventually begins to oscillate. 

• Barkhausen’s Criteria : The circuit may oscillate at frequency 

o 

( jω 

) = ∠βH 

( jω 

) = 

βH 180 

1 

1 

1 

− 

• The total phase shift around the loop at ω 1 is 360 o . (180 o from negative feedback) 

• The feedback signal add in phase to the original noise to allow oscillation buildup. 

ω 1 

if 

Analog IC Analysis and Design 10- Chih-Cheng Hsieh 

3

Bode Plot for Stability Analysis 

• Assume that β is less than or equal to unity and does not depend on the 

frequency. 

• The worst case stability corresponds to β = 1. 

• We often analyze the magnitude and phase plots for βH = H . 

Unstable 

Stable 

• In a stable system 

Gain 

20log βH 

Margin = −20log 

βH 

o 

( jω 

o ) < 0 ∠βH 

( jωunity 

) > −180 

180 

o 

( jω 

o ) Phase Margin = 180 + ∠βH 

( jω 

) 

180 


4 

unity

Time Response vs. the Position of Poles 

• Expressing each pole frequency as 

s 

p 

= 

jω + σ 

p 

p 

• The impulse response of the system 

includes a term 

exp( 

jω + σ )t 

p 

p 


5

Stability and Pole Location 

http://en.wikipedia.org/wiki/Laplace_transform 

Consider an amplifier with a complex-conjugate pole pair : s = σ0 ± jωn 

σ0 0 

( ) t [ j ωn 

t − j ωn 

t σ 

] 2 t cos( ) 

vt = e e + e = e ω t 

A sinusoidal signal with an envelope 

n 

0 

e σ t 

Stable system 

Left-plane pole 

σ 

0 

< 0 

: Oscillation decays exponentially to zero. 


Chih-Cheng Hsieh

Stability and Pole Location 

Right-plane pole 

σ 

0 

> 0 

: Oscillation grows exponentially to nonlinear. 

Unstable system 

jw axis pole 

σ 

0 

= 0 

: Oscillation sustained. 

Oscillation system 

The existence of any righ-half-plane poles results in instability 


Chih-Cheng Hsieh

Bode Plots of Loop Gain 

• Consider an one pole amplifier 

A0 

H() 

s = ⎛ 

1 s ⎞ 

⎜ + 

ω ⎟ 

⎝ 0 ⎠ 

A0 

Y 1+ 

β A 

() s = 

X 

1+ 

s 

ω 

0 

( 1+ 

βA 

) 

0 0 

• A single pole cannot contribute a phase shift greater than 90 o , and the system is 

unconditionally stable for all non-negative values of β. 


8

Two-pole Systems 

• Consider a two-pole system 

H() 

s 

0 

= ⎛ 

1 s ⎞⎛ 

1 s ⎞ 

⎜ + 

ω ⎟⎜ + 

P1 ω ⎟ 

⎝ ⎠⎝ P2⎠ 

Y 

A0 

() s = 

X ⎛ s ⎞⎛ s ⎞ 

1+ 1+ + β A 

⎜ 

⎟⎜ ω ⎟⎜ ω ⎟ 

A 

⎝ p1⎠⎝ p2 

⎠ 

0 

• Maximum phase shift = 180 o 

• As the feedback becomes weaker, the gain crossover point moves toward the 

origin. 

• The stability is obtained at the cost of weaker feedback. 


9

Multipole Systems 

• Consider a 3-pole system 

• Maximum phase shift = 270 o 

• The phase begins to change at 0.1 of pole 

frequency whereas the magnitude begins to 

drop only near the pole frequency. 

• Additional poles(and zeros) impact the phase to 

a much greater extent than magnitude. 

• The stability is obtained at the cost of weaker 

feedback. 


10

Outline 







Chih-Cheng Hsieh

Phase Margin 

• If 

o 

o 

( u ) 175 , ( 1) 1 exp( 175 ) 

∠ βH jω =− βH jω 

= × −j 

1 exp 175 

Y H β 

1 −0. 9962 − j 0. 

0872 

( jω1 

) = = = ⋅ 

X 1+ βH j ω 1+ exp − 175 β 0. 0038 − j 0. 

0872 

( jω1 

) 

( ) 

1 

o 

( − j ) 

o 

( j ) 

Y 

X 

( jω 

) 

1 1 

⋅ 

β 0.0872 

1 

= ≈ 

11.5 

β 

Y 

1 

• Since at low frequencies, ≈ 

X β , the closed-loop frequency response 

exhibits a sharp peak in the vicinity of ω = ω 1 

• The closed loop system is near oscillation and its step response exhibits a very 

underdamped behavior. 


12

Closed-Loop Freq. and Time Response 

‣ For a small phase margin 

log βH 

( ω) 

‣ For a large phase margin 

log βH 

( ω) 

∠βH ( ω) 

∠βH ( ω) 

• The greater the phase margin, the more stable the feedback system. 


13

CL Freq. Response with 45 o PM 

• For PM = 45 o , 

∠βH 

o 

( ω ) = − βH 

( ω ) 1 

1 

135 

1 

= 

Y 

X 

H 

1 

= 

1+ 

1× 

exp( − j135 

( jω 

) H ( jω 

) 

o 

) 

1 

= 

0.29 − 0.71j 

Y 

X 

= 

1 

⋅ 

β 

1 

0.29 − 0.71j 

1.3 

≈ 

β 

– The feedback system suffers from a 30% peak at ω = ω 1 

• For PM = 60 o , the peaking is negligible 

1 

= 

β 

• The concept of phase margin is well-suited to the 

design of circuits that process small signals, not large 

signals. 

Y 

X 


14

Outline 







Chih-Cheng Hsieh

Frequency Compensation 

• Frequency compensation : their open loop transfer function must be modified such that 

the closed-loop circuit is stable and the time response is well behaved. 

• Frequency compensation can be achieved by 

– Minimizing the overall phase shift, thus pushing the phase crossover out. 

– Dropping the gain, thereby pushing the gain crossover in. 


16

Single-Ended Telescopic OP Amp. 

• Frequency compensation : their open loop transfer function must be modified 

such that the closed-loop circuit is stable and the time response is well behaved. 

• Path 1 : a high-frequency pole at the source of M 3 , a mirror pole at node A, 

another high-frequency pole at the source of M 7 , and a pole at the output. 

• Path2 : a high-frequency pole at the source of M 4 , and a pole at the output. 

• Dominant pole (d.p.) : ω p,out usually sets the open loop 3-dB bandwidth. 

• Non-dominant pole : the closet pole to the origin after the d.p. is at node A. 

C = C + C + C + C + C + C 

A 

GS 5 GS 6 DB5 

2 

GD6 

DB3 

GD3 


17

Bode Plots of Telescopic OP Amp 

• The mirror pole typically limits the phase margin. 

• The circuit contains a zero at 2ω p,A . (ignore it now for simplicity) 

• Force the loop gain to drop the gain crossover point moves toward the origin. 

• The dominant pole can be pushed to a lower freq. by increasing the load cap. 

• The unity-gain bandwidth of the compensated op amp is equal to the frequency 

of the 1 st non-dominant pole. 


18

Bode Plots of Loop Gain for Higher R o 

• Translating the dominant pole toward the origin for 45 o phase margin, the unity 

gain frequency is equal to the frequency of the first non-dominant pole. 

• Increasing R out does not compensate the op amp. 

• To achieve a wideband in a feedback system, the 1 st nondominant pole must be 

as far as possible The mirror pole is undesirable. 


19

Fully Differential Telescopic Op Amp 

• This topology avoids the mirror pole, thereby 

exhibiting stable behavior for a greater bandwidth. 

• One dominant pole at the output node and only 

one nondominant pole arising from node X (or Y). 

• Capacitance at node N, C N = C GS5 + C SB5 + C GD7 + 

C DB7 , shunts the output resistance of M 7 at high 

frequencies, thereby dropping the output 

impedance of the cascode. 

• The pole in the PMOS cascode is merged with the 

output pole, thus creating no additional pole. 


20 

Z 

out 

( ) 

− 

( ) 

Z = 1 + g r Z + r , 

out m5 O5 N O5 

Z = r || C s , 

N O7 

N 

1 

( 1 ) 

Z ≈ + g r 

out m5 O5 

1 

|| 

sC 

L 

= 

rO 

7 

1+ 

sr C 

O7 

( 1+ 

gm5rO 

5 

) rO 

7 

[( 1+ 

g r ) r C + r C ] s + 1 

m5 

O5 

O7 

N 

L 

O7 

N

Outline 







Chih-Cheng Hsieh

Compensation of Two Stage Op Amps 

• We identify three poles : a pole at X (or Y), another at E (or F), and a third at A 

(or B). 

• Since the poles at E and A are relatively close to the origin, the phase 

approaches -180 o well below the third pole. 

• If the magnitude of ω P,E is to be reduced, the available bandwidth is limited to 

approximately ω P,A , a low value. A very large compensation capacitor is 

required. 


22

Miller Comp. of a Two-Stage Op Amp 

• Create a large capacitance at node E, equal to (1 + A v2 ) C C . 

• The corresponding pole is moved to 1/ Rout1⎡⎣CE + ( 1+ 

Av 2) 

CC 

⎤⎦ 

• Miller compensation 

– Lowering the required capacitor value. 

– Moves the output pole away from the origin. 

• If R S denotes the output resistance of the first stage, R L = r O9 || r O11 . 

ω

Pole Splitting 

• Before compensation, ω p1 and ω p2 are 

of the same order of magnitude. 

• Miller compensation increases the magnitude of the output pole by roughly a 

factor of gm 9 

R L . 

• At high frequencies, C C provides a low impedance between the gate and drain of 

−1 

−1 

M 9 . The resistance seen by C L becomes R || g 

9 

|| R ≈ g 

9 

. 


24 

ω ≈ 1/[ R( C + C )] ≈1/ 

RC 

p2 S E GD9 

L L 

• After compensation , 

gm gm 

If CC + CGD9 >> CE , ω 

p2 

≈ ≈ 

C + C C 

S 

m 

L 

m 

9 9 

E L L

Effect of Right Half Plane Zero 

ω < ω < ω 

p1 z p2 

• A right-half plane zero at 

• Numerator is 

ω = 

• The phase shift is − tan −1 

negative 

• Zero slow down the drop of magnitude. 

• The stability degrades considerably. 

( C C ) 

/ m9 C GD9 

• The right half–plane zero in a two-stage 

CMOS op amps is a serious issue because 

gm is relatively small and C C is usually large. 

• A series resistor to eliminate the right-half 

plane zero. 

ω ≈ 

z 

C 

C 

• The output stage now exhibits three poles. 

• To cancel the first non-dominant pole 

1 

C g R 

−1 

( 9 

− ) 

C m Z 

( 1 − s / ωz 

) 

1 

z 

( ) 

ω /ω z 

−1 

( g − R ) 

m9 

Z 

g + 

− g C + C + C C + C 

= , R = ≈ 

C C g C g C 

m9 

L E C L C 

Z 

L 

+ 

E m9 C m9 

C 


25

Resistor for Miller Compensation 

• Typically the feedback R Z is realized by a MOS transistor in the triode region. 

• R Z changes substantially as output voltage excursions are coupled through C C to 

node X, thereby degrading the large-signal settling response. 

• Let 

( ) ( ) 

• For pole zero cancellation to occur 

( ) ⎛ 14 

C ⎞ 

− L 

m 

m 

( ) 

− ⎜ ⎟ 

( ) ( ) 

14 

( W L 

−1 ) 

14 

( W L) 

V = VμC , Wthen L V V = V V , g = − , 

GS13 GS 9 GS15 GS14 m14 p ox GS14 TH14 

RμC = ⎡ 

⎣W L V V − R⎤ 

⎦ , g = 

on15 p ox 15 GS15 TH15 on15 m14 

W L ID CC 

g = g 1 + , W L = W L W L 

-1 

( ) ( ) ( ) 

1 1 9 

14 9 15 14 9 

W L 

15 ⎝ CC ⎠ 

ID 14 

CC + CL 

15 


26

Defining g m9 with respect to R S 

• Use a simple resistor for R Z and define g m9 as a resistor that closely match R Z . 

• Incorporating M b1 -M b4 along with R S to generate 

I b 

∝ R 

−2 

S 

g 

m9 

∝ 

I 

D9 

∝ 

I 

D11 

∝ 

R 

−1 

S 

• Proper ratio-ing of R Z and R S with temperature and process variations. 

• Short channel effects may deviate from the square low regime and create errors. 


27

Increased Load Cap. on Step Response 

• In a two-stage op-amp, a higher load capacitance presented to the second stage 

moves the second pole toward the origin, degrading the phase margin. 

– The response approaches an oscillatory behavior if the load capacitance 

seen by the two-stage op amp increases. 

• In one-stage op amps, a higher load capacitance brings the dominant pole closer 

to the origin, improving the phase margin. 

– Making the feedback system more overdamped. 


28

Slewing in Two-Stage Op Amps 

V ≈ I t C , SR = I C 

out SS C SS C 

• For the positive slew rate, M5 must provide two currents : I SS + I 1 

– If M5 not wide enough to sustain I SS + I 1 in saturation, then V X drops 

significantly, possibly driving M 1 into triode region. 

• For the negative slew rate, I 1 must support both I SS and I D5 . 

– If I 1 = I SS , V x rises so as to turn off M5. 

‣ C C is charged by a constant I SS if parasitic cap. 

at node X are negligible. 

‣ The gain of the output stage makes node X a 

virtual ground. 

– If I 1 


29

Outline 







Chih-Cheng Hsieh

Other Compensation Techniques 

• The feedforward path formed by the comp. 

capacitor causes a right-half plane zero. 

• If C C could conduct current from the output 

node to node X but not vice versa, then the 

zero would move to a very high frequency. 

– Inserting a source follower in series with 

the capacitor. 

−V 

− g V = V R + C s V = + sR C 

−1 

out 

( ), ( 1 ) 

m1 1 out L L 1 

L L 

gm 

1RL 

V −V V 

+ I = 

(1 / g ) (1 / sC ) R 

out 1 1 

in 

m2 

+ 

C S 


31

Compensation Using Source Follower 

• We have 

V 

I 

out 

in 

= 

R 

L 

C 

C 

C 

L 

− gm 

1RLRS 

( gm2 

+ sCC 

) 

2 

( 1+ 

gm2RS 

) s + [( 1+ 

gm 

1gm2RLRS 

) CC 

+ gm2RLCL 

] s + gm2 

ω 

p1 

≈ 

g 

m1 

( 1+ 

g g R R ) 

If 1+ 

gm 

RS 

>> 1 

m1 

m2 

L S 

CC 

>> gm2R 

g 

2 L L 

gm2 

1 

gm 

1gm2RLRSCC 

gm 

1 

≈ 

ω 

p2 

≈ 

≈ 

m2RLRSCC 

gm 

1RLRSCC 

RLCLCC 

gm2RS 

CL 

C 

• Thus, the circuit contains a zero in the left half plane, which can be chosen to 

cancel one of the poles. 

• The new values of ω , p 1 

ω 

p2 

are similar to those obtained by simple Miller 

approximation. 

• The source follower limits the lower end of the output voltage to V 

GS 2 

+ VI 

2 . 

• It is desirable to utilize the compensation capacitor to isolate the DC levels in the 

active feedback stage from that at the output. 


32

Compensation Using a CG Stage 

• C C and the common gate stage M 2 converts the output voltage swing to a 

current, returning the result to the gate of M 1 . 

• If V 1 changes by ∆V 

, V changes by A v 

∆V 

out 

• The current flows through the capacitor is nearly equal to Av∆VCCs 

because 

1/g m2 can be relatively small. 


33

Compensation Using a CG Stage 

g V sC 

, 

m2 2 

C 

Vout 

+ =− V2 V2 

=−Vout 

sCC sCC + gm2 

g V + V 

⎛ 1 

+ sC 

⎞ 

= g V , I 

V 

= + g V 

⎝ ⎠ 

1 

ω 

p1 

≈ 

V 

− g 

out 

m1RSRL( gm2 

+ sCC) 

gm 1RRC 

L S C 

= 

2 

Iin RCCs 

L C L 

+ ⎡⎣( 1+ gm 1RS) 

gm2RC L C 

+ CC + gm2RC L L⎤⎦s + g 

g 

m2 

m2Rg 

S m1 

ω 

p2 

≈ 

C 

• Contains a left-plane zero 

• The second pole has considerably risen in magnitude – by a factor of g m2 R S . 

– At very high frequencies, the feedback loop consisting of M 2 and R S lowers 

the output resistance by the same factor. 

– If the capacitance at the gate of M1 is taken into account, pole splitting is 

less pronounced. 

– This technique can potentially provide a high bandwidth in 2-stage op-amp. 


34 

1 

m1 1 out ⎜ 

L ⎟ m2 2 in m2 2 

RL 

RS 

L

Slew Rate Analysis 

I ≥ I + I 

I2 ≥ ISS 

+ ID2 

1 SS D1 

• For positive slewing, M2 and hence I 1 must support I SS , requiring that 

– If I 1 

– If I 1 

yielding a slew rate equal to I 1 /C C . 

• For negative slewing, I 2 must support both I SS and I D2 . 

– As I SS flows into P → V GS1 ↑ → I D1 ↑ → V X ↓ → P is a virtual ground node. 

– I 3 and I 2 must be as large as I SS , raising the power dissipation. 


35

Alternative Comp. of 2-Stage Op-Amp 

• The zero appears at 

( g R )( g / C ) 

m4 

eq 

m9 

C 

• The dominant pole is located at approximately 

• The first non-dominant pole is given by 


36 

gm4gm9 

C 

L 

R 

eq 

Reqgm9 

1 

R 

L 

C 

C

Download

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?