Lecture 20:
Stability and Frequency Compensation
for CMOS Amplifiers
Gu-Yeon Wei
Division of Engineering and Applied Sciences
Harvard University
guyeon@eecs.harvard.edu
Wei
ES154 - Lecture 20
1
Overview
•
•
Wei
Reading:
– S&S Chapter 8.8~8.11
Background
– When we apply feedback, it is important to also consider
stability. Sometimes, we need to frequency compensate
amplifiers to ensure stability.
ES154 - Lecture 20
2
Feedback Stability
•
Consider the feedback circuit where β is constant and the closed-loop transfer
function is:
Y
(s ) = A(s )
1 + β A(s )
X
X(s)
Σ
A(s)
Y(s)
β
– As βA(s=jω) Æ -1, the gain goes to infinity and the amplifier can oscillate
– In other words, oscillation can occur when
β A ( jω 1 ) = 1
∠ β A ( j ω 1 ) = − 180 °
Wei
ES154 - Lecture 20
3
Stability Analysis with Bode Plots
•
Bode plot for loop gain Aβ
– Gain margin is an indication of
excess gain before instability
– Phase margin is an indication of
excess phase before -180° phase
shift at unity gain
Wei
ES154 - Lecture 20
4
Stability Analysis using Bode Plot of |A(s)|
•
Can use the Bode plot of |A(s)| to
determine β for stable loop gain
– Larger values of β leads to instability
Wei
ES154 - Lecture 20
5
Nyquist Plot
•
Wei
Polar plot of loop gain where radial distance is |Aβ| and angle is the phase angle
– Unstable if Nyquist plot encircles (-1,0)
ES154 - Lecture 20
6
Plotting Poles in S-Plane
•
•
Wei
For an amplifier to be stable, poles must lie in the left half of the s-plane
– A pair of complex poles on the jω-axis Æ sustained oscillations
– Poles that are right of jω-axis Æ growing oscillations
– Poles that are left of jω-axis Æ stable
Consider pole pair at σp±jωp
jω t
σ t
− jω t
σ t
– Results in transient response v(t ) = e p e p + e p = 2e p cos ω p t
(
ES154 - Lecture 20
)
( )
7
Effect of FB on Poles (Amp with Single Pole)
•
•
For a closed loop transfer function, the poles are found by solving for
1 + Aβ = 0
Also known as the characteristic equation of the feedback loop
– Assume only poles in the system (zeros at infinite frequency)
– Assume β is independent of frequency (to simplify analysis)
Consider an amp with open-loop transfer function A(s) characterized by a single
pole… A(s) = A0/(1+s/ωp)
A f (s ) =
A0 (1 + A0 β )
1 + s ω p (1 + A0 β )
– Feedback moves the pole to
ω pf = ω p (1 + A0 β )
Wei
ES154 - Lecture 20
8
•
Wei
See pole movement in S-plane and Bode plot
ES154 - Lecture 20
9
Amplifier with Two-Pole Response
•
Now consider an amplifier with open-loop transfer function with two real poles
A(s ) =
•
A0
(1 + s ω p1 )(1 + s ω p 2 )
Closed loop poles come from 1+Aβ=0
s 2 + s (ω p1 + ω p 2 ) + (1 + A0 β )ω p1ω p 2 = 0
s = − 12 (ω p1 + ω p 2 ) ± 12
(ω
+ ω p 2 ) − 4(1 + A0 β )ω p1ω p 2
2
p1
– From the equation, as β increases, the poles move closer together and then
when they are equal, they become complex
– Complex poles cause peaking in the frequency response depending on Q
(see Figure 8.33)
s2 + s
Wei
ω0
Q
+ ω0 = 0
2
ES154 - Lecture 20
10
Frequency Compensation
•
•
•
Wei
Typical op amps have many poles. In a folded-cascode topology, both the
folding node and the output load contribute poles. Due to having multiple poles,
op amps usually have to be compensated – their open-loop transfer function
must be modified in order for the closed-loop circuit to be stable.
– Need compensation b/c Aβ not <1 well before –180 phase shift
Stabilize by
– Minimizing overall phase shift (zero compensation)
– Dropping gain or pushing dominant pole towards origin
Watch out for right-half plane zeros
– Magnitude response increases +20dB/dec but…
– Causes negative phase shift
ES154 - Lecture 20
11
Compensating Two-Stage Op Amp
•
Consider the following two-stage op amp:
vb3
vb2
E
F
X
Y
vb1
Vout1
A
CL
B
Vout2
CL
vin
– Pole at node X (and Y) at high frequency
– Poles at nodes A (and B) and E (and F) can be close to the origin b/c output
resistance at E is high, CL can be large (even though Rout of 2nd stage is
small) Æ two dominant poles
– Will the amplifier be stable in a unity-gain feedback configuration?
Wei
ES154 - Lecture 20
12
•
The resulting Bode plots…
•
•
Assume β=1 and notice that at unity gain, phase shift is <-180°
Move one of the dominant poles toward the origin to place gain crossover well
below phase crossover frequency
Wei
ES154 - Lecture 20
13
Miller Compensation
•
Create a large capacitance at node E (and F) by relying on the moderate gain of
the 2nd stage and the high Rout of the first stage
CC
E
Av1
A
Av2
Rout
– Create a large capacitance at node E, (1+Av2)CC
– Pole associated with node E now becomes
1
Rout1 [C E + (1 + Av 2 )CC ]
– Miller multiplication enables us to use a smaller capacitor size versus just
adding a capacitor to Vdd or Gnd (small-signal gnd)
– This also has the effect of moving the output pole (at A and B) away – also
known as pole splitting
Wei
ES154 - Lecture 20
14
•
Using the simplified circuit of the two-stage amp
– From Lecture 17 (HF response of CS amp)
we get (w/ CC=Cgs, no explicit CC, and
CE=Cgs)
ω P1 =
ωP 2 =
1
Rs C gs + Rs C gd (1 + g m RL ') + Rs C gd RL ' Rs
C gs + C gd (1 + g m RL ') + C gd RL ' Rs
C gs C gd RL '
≅
gm
C gs
– With CC added… (replace Cgd w/ CC+Cgd)
ω P1 =
ωP 2 =
Av1
Av2
RS
CE
CC
Vout
CL
RL
Simplified circuit model of twostage amp w/ Miller Compensation
1
Rs C gs + Rs (C gd + CC )(1 + g m RL ') + Rs (C gd + CC ) RL ' Rs
C gs + (C gd + CC )(1 + g m RL ') + (CC + C gd ) RL ' Rs
C gs (C gd + CC )RL '
– So moves ωp1 to lower frequency and ωp2 to
higher frequency
Wei
ES154 - Lecture 20
15
+Miller Compensation with Zero
•
•
One of the drawbacks of Miller compensation comes from the zero that it
introduces in the right-half plane (right of jω axis)…
– While it increases gain, it causes additional phase shift
To combat this effect, add in a series resistor Rz
Av1
Av2
RS
CE
Rz
CC
RL
Vout
CL
– Then we get a zero frequency at
ωz ≈
1
CC (1 g m − Rz )
– If Rz > 1/gm then ωZ < 0 and get a left-half plane zero
Wei
ES154 - Lecture 20
16