Feedback. Operational amplifiers invariably are incorporated within a circuit with negative feedback. Consider the inverting amplifier configuration :

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1 Feedback Operational amplifiers invariably are incorporated within a circuit with negative feedback. Consider the inverting amplifier configuration : Vt t fz V2 1 -t `,i 2z my rtmg amplifier The "loop-gain" A,g., is a measure of the amount of feedback in the amplifier-feedback network. It is equal to the transmission between the inverting input terminal, v_, and the network which drives this terminal if the loop were opened and the signal source was 0. Notice that the equations can be formulated to explicitly include A, e. If the loop-gain is much larger than 1 then the closed-loop gain is approximated by Z2/Z1 This is the usual design goal. This goal usually is not met at high frequencies. It is useful to separate the effects of the gain of the op-amp from the effects of the passive components in a feedback configuration. The quantity (A,g/l + A,g) is a measure of the effectiveness of feedback. You'd like it to be equal to 1. Then the gain would depend only on the feedback network...not the amplifier. We can express the gain for the inverting op-amp configuration in these terms : the overall gain is (A,/1 + A, g) (Z2/Z,) ~Z ~.2 Vt inverting amp non-inverting amp differential amp For the non-inverting configuration the gain is (A,~1 + A, e )(1 + Z2/Z,). For the differential configuration (A,~1 + A,g)(ZZ/Z,). In each case the term (A,g/1 + A,g) appears as a multiplicative weighting, which we would like to be equal to 1. We can examine the "sensitivity''and the "stability" of (A,V1 + A,,) and predict the performance limitation for all three configurations. Sensitivity : The "sensitivity" of a factor is essentially a measure of the change in that factor due to a change in another factor. Remember that the fundamental tenant of a linear system is that "if you kick the system a little bit, it will jump a little bit, and if you kick it twice as hard, it will jump twice as high". Then "sensitivity" is the constant of proportionality which says how high it will jump for how hard you kick itl. The sensitivity of the term (A,~l + A,,) to the loop gain, A,g, is S = -1/(1 + A, g). Whenever the loop-gain is much greater than 1, the overall gain is not very

2 sensitive. Note however that when A, g = -1, it gets very sensitive indeed...in fact it becomes infinitely sensitive...perhaps unstable. Of course the real physical meaning of an infinite gain term is that you get an output with no input...a manifestation of the presence of energy in the system you are describing. If the output tends to increase with time it is called "unstable". Stability : When 1 + A,g(s) = 0 the "gain" of the system becomes infinite...which simply means that you get an output for no input at all! A "pole" of the gain is located frequencies at which 1 + A,g(s) = 0 If this "output for no input" grows with time, then the system is "unstable". If we represent signals in terms of LaPlace transforms (functions of es`) then growing signals correspond to Re[s] > 0...so stability demands poles in the left-half-plane Model of a real op-amp The gain of a real amplifier is finite and is a decreasing function of frequency. The frequency dependence of the amplifier, A(s) introduces phase shift which can make negative feedback become positive. The frequency response of op-amps can often be adequately modeled as a system incorporating one or more resistor-capacitor (R-C) filters. Thus you can extend your intuition of the performance of simple RC network to include the performance of op-amps. real op-amp idealized op-amp with frequency limitation This model is easy to analyze. It explains a typical operational amplifier but the physical mechanism which limits the op-amps frequency response is not a series of R-C low-pass filters.. Typically an opamp has three internal stages : an input differential amplifier, a high gain voltage amplifier and a high current output stage. Capacitance associated with the high gain voltage amplifier is the dominant frequency limiting mechanism. In fact, capacitance is often added to reduce the frequency of the first pole so that its frequency response looks like that limited by a single pole until the gain is less than 1..This contributes to stability for resistive feedback. The frequency response (Bode diagram) is plotted for the LMC6484 in the following figure w * 1 C "C3 a..~.~ s 0 kq :+u

3 frequency response (Bode diagram) is plotted for the LMC6484 The response asymptotically approaches two limits ; the frequency independent DC gain and a gain which decreases as 1/f and equals unity at the so-called "gain-bandwidth product." of the amplifier. These parameters can be readily estimated and used in calculations. There are several mechanisms which will introduce additional phase shift in real feedback systems. The output impedance of the amplifier is not zero... it is about 350 ohms for the LMC6484 powered with 5 volts. External load capacitance will introduce an additional high frequency pole. The actual output impedance is not easily predicted. It is a function of output load, the particular device, supply voltages, etc. Often an additional resistance is placed in series with the output terminal to isolate the external capacitance and a small capacitor added between the output terminal and the inverting input terminal to "lead" compensate the amplifier. High values of the feedback impedance can cause problems as well due to the input capacitance of the amplifier (2 pf or so...plus the effects of wiring). The LMC6484 is an example of a "fully compensated" op=amp. It is unconditionally stable for any resistive feedback network. Resistive feedback networks provide a frequency-independent attenuation. All of the frequency dependence comes from the amplifier. Fully compensating the amplifier guarantees no more than -90 of phase shift when the gain around the feedback path (loop gain) is greater than one. A few thoughts about stability of feedback system are in order. Let's take a look at the closed-loop gain equation closed-loop gain equation This "gain" can be infinite. This occurs when the denominator is zero. This corresponds to a finite output with no input. It comes from energy within the network. If that finite output grows or lasts forever, then the system is not stable. If it diminishes with time, then the system is stable. The waveform reflecting energy within the network has the form V e` = V e Re[s]t + j tm(sjt The problem then is to design systems in which the poles (values of s where I + A,g = 0 occur in the left half plane...where Re[s], 0 These occur when A,g = -1) stability criteria s-plane to A-plane If the system is stable (bounded output for bounded input) then it's poles must lie in the left-half plane. Poles occur when the loop gain, A,g = -1. There is a simple graphic technique which uses A to map the r.h.p. A conformally maps s ; everything inside a closed boundary in the s-plane will be mapped

4 into a closed boundary in the A plane if A is analytic on the periphery. If the point -1 is in the region which includes"the r.h.p. of s, then the system output may be unbounded... and is unstable. The quantity A,g is the loop-gain of the feedback system. If we assume minimum phase-shift networks, then we can readily relate a Bode diagram (log gain vs log frequency) to a polar diagram (guessing the phase response by associating a phase shift of -90 with a slope of - 6 db/octave (-20 db/decade), -180 with -12 db/octave (-40 db/decade) and so forth. 1 ", - k-1 6, Wc'? r _~Ce ~tr3/ - 406B(d e-, -To., C k{(s2 1st order system 2nd order system 3rd order system bode and polar plots of loop gains ~Acs~ -- %.C> S.:Ṣ -Sty L/ VLAa to / v LLlti~ I,- 4

5 Let's consider several feedback systems and try and predict if they will be stable. Notice that IA,, I < I A ` \c~ t A t A 1 i _ A,,~ ; A - a2\,ioll loc,4.-,. u VA U (2 1 resistive For a compensated amplifier, like the LMC6484, the system will always be stable. It always looks like a first-order system ( a maximum of -90 of phase shift). A Cv) -_ VZ f v~ V Fti' : p^ lv`~ 4\ r!. integrator The integrator essentially reduces the gain-bandwidth product of a compensated amplifier to 1/2 RC. An integrator can be implemented with an amplifier which would otherwise not be stable with feedback. The sketch on the right shows the closed-loop response of an op-amp set up as an integrator. Notice that it has the same shape as the gain of the compensated op-amp. 'FS Fif 2Z 1' S t t tt~+flz GC Zd. I st I ~~ 1 ~ 1~ 1~= capacitive A capacitive load introduces additional negative phase-shift (lag) into the loop and makes the loop potentially unstable. For the LMC6484 powered at 5 volts, the output resistance, modeled here as R3, is about 350 ohms....a 500 pf capacitive load (from long-leads, for example) produces a pole at 1/2 RC = 0.91 mhz.. So you. conclude that unity-gain followers incorporating LMC6484 cannot drive highly capacitive loads.

6 t. capacitive load with lead compensation. Capacitive loads are often the lot of operational amplifiers...especially those in power-supplies or applications which constrain potentials. The amplifier can be compensated (made to work with the desired feedback) by reducing the amplifier's gainbandwidth product so the loop gain is less than 1 when the phase shift approaches 180. But reducing the amount of feedback decreases the benefits of feedback. An alternative approach is to tailor the frequency response of the feedback (loop-gain) so that part of the loop's response is increasing while the unavoidable effect occurs...like the more rapid decrease in response because of a capacitive load. Lead compensation does this. In the example, the decrease in response because of the capacitive load begins at fl = 1 /2 R C. We can make the loop-gain flatten-out at fl or below...call it fc with R and C. Resistor R5 provides isolation between the op-amp output and the system output (capacitive load) -active ground A large common-mode voltage characterizes the difference between potentials on the body surface and our measurement system. ' We can equalize...or a least reduce the difference by connecting the subject to the amplifier reference. But we must limit the current possible in this connection and the reduction is limited by the lead and skin resistance. An active "ground" or feedback system which sense the common-mode voltage, compares it with our reference and generates the necessary current to equalize the common-mode to the reference. This can reduce the commonmode potential by approximate the gain of the amplifier. But it will oscillate without compensation...and simple gain-bandwidth reduction reduces the gain substantially at important frequencies.