Lecture 20: Stability and for CMOS

Gu-Yeon Wei Division of Engineering and Applied Sciences Harvard University [email protected]

Wei ES154 - Lecture 20 1 Overview

• Reading: – S&S Chapter 8.8~8.11 • Background – When we apply , it is important to also consider stability. Sometimes, we need to frequency compensate amplifiers to ensure stability.

Wei ES154 - Lecture 20 2 Feedback Stability

• Consider the feedback circuit where β is constant and the closed-loop transfer function is:

X(s)Σ A(s) Y(s) Y A(s) ()s = X 1+ βA()s β

–As βA(s=jω) Æ -1, the gain goes to infinity and the can oscillate – In other words, oscillation can occur when β A(ωj ) = 1 β 1 ∠ A()jω 1 = − 180 °

Wei ES154 - Lecture 20 3 Stability Analysis with Bode Plots

• for Aβ – Gain margin is an indication of excess gain before instability – 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

• Polar plot of loop gain where radial distance is |Aβ| and angle is the phase angle – Unstable if Nyquist plot encircles (-1,0)

Wei ES154 - Lecture 20 6 Plotting Poles in S-Plane

• 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 growingσ oscillations – Poles that are left of jω-axis Æ stable ω ω • Consider pole pair at σp±jωp pt j pt − j pt σ pt – Results in transient response v(t)= e (e + e )= 2e cos(ω pt)

Wei 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)

A0 ω(1+ A0β ) Af ()s = 1+ s p ()1+ A0β

– Feedback moves the pole to

ω pf = ω p (1+ A0β )

Wei ES154 - Lecture 20 8 • See pole movement in S-plane and Bode plot

Wei ES154 - Lecture 20 9 Amplifier with Two-Pole Response

• Now consider ωan amplifier with open-loop transfer function with two real poles ω A ω A()s = 0 ()1+ s ()1+ s ω ω β p1 p2 ω ω ω • Closed loop poles come from 1+Aβ=0ω 2 ω s + s()p1 + p2 + (1+ A0 ) p1 p2 = 0 β 1 1 2 ω s = − 2 ()p1 + p2 ± 2 ()p1 + p2 − 4()1+ A0 p1ω p2

– 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) ω s2 + s 0 +ω 2 = 0 Q 0

Wei ES154 - Lecture 20 10 Frequency Compensation

• 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

Wei ES154 - Lecture 20 11 Compensating Two-Stage Op Amp

• Consider the following two-stage op amp:

vb3

vb2

EF

vb1

Vout1 ABXY Vout2

C C L vin L

– 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 nd resistance at E is high, CL can be large (even though Rout of 2 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 nd the 2 stage and the high Rout of the first stage

CC

E A

Av1 Av2

Rout

– Create a large capacitance at node E, (1+Av2)CC – Pole associated with node E now becomes 1

Rout1[]CE + ()1+ Av2 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 Av1 Av2 – From Lecture 17 (HF response of CS amp) we get (w/ C =C , no explicit C and C gs C, RS CE CE=Cgs) C ω 1 C P1 = Vout RsCgs + RsCgd ()1+ gm RL ' + RsCgd RL ' Rs RL CL C + C ()1+ g R ' + C R ' R g ω = gs gd m L gd L s ≅ m P2 C C R ' C gs gd L gs Simplified circuit model of two- stage amp w/ Miller Compensation –With CC added… (replace Cgd w/ CC+Cgd)

ω 1 P1 = RsCgs + Rs ()Cgd + CC ()1+ gm RL ' + Rs ()Cgd + CC RL ' Rs

Cgs + ()Cgd + CC ()1+ gm RL ' + ()CC + Cgd RL ' Rs ωP2 = Cgs ()Cgd + 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

Vout Rz CC RL CL

– Then we get a zero frequency at 1 ωz ≈ CC ()1 gm − Rz

– If Rz > 1/gm then ωZ < 0 and get a left-half plane zero

Wei ES154 - Lecture 20 16