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1 ECEN 4228 Notes 2. Feedback in Electronic Circuits: An Introduction cæ 1997 Dragan Maksimoviçc Department of Electrical and Computer Engineering University of Colorado, Boulder There are several reasons why feedback is applied in electronic circuits: æ Circuit characteristics, such as gain, can be precisely controlled, and made relatively independent of wide variations in active device parameters; æ Circuit characteristics can be made relatively independent of operating conditions such as supply voltages or temperature; æ Signal distortion, which is a result of the nonlinear nature of active devices, can be signiæcantly reduced; æ Frequency response and the gainèbandwidth tradeoæ can be controlled. Feedback is a powerful tool that allows us to design highperformance circuits using components with relatively poor or uncertain parameter values. Two types of feedback are applied in electronic circuits: positive and negative. In positivefeedback circuits, devices are usually driven up to their extreme operating modes èsuch as cutoæ or saturation of bipolar transistorsè. Applications include voltage comparators, æipæops, oscillators and timing circuits. In negativefeedback circuits, devices are usually in their linear operating modes èsuch asactive mode of bipolar transistorsè. Applications include ampliæers, linear voltage regulators and ælters. The purpose of this part of the notes is to present basic tools for analysis and design of feedback circuits. In Section 1, we ærst look at simple opamp applications with negative and positive feedback. A simple and general approach to evaluate and design circuits with negative feedback is based on the concepts of loopgain and ideal closedloop gain introduced in Section 2. In this part of the notes, we neglect dynamic response limitations in electronic circuits. Eæects of feedback on the frequency response, including: gainbandwidth tradeoæs, stability limits, and compensation techniques will be discussed later. 1 Examples of negative and positive feedback circuits Consider the operational ampliæer in Fig. 1. In this section, we will assume that the opamp has ideal characteristics: its openloop gain is very large: A OL = v o v e!1; è1è 1

2 VCC ie VEE Rin Rout A OL (b) Figure 1: Operational ampliæer èaè and its simpliæed model èbè. its input resistance is very large, R in = v e i e!1; and its output resistance is very small R out ç 0. Fig. 1èbè depicts a simple opamp model that includes the parameters A OL, R in and R out. This model is a crude, but in many cases adequate approximation to actual opamp characteristics, as long as the opamp output voltage v o stays within limits determined by the opamp internal structure and the supply voltages. The opamp outputs cannot exceed the supply voltages V CC and,v EE. The actual limits èpositive V O and negative V, O è are usually more restrictive, but close to the supply voltages:,v EE év, O çv oçv O év CC : è3è è2è The opamp is used in simple feedback applications as shown in Fig. 2: the input v i is applied to one of the opamp inputs, while the output v o is fed back to the other input via two resistors, and R b. In Fig. 2èaè, the output is returned to the, input, while in Fig. 2èbè, the output is returned to the input. Our goal is to determine the properties of the two closedloop application circuits, i.e. to determine how the output v o depends on the input v i. Before we ænd v o èv i è for the two circuits in Fig. 2, consider ærst what happens qualitatively in the two circuits when the input voltage is nulled, as shown in Fig. 3. With no input, v i =0, one would expect that v o = 0, but in one of the two circuits this is not the case. Even with zero input, small timevarying noise signals do exist in any circuit. Suppose that there is a very small positive voltage change at the outputs of the opamps in Fig. 3èaè and 2

3 VCC VCC Vi V1 VEE Vi V1 VEE (b) Figure 2: Two opamp application examples. èbè. This is indicated by the arrow pointing upwards. Because of the feedback connection, this small output voltage change is returned back to the ampliæer input through the voltage divider, R b. A positive change in the output will in both circuits result in a positive change at the input where the output signal is fed back. Achange at the input will produce a change in the output voltage, negative in the circuit of Fig. 3èaè, and positive in the circuit of Fig. 3èbè. Because the initially assumed change produces opposite results when propagated through the feedback loop, the behavior of the two circuits is very diæerent. In the circuit of Fig. 3èaè, the feedback produces a signal that opposes and tends to cancel the originally assumed change. As a result, the steadystate signal output with zero signal input is indeed zero. Furthermore, since the opamp openloop gain A OL is very large, the opamp input voltage v e between the and the, inputs must be equal to zero, with or without the input voltage v i applied. This is acharacteristic of a negative feedback circuit. In the circuit of Fig. 3èbè, however, the feedback produces a change that tends to amplify the originally assumed change. As a result, the output v o in the circuit of Fig. 3èbè will move away from zero until it hits the upper limit v o = V O. This is a characteristic of a positive feedback circuit. It is important to note that the circuit of Fig. 3èbè may also end up in the negative saturation limit, v o = V, O. In fact, based on the available information, it is not possible to say whether v o = V O or v o = V, O, but in any case the output is saturated at one of the limits, away from zero, although no input voltage is applied. It is also important to note that in the positive feedback circuit, the opamp input voltage v e is not zero. We can now determine the closedloop v o èv i ècharacteristics of the two application circuits. In the circuit of Fig. 2èaè, because of the negative feedback, and because the opamp openloop 3

4 V1 V1 feedback loop feedback loop Figure 3: Feedback loops in the examples of Fig. 2. gain is very large, the opamp input is zero, v e = 0, which yields: v e = v i, v 1 = v i, v o =0; R b è4è A CL = v o v i = R b : è5è Therefore, as long as the output is within the saturation limits, the circuit is a noninverting ampliæer with the closedloop gain equal to è R b è=. The complete v o èv i ècharacteristic, including the saturation limits, is shown in Fig. 4èaè. In the circuit of Fig. 2èbè, because of the positive feedback, and because the opamp openloop gain is very large, the output is always in one of the two saturation limits, v o = V O, or v o = V, O, depending on both v i and v o. Suppose that initially v i is a large negative voltage so that v o = V O. The input is positive at v 1 = V O =è R b è. The output stays in the positive saturation limit as long as v e = v 1, v i is greater than zero. Once v i exceeds v 1, v e becomes negative, and because the opamp has very large openloop gain, the output jumps abruptly to V, O. As a result, v 1 jumps to the negative value v 1 = V, O =è R b è. The output stays at the negative saturation limit as long as the opamp input v e = v 1, v i is negative. The input voltage must drop below v i for the output to jump back to V O. The complete characteristic of the positivefeedback circuit is shown in Fig. 4èbè. The characteristic is distinctly diæerent from the ampliæer characteristic in Fig. 4èaè: the output jumps from one saturation limit to the other at two diæerent points, depending on the previous state of the output. The characteristic with hysteresis is useful in voltagecomparator applications where the circuit can be used to decide a logiclevel èhigh or lowè depending on the analog voltage input. The hysteresis provides noise immunity in the decision process. 4

5 Vi Vi (b) Figure 4: v o èv i ècharacteristics of the application examples in Fig Analysis of negative feedback circuits Consider again the negativefeedback ampliæer of Fig. 2èaè. Assuming ideal opamp characteristics, we have found that the closedloop gain is given by è5è. Suppose now that the opamp openloop gain A OL is ænite. Keeping the assumptions about ideal input and output resistances, we can ænd the closedloop gain A CL = v o =v i as follows: v e = v i, v 1 = v i, R b v o ; è6è Solve for v o =v i, or, where A CL = R b is the ideal closedloop gain, and v o = A OL v e ; è7è ç v o = A OL v i, R ç a v o : è8è R b A CL = v o v i = A OL 1A OL R b ; è9è A OL R b 1A OL A T!1 = R b R b = A T!1 T T = A OL R b 1T ; è10è è11è è12è 5

6 is the loopgain of the feedback circuit. The ænal expression è10è for the feedback circuit of Fig. 2èaè introduces two important concepts in the analysis of feedback circuits: loop gain T and ideal closedloop gain A T!1. In general, the loop gain shows how much the signal is ampliæed when propagated through the feedback loop. The concept of ëpropagation through the loop" was used in Section 1 to examine qualitatively whether the feedback is positive or negative. T is the quantity that tells us not only whether the feedback is negative èt é0è or positive èt é0è but also how ëstrong" the feedback is. Referring to the example of Fig. 3èaè, assuming a small change in v o, and propagating this change through the loop, we have that the change is ærst attenuated by the resistive voltage divider, v 1 = v o ; è13è R b and then ampliæed through the opamp, v o =,A OL v 1 è14è Thus the initially assumed change in the output voltage is ampliæed through the feedback loop by the total gain equal to ç çç ç ç ç v1 vo = è,a OL è è15è v o v 1 R b The loop gain is the negative of this total gain, so that for negative feedback wehaveté0, ç ç T = èa OL è : è16è R b When the feedback is very strong èt! 1è, the closedloop gain assumes the ideal value A T!1. In the example of Fig. 2èaè, this is the value found in è5è, where we assumed that the opamp had inænite openloop gain. 2.1 Finding the loop gain T and the ideal closedloop gain A T!1 The expression A CL = A T!1 T 1T ; è17è which was derived for a particular circuit example, is very useful and important because it holds in general. Although feedback circuits can be analyzed using any standard circuitanalysis techniques èfor example writing and solving a system of nodal equationsè, the analysis of feedback circuits based on è17è has a number of advantages: 1. In most cases, both A T!1 and T can be obtained by inspection or with very little amount of èmessyè algebra. 2. A T!1 shows what the circuit is ideally supposed to do, it reveals the function and the purpose of the circuit, and is therefore valuable in understanding the circuit conæguration and its properties. The factor T=è1 T è shows how close the actual circuit performance is to the ideal. 6

7 Vt y u ue K ue Vy Vx Figure 5: A general block diagram of a feedback circuit. 3. The expression A T!1 T leads directly to the circuit design: one constructs the feedback 1T so that A T!1 has the desired value, and than makes sure that T is large enough èin the frequency range of interest and subject to stability conditions discussed laterè so that A CL ç A T!1. 4. The loop gain T is the main quantity used to determine frequencyresponse and stability of feedback circuits, which will be discussed later. The approach based on è17è requires knowledge of the ideal closed loop response A T!1, and the loop gain T. We now discuss how to determine A T!1 and T for an arbitrary feedback circuit. A general block diagram of a feedback circuit is shown in Fig. 5. If the circuit is nonlinear, as is essentially always the case, we assume that the steadystate dc operating point has been found, and that the block diagram of Fig. 5 corresponds to the linear smallsignal model of the feedback circuit. The input is u, and the output is y. At some point, the feedback loop contains a gain block where the output behaves as a voltage source Ku e, controlled by the input u e. The other parts of the feedback circuit may contain other gain stages or passive components. For the purpose of ænding A T!1 and T, a test signal v t is introduced in series with the controlled voltage source Ku e. The test source is needed only to produce the signal v x that is propagated through the loop in order to ænd the loop gain T. It is not a part of the feedback circuit. With the signals deæned as in Fig. 5, we have: æ The loop gain shows how much the signal v x is ampliæed when propagated through the loop èwhile the input signal is nulled, u = 0è. T = v y v x æ æææu=0 è18è 7

8 Note that v x is the signal that enters the loop, and v y is the signal that comes back through the loop. The polarity of v y is selected so that T é 0 corresponds to negative feedback. æ The ideal closed loop gain shows the gain of the circuit when the loop gain is very large, i.e., when u e = v y =0, A T!1 = y æ è19è u æ vy=0 In opamp applications, one usually gets A T!1 simply by solving the circuit where all opamps are assumed ideal. An ideal opamp has inænite gain, and when negative feedback is applied around the opamp, the voltage between the èè and the è,è input terminals is nulled. Therefore, the ideal A T!1 can be obtained starting from the condition that vèè = vè,è at the opamp inputs. For a real opamp, the openloop gain is ænite, and vèè 6= vè,è. The diæerence v e = vèè, vè,è can be considered an error signal for the feedback circuit, and the ideal response A T!1 corresponds to zero error. In general, to determine A T!1 for an arbitrary feedback circuit, it is necessary to identify the error signal u e in the circuit. Once the location of the error signal is identiæed, nulling the error yields the ideal response A T!1. æ Finally, the closedloop gain is the gain of the original circuit where the test source is zero, A CL = y æ = A u T!1 T è20è 1T æ vt=0 The setup of Fig. 5 and the results è1820è are the basis of our approach to feedback circuit analysis and design. 2.2 Examples As an example, consider another simple opamp application shown in Fig. 6èaè. ry large openloop gain of the opamp èequivalent tovery large T è, and negative feedback imply that the opamp input is v e = 0. As a result, the, input is at the ground potential èëvirtual ground"è and the ideal closedloop gain is èby inspectionè: A T!1 =, R b : è21è The circuit functions as an inverting ampliæer with gain set by, R b. If the open loop gain of the opamp is ænite, the closed loop gain can be found using the general feedback results. The setup for ænding the loop gain is shown in Fig. 6èbè. Note that the input is zero, v i = 0, and that the test source is inserted in series with the opamp output, which behaves as a controlled voltage source A OL v e. The loopgain can be found as follows: T = v y v x = v 1 v x v y v 1 = R b A OL : è22è 8

9 Vy Vt Vx Vi v() feedback loop v() feedback loop (b) Figure 6: Another feedback circuit example: inverting ampliæer. It is interesting that the expression for the loop gain is exactly the same as for the noninverting ampliæer of Fig. 2èaè. Finally, the closedloop gain is A CL = A T!1 T 1T =,R b A OL R b 1A OL R b : è23è Fig. 7 shows another negativefeedback circuit example. Note that in this case the feedback loop is closed through an active device Q. The ideal èlargesignalè response is obtained assuming that the opamp has very large gain so that v e = 0, i.e. vèè = vè,è =0: I=I D =V g =R ; V o =,V GS =, p KI, V t : è24è è25è The ideal smallsignal response is obtained by using the smallsignal model for the MOS device èassume r ds!1è, v g =R = i = i d = g m v gs =,g m v o ; è26è A T!1 = v o v g =, 1 g m R : The loopgain T is found as shown in Fig. 7èbè: the input v g is nulled èshorted to zeroè, and the test source v t is inserted in series with the opamp output which behaves as a controlled voltage source A OL v e. The device is replaced with the smallsignal model. è27è T = v y v x = v y vè,è i d =èa OL èè,rèè,g m è=g m RA OL : vè,è i d v x è28è The fact that the loop gain T = g m RA OL is positive is an indication that the feedback is indeed negative. The smallsignal response of the feedback circuit can now be determined using the general result è17è. 9

10 Vi i R v() id Q VGS feedback loop R id v() gm Vgs (b) Vgs Vy Vt Vx feedback loop Figure 7: A feedback circuit example with an active device Q in the feedback loop. 10