feedback2_s08.pdfcompensation1_w03Feedback Amplifiers_new intro.pdf0BFeedback Amplifiers: One and Two Pole cases1BNegative Feedback:2BFrequency Response3BBandwidth of feedback amplifiers: Single Pole case4BS - plane5BBode Plot6BStability condition:Compensation.pdf1. Dominant Pole Compensation3. What if we reduce p1 and increase p2 at the same time!Feedback Amplifiers: One and Two Pole cases Negative Feedback: afΣ_afΣ_ There must be 180o phase shift somewhere in the loop. This is often provided by an inverting amplifier or by use of a differential amplifier. Closed Loop Gain: 1aAaf=+ When a >> 1, then 1aAaf f≅= This is a very useful approximation. The product af occurs frequently: Loop Gain or Loop Transmission T = af afafAt low frequencies, the amplifier does not produce any excess phase shift. The feedback block is a passive network. But, all amplifiers contain poles. Beyond some frequency there will be excess phase shift and this will affect the stability of the closed loop system. Frequency Response Using negative feedback, we have chosen to exchange gain a for improved performance Since A = 1/f, there is little variation of closed loop gain with a. Gain is determined by the passive network f But as frequency increases, we run the possibility of • Instability • Gain peaking • Ringing and overshoot in the transient response We will develop methods for evaluation and compensation of these problems. Bandwidth of feedback amplifiers: Single Pole case Assume the amplifier has a frequency dependent transfer function 1() ()()1() 1(()1/oas asAsas f Tsandaassp==++=−) where p1 is on the negative real axis of the s plane.With substitution, it can be shown that: 11()11/[(1oooaAsaf s p af⎛⎞⎛⎞⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠=+−+)] We see the low frequency gain with feedback as the first term followed by a bandwidth term. The 3dB bandwidth has been expanded by the factor 1 + aof = 1 + To. xxp1p1(1+To)xxp1p1(1+To)S - plane Bode Plot Bode Plot 20 log |a(jω)|∠a(jω)0-90-18020 log |A(jω)|= 20 log (1/f)|p1||p1(1+To)|x20 log |a(jω)|∠a(jω)0-90-18020 log |A(jω)|= 20 log (1/f)20 log |a(jω)|∠a(jω)0-90-18020 log |A(jω)|= 20 log (1/f)|p1||p1(1+To)|xNote that the separation between a and A, labeled as x, 20log(| ( )|) 20log(1/ ) 20log(| ( ) |) 20log( ( ))xaj f aj f Tjωωω=−= = Therefore, a plot of T(jω) in dB would be the equivalent of the plot above with the vertical scale shifted to show 1/f at 0 dB. We see from the single pole case, the maximum excess phase shift that the amplifier can produce is 90 degrees. Stability condition: If |T(jω)| > 1 at a frequency where ∠T(jω) = -180o, then the amplifier is unstable. This is the opposite of the Barkhousen criteria used to judge oscillation with positive feedback. In fact, a round trip 360 degrees (180 for the inverting amplifier at low frequency plus the excess 180 due to poles) will produce positive feedback and oscillations. This is a feedback based definition. The traditional methods using T(jω) • Bode Plots • Nyquist diagram • Root – locus plots can also be used to determine stability. I find the Bode method most useful for providing design insight. To see how this may work, first define what is meant by PHASE MARGIN in the context of feedback systems.Two pole frequency and step response. Low pass. No zeros. Frequency Response 3 dB bandwidth 224312 24 4dB nωωζ ζζ=−+−+ Gain peaking 210.70721PMζζζ=<− Step Response Risetime (10-90%) 32.2/rdBtω= Overshoot (%) 2100 exp1πζζ⎛⎞−⎜⎟⎜⎟−⎝⎠ Ringing frequency 21dnωωζ=− Settling time 1%ln100Sntζω⎛⎞=−⎜⎟⎝⎠Zero in Loop Gain, T(s). There will be some cases where we will want to add a zero to the loop gain T(s). How does this zero affect the transient response? Let’s locate the zero frequency relative to the real part of the closed loop pole location using α as a proportionality factor s = - αζωn . ()2()1()2/ 1nnsHsssαζωζω+=++ A large α will place the zero far to the left of the poles. Let’s normalize ωn = 1. Then, 21()21sHsssαζζ+=++ Split this into 2 equations. 2211()21 21sHsssssαζζζ=+++++ We see that the second term is the derivative of the first term (first term is multiplied by s times a coefficient). This can produce a bump in the step response. See the next figure from G. Franklin, et al, “Feedback Control of Dynamic Systems,” 3rd edition, Addison-Wesley, 1994. Ho(s) is the first term; Hd(s) is the derivative term. We see that if α is close to 1, we get a big increase in the overshoot.Ref. Franklin, op.cit.Compensation Question: What happens when the phase margin is too small or negative for the particular value of feedback required for an application. • The transient response will ring, • gain will peak, • or possibly oscillation. If you are building an oscillator, that might be good, but if you intended it as an amplifier then you must modify its frequency response to make it useable. Compensation is a technique that accomplished this, albeit at the expense of bandwidth. Many techniques are available: • Add a dominant pole • Move a dominant pole • Miller compensation • Add a zero to the closed loop gain 1. Dominant Pole Compensation Add another pole that is much lower in frequency than the existing poles of the amplifier or system. This is the least efficient of the compensation techniques but may be the easiest to implement. • Reduces bandwidth, • but increases the phase margin at the crossover frequency. • |T| is reduced over the useful bandwidth, Thus, some of the feedback benefits are sacrificed in order to obtain better stability Extrapolate back from the crossover frequency at 20 dB/decade The frequency, PD, where the line intersects the open loop gain Ao is the new dominant pole frequency. In this case, |T(jω)| = 1 at ω = |p1| 45o phase margin Note that we are assuming that the other poles are not affected by the new dominant pole. This is not always the case, and computer simulation will be needed to optimize the design. Nevertheless, this simple method will usually help you get started with a solution that will generally work even though not optimum.For example, you could add C to produce a dominant pole at 1212( || )DLipCR R= This approach may require a large