Chapter 10 Feedback ECE 3120 Microelectronics II Dr. Suketu Naik

Transcription

1 1 Chapter 10 Feedback

2 Operational Amplifier Circuit Components 2 1. Ch 7: Current Mirrors and Biasing 2. Ch 9: Frequency Response 3. Ch 8: Active-Loaded Differential Pair 4. Ch 10: Feedback 5. Ch 11: Output Stages

3 Feedback 3 Two Stage Op Amp (MOSFET)

4 Example: Non-inverting Amplifier 4 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 -

5 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 5 Resistor R ref here i i i V 1 V s V f V 0

6 How does the feedback work in this circuit? i =instantaneous current=small ac current around DC Stage 1 6 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

7 i =instantaneous current=small ac current around DC Drain current i 3 increases Stage 1 7 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

8 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 8 Resistor R ref here V s + Δ V i + Δ i i + Δ i i V 1 V f V 0

9 i =instantaneous current=small ac current around DC Stage 1 V 1 decreases Q4 will copy the change 9 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

10 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 decrease since gate voltage has decreased Stage 2 10 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

11 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 litle since V 0 = V DD - i 6 ro 6 Stage 2 11 Resistor R ref here V s + Δ V i + Δ i i + Δ i i V 1 - ΔV V f V 0 - ΔV

12 V f increases 12 V o affects the feedback voltage V f through voltage divider law: V o increases, V f increases + + V i - V f -

13 Drain current i 2 increases and change is countered i =instantaneous current=small ac current around DC Stage 1 Q4 will copy the change 13 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

14 Can you explain how the FB works in this circuit? 14 Stage 1 Stage 2 Resistor R ref here

15 15 Stability

16 10.10 The Stability Problem In a feedback amplifier, the open loop gain (A) is generally a function of frequency. 16 It is called open-loop transfer function A(s). Question: What happens to gain at higher frequencies? Frequency response determines stability of the amplifier.

17 (10.83) loop-gain: The Ideal Case (10.81) closed-loop gain t-function: A (10.82) closed-loop gain t-function: A s f f j A 1 A 1 s sβs A j A jβ j L j A j β j A j β j e angle jφ w 17 magnitude of gain It is the manner in which the loop gain varies with frequency that determines the stability or instability of the feedback amplifier

18 Loop gain and Amp Stability at High Frequencies 18

19 Nyquist Plot (Loop Gain with Varying Freq) Figure 10.34: The Nyquist plot of an unstable amplifier 19 1) At ω=ω 180, the feedback becomes positive 2) If the loop gain at ω=ω 180 crosses the x-axis to the left of (- 1,0), the amplifier will be unstable because Aβ < -1: oscillations will grow with nonlinearity 3) If the loop gain at ω=ω 180 crosses the x-axis exactly at (-1,0), the amplifier will be unstable because Aβ = -1: sustained oscillations 4) If the loop gain at ω=ω 180 crosses the x-axis to the right of (- 1,0), the amplifier will be stable 5) If the Nyquist plot encircles (- 1,0), then the amplifier will be unstable

20 The Ideal Case 20 (10.84) instantaneous voltage: (10.85) feedback-ampflier pole constraint: (10.86) open-loop transfer function: (10.87) closed-loop transfer function: A (10.88) pole: 1 A Pf (10.89) closed-loop transfer function: A P 0 0t nt nt 0t t 2 nt 1 Asβs 0 v e e e e cos A s f f A0 1 s / s s A 0 s P P A0 /1A0 1 s/ 1 A P A s 0

21 Effect of Feedback on the Amplifier Poles 21 STABLE UNSTABLE (railto-rail oscillations) UNSTABLE (sustained oscillations)

22 10.12 Stability Study Using Bode Plots 22 Since the loop gain A(s)β = 1 at low frequencies, we define A(s)β= 1e jθ, where 1) β= feedback factor at low frequencies 2) θ=180-phase margin (PM) At low frequencies closed-loop gain=(1/β) At phase margin=70, closed-loop gain=0.87(1/β) At phase margin=45, closed-loop gain=1.3(1/β) Trade-off PM 1 BW The stability of the feedback amplifier reduces as the phase margin reduces

23 STABLE UNSTABLE UNSTABLE Stability Study Using Bode Plots The stability of the feedback amplifier can be determined directly from the plot of A(s) 23 After plotting A(s), we look at the phase at 1/β, since A(s) = (1/β)e jθ, phase margin (PM) = -180-phase of A(s) If the phase < -180deg: amplifier will be unstable If the phase is very small: amplifier will be stable but the BW will be small If the phase is about deg: stable with acceptable BW

24 10.13 Miller Compensation and Pole Spitting 24 Problem: open-loop response A(s) shows instability Solution: shift the response to the left so that the phase angle is positive and lies between deg While shifting, we end up reducing the BW and desired DC gain. To address this issue, we will compromise. We can shift the pole at the intersection of 1/β and A(s) curve to the right by introducing compensation capacitor C f

25 10.13 Miller Compensation and Pole Spitting 25 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 and make it dominant (eq ), 2) shift ω p2 (=1/(R 2 C 2 )) to far right and make it insignificant (eq )

26 Compensation Capacitor in Two-stage BJT Op-amp 26

27 Compensation Capacitor in Two-stage CMOS Op-amp 27

28 List of Problems Feedback and Stability p10.82: stability of op amp with feedback p10.92: phase margin of op amp p10.99: Miller capacitance compensation 28

29 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 29 The advantages above are obtained at the expense of a reduction in gain and at the risk of the amplifier becoming unstable (that is, oscillating). The latter problem is solved by careful design For each of the four basic types of amplifier, there is an appropriate feedback topology. The four topologies, together with their analysis procedures, are summarized in Table 10.1.

30 Summary The key feedback parameter are the loop gain (A. ), which for negative feedback must be a positive dimensionless number, and the amount of feedback (1+A. ). The latter directly determines gain reduction, gain desensitivity, bandwidth extension, and changes in input and output resistances 30 Since A and are in general frequency dependent, the poles of the feedback amplifier are obtained by solving the characteristic equation 1+A(s)(s) = 0 For the feedback amplifier to be stable, its poles must all be in the lefthand side of the s-plane.

31 Summary Stability is guaranteed if at the frequency for which the phase angle of A is 180 O, A 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 A = 1, the phase angle is less than 180 O, the difference ifs the phase margin 31 The stability of a feedback amplifier can be analyzed by constructing a Bode plot for A and superimposing it on a plot for 1/. Stability is guaranteed if the two plots intersect with a difference in slope no greater than 6dB/decade.

32 Summary 32 To make a given amplifier stable for a given feedback factor, the open-loop frequency response is suitably modified by a process known as frequency compensation. A popular method for frequency compensation involves connecting a feedback capacitor across an inverting stage in the amplifier. This causes the pole formed at the input of the amplifier stage to shift to a lower frequency and thus become dominant, while the pole formed at the output of the amplifier stage is moved to a very high frequency and thus becomes unimportant. This process is known as pole splitting.