Chapter 10 Feedback. PART C: Stability and Compensation

Transcription

1 1 Chapter 10 Feedback PART C: Stability and Compensation

2 Example: Non-inverting Amplifier We are analyzing the two circuits (nmos diff pair or pmos diff pair) to realize this symbol: either of the circuits can be used V i - V f -

3 How does the feedback work in this circuit? i =instantaneous current=small ac current around DC Stage 1 1) DC bias points are established 2) Small AC signal (V s ) is amplified Stage 2 3 Resistor R ref here i i -i V 1 V s V f V 0

4 How does the feedback work in this circuit? i =instantaneous current=small ac current around DC Stage 1 4 Now suppose there is an abrupt positive change in V s, how does FB counter it? Stage 2 Resistor R ref here V s + Δ V i i -i V 1 V f V 0

5 i =instantaneous current=small ac current around DC Drain current i 3 increases Stage 1 5 Drain current i will increase a little since gate voltage increased Stage 2 Resistor R ref here V s + Δ V i + Δ i i -i V 1 V f V 0

6 i =instantaneous current=small ac current around DC Drain current i 4 increases Stage 1 Q4 will copy the change 1) Drain current i 4 will increase a little 2) FB hasn t happened so no change in drain current i 2 yet Stage 2 6 Resistor R ref here V s + Δ V i + Δ i i + Δ i -i V 1 V f V 0

7 i =instantaneous current=small ac current around DC Stage 1 V 1 decreases Q4 will copy the change 7 Assuming all transistors are in saturation Voltage V 1 will decrease since V 1 = V DD - i 4 ro 4 Stage 2 Resistor R ref here V s + Δ V i + Δ i i + Δ i -i V 1 - ΔV V f V 0

8 i =instantaneous current=small ac current around DC Drain current i 6 decreases Stage 1 Q4 will copy the change Drain current of i 6 will increase since gate voltage has decreased (PMOS current flow) Stage 2 8 Resistor R ref here V s + Δ V i + Δ i i + Δ i -i V 1 - ΔV V f i 6 + Δ i V 0

9 i =instantaneous current=small ac current around DC Stage 1 V o increases Q4 will copy the change Assuming all transistors are in saturation Voltage V 0 will increase a little since V 0 = -V SS + i 6 ro 7 Stage 2 9 Resistor R ref here V s + Δ V i + Δ i i + Δ i -i V 1 - ΔV V f V 0 + ΔV

10 V f increases 10 V o affects the feedback voltage V f through voltage divider law: V o increases, V f increases + + V i - V f -

11 Drain current i 2 increases and change is countered i =instantaneous current=small ac current around DC Stage 1 Q4 will copy the change 11 Drain current i 2 will increase since gate voltage has increased: This is the required counter action so there is no net change in current or voltage out of stage 1 Stage 2 Resistor R ref here V s + Δ V i + Δ i i + Δ i -i + Δ i V 1 - ΔV V f V 0 + ΔV

12 Can you explain how the FB works in this circuit? 12 Stage 1 Stage 2 Ignore Resistor R ref here

13 13 Stability

14 10.10 The Stability Problem In a feedback amplifier, the open loop gain (A) is generally a function of frequency. Open-loop transfer function: A(s), where s=jω=j(2πf) Question: What happens to gain at high frequencies? Stability of the amplifier is affected A f (jω) = A(jω) 1 + A(jω)β(jω) L(jω) = A jω β jω = A jω β jω e jφ(ω) 14 Amplitude Angle Loop gain L(jω) determines the stability or instability of the feedback amplifier

15 Loop gain and Amp Stability at High Frequencies A f (jω) = A(jω) 1 + A(jω)β(jω) Loop gain must be positive for negative feedback L(jω) = A jω β jω = A jω β jω e jθ(ω) 15 Problem: Loop gain will be negative at (ω)=180 deg at some ω 1) If A(jω)β(jω) =1, then A f will be infinite -Unstable, Unbounded (Rail-to-rail) oscillations 2) If A(jω)β(jω) >1, then A f will be negative -Unstable, Sustained oscillations (oscillator design) 3) If A(jω)β(jω) <1, then A f will be positive -Stable, A f >A at that ω, feedback will work

16 10.11 Effect of Feedback on the Amplifier Poles 16 1) UNSTABLE (rail-to-rail oscillations) 2) UNSTABLE (sustained oscillations) 3) STABLE

17 Stability Study Using Bode Plot of Aβ 1) Note that β(s)= β since resistor network is used to realize β 2) A(s)β(s)=A(s)β 3) Plot A(s)β: amplitude and phase responses 4) Gain Margin ( Aβ plot) -Amount by which loop gain can be increased while maintaining stability -Accounts for changes in loop gain with temperature, bias drift over time, etc. 5) Phase Margin (Angle Aβ plot) -If phase >180 then unstable -If phase <180 then stable -PM=180- -Trade-off between PM and BW PM 1 BW Amplitude Response Phase Response Higher the PM, More stable the Amplifier 17

18 STABLE UNSTABLE UNSTABLE Step1: Prediction of Stability from A(s) 18 1) 20log 10 A(s)β = 20log10 A(s) -20log10 1/β 2) 20log10 A(s) plot will shift down by 20log10 1/β : so we only need to look at the phase at this intersection a) If the > 180deg: amplifier will be unstable b) If the phase is very small: amplifier will be stable but the BW will also be very small c)if the is about : stable with acceptable BW phase margin (PM) = 180- =60 to 70 is ideal for stability

19 Step2: Frequency Compensation 19 Problem: open-loop response A(s) shows instability at desired 1/β Solution: Pick another 1/β and shift the response to the left so that the phase angle lies between deg 1) While shifting, we end up reducing the BW and desired DC gain. To address this issue, we will compromise. 2) We can shift the pole at the intersection of 1/β and A(s) curve to the right by introducing compensation capacitor C f

20 10.13 Miller Compensation and Pole Spitting 20 C 1 and C 2 include the Miller component due to Cμ R 1 and C 1 = total resistance and capacitance at the input R 2 and C 2 = total resistance and capacitance at the output C f = compensation capacitor C f, the compensation capacitor will 1) shift ω p1 (=1/(R 1 C 1 )) to left (eq ) 2) shift ω p2 (=1/(R 2 C 2 )) to far right (eq )

21 Compensation Capacitor in Two-stage CMOS Op-amp 21 Stage 1 Stage 2 Resistor R ref here

22 List of Problems Feedback and Stability p10.82: stability of op amp with feedback 22 p10.92: phase margin of op amp p10.99: Miller capacitance compensation

23 Summary Negative feedback is employed to make the amplifier gain less sensitive to component variations; to control input and output impedances; to extend bandwidth; to reduce nonlinear distortion; and to enhance signal-to-interference ratio 23 The advantages above are obtained at the expense of a reduction in gain and at the risk of the amplifier becoming unstable (oscillations). The latter problem is solved by careful design. The key feedback parameter are the loop gain (Ab. ), which for negative feedback must be a positive dimensionless number, and the amount of feedback (1+Ab. ). This factor determines gain reduction, gain de-sensitivity, bandwidth extension, and changes in input and output resistances.

24 Summary Stability is guaranteed if at the frequency for which the phase angle of Ab is 180 O, Ab is less than unity; the amount by which it is less than unity, expressed in decibels, is the gain margin. Alternatively, the amplifier is stable if, at the frequency at which Ab = 1, the phase angle is less than 180 O. We look at phase margin (PM). PM = is ideal. 24 The stability of a feedback amplifier can be analyzed by constructing a Bode plot for A and superimposing it on a plot for 1/ b. Stability is guaranteed if the two plots intersect with a difference in slope no greater than 6dB/decade. To make amplifier stable for a given feedback factor b, the open-loop frequency response is modified by frequency compensation:connect a feedback capacitor across an inverting stage in the amplifier.