1. INTRODUCTION. Now we have to consider that: Reliability is inversely proportional to the number of contacts in the circuit;

Transcription

1 1. INTRODUCTION Versione del:venerdì 9 giugno 2017 The design of amplifiers has changed from the beginning as the technology grew. The initial standard way to design was connecting transistor by transistor, resistor by resistor, etc.; namely brick by brick. We can imagine the cost and, more important, the reliability of such a way to proceed, if we consider that an amplifier to work, needs at least a few tens of transistors, every transistor has 3 legs to connect, and so on. Space occupation is another very important issue. Now we have to consider that: Reliability is inversely proportional to the number of contacts in the circuit; The test of the circuit, done automatically on large series, takes a time duration that is proportional to the measurement nodes. In conclusion, the development time rises with the number of components to be used and efforts have been made so far in trying to improve this. Amplifiers and feedback gpessina 1

2 . introduction 2 We are now living the age of complex integrated circuits. These devices have a large number of connecting pins and are composed of several transistors, or bricks, already connected to form circuits with specific functions. A few external components can then be added from outside to personalize several of their functions. Two categories of such circuits exist: digital circuits (microcontrollers, microprocessors, etc.) capable to contain millions of bricks per square mm and analogue circuits that, strategically, contain a few hundred of transistors as their functionality are limited. It is interesting here a brief historical review of the technological evolution of the solid state electronics. The first form of transistors were the valves (components in 3D) that worked on a different operating principle: thermionic emission of charge from metal. Valves generate heat by default and are totally incompatible with the modern electronics. To give the order of magnitude: the implementation of the microprocessor of our smartphones with valves takes probability the volume of this building. Amplifiers and feedback gpessina 2

3 . introduction 3 At the end of the 40 of the last century it started the study of semiconductors, mainly Germanium and Silicon, Three researchers that played a big role in this business were William B. Shockley, John Bardeen and Walter H. Brattain, Nobel Prize in 1956 for this subject. The first working transistor, a bipolar transistor, was in Ge and was developed at the Bell Labs. Since then transistor manufacturing became planar, namely 2D and Si was the standard substrate for it. Bonding wire Power transistor ( mm 2 ) Amplifiers and feedback gpessina 3

4 . introduction 4 Pretty soon the transistor evolved and the valves retired. The change was dramatic: the space occupation was reduced by order of magnitude (circuits changed from 3D to 2D ) and, more important, the power consumption became negligible in comparison to valve since the operating principle of transistors is not thermionic. One of the first commercial transistor in its metallic package. The technology evolution was not only in the semiconductors, but also in packaging and the managing of the semiconductorchip inside. Several are the constraints to be followed: compactness, stress resistance, power dissipation, hence thermal conductance, Amplifiers and feedback gpessina 4

5 . introduction 5 Having the transistors available, amplifiers can be implemented combining several of them with resistors, capacitors and inductances. At this moment of our study we do not know which way an amplifier is composed and we can consider it as a mess of components for the moment: several transistors are needed to compose an amplifier just as many bricks are needed to construct buildings. Having available only one transistor per package, the only way to design amplifiers was by connecting packages to packages. This way to proceed is adopted also nowadays, but a new approach made available by the evolution of the technology was possible. For quite a long time one packages included only one transistor and there was an evolution in packaging from metal can, throughhole (TO) to surface mount devices having much smaller dimension. Amplifiers and feedback gpessina 5

6 . introduction 6 Throughhole means that the connecting pins of components are inserted in corresponding holes and soldered on the other side of the Printed Circuit Board, PCB, the substrate support where copper tracks are printed for the electrical connection of the pins of the components: There is a limit in the size of and the distance between holes, so the compactness of this circuits is limited. Surface mount devices, or SMD devices, are soldered on the same side where the devices are located and take advantage from the fact the soldering pads can be closer and smaller: these circuits are much more compact. The assembling of SMD circuits is more expensive compared to throughhole circuits and this is a parameter of decision about which technology to use: if space is not an issue throughhole is the choice. Amplifiers and feedback gpessina 6

7 . introduction 7 It could be considered strange, but the idea of trying of implementing more than one transistor in a package was not immediate (a similar, much longer, evolution was the introduction of the zero with numbers ). At the end of the 50 of the last century the first integrated circuits as we know now came. The idea and implementation were developed in parallel at Texas Instruments and Fairchild semiconductor mainly by Jack S. Kilby and Robert N. Noyce Nobel Prize in 2000 For the first time resistors, capacitors and transistors share space on the same substrate (Ge). An oscillator A flipflop (memory) Amplifiers and feedback gpessina 7

8 . introduction 8 Monolithic technology grew very fast since the first prototypes, and now the number of transistors that can be integrated on a singe piece of Si and housed in a single package have practically no limit, or the limit is approaching that of atom volume: Concerning our interest about analogue circuits, there is a very very wide choice of the so called Operational Amplifiers, OA, available with a large choice of different characteristics, therefore able able to satisfy a very wide range of applications. The space occupied by the package is not always related to the dimension of the Si on its inside. This is a single transistor in SOT23 package. And this is an amplifier with a few tens of transistors contained in its inside: same package above, but with 2 additional pin connections. The name is called SOT235. Amplifiers and feedback gpessina 8

9 . introduction 9 Now, this very large number of different kind of monolithic amplifiers are not intended to satisfy any possible application. Namely, things do not work such that an amplifier with X gain, Y bandwidth, etc., needed for our application, can be found oftheshelf. Operational Amplifiers are sold rough and, with the help of a few components added externally, the final needs can be pursued, generally with great precision: This extended range of obtainable possibilities are all based on the exploitation of the negative feedback, which needs to be managed with care to obtain the final goal. We will study deeply the theory of negative feedback with operational amplifiers, as this technique does not reduces merely to the addition of a few components around an operational amplifier, since other constraints must be met, such as dynamic, frequency, noise. etc. Then, the physical principle of operation of transistors will be also part of our study. But, because of transistors are themselves feedbacked amplifiers, they will be considered a class of OAs and several of the rules valid for operational amplifiers will maintain also for them. Amplifiers and feedback gpessina 9

10 . introduction 10 Let s briefly summarize our next steps in our study. First we will study the use of amplifiers as blackboxes, namely, we assume that the Operational Amplifier, OA, is available and we can manage it by adding some components around: OA as a black box, a mathematical tool. Then, we will enter the black box to try to understand the OA, or Amplifier structure: This is the inside of the black box above. Amplifiers and feedback gpessina 10

11 . introduction 11 We start first with the introduction of a few simple electrical properties that will be necessary afterwards. Then, we will enter deeply in the study of amplifiers with feedback. The amplifier will be modeled as a device with some characteristics. The last step will be the introduction of transistors and their use in the implementation of amplifiers. This way to proceed is in some sense not standard, as normally transistors are introduced first with the aim of their use in the realization of amplifiers. Experience shows that facing 2 new concepts at the same time, transistors and the design of amplifiers with transistors, leads to some difficulties in the comprehension from one side, and does not follow the way circuits are designed nowadays for which, at the startup of a new design, the answer to the following question is at the basis: Does it exist an OA suitable to the application? 1) If yes (almost 95% of cases), use it with a handful of passive components (resistors and capacitors); 2) If not, the amplifier is designed with the use of transistors. Actually there is a further inbetween solution consisting in merging OAs with a few transistors. This is often necessary when noise is an issue (we will see this). Amplifiers and feedback gpessina 11

12 2. THE BASIC PASSIVE ELECTRICAL ELEMENTS TO KNOW RESISTORS: or R v = Ri The current is linearly dependent on voltage through a resistor. If the current is zero the voltage is zero: this means that the energy given to resistor is promptly dissipated into heat. CAPACITOR OR CAPACITANCE: t C v(t) = 1 C 0 i(t) Current and voltage are not linearly dependent. If a short current pulse, namely a charge, is loaded on a capacitor a voltage drop is created that remains indefinitely: the energy spent for that is not dissipated, but stored in the capacitance. INDUCTOR OR INDUCTANCE: L v(t) = L di dt Current and voltage are not linearly dependent. If a short voltage pulse, namely a magnetic field, is applied to an inductor, a current is created that circulates indefinitely: the energy spent for that is not dissipated, but stored in the inductor. Amplifiers and feedback gpessina 12

13 . the basic electrical elements to know 2 IDEAL VOLTAGE SOURCE: An ideal voltage source is able to maintain a dropout (constant or variable with time) that does not depend on what it is connected across its terminals. When this is the case we say that it has zeroseries impedance and is ideal: we will show why in a while. NONIDEAL VOLTAGE SOURCE: V G R G V O A nonideal voltage source is modelled with an ideal voltage source having a resistor R G in series. The resulting signal at the load is then: V O = R G V G R G 0 V G (see next page for evaluation) According to the model above, a nonideal voltage source tries to be close to an ideal voltage source as soon as its series impedance tends to zero, or it is << smaller than the load impedance. ZEROVOLTAGE SOURCE: To null a voltage source means that the source itself generates a zeroamplitude voltage. A voltage source with zeroamplitude can be modelled with a short circuit. V O V G V G =0 V Amplifiers and feedback gpessina 13

14 . the basic electrical elements to know 3 R G V O V G I G Let s evaluate the output voltage: The current I G flows through the series of the 2 resistors: I G = V G R G Now V O is the voltage drop across : Therefore: V O = V O = I G R G V G We can arrive at the same result by considering the voltage V G is linearly shared between both resistors, while the current is the same: V G : R G = V O : Amplifiers and feedback gpessina 14

15 . the basic electrical elements to know 4 IDEAL CURRENT GENERATOR An ideal current source is able to drive a current (constant or variable with time) that does not depend on what it is connected across its terminals. When this is the case we say that it has an infiniteparallel impedance and is ideal: we will show why in a while. NONIDEAL CURRENT GENERATOR I O I G R G I O I O = A nonideal current source is modelled by an ideal current source having a resistor R G in parallel. The resulting signal at the load is then: According to the model above, a nonideal current source tries to be close to an ideal current source as soon as its parallelimpedance tends to infinite, or it is >> larger than the load impedance. ZEROCURRENT SOURCE: R G R G (see next page for I R G IG G evaluation) To null a current source means that the source itself generates a zeroamplitude current. A current source with zeroamplitude can be modelled with an open circuit. I G I G =0 A Amplifiers and feedback gpessina 15

16 . the basic electrical elements to know 5 V O I O I G R G Let s evaluate the output current: The voltage V O is common to both resistors: Hence: The current through is: Finally: I G = V O R G V O V O = R G R G I G I O = V O I O = R G R G I G We can arrive at the same result by considering the current I G is linearly shared between both resistors, while the voltage is the same: I G : 1 1 R G = I O : 1 Amplifiers and feedback gpessina 16

17 . the basic electrical elements to know 6 CURRENTCONTROLLED/VOLTAGECONTROLLED SOURCE: There is a class of voltage and current generators that are at the basis of the modelling of amplifiers and transistors: the voltagecontrolled and currentcontrolled sources. These sources generate a voltage (current) that is dependent on the voltage or current of a node or branch of the circuit or on the voltage or current of other generators. Here an example: V 1 I 3 N R 3 V 2 =BV N As it can be seen voltage generator V 2 generates a voltage that is linearly proportional, with coefficient B, to that at node N. Amplifiers and feedback gpessina 17

18 3. SUPERIMPOSITION PRINCIPLE THEO: The behaviour of a linear network containing voltage and/or current sources can be calculated by solving a number of networks equal to the number of sources, where each network differs from the other for having all the sources nulled but one (dependent sources are considered independent during calculation, their dependence being considered at the end of the process, once the nodes to which dependences are appended are highlight); then the algebraic sum of every individual elaboration is the final answer. NOTE: this theorem is very useful when dealing with noise networks. Do Not forget: a nulled ideal voltage source is replaced by a short circuit; a nulled ideal current source is replaced by an open circuit. ES. V I 3 1 R 3 V 2 I 3 =???? V 1 =6 V V 2 =14 V =1K Ω =9K Ω R 3 =2K Ω CASE a) let s null V 2 : V 1 I A V A R 3 I A = (see next page for evaluation) This first result is: Amplifiers and feedback gpessina 18 R 3 R 3 V 1

19 . superimposition principle 2 Let s write the solving eq for the network: V A V 1 I A R 3 It is convenient to start from one side and stop at the first node. Then write the balance of the currents at the node and verify if the eq has only one unknown. If no, then we must proceed to the following node for adding a new equation. This process will end when the number of written equations equals the number of the unknown. For the network above the balance at node V A is: V 1 V A = V A R 3 V A With only one unknown and: I A = V A R 3 Amplifiers and feedback gpessina 19

20 . superimposition principle 3 CONSIDERATION: Let s see why the voltage, common to all, at potential Can be assumed equal to 0 V. Let s call it V B and see what happens: V A V 1 I A R 3 V B V 1 V B V A = V A V B R 3 V A V B Now, if we call: V X = V A V B We obtain: V 1 V X = V X R 3 V X Which is the same equation of the previous page. Amplifiers and feedback gpessina 20

21 . superimposition principle 4 CASE b) let s null V 1 : V B I B R 3 V 2 The second result is: I B = R 3 R 3 V 2 Finally (please, take care at the arrows that indicate the direction of the current flows: currents having the same direction must be summed with the same sign opposite to that of the currents having opposite direction): I 3 = I A I B = V 1 V 2 R 3 R 3 = 9K 6 1K ( 14) 9K 2K 1K 2K 1K 9K = ma Amplifiers and feedback gpessina 21

22 4. THEVENIN THEOREM With respect to a pair of its terminals, a netwtork can be modelled by an ideal voltage source, V TH, having in series a resistor, R TH. The voltage V TH is the voltage measured at the terminals when the load impedance is, while R TH is the impedance that can be measured once that all the independent sources of the network are nulled. ES. V A R 3 R TH V 1 R 4 V TH VTH With respect to the 2 terminals above V TH results in: V TH = R 4 R 3 R 4 V 1 (see next page for evaluation) To calculate R TH let s proceed by applying a test voltage, V T, at the nodes of interest and measuring the resulting current, just as if we were using a measuring instrument. The ratio of the applied voltage and the driving current being the wanted impedance R TH. Before doing this all the independent voltage sources must be nulled: V B R 3 I T V 1 =0 V R 4 V T The result is: R TH = V T I T = R 4 R 3 R 3 R 4 R 3 R 3 Amplifiers and feedback gpessina 22

23 . Thevenin theorem 2 Network evaluation: V A R 3 V 1 R 4 V TH Starting from the left we encounter the first node V A : V 1 V A = V A V A V TH R 3 We see now that the above eq has 2 unknown, V A and V TH, and we need to add an eq. For that we move further on the right: V A V TH R 3 = V TH R 4 This new eq does not add any new unknown and the 2 eq above allow to solve the network. CONSIDERATION: Why the voltage V TH is the voltage measured at the given terminals is soon understood if we remember that the output voltage V O of the network below equal V G only and if only equals every time R G has a finite value. V G R G V O V O = R G V G VG Amplifiers and feedback gpessina 23

24 5. NORTON THEOREM With respect to a pair of its terminals, a netwtork can be modelled by an ideal current source, I TH, having in parallel a resistor, R TH. The current I TH is the current measured at the terminals of interest when the load impedance is 0, a short, while R TH is the impedance that can be measured once that all the independent sources of the network are nulled. ES.: V 1 V A R 3 R 4 I O I NO R NO The impedance seen at the nodes of interest is the same as that of the previous example (the network is the same). To evaluate the current I NO we consider the following: V A R 3 I NO V 1 R 4 From which we obtain: I NO = V A R 3 = R 3 R 3 V 1 IMPORTANT: if we apply the Norton Theo to a network to which we have applied the Thevenin Theo, or vice versa, it can be shown that: V Th = I No R No, R Th = R NO Amplifiers and feedback gpessina 24

25 . Norton theorem 2 Let try to understand why the Norton current I NO is evaluated this way: V A R 3 I NO V 1 R 4 I O I NO R NO Interpretation: we know that current I O equals I NO if and only if is zero if R NO has a finite value: I O = R NO R NO I G 0 I G Amplifiers and feedback gpessina 25

26 6. A NOTE ON THE TEXT COLOUR OF EQUATIONS In the following a formula coloured like this: 1 = R of βa R o Aβ RsR oβ Represents a standard result of an evaluation. Equations coloured like this: T AβR = i 1 R ool RsR oβ R o represents an intermediate passage of an evaluation and can be temporarily skipped. Equations coloured like this: T= Aβ RsR oβ βa βa R oo Indicates a result that is widely used around. Amplifiers and feedback gpessina 26

27 7. AMPLIFIERS: SYMBOLS The simplest form for representing a voltage amplifier is the following: V O = AV i v i v O The above is an ideal amplifier since there is no indication of the characteristics of its input and output and its frequency behaviour. Dealing with an ideal model we can introduce some useful properties when driving it with a nonideal voltage source: R S v i A v O I i The amplifier, being ideal, has no current that enter its input, so there is no current that flow through R S. The consequence is that the voltage V i is equal to V S. Being ideal, the output voltage does not depends on the load impedance. As a consequence: V O =AV S The result is independent from the source resistor R S and load resistor. In this latter respect the output of the ideal amplifier behaves as an ideal voltage source. Amplifiers and feedback gpessina 27

28 . amplifiers: symbols 2 R S v i A v O V O =AV S Moreover, there is another important consequence: The input power, or energy, is negligible, being: The output power is instead: P i = V S I i ~0 P o = V o 2 P L The amplifier is a transducer able to bring a weak signal, from a weak source, amplifying it for generating a strong signal across a load; the output delivered energy being larger than the input energy. The amplifier is not able to generate energy by itself and, for taking into account the actual balance of energy, we must consider that the amplifier is not isolated and some voltage sources are needed to provide the energy: v i R S I DC A V CC V EE The amplifier merely takes energy from the supply voltages, bringing it to the load. In this respect the output cannot exceed the supply rail. Amplifiers and feedback gpessina 28

29 . amplifiers: symbols 3 There is another common, actually the most widely diffused, amplifier configuration: the differential amplifier or Operational Amplifier, OA: R S V v O =A(V V ) i The advantage of such a configuration will be soon evident in connection with feedback, but also in several other applications. OAs are able to amplify signals floating from a common potential, in contrary to single ended input amplifiers. OAs have only a disadvantage in some situations concerning noise. The concept of the OA as it is shown above is very useful and widely exploited in modern electronic design. We will start from the above ideal model and, as soon as we will acquire confidence, we will complete it with its several actual characteristics. Amplifiers and feedback gpessina 29

30 . amplifiers: symbols 4 The model, actual or ideal, is not an abstraction as OAs exist and are widely used. The knowhow we will acquire will find practical applications. Vcc In Vcc Out In In Vee Out Vcc Out Vee In In InVee We can consider our OA as a black box having some specifications which links its output to the difference of its inputs with a certain proportionality, or gain, A that is a more or less complicated function of frequency or time. The factor A we are referring is generic, but must assume different values, which depend on the applications and we should find the method to personalize it. Amplifiers and feedback gpessina 30

31 8. THE NEGATIVE FEEDBACK CONCEPT Now let s try to enter in more details in the actual word. Let s start again with our model and requirement: we want an amplifier with gain A: R S V v O =A(V V ) i The idea is to find an amplifier with such a gain offtheshelf. This is not trivial for 2 reasons: (a) the number of applications is so wide that it would be very difficult for a constructor to face all the possible values of gain; (b) the developing of an amplifier with gain A and good stability against temperature, process parameters and the like is very difficult (we will see why soon). To approach the problem the philosophy used about the gain of the OA is completely different. Let s consider only amplifiers with a voltage gain very very large, let s exaggerate (the actual OAs work also this way) and consider V/V of gain, without any requirements concerning precision and stability. R S V v O =10 8 (V V ) i Amplifiers and feedback gpessina 31

32 . the negative feedback concept 2 V v O =10 8 (V V ) i If the gain is so large the allowed input excursion necessary to obtain a finite value of voltage at the output is very small, as the output will saturate toward the rails (the positive or negative supply voltages) in a while. As an example, if the positive rail is 10 V an input voltage larger than 0.1 µv is able to saturate the output. Under this point of view this class of OAs is quite useless. But we can see the situation in an other way: if the output has a finite excursion, let s say a few V at most, the differential input needs a negligible difference to take care of it. This is an important consideration to remember: if the OA gain is very very large a negligible change of the input voltage is needed to obtain an output level within the rails. Now we can start to exploit this idea for our purposes. Amplifiers and feedback gpessina 32

33 . the negative feedback concept 3 V v O =10 8 (V V ) i Take a fraction β of V O and subtract it from the input. Suppose now to be able to take a fraction β of the output signal V O and subtract it from the input, at the inverting node. This way we are adding negative feedback: we force the inverting input to follow the output, but the gain is large, so we have that: From A then V V, but V = V S and V = βv O therefore we obtain: V S βv O We can also be a bit more precise in considering A< : V O = A V V = AV S AβV O V O 1 Aβ = AV S A VV OO = V S 1 Aβ VV SS ββ This is a very important result: the voltage gain does not depend on the OA gain as long as it is very very large. For instance, if β is 0.1, then the output becomes ten times the input and the voltage excursion of the input can be much larger than with the unfed backed OA. A number of other useful consequences follow. Amplifiers and feedback gpessina 33

34 . the negative feedback concept 4 Before entering into more details let s move to consider a more real model of an OA: i V A =A(v v ) Ro i An actual OA has an input impedance, normally of large value, an output impedance, normally of small value and a link between the input and the output given by the voltage controlled voltage source V A, a voltage generator whose value is proportional to the voltage present at the input nodes according to the rule given above. The circuit of the previous page becomes now: R S VA=10 8 (v v ) i Ro i Take a fraction β of V O and subtract it from the input. Amplifiers and feedback gpessina 34

35 . the negative feedback concept 5 R S VA=10 8 (v v ) i i Ro Being present a finite value of the input impedance we expect that a current would flow through it. This current is given by: Take a fraction β of V O and subtract it from the input. I = I = V V 0 = 0 The consequence of the large OA gain and the feedback which forces V to be equal to V makes negligible the input current. This means that there is no current through R S and, as a consequence, V S equals V. And again, given that V V and also that I =I 0 we can confirm the result of the previous pages: V O = A V V = AV S AβV O VV OO = V S A 1 Aβ VV SS ββ Amplifiers and feedback gpessina 35

36 . the negative feedback concept 6 R S VA=10 8 (v v ) i i Ro We have another very important result: the gain so far obtained is not dependent on the parameters, or characteristics, of the OA as long as its gain is very very large. Another very important result is that we did not make any requirement on the feedback implementation: it needs only to be a fraction of the output voltage. A practical consideration: since β is smaller than 1 for obtaining a final gain larger than 1, it can be easily implemented with passive components, resistors, capacitors, inductors, that are very inexpensive and precise at the same time. Let s see an example. Take a fraction β of V O and subtract it from the input. VV OO VV SS ββ Amplifiers and feedback gpessina 36

37 9. THE NEGATIVE FEEDBACK CIRCUIT EXAMPLES R S V A =A(v v ) I =0 Ro Let s apply the rules: 1) A = 2 V V, According to (1) and (2): 3)I = I = 0 4 Vo = 1 β V s From (2): v = v Therefore the currents in and satisfies, with the position (3): v = v = v R 2 = v The final result has been obtained very quickly without the need of modelling the feedback by mean of a 2ports network. and = vs Finally, from (4): β = Amplifiers and feedback (See also Appendix A) gpessina 37

38 . the negative feedback circuit examples 2 R S V A =A(v v ) I =0 Ro vo = vs hence: β = In obtaining the above result we did not apply any considerations about the input/output impedance of the fed backed OA, nor the direct/indirect transmission across the feedback, provided that the gain can be considered large. Moreover, under the approximation we have carried out, the result is independent from the source impedance R S and the load impedance. We are dealing with negative feedback: a fraction of the output is taken to the inverting input, namely, is subtracted from the input. We will see soon that when the OA is assumed not ideal and all the elements are taken into consideration, the above result remains valid and only terms are added, or summed. Amplifiers and feedback gpessina 38

39 . the negative feedback circuit examples 3 Let s come back to our original result: VV OO = A 1 Aβ V S VV SS ββ And elaborate a bit: VV OO = V v O =10 8 (V V ) i A 1 Aβ V S = 1 β Take a fraction β of V O and subtract it from the input. Aβ 1 Aβ V S = 1 β T 1 T V S In this new look the overall gain is put into evidence as a product of the gain term, 1/β, with a correction factor which measures the departure of the gain from ideal, T/(1T). The quantity T=βA is called loop gain and describes the path of the signal that starts at the a whatever point in the feedback loop and return to it, running across the feedback loop. We introduce the parameter T because in the actual case in which the amplifier is not ideal, the loop gain will result: T=βAf(β), f(β) being dependent on the elements around, and it will be easier to determines T, rather than the 3 factors, individually. This view puts into evidence that the parameter to consider is the product of the OA gain and the return factor β and we know that βa is smaller than A. Amplifiers and feedback gpessina 39

40 . the negative feedback circuit examples 4 VV OO = 1 β T 1 T V S The larger is the value of T, the closer to ideal is the behaviour. This is what it happens in most of the application and one could be feel happy of it. Nevertheless, in some situations, and especially when the frequency behaviour is considered, an estimation of the correcting factor is necessary and now we will try to set the recipe for its elaboration. We will see that T will be not equal to βa an other correcting factors are present which are dependent on the network around. As we can see from the above result, we are constructing the relation between the output and the input signals by successive approximation, so that we can stop as soon as we feel satisfied. In our network below we have the input signal and the voltage controlled voltage source. We can take profit from the fact that the ideal generators (current or voltage) can assume arbitrary values for a while without affecting the morphology of the network. R S V A =A(v v ) I =0 Ro Amplifiers and feedback gpessina 40

41 . the negative feedback circuit examples 5 The easiest way to understand what happens is to superimpose the signals; namely to null one source at a time. The first step is to consider V S =0 and breaking the feedback loop by assuming the voltage controlled voltage source independent and equal to a test voltage V T for a while: V s =0 R S I =0 V A =V T Ro The circuit above is equivalent to this: R S V A =V T I =0 Ro Amplifiers and feedback gpessina 41

42 . the negative feedback circuit examples 6 R S V A V A V A =V T =generic variable A Ro We can solve the network: V A V OA R O = V OA V A V OA V OA V A = V A V A V A V A V A = V A R SS The syntax V xa indicates that we are considering the case V S is 0. The solution of the network gives: V OA = αv A V A V A = γv A We do not care about the expressions of α and γ: we stay general as the following considerations are intended to be valid for every fed backed network, as the expression at this right will be still valid with different values of α and γ, in case a different feedback is considered. Amplifiers and feedback gpessina 42

43 . the negative feedback circuit examples 7 Let s come back now to our original network and suppose that our OA has a very small gain for a while, namely A=0. Then: R S v V s s V s s Ro We can solve the network: V S V S R S V S V S V S V OS = V S V S = V S V S V OS = V OS R O V OS The result of the system of equation is: Where with V xs is intended voltage node evaluated when A=0. V OS = A DIR V S V S V S = ξv S Here A DIs the direct transmission of the signal through the feedback and take into account that the feedback is not an unidirectional network, in general. Amplifiers and feedback gpessina 43

44 . the negative feedback circuit examples 8 R S V A =A(v v ) I =0 Ro The OA and the source are back to their original function. We remember that: V OA = αv A V A V A = γv A V OS = A DIR V S V S V S = ξv S For obtaining: V O = V OS V OA =A DIR V S αv A V V = V S V S V A V A = ξv S γv A Now we have to consider the link between the input and the output through the OA, to be merged within the last equation: V A = A V V = AξV S AγV A Amplifiers and feedback gpessina 44

45 . the negative feedback circuit examples 9 R S V A =A(v v ) I =0 Ro From which: V A = A V V = AξV S AγV A V A = ξ A 1 Aγ V S Now we have to include this result in the first eq of the system: V O = A DIR V S αv A V O = A DIR V S αξ A 1 Aγ V S That we can rewrite as: V O = A DIR αξ γγ Aγ 1 Aγ V S Amplifiers and feedback gpessina 45

46 . the negative feedback circuit examples 10 R S V A =A(v v ) I =0 Ro V O = A DIR αξ γγ Aγ 1 Aγ V S Now we can arrive at a compact solution if we consider that the above result is valid for whatever value of A in the range 0 A, and we have shown that: 1 V O = A β V S Since A DIR does not depend on A we have: lim A So that, from above: A DIR αξ γγ Aγ 1 Aγ = A DIR αξ γ A DIR αξ γ = 1 A β αξ γγ = A DIR 1 β Therefore: Aγ V O = A DIR A DIR 1 Aγ 1 Aγ β 1 Aγ V S that is: V O = A DIR 1 Aγ 1 Aγ β 1 Aγ V S Amplifiers and feedback gpessina 46

47 . the negative feedback circuit examples 11 R S V A =V T What does represent V A V A? I =0 Ro V V A V A is the signal that is generated by V A which travels the path arriving at the input. Then, if V A is made a voltage dependent again, this signal return to R 1 the starting point, once multiplied by the OA gain A. Indeed, if, from the previous page, we define: T = Aγ We can set: V A V A V A = γ = T A It is very easy to show that if the OA is made ideal, namely R o =0 and =, γ reduces to β Amplifiers and feedback gpessina 47

48 . the negative feedback circuit examples 12 Considering what we obtained by setting A=0 we have the 3 rules to apply: 1) A = V V and I = I = 0 Vo = 1 β V s 2) Null the input source and make the input node controlled source (V A, the gain) independent for a while, then, solving the network it results that: V A V A = T V A A 3) Null the gain then, solving the network it results that: V OS = A DIR V S Now the last step to carry out is to superimpose the results obtained considering nulled the source, first, then nulled the gain From 2) above we must remember the following: T is always proportional to the gain A. As a consequence, the limit for A 0 of the ratio T/A is a finite value independent from A. Again, if β 0 then T 0 as well, but T/β A something. Amplifiers and feedback gpessina 48

49 . the negative feedback circuit examples 13 vo= 1 β T 1 T A DIR 1 T We see that if T is very very large or, as we know, A is very very large, the first term reduces to 1/β and the second term disappears (also from the fact that A s already small, in general). We can then consider 3 levels of accuracy: 1. T is very very large: reduces to 1/β; 2. T is large, but we would be a bit more accurate: is 1/β [ T/(1T)]; 3. T is large, or even not too large. We can be very accurate and consider both terms in the expression above for. Whatever is the level of approximation considered, we can improve it simply adding terms that can be easily evaluated. At the maximum level of accuracy the result will be not approximated any more: remember, if we consider both terms above the result is not an approximated estimate of V O, but its exact expression. Amplifiers and feedback gpessina 49

50 10. FEEDBACK PROCEDURE Let s summarize still another time. Step #1: vo= 1 β T 1 T A DIR 1 T The only way to obtain the gain 1/β is to consider A=, even if it is not: if A= in the expression for all terms hide, except 1/β. R S V A = (v v ) 1) A = 2 V V, I =0 Ro 3)I = I 0 4 Vo = 1 β V s If A= the 4 rules above apply and we can easily arrive at the result. Once 1/β has been found we can decide if we need or not to proceed to find the other terms. The following step is to calculate T and see if it is worth to consider T/(1T) or not (the term βa is the upper limit for T and gives a first guess for the necessity of calculating the actual value of T). Amplifiers and feedback gpessina 50

51 . feedback procedure 2 Step #2: vo= 1 β T 1 T A DIR 1 T To evaluate the loop gain T we null the input source and let v A to be generic and not dependent on the input nodes. From the circuit on the left: v v = γv A then: R S V A =V T Ro T =γa Once that we know T we can decide if T/(1T) can be considered well approximated with 1 or not. Amplifiers and feedback gpessina 51

52 . feedback procedure 3 Step #3: vo= 1 β T 1 T A DIR 1 T Finally, once we know β and T we can calculate A DIR by nulling the amplifier gain A. R S vo=a DIR vs Ro Combining all together the nonapproximated result is so far obtained. The application of the 3 steps to the practical case is straightforward and we do not need to model the feedback with a 4terminals block. Amplifiers and feedback gpessina 52

53 11. A FURTHER BIT OF HISTORY The first high gain amplifiers introduced were not monolithic, but based on valves. They were the K2W and 12AX7 in the of the last century. and then it came Bob Widlar with the µa700 in 1963: Amplifiers and feedback gpessina 53

54 12. THE NEGATIVE FEEDBACK AND STABILITY At a first glance it could be that the precision of the final gain would not be good, as the gain A changes with the environment (temperature, humidity, ) and from sample to sample. We can verify if this is true or nature is kindness with us. Let s start from: A f = = 1 Aβ β 1 βa and calculate: ΔA f = 1 β β 1 βa β 2 A 1 βa 2 ΔA ΔA f = 1 β ΔA f = 1 β ΔA f = 1 β β 1 βa 2 ΔA βa ΔA 1 βa 2 A Aβ 1 ΔA 1 βa 1 βa A ΔA f = A f 1 1 βa ΔA A ΔA f = 1 ΔA A f 1 βa A The above result says that the percentage of change of the final gain, A f, equals the percentage of change of the gain A, but attenuated by the factor 1βA. Amplifiers and feedback gpessina 54

55 . the negative feedback and stability 2 ΔA f = 1 ΔA A f 1 βa A Just for having an idea. If β=0.01, or the gain 1/β is 100, and A=10 7 V/V, the factor βa results in 10 5, and even a 100% of change of the gain A results in only 10 3 % of change in A f. This effect can be interpreted graphically. The response to the input signal of the openloop OA, namely, the OA with no feedback, or with β=0 is very steeply and gets saturation towards the rails as soon as the input move away from zero: A V CC V i Narrow linear region: the gain is 10 7 V/V V EE The gain A f is much smaller than A and the input voltage can spans larger values before the saturation toward the rails occurs: V CC A f V EE V i The gain slope A f is smaller and the input can span a widely region before saturation occurs Amplifiers and feedback gpessina 55

56 . the negative feedback and stability 3 Another example: i 0 1) A = 2 V V, v i s R 3 v A 3)I = I 0 4 Vo = 1 β V s Again a fraction of the output is taken at the inverting input, where it is subtracted: once again the feedback is negative. v v = 0 Apply (1) and take care: here v results 0 just because v is connected to ground. is = v v A R 3 = 0 v A R 3 = v A R 3 Apply (3) v A v A R 3 = v A v A i s = v A = i v 1 R A 2 = i R 1 R 3 i s 2 Again we have obtained the link between the output and input. = R 3 i 2 Amplifiers and feedback gpessina 56

57 . the negative feedback circuit examples 4 i 0 1) A = 2 V V, v i s R 3 v A 3)I = I 0 4 Vo = 1 β V s = R 3 i s Again, (2) is valid, therefore: β = R 3 Here we have found other 2 properties: The output is a voltage, while the input is a current: when to disentangle between the input/output variable type? The consequence of the difference in output and input variables type is that the gain 1/β results dimensional. This seems to conflict to the fact found that βa must be dimensionless. The answers to both of the above questions will be soon given. Amplifiers and feedback gpessina 57

58 . the negative feedback circuit examples 5 R F V 1 i s V 1 v A A 2 A 1 i o V 2 V 2 Here we have a first OA A 1, for which across its feedback path there is another feedbacked OA, A 2. Note that since A 2 is inverting the feedback at the input of A 1 is taken at the noninverting node, for negative feedback. Again: 1) A = 2 V V, At the input node: 3)I = I 0 4 Vo = 1 β V s V 1 V 1 = 0 From (2) I s = V V A R F = V A R F From (3) Now from the output, merging the input: V 2 V 2 = 0 I o = V 2 V A = V A i o = v 2 = We see that input and output have the same dimension, but are currents. I o = I sr F = R F I s From (4): β = R F Parameter β is dimensionless. Amplifiers and feedback gpessina 58

59 . the negative feedback concept 6 So far we have became familiar in solving fedbacked networks in case the OA gain A is close to. In doing this, some question arose. One of the results obtained is that the output and the input signal can differ in type. Actually this is the case and any combination of the output over the input can be obtained. Here the list of all cases. Voltage output over voltage input: v i β RL The output voltage is read, and a fraction of it is subtracted from the input. Current output over voltage input v i β o i o The output current is read, and a fraction of it is multiplied by a coefficient having proper dimension and subtracted from the input. Amplifiers and feedback gpessina 59

60 . the negative feedback concept 7 Current output over current input i s The output current is read, and a fraction of it is subtracted from the input. β i i o i o Voltage output over current input i s v i β G i o The output voltage is read, and a fraction of it is, after being multiplied by a coefficient having proper dimension, subtracted from the input. Amplifiers and feedback gpessina 60

61 . the negative feedback concept 8 A further consideration is how to understand which is the variable, output or input, worked out by the feedback. The answer is tied on what is our interest: we would like that the amplified signal is as much as possible independent from the source and load impedances, so approaching the ideal situation. In this respect, rather than thinking at a variable that is managed we can think at our final aim. Let s see the practical cases: =A(v v ) I =0 v We have found that: vo = vs The output voltage is independent from the load impedance, under the assumption that the gain is very very large, whereas the output current is: io = = R vs 1 We can see that the output current is fully dependent on the load impedance: the output variable in this configuration is therefore the voltage, or reading the output current is meaningless as it depends on the load. Amplifiers and feedback gpessina 61

62 . the negative feedback concept 9 vs Rs v I=0 R1 vo=a(vv) vo RL R2 From our approximation we know that the input current is negligible; the consequence is that the voltage drop across is negligible and V s =V. Indeed our result is: vo = So the final result does not depends on the input source resistor and we can consider the voltage the input variable. Let s consider the case the input is a current generator with its source parallel resistor. vs I =0 v =A(v v ) i s We have now that: v = I s and v = Therefore: vo = RsI s We can see that the output voltage is fully dependent on the source impedance in this latter case: the input variable in this configuration is therefore the voltage. Amplifiers and feedback gpessina 62

63 . the negative feedback concept 10 Let s consider the second of our examples. i 0 v i s R 3 v A We have found that the gain is: = R 3 That does not depends on the load : the output variable is therefore a voltage. We now consider the source resistor in parallel to the input current source. Since v =0 V and v v, then v 0 V and the current through is 0 A: the output signal will not depend on, as it happens, actually. i s Let s consider the case the input is a voltage generator with its source series resistor. Now we have that: i s i 0 v R 3 v A i s = So that: = R 3 The output becomes dependent on the source impedance in this case: the amplification is about current. Amplifiers and feedback gpessina 63

64 . the negative feedback concept 11 Let s consider the last of our examples. i s V 1 V 1 R F v A We know that V 2 is about 0 V so that: I o V 2 V 2 We have found that the gain is: I o = R F I s The input follows the same consideration of the previous case: v 1 is at ground and the current in is nulled. = I o and: = I o = R F I s The output voltage depends on the load impedance, and the output signal is, therefore, a current. Amplifiers and feedback gpessina 64

65 . the negative feedback concept 12 There is another approach for understanding which kind of feedback is working with respect to the input variable. i s V 1 V 1 R F v A V 2 V 2 In the circuit on the left one of the input, V, is held at constant potential. As a consequence, the other terminal, V, is held close at constant potential as well. This suggest that the input is at low impedance since it does not change by injecting current. The conclusion is: every time one of the inputs is held at constant potential, or when the feedback insists on the same node where the source is connected, then the input is a current. vs Rs vo=a(vv) v The opposite is true for the voltage vo at the input: when the 2 inputs are I=0 floating and one of the input is RL connected to the source, or when R1 R2 the feedback does not insist on the same node as that of the source, then the input variable is a voltage. Amplifiers and feedback gpessina 65

66 . the negative feedback concept 13 There is another approach for understanding which kind of feedback is working with respect to the output variable. vs Rs vo=a(vv) v vo I=0 RL R1 R2 When the feedback reads the same node of that of the OA output, or one of the nodes of the load impedance is held at constant potential, then the voltage is the output variable. V 1 V 1 i s R F v A V 2 V 2 If the load impedance has it legs connected to floating nodes, then a current is the output variable. A current is read also in case the feedback does not connect the load. See the example below. Amplifiers and feedback gpessina 66

67 13. EXAMPLES OF COMPLETE FEEDBACK EVALUATION Let s consider some examples. i V A = (v v ) i Ro Step #1: 1) A = 2 V V, 3)I = I 0 4 Vo = 1 β V s If (1) is valid then (2) is valid, too, and v v = therefore: = From which we obtain the same result of the ideal case: or: vo = 1 β = Amplifiers and feedback gpessina 67 vs

68 . examples of complete feedback evaluation 2 i V A is generic The input is nulled i vo Ro Step #2: v v =T v A A Let s solve the network as the feedback is not present (v A considered independent) for finding the loop gain T: v A R o = v v = v v v v = Let s define: v R v = v i R i 1 = 1 1, namely: = From the second eq: v = v = 1 1 v R = v 2 R 1 v = v R o 2 Amplifiers and feedback gpessina 68

69 . examples of complete feedback evaluation 3 v A R o = v v = v = v v v v = v Now the first eq: v A R o = R o 1 v A R o = R o 1 v A R o = R o 1 With the position: 1 = 1 1, namely: = v A R o = R o v A = R o = R o v AA And: v = R o v AA Amplifiers and feedback gpessina 69

70 . examples of complete feedback evaluation 4 i V A is generic The input is nulled i vo Ro Remembering that: T v A A = v v = v = R o v AA We, finally, obtain: T = A R o Remember that in a negative feedback loop T must be negative (at DC). If you do not find such a result two possible drawbacks are: the feedback path was set positive or an error happened in the solution of the system of equations. Amplifiers and feedback gpessina 70

71 . examples of complete feedback evaluation 5 i V A =0 Ro i Step #3: vo=a DIR vs We null the gain source: v = v v v From the second: = v 1 o We put: = 1 1 R 0 R o R 0 v = R 0 v = R 0 R 0 = R 0 R 0 v From the first: = v v v 1 R 0 v R 0 = v v v 1 R 0 R 0 Amplifiers and feedback gpessina 71

72 . examples of complete feedback evaluation 6 v = v v v = R 0 = v v v 1 R 0 R 0 Again, for convenience, we can put: 1 = 1 1 R 0 = v v = v v = Finally: s = R 0 R 0 v = R 0 R 0 s Or: A DIR = = R 0 R 0 Amplifiers and feedback gpessina 72

73 . examples of complete feedback evaluation 7 i V A =A(v v ) Ro i Putting all together: = And: T 1 T R 0 R T T = A R o Amplifiers and feedback gpessina 73

74 One consideration We learned to find the value of T starting by solving an appropriate network. The result was: Where γ is the result of the linear system of equations. The term γ accounts on the fact that the fed backed network include non ideal terms and the OA itself is not ideal, having an input and output impedances of finite values. Nevertheless, in case of an ideal amplifier having an input impedance of value and an output impedance of zero value we have found that: As a consequence: T = γa T = βa lim γ = β, R o 0 In practice, we can consider γ as expressed with the form: Namely: γ = βf(β) T = βf(β)a (f(β) 1 when and 0) Amplifiers and feedback gpessina 74

75 and another consideration V B v i v B Av i v B = v B R o At first it might seem a complicated procedure the following of the 3 steps. The alternative way is to solve the network as one were not be aware of the feedback, as it follows: This is the system of equations to be written to solve the network as if we did not know the feedback A v B R o = v B v i = v B 1 = v R B 1 1 i A R o = v B A 1 1 v R o R o 1 2 R o To come to the final solution several intermediate steps are needed. = A R o v R o A 1 i A v A 1 v i cvd! 1 At the end the final result does not distinguish the different contributions. Amplifiers and feedback gpessina 75

76 14. INPUT IMPEDANCE OF A FED BACKED OA Av i R o The source and load impedances introduce signal partitions, in a real OA, at the input and output across which part of the signal is lost. Nature of feedback is kindly in making the input and output of the fed backed OA close to an ideal condition. This was implicit in the results we have obtained in our calculation as we have taken into account the presence of the source and load impedance, obtaining little effects. We can try to estimate the input and output impedances of the fed backed OA in view of a final modeling, as we will see later. Again, we will try to exploit the parameters we are learned to estimate. Amplifiers and feedback gpessina 76

77 . input impedance of a vtov fed backed OA 2 vtov= voltage input and voltage output amplifier i V A =A(v v ) Ro i The input impedance of the network is that seen at the input source node, and is given by: f = i s The current i s is flowing through and also thought and through this latter satisfies: i s = v v or: i s = v v = v A A Amplifiers and feedback gpessina 77

78 . input impedance of a vtov fed backed OA 3 i v A = 1 β T 1 T vo = 1 β T 1 T Ro i v p We now will try to verify if v A can be obtained exploiting the parameters we have just considered. Node v A has not special properties with respect node or any other node of the network; therefore we can write: v A = 1 β T 1 T A DIR 1 T Here β and A R refer the fact that node v A is different than node, but T is the same, as expected all around the loop. The expression above is expected to be valid as the evaluation made for node was completely general, not base on a particular assumption, so that the same procedure can be applied at every other node of the network. Amplifiers and feedback gpessina 78

79 . input impedance of a vtov fed backed OA 4 i V A = (v v ) Ro i v p =present evaluated feedback node Although not necessary for what is our intended goal, let s determine β, just to practice: β = = β = = 1 With the position above: v A R o = v A R o = 1 R o 1 1 or: v A = 1 R o 1 R o namely: v A v S = 1 β = 1 R o 1 R o Amplifiers and feedback gpessina 79

80 . input impedance of a vtov fed backed OA 5 i i Ro Now, if the gain is reduced to 0 we see that node v A cannot change for direct transmission as, being an ideal voltage source, it has a zero impedance and A R results to be 0: v A = 1 β T 1 T Therefore, coming back to the restored amplifier: v v = v A A = 1 Aβ T 1 T = 1 Aβ v v = γ β 1 T =present evaluated feedback node The loop gain T is proportional to A, and we have found to depend on the components of the network. It can be put in the form: T = γa, (γ < 0) γa 1 T i s = v v = γ β v 1 T i Amplifiers and feedback gpessina 80

81 . input impedance of a vtov fed backed OA 6 i V A =A(v v ) Ro i i s = v v = γ β v 1 T i =present evaluated feedback node Finally: f = i s = γ β 1 T The feedback multiplies the parameter /( γ β ) by 1T, with the aim of increasing it, obtaining a condition closer to the ideal case. Nevertheless, if we set T to 0 by nulling the OA gain, as it is the case below, it results evident that the parameter /( γ β ) represents the input impedance that is measured in open loop condition, or when A is set to 0 Namely: ol = γ β Amplifiers and feedback gpessina 81

82 . input impedance of a vtov fed backed OA 7 i i Ro Therefore, we can write that: f = i s = ol 1 T where ol (here iol is for input, open loop) is the impedance that is measured when the OA gain is set to 0, in order to obtain T=0. This measurement is easily done with the network above: ol = with: R 1 = and: R 2 = R o A practical example is given in Appendix A. Amplifiers and feedback gpessina 82

83 . input impedance of a vtov fed backed OA 8 A consideration (a): In determining the input impedance the source of test signal was a voltage. That was a good choice as in this way we were sure that the driving current was that flowing through f. Namely, in this way the feedback is not perturbed when we apply the A= rule: when A= the current thorough drops to a negligible value and is managed from the feedback. i i V A =A(v v ) Ro Moreover, if the testing voltage source is not ideal, namely is not completely negligible, the measured input impedance value is not contributed, or is marginally contributed, by itself. Amplifiers and feedback gpessina 83

84 . input impedance of a vtov fed backed OA 9 A consideration (b): If a current source were used for the characterization we would perturb the feedback, acting on the actual value of T. We see the way this perturbs T: V A =A(v v ) v I o s Ro i i i V A =A(v v ) T= v I o s Ro Aβ RsR oβ βa Ro 1 βa R o i We see that if because we use a current source, than T 0, therefore. The direct gain was omitted for brevity: = 1 β = T 1 T = 1 T β 1 T i s AR L βa 1 1 RsR oβ Ro βa R o 1 T i s AR βa 1 Rs i i Ro βa o Amplifiers and feedback Namely, the gain depends on A and not on 1/β only: we are unhappy of this.. gpessina 84

85 . input impedance of a vtov fed backed OA 10 A consideration (c): In case a current source were used anyway, it would help to consider a source impedance,, in parallel to it. Remembering the Thevenin theorem, this would be equivalent to consider a voltage source with a series resistance in series to it. V A =A(v v ) v I o s Ro i i GENERAL RULE: In characterizing the input impedance use a voltage source to measure large impedance and a current source to measure small impedances; namely, if the feedback reads a voltage, then use a voltage source; if the feedback read a current use a current source. Said in another way: a voltage amplifier tries to show a large input impedance, then use a source with a small impedance for its characterization; a current amplifier tries to show a small input impedance, then use a source with a large impedance for its characterization. The golden suggestion: look at the found loop gain and see what happens if the source impedance is very big or very small: only in one case T will be 0; than, take a current source if T 0 when 0, or take a voltage source if T 0 when Amplifiers and feedback gpessina 85

86 . input impedance of a vtov fed backed OA 11 A consideration (d): Using the configuration on the figure is anyway possible, provided that some considerations are done. It has to be considered that: i s = v v v V A =A(v v ) v I o s Ro i i Where, as we have already shown: v A = 1 T β 1 T = 1 T β 1 T i s i s = v i s And: 1 γ β R 1 T i s i γ β R 1 T = v i and v v = v A A fp = v i s = 1 = 2 1 γ β 1 2 R 1 T = 1 i f 1 fo = fo fo = fo fo And f0 is the feedback input impedance measured when is set to 0 in the network, see the next consideration for detail. Amplifiers and feedback gpessina 86

87 . input impedance of a vtov fed backed OA 12 A consideration (e): An useful utility is to disentangle the source impedance,, from the amplifier input impedance. In practice the impedance so far evaluated, f, we would like to be express as: i s V A =A(v v ) Ro f = f0 Starting from the zero gain, or open loop, input impedance: ol = ol0 We show that it is indeed valid the following expression: f = ol0 1 T 0 Where ol0 and T 0 refer to ol and T evaluated when =0 Ω. Amplifiers and feedback gpessina 87

88 . input impedance of a vtov fed backed OA 13 i Ro i In the above circuit we have that: ol = Where is the impedance seen to the right of node V. Thev. i s V A =A(v v ) Ro By applying the Thevenin th. to node V we obtain that the impedance seen is, while αv A will be the voltage, α being the fraction of V A measured across in the circuit on the left. Amplifiers and feedback gpessina 88

89 . input impedance of a vtov fed backed OA 14 i s V A =αv A =αa(v v ) Pay attention, this node is not V o! In the new equivalent circuit above both V s and V A see the same impedance: ol = If now we take V s =0 we can write, considering that the dropout across a resistor in a series of resistors is proportional to the resistor value, according to the following proportion: We have: V V = V A ol V V = V A ol0 (R S = 0) ol V V = V A ol0 V V = V A 0 But from the last eq: = ol V V V A = ol V V αv A = ol T αa and: Hence: V V 0 V V 0 T 0 = ol0 = R V iol0 = R A αv iol0 A αa ol T = ol0 T 0 Amplifiers and feedback gpessina 89

90 . input impedance of a vtov fed backed OA 15 ol T = ol0 T o Now let s elaborate: T = ol0 ol T 0 So that: f = ol 1 T =ol ol T =ol ol0 T 0 = ol0 ol0 T 0 = ol0 1 T 0 CVD! Amplifiers and feedback gpessina 90

91 15. OUTPUT IMPEDANCE OF A VTOV FED BACKED OA The output impedance is the impedance seen from the load and is characterized as shown below: i s R 1 V A =A(v v ) Ro I T (the practical standard way to characterize a voltage source is with a current source, as in this case there are no conflicts and the network remains unperturbed. The vice versa also applies) The source of the signal is nulled and the test signal is applied at the node where the impedance has to be measured. Note that this mathematical procedure reflects what it is done in the lab, with a standard instrumentation. Also for this case we would try to apply our known procedure as the node where the signal is applied now has not a particular property with respect to V s, provided that the source does not perturb the network, hence the loop gain T. Again, we expect that: = 1 β" T 1 T i T A DIR" 1 T i T Parameters β and A DIR are only to indicate that is excited by a source other than, and are different from the respective β and A DIR. Amplifiers and feedback gpessina 91

92 . output impedance of a vtov fed backed OA 2 i s R 1 i 2 V A =A(v v ) Ro = 1 β" I T T 1 T i T =present evaluated feedback node Let s begin with considering T= : 0 = v = v i 2 = 0 therefore: = v i 2 = 0 I T Therefore results 0, that means: β = and we remain with: So that: = A DIR" 1 T i T R ofp = i T = A DIR" 1 T With T=0 R ofp reduces to the output impedance, R ool (R output, measured in open loop), measured when the gain is zero, hence: R ofp = R ool = A DIR" T=0 Amplifiers and feedback gpessina 92

93 . output impedance of a vtov fed backed OA 3 i s V A =A(v v ) Ro I T =present evaluated feedback node So we can write: R ofp = R ool 1 T The feedback works in lowering the output impedance by a factor of 1T, making the output of the network more close to an ideal voltage source. Of course this behaviour is made more and more effective if the loop gain T is large. Amplifiers and feedback gpessina 93

94 . output impedance of a vtov fed backed OA 4 i s Ro I T Just to exercise, now we have that: i T = R o v =present evaluated feedback node v = v v Or, simply: We have: v = i T = R o And: R ool = i T = 1 R o Amplifiers and feedback gpessina 94

95 . output impedance of a vtov fed backed OA 5 A consideration (a): i s V A =A(v v ) Ro vo I T We expect to evaluate a very small output impedance, then, for not perturbing the loop, it is recommended to use a current source for the test signal. Below, we use a voltage source. See what happens. V A =A(v v ) i s R 1 vo R T Ro V T It is easy to imagine that if the voltage source is close to ideal, namely R T close to 0 Ω, the output node is shorted and the loop does not work. See next slide for details. The golden suggestion: To measure a small impedance use a source with a large impedance, a current source; to measure a large impedance use a source with a small impedance, a voltage source. Amplifiers and feedback gpessina 95

96 . output impedance of a vtov fed backed OA 6 A consideration (b): V A =A(v v ) i s vo R T Ro V T Then: Let s evaluate the loop gain. For convenience we define: = R T RT 0 R T = T= A Ro T RT 0 0 In this case, obviously, again neglecting the direct transmission: = 1 β = T 1 T AR L βa 1 1 RsR oβ Ro βa R o 1 T 0 RT 0 The golden suggestion: look at the found loop gain and see what happens if the load test impedance is very big or very small: only in one case T will be 0; than, take a current source if T 0 when 0, or take a voltage source if T 0 when. Amplifiers and feedback gpessina 96

97 . output impedance of a vtov fed backed OA 7 A consideration (c): nevertheless i s R 1 Ro vo i T V T The test signal in the position shown does not affect the loop and can be used for testing. Also in this case we have only direct transmission, given by: = = R o R o R o v T When A=0 Therefore, finally, wen A is its value: = R o v T R o R o 1 T = A DIR 1 T v T Amplifiers and feedback gpessina 97

98 . output impedance of a vtov fed backed OA 8 A consideration (d): nevertheless i s vo Ro i T V T = R o v T R o R o 1 T = A DIR 1 T v T Therefore: i T = v T = 1 A DIR 1 T v T R ofs = R of = v T i T = 1 1 A DIR 1 T It can be shown that the result obtained in this measurement configuration for R of and in the previous configuration with a current source as test signal are the same. The adoption of the test current source approach is a bit more physical as it links the impedance measured in open loop condition with the impedance measured in closed loop condition. Amplifiers and feedback gpessina 98

99 . output impedance of a vtov fed backed OA 9 A last consideration : An useful modelling is obtained when the load impedance,, is disentangled from the amplifier output impedance. In practice the impedance so far evaluated, R of, we would like to be expressed as: R of = R of i s V A =A(v v ) Ro I T Starting from the zero gain condition, or open loop, output impedance: R ool = R o = We can write, obviously: R ool = R ool We will show that it is possible to write: R of = R ool 1 T Where R ool and T refer to R ool and T evaluated when = Ω. Amplifiers and feedback gpessina 99

100 . output impedance of a vtov fed backed OA 10 i s V A =A(v v ) Ro Norton I T i s I A =V A /R o = A/R o (v v ) R o This way the impedance seen from I T and I A are the same and equal to: R ool = R o I T = I R Assuming that I T =0 the current I s a fraction of the total current weighted on the inverse of the impedance on which it flows: I R = I A I R 1 1 R ool If we set =, then: R ool = R o But, from above: I R RL = = I A R ool R ool = I A Amplifiers and feedback gpessina 100

101 . output impedance of a vtov fed backed OA 11 i s I A =V A /R o = A/R o (v v ) R o I T I R Now it is evident from above that we can write: V V = ρi R V V = ρ I R RL = RL = So that: I R = V V R O = T I A ρρ V A A R I O R RL = ρv A I A = V V RL = R O = T ρ V A A R O ρv A So that from the result of the previous slide: I R R ool = I A 1 R ool I R I A = 1 I R RL = R ool = I A 1 I R RL = = 1 R ool I A We can write: 1 T R O = 1 T R ool A ρv A R ool A R O ρv A T R ool = T R ool Amplifiers and feedback gpessina 101

102 . output impedance of a vtov fed backed OA 12 T R ool = T R ool Let s elaborate: 1 = 1 T R of R ool = 1 R ool = 1 R ool Namely: T R ool T R ool = 1 1 R ool = 1 1 R of T R ool R of = R of Amplifiers and feedback gpessina 102

103 16. INPUT/OUTPUT IMP. SUMMARY OF A VTOV FED BACKED OA Summary about the input/output impedance of a vtov. Input impedance measured with a voltage source having a series resistance in series: fs = ol 1 T fs = ol0 1 T o = fs0 ol T = ol0 T 0 Where olo and T o are ol and T measured when is set to 0 Ω. If the input impedance is measured using a current generator with in parallel a source resistor we have: fp = fso oooooooo Output impedance measured with a current source having in parallel the load resistor : R ofp = R ol 1 T R ofp = R ol 1 T = R ofp T R ool = T R ool Where R ool and T are R ool and T measured when is set to Ω. If the output impedance is measured using a voltage generator in series to the load we have: R ofs = R ofp Amplifiers and feedback gpessina 103

104 17. GAIN OF A CTOC FED BACKED CURRENT OA ctoc= current input and current output amplifier i A =Ai i i i vo R o i o v i s R F v B V 2 i F V 2 Just for exercise we can evaluate the loop gain and direct transmission. For the loop gain we put i s =0 A, namely, we open i s, since it is a current generator. To simplify let s consider the OA having in feedback ideal otherwise, being a feedback in a feedback, it would be recommended to replace it with its model, to avoid to introduce errors (see later for this). i A =is independent i i R o i o v i s R F v B V 2 i F V 2 Amplifiers and feedback gpessina 104

105 . gain of a ctoc fed backed current OA 2 i A =is independent i i R o i o v i s R F v B V 2 i F V 2 With: R F = R F R S and: R S = R S It is: i A = 1 R o 1 and v B = v = R S v R S R B = R S 1 F R S R F 1 1 R o i A i i = v = 1 R S 1 R S R F 1 1 R o i A To obtain: T = A i i = A S RS R F 1 i A R S R F 1 1 = γa, (γ < 0) R o Amplifiers and feedback gpessina 105

106 . gain of a ctoc fed backed current OA 3 i i i A R o i o v i s R F v B V 2 i F A 2 V 2 Now for the direct transmission we null the current dependent current generator, i A : note that it is left an open circuit since it is a current generator. For this particular case the fed backed amplifier A 2 shows a very small transmission between its output and its input, being fed backed. Anyway, let s call α this small term: Exercise: evaluation of α α = i o i F i s v i F R F i A v B R o V 2 A 2 R o2 i o Let s suppose,, and 2 very large to simplify. We see that if A 2 =, then V 2 is 0 and 1/β 2 =0. V 2 Amplifiers and feedback gpessina 106

107 . gain of a ctoc fed backed current OA 4 i s v i F If R o2 were 0 the transmission would be 0. Otherwise: i o = R o2 i R F R o2 R o2 i o R F o R F i A v B R o R o2 A 2 V 2 V 2 i o Now the loop gain of A 2 : R F = R F R o2 R o i A R o i o i s v R F v B R o2 V 2 Very simple ( ): i F V T V 2 v 2 R F R o R F R o R o2 R o R o V T v 2 V T Therefore: T 2 A 2 and: α R o2 R o 1 1 A 2 Amplifiers and feedback gpessina 107

108 . gain of a ctoc fed backed current OA 5 i A = i i i i vo R o i o v i s R F v B V 2 Finally, let s determine 1/β, considering A= : i F V 2 Remember, if A= it results i i 0 and v v, but in this case v =0, then v =0, too, and: i s = v B R F, i o = v B i o = R F i s As it can be seen the gain does not depends on both the input source and output load impedances, that suggests that the above is a current output over current input amplifier. Finally: = 1 β T 1 T i s A DIR 1 T i s i o = R F T 1 T i s αα 1 T i s T = A i i = A R S 1 i A R S R F 1 R 1 o α R o2 R o 1 1 A 2 Amplifiers and feedback gpessina 108

109 18. INPUT IMP. OF A CTOC FED BACKED CURRENT OA i A =Ai i i i vo R o i o v i s R F v B V 2 i F V 2 The procedure we developed for measuring the input/output impedance for the voltage amplifier remains valid also for this configuration, as it was determined using a general approach. Being amplified a current we expected that: f = ol 1 T ol being the impedance measured when the gain is set to 0 and a test current is injected at the input. Since the output is a current, then: R of = R ool 1 T Again, R ool is the impedance measured at the output, by injecting a test voltage in series to the load, when the gain is set to 0. Amplifiers and feedback gpessina 109

110 . input impedance of a ctoc fed backed current OA 2 =present evaluated feedback node i i i A = 1 β T 1 T i s R o i o i s v R F v B V 2 Previous feedback branch i F V 2 Just to exercise let s verify the positions given in the previous slide. From the circuit above we expect that: f = v i s = i i i s = i A Ai s Now if we consider i A the branch where to consider the feedback: i s v i F i i i A i A = 1 β T 1 T i s A DIR 1 T i s We see that if A=0 than i A cannot be different than 0, since the current source is replaced by an open. We have seen that I o =αi s, but I A remains 0 since the impedance in the branch is. Amplifiers and feedback gpessina 110 R o R F v B A 2 V 2 V 2 i o

111 . input impedance of a ctoc fed backed current OA 3 =present evaluated feedback node i i i A = i i vo R o i o v i s R F v B V 2 So we remain with: i F i A = 1 β T 1 T i s V 2 Now let s consider A= : i A = R o and = I o = R F i s = R F i s i A = R o 1 R F i s = 1 β i s Therefore: f = i A Ai s = R o 1 R F T A 1 T But we know that T is proportional to A: T = γa, (γ < 0) Finally: f = v i s = i i i s = i A Ai s = A 1 β γa 1 T = ( γ β ) 1 T = ol 1 T CVD! Amplifiers and feedback gpessina 111

112 . input impedance of a ctoc fed backed current OA 4 A consideration (a): Likewise with the voltage amplifier we would disentangle the input impedance of the fed backed OA from. Or: f = ol 1 T i A =Ai i i i vo R o i o v i s R F v B V 2 Norton V 2 Voltage nulled i s i i v R o R F v B Norton v B 0 V 2 V 2 i o If i A =0 then v B =0. Therefore the Norton impedance reduces to R F. Impedance, R N Amplifiers and feedback gpessina 112

113 . input impedance of a ctoc fed backed current OA 5 Now the Norton current, namely, the current in the short set at the node: i i i A =Ai i vo R o i o v i s R F v B V 2 Norton I N V 2 That results in: i o = R o R o i A v B = i o i N = v B R F Or: i N = R o R O R F i A = αi A = αai i i N =αai i i i v R F i R i s Amplifiers and feedback gpessina 113

114 . input impedance of a ctoc fed backed current OA 6 i N =αai i i i v R F i R i s Now we have that, in case i s =0: i i = i N i i 1 1 ol But: ol = i N i i Rs = = 1 i N i i = 1 ol ol = i N i i i N = We obtain: i i αi A = i i = i N T ol ol αa T αa i i Rs = = i N αi A i i = T = αa ol i N = T αa i i Rs = = i N T ol ol αa T ol = αa i N From which: T ol = T ol Amplifiers and feedback gpessina 114

115 . input impedance of a ctoc fed backed current OA 7 The open loop input impedance results: v i i i A R o i o i s R F v B V 2 ol = R F i F A 2 (let s assume negligible the output impedance of A 2 with its feedback) V 2 An we have just shown that: T ol = T ol (here ol is ol evaluated when.) From which: 1 = 1 T f ol = 1 ol T ol = 1 1 ol = 1 1 T ol = 1 1 f T ol CVD! Amplifiers and feedback gpessina 115

116 . input impedance of a ctoc fed backed current OA 8 A consideration (b): i A =Ai i i i vo R o i o v I s R F v B V 2 i F V 2 As with the vtov configuration we see here that it is not convenient to adopt a voltage source at the input to measure the input impedance. If we short the loop gain is the same found previously: T = A i i = A R S 1 i A R S R F 1 1 R o R S = R S We see that T 0 if 0, and this suggest to use a current source. Insisting anyway in using a voltage source we obtain (Norton equivalent): i o = 1 T β 1 T i s A DIR 1 T i s = 1 T A DIR β 1 T 1 T T i o A Rs 0 β R DIR s i o Rs 0 R F A 1 1 R F 1 R 1 o R o2 R F R o 1 1 A 2 Amplifiers and feedback gpessina 116

117 . input impedance of a ctoc fed backed current OA 9 i A =Ai i i i vo R o i o v I s R F v B V 2 i F So that it is not recommended to adopt this configuration if has a negligible value. Nevertheless, we will see that this topology is useful, provided that some precautions are taken. In case has a finite value we can consider to measure the impedance seen by the voltage source V s : V 2 i s = v = i i = i A A i A = 1 γa β 1 T = A γ β 1 T = Av i s = v = γ β 1 T T = γa, (γ < 0) i s = 1 ( γ β ) 1 T or: fs = i s = 1 ( γ β ) 1 T This input impedance results slightly larger than, suggesting that results in series with a small impedance. This small impedance results f as it is shown in the following pages: fs = i s = f Amplifiers and feedback gpessina 117

118 . input impedance of a ctoc fed backed current OA 10 i A =Ai i i i vo R o i o v I s R F v B V 2 To prove that: i F fs = i s = f V 2 We start by considering that: i A = 1 T β 1 T = A γ β 1 T While, in case the source is a current: i A = 1 T β 1 T = A γ β 1 T i s The 2 configurations are linked by the Norton theorem: i i i A =Ai i vo R o i o = i s v i s R F v B V 2 Norton V 2 I s Amplifiers and feedback gpessina 118

119 . input impedance of a ctoc fed backed current OA 11 i A =Ai i i i vo R o i o v I s R F v B V 2 i A = A γ β, and 1 T i A = A γ β 1 T i s Results in: We have shown that: So that: i F = β β or f = γ β 1 T 1 β = 1 β V 2 = i s fs = 1 γγ β 1 T By continuing: = 1 1 γ β 1 T = 1 f 1 f R f fs = s 1 R = 1 if Amplifiers and feedback gpessina 119 f 1 f = 1 fs = f = f CVD! f

120 19. OUTPUT IMPEDANCE OF A CTOC FED BACKED CURRENT OA i A =Ai i v i i vo R o i o v T i s R F v B V 2 where R ool is the impedance measured at the output, by injecting the test voltage in series to the load, see above, when the gain is set to 0. By applying v T we expect: But, at the input mesh, with A= is i i =0, hence v =0, i F =0 and: And we reduces to: i F We now want to show true what we introduced in the previous chapter: R of = R ool 1 T i o = 1 β T 1 T v T A DIR 1 T v T i o = 0 i o = A DIR 1 T v T Or (take care of the direction chosen for the current) and that V 2 =0: R ofs = v T i o = 1 T A DIR = R ool 1 T now let s determine A DIR that is our 1/R ool Amplifiers and feedback gpessina 120 V 2

121 . output impedance of a ctoc fed backed current OA 2 i s i i v R o R F v B i o v T V 2 Assuming that V 2 is almost 0 it is: i F V 2 i o = v T R o = A DIR v T But it is evident that: R ool = v T i o = R o So that it is proved the last position of the previous page. Amplifiers and feedback gpessina 121

122 . output impedance of a ctoc fed backed current OA 3 Consideration (c): i A independent Thevenin v i i R o i L v T i s R F v B V 2 i F V 2 Adopting the usual procedure we can show that it is valid that: R of = R ool0 1 T 0 To arrive at this we must put both v T and i A to see the same impedance R ool or R ool0 : i i V A =R o i A vo R o i L So now if v T is 0 we have that: i i = α R ool v A v v T R ool i i = α v A i s R F v B V 2 But also: i F V 2 R ool0 i i = αr RL 1v A =0 Amplifiers and feedback gpessina 122

123 . output impedance of a ctoc fed backed current OA 4 V A =R o i A i s i i v vo R o R F v B i L v T V 2 Nevertheless: i i v A = i F i i R o i A = T R o A V 2 i i RL =0 = i i =0 = T 0 v A R o i A R o A Therefore, from the previous slide: R ool i i = α v A R ool0 i i = αr RL 1v A =0 We have: R ool T = R ool0 T 0 Amplifiers and feedback gpessina 123

124 . output impedance of a ctoc fed backed current OA 5 V A =R o i A i s i i v vo R o R F v B i L v T V 2 i F V 2 R ool T = R ool0 T 0 R ool = R o Therefore: R of = R ool 1 T = R ool R ool T = R ool0 R ool T = R ool0 R ool0 T 0 = R ool0 1 T 0 = f0 where R ool0 and T 0 are R ool and T when is 0 Ω. Amplifiers and feedback gpessina 124

125 20. INPUT IMPEDANCE OF A CTOC FED BACKED OA In the previous configuration studied we have used a current OA for implementing a fed backed current amplifier. This choice is the natural, but current OA are not easy to be found offtheshelf or, if they are available, are expensive. Normally, the fed backed current amplifier is implemented with a standard OA. We will study now also this configuration and put into evidence the differences. v A =A(v v ) v i i R o i R o L i s R F v B V 2 i F V 2 The evaluation procedure is rather similar to that seen before: f = v i s where, from the scheme above: and we need to elaborate v A. v = v A A Amplifiers and feedback gpessina 125

126 . input impedance of a ctoc fed backed OA 2 v A =A(v v ) v i i R o i R o L i s R F v B V 2 i F V 2 Continuing: v A = 1 β T 1 T i s A DIR 1 T i s from the scheme above we see that the current i o cannot change the voltage at node v A when the source is nulled and, as a consequence, A R =0: v A = 1 β T 1 T i s 1 β = Ω Let s determine β Amplifiers and feedback gpessina 126

127 . input impedance of a ctoc fed backed OA 3 v A = (v v ) v i i R o i R o L i s R F v B V 2 i F V 2 Let s take A=, which gives v =0. We know that i o is given by: i o = 1 β i s and: therefore: v A = R o i o v A = R o β i s = 1 β i s And from: f = v A = 1 T Ai s Aβ 1 T The final: f = γ β 1 T = ol 1 T Amplifiers and feedback gpessina 127

128 . input impedance of a ctoc fed backed OA 4 Let s compare the difference of the results obtained for the 2 types of OAs. In both cases the expression for the input impedance is the same, namely the ratio between the open loop input impedance divided by the 1T term. What is remarkable different is the open loop input impedance. If we consider negligible the output impedance of the second OA, for both circuits it results that: ol = R F For the current OA is of a few Ω and: ol_current_oa For the standard OA can be very large, or a few MΩ at least and the above approximation does not applies and ol could be limited by R F, a few KΩ. Amplifiers and feedback gpessina 128

129 . input impedance of a ctoc fed backed OA 5 Just for giving the order of magnitude let s suppose that our OAs have a voltage/current gain of the order of 10 7, a term close to 1 for f(β ) and the close loop gain is 100, then T results of the order of As a result: f ol currentoa 1 T = Ω f olstandardoa 1 T = 1 MΩ Ω That is the difference. Amplifiers and feedback gpessina 129

130 21. OUTPUT IMPEDANCE OF A CTOC FED BACKED OA =present evaluated feedback node v A =A(v v ) i s v i F i i i o = 1 β R F T 1 T v T A DIR 1 T v T But, at the input mesh is, if A= : v =0, then i F =0, then, again, v B =0 and, finally: R o v B V 2 V 2 i o v T And we reduces to: i o = 0 i o = A DIR 1 T v T Where, by setting A=0, and supposing v 2 =0 : And finally: i o = R ofs = v T i o = 1 T A DIR = R o 1 T R ofs = R ool 1 T v T R o = A DIR v T Where R ool is the impedance that is measured when the gain is set to 0, or in the open loop condition. Amplifiers and feedback gpessina 130

131 . output impedance of a ctoc fed backed OA 2 v A =independent v i i R o i R o L i s R F v B V 2 i F V 2 Let s determine T: T = A v R S = A v A R S R F R o = In this case can be >> R F and >>R o : T A We can compare this loop gain with that obtained when the OA is a current amplifier: T = A i i = A R S 1 i A R S R F 1 1 R o For which is << and R o >> : T = A i i i A = A R F So that the loop gains in the 2 cases are similar if R F and are comparable. Amplifiers and feedback gpessina 131

132 . output impedance of a ctoc fed backed OA 3 Consideration: As with the vtov amplifier we can try to measure the output impedance using the complementary test source, a current source in this case. We have to take care the way we connect the source. The connection shown below does not allow to measure the output impedance. v A =A(v v ) i i R o i o v i T i s R F v B V 2 i F V 2 Indeed, considering that 1/β =0: A R = i o = 1 T i T R o 1 = R o 1 T i T R o_wrong = i T = R o 1 R o 1 T A very small value: that is correct as the feedback wish to force a negligible current into and, as a consequence, its dropout voltage is negligible: the current i T flows into v A. Amplifiers and feedback gpessina 132

133 . output impedance of a ctoc fed backed OA 4 The impedance must be characterized in parallel to the load as it is shown below: the terminal of i T previously connected to ground is now connected to the second terminal of, see the red connection. v A =A(v v ) i i R o i o v i T i s R F v B V 2 i F V 2 This is an interesting application of the A= case. Considering A= it results that v =0, then i F =0 and therefore v B =0 and the consequence is that the current in is 0, so it must results that, at node V 2 : i o = i T v i i R o i o i T For what concern the direct transmission: i o = R o R o i T i s R F v B V 2 i F V 2 Amplifiers and feedback gpessina 133

134 . output impedance of a ctoc fed backed OA 5 v A =A(v v ) v i i R o i R o L i T i s R F v B V 2 i F V 2 Putting together: i o = T 1 T i T R o R o 1 1 T i T Now, the impedance seen by the current generator satisfies: R ofp = v 2 i T = i o i T = i T T R o R o = R o 1 T T R o R o = R o 1 T R ofs = R ofs = R ofs0 R ofs0 = R ofs0 T 1 T R o R o 1 1 T i T Amplifiers and feedback gpessina 134

135 22. INPUT/OUTPUT IMP. SUMMARY OF A CTOC FED BACKED OA Summary about the input/output impedance of a ctoc. Input impedance measured with a current source with a parallel resistance : fp = ol 1 T fp = fp T ol = T ol Where ol and T are ol and T measured when is set to (open) Ω. If the input impedance is measured using a voltage generator with in series a source resistor we have: fss = fp oooooooo Output impedance measured with a voltage source with in series the load resistor : R ofs = R ol 1 T R ofs = R ofs0 (R ool T = R ool0 T 0 ) Where R ools0 and T 0 are R ool and T measured when is set to 0 Ω. If the output impedance is measured using a current generator in parallel to the load we have: R ofp = R ofs0 Amplifiers and feedback gpessina 135

136 23. OTHER CONFIGURATIONS So far we have studied the vtov and ctoc configurations. There are other 2 possible and affordable feedback networks: the vtoc (voltage input, current output) and the ctov (current input, voltage output). There is no reason why these configurations should not behave as the two already studied. Indeed, this is what it happens and we will investigate this shortly. Amplifiers and feedback gpessina 136

137 24. THE VTOC FED BACKED OA Having already shown that the kind of OA does not have impact on the feedback loop we consider only voltage to voltage OAs from now on. v v A =A(v v ) R o v i o i B i B R A R B R C v B As usual we expect that: i 0 = 1 β R T 1 T A DIR 1 T Where we must not forget that T is dimensionless, but, this time, β R has dimension: Ω, as well as A DIR : Ω 1. Now, if we assume A= we can evaluate 1/β: (v v ) v = v B R B = R A v B = R A R B R A (i i 0) i o = R A R B v R A R C A = R C R A R B R C R A Amplifiers and feedback gpessina 137

138 . the vtoc fed backed OA 2 v v A =A(v v ) R o v i o v B R A R B R C If we consider the output voltage rather than the output current we have: = i o v B = R C R A R B R C R A R A R B R A So we see that the output voltage depends on the load impedance, that suggest that the more suitable output variable is the current. The output does not depends on the source resistor and this tells us that the input variable is a voltage. Amplifiers and feedback gpessina 138

139 . the vtoc fed backed OA 3 and now A DIR : v v A =0 R o v i o v B R A R B R C Introducing: = We can write: R A = R A R B R C R C = R C R o v = R A R A v B = R C R C R B v i o = 1 R o v B Or: i o = 1 R o R C R C R B R A R A A DIR = i o 1 R C R A = R o R C R B R A Amplifiers and feedback gpessina 139

140 . the vtoc fed backed Oa 4 Let determine T: v v A =generic R o v i o v B Again, introducing: R A R B R C R o = R o RC" = R C R B R A" R A" = R A We have: v B = R C" R C" R o v A v = R A" R A" R B v B v v = v Therefore: And: T = A v v v A = A R A" R A" R B R C" R C" R o i o = R C R A R B R C R A T 1 T 1 R o R C R C R B R A R A 1 1 T REMEMBER: the contribution of A DIR has not necessarily the same sign of the fed backed gain 1/β. Amplifiers and feedback gpessina 140

141 . the vtoc fed backed OA 5 Input impedance: v v A =0 R o v i o v B R A R B R C Considering the definitions already given: = We can write: R A = R A R B R C R C = R C R o Of course: ol = R A f = ol 1 T = R A 1 T Amplifiers and feedback gpessina 141

142 . the vtoc fed backed OA 6 Output impedance: v v A =0 v i o R o v T B R A R B R C Again, introducing: R o = R o RC" = R C R B R A" R A" = R A R ool = R o R C" Therefore: R ofs = R ool 1 T = R o R C" 1 T Amplifiers and feedback gpessina 142

143 25. THE CTOV FED BACKED OA v v A =A(v v ) R o v i o i s The expect result: R F v 0 = 1 β G T 1 T i s A DIR 1 T i s Again, we must not forget that T is dimensionless, but, this time, β G has dimension: Ω 1, as well as A DIR : Ω. Setting A= we can get 1/β G : v =0, then the current i s flows all into R F and: Or: i s = v R F = R F i s = 1 β G i s Note that does not depends on and, proving that the input variable is a current while the output a voltage. Amplifiers and feedback gpessina 143

144 . the ctov fed backed OA 2 v v A =Generic R o v i o i s R F Loop gain T calculation: Be: = and: R F = R F We have: = R F R F R o v A v = R F = R F To give: v = R F R F R F R o v A And: T = A v v v A = A R F R F R F R o Amplifiers and feedback gpessina 144

145 . the ctov fed backed OA 3 v v A =0 R o v i o i s Direct transmission: R F Starting from: v = i s v R F R o = R o v o R F R o We obtain: = R o i s = A DI s R F R o R F R o In the end: = R F T 1 T i s R o 1 1 R F R o R F R o 1 T i s Amplifiers and feedback gpessina 145

146 . the ctov fed backed OA 4 v v A =0 R o v i o i s R F Input impedance: ol = R F R o 1 And: of = R F R o T Amplifiers and feedback gpessina 146

147 . the ctov fed backed OA 5 v v A =0 R o v i o i T i s R F Output impedance: R ool = 1 R o 1 1 R F 1 Therefore R ofp = 1 R o 1 R F T Amplifiers and feedback gpessina 147

148 26. QUADRUPOLE MODEL OF AN AMPLIFIER We are now in a condition of giving a simpler representation of our fed backed amplifier, useful in several applications and particularly important when we design a system for a customer. The model we would like to obtain is a quadrupole: The model we would like to obtain is a quadrupole: R oq V i q q =a q v i Such that: = q a q R q v s oq Where q, R oq and a q parameters that depend on our previous characterization of the fed backed network. At minimum we need 3 parameters. If we have to consider that the feedback network is not ideal, but bidirectional, some additional parameter comes. Amplifiers and feedback gpessina 148

149 . quadrupole model of an amplifier 2 The simplest approximation is like this: v V i s f0 q =A f v i R of Such that: = f0 A f0 R f v s of Where, if we represent T in the form: T=T(, ): = f0 1 f0 β T(0, ) 1 T(0, ) R of (let s omit for a while the term given by the direct transmission) Amplifiers and feedback gpessina 149

150 . quadrupole model of an amplifier 3 Unfortunately, because the feedback is not unidirectional, we expect that: f0 = f0 ( ), R of = R of ( ), A DIR =A DIR (, ). We have to remember that: f = ol0 1 T 0 = f0 R of = R ool 1 T = R of v β = 1 β T 1 T A DIR 1 T T(, ) = T 0, T(, ) = T, ol0 ol0 R ool Amplifiers and feedback gpessina 150

151 . quadrupole model of an amplifier 4 A first step is to disentangle, for instance, the source from the input: R of V i f0 q =A f v i We know from the theory developed: = 1 T β 1 T = 1 T 0, f0 Proved below v β 1 T 0, f0 s Remember: f0 =f0 ( ). Proof: T 0, 1 T 0, f0 f0 = T 0, 1 T 0, f0 f = T 0, 1 T 0, ol0 1 T 0, ol 1 T 0, But we have already shown that it is valid: Therefore: T 0, ol0 ol 1 T ol T = ol0 T 0, = olt ol 1 T = T 1 T CVD! Amplifiers and feedback gpessina 151

152 Now to the output:. quadrupole model of an amplifier 5 R of V i f0 q =A f v i From the model above: q = = 1 T 0, RL = β 1 T 0, f0 RL = v f0 RL = s With: R of = R of (Rs)= R of Rs =0 1 T 0, l0 l0 f0 = RL = l0 RL = 1 T 0, The 2 critical parameters are: R of Rs =0 and f0 RL =. Things simplifies if l0 is >> and R of and f0 are tiny dependent on and respectively. Under this assumption we can give the parameters f0, R ol and A f =1/β(T(0, )) /(1 T(0, ) ), to give: q = 1 T 0, β 1 T 0, v i = A f v i f0 vo = Af v f0 R of L Amplifiers and feedback gpessina 152

153 . quadrupole model of an amplifier 6 A similar evaluation can be done for A DIR. A similar approach can be used when the fed backed OA is a current amplifier, or transimpedance amplifier or transconducatnce amplifier. Amplifiers and feedback gpessina 153

154 27. LOOP GAIN EVALUATION OF CASCADED STRUCTURES Sometimes the fed backed amplifier is composed by the cascade of several amplifiers. In this case there are 2 possible configurations. The first is when every stage is an open loop stage itself. In this case there is no need for a particular precaution in choosing the source where to open the loop and find the loop gain. Here an example: v A =is generic now R o β Some precautions are instead to be taken when inside the loop there are one or more closed loop amplifiers. In this case it is not recommended to choose the controlled source to be made independent at a nested fed backed amplifier: that amplifier is subjected to a double feedback and the procedure we developed so far does not apply to it, unless some other rules are taken. v A =is generic now R o Not recommended here! β β Amplifiers and feedback gpessina 154

155 . loop gain evaluation of cascaded structures 2 To avoid any possible misleading it is a good choice to make generic a source at a stage not fed backed, first: v A =is generic now R o β β Then, a second useful step is to replace the inner local closed loop amplifier(s) with its model: v A =is generic now R o β Amplifiers and feedback gpessina 155

156 28. BIBLIOGRAPHY La storia del Silicio, F S Norman e G Einspruch, Boringhieri Sergio Franco Amplificatori operazionali e circuiti integrati analogici : tecniche di progetto, applicazioni, U. Hoepli, c1992. P.R.Gray, R.G.Meyer, Analysis and Design of Analog Integrated Circuits, John Wiley & Sons; Sergio Franco Electric Circuits Fundamentals Oxford University Press, USA,1994, C. Biblio FRAS.ELE/1995 Sergio Franco Design with Operational Aplifiers and Analog Integrated Circuits McGrawHill, 2002, C. Biblio FRAS.DES/2002. Amplifiers and feedback gpessina 156

157 APPENDIX A (1): voltage amplifier The aim here is to measure the input impedance f and the output impedance R Of of the below network, for which we already studied the gain: A=10 4, =10 6, =R O =100 Ω, =1000 Ω, =9 KΩ, =200 Ω. =0 v i v A =is generic now Ro Let s study T. Let s write the node equations: v A R o = v v A R o = 1 R o 1 1 v v = v v = v v i = v v v i = v v With: = ( )=999 Ω v A R o = 1 R o 1 1 v v = v i = v v Amplifiers and feedback gpessina 157

158 . Appendix A (2) =0 v i v A =is generic now Ro v A R o = 1 R o 1 1 v v A R o = 1 R o v = v = v i = v v v A R o = 1 R o 1 v = 1 v i = v v In addition: = ( )=999 Ω e R E = ( ). v i = v v va R o = 1 R o 1 R E = R E R E R o v A v = v = v i = v v v i = v v Amplifiers and feedback gpessina 158

159 . Appendix A (3) =0 v i v A =is generic now Ro = R E R E R o v A v = v i = v v = R E R E R o v A v = v i = v v = R E R E R o v A v = v i = v Therefore: And then: v v = R E v R E R A = βf(β)v A o T = A v v v A = A R E = A R E R R β = o A f = 1 β T 1 T = 1 β with an error of the order of 0.2% Amplifiers and feedback gpessina 159

160 . Appendix A (4) =0 v i v A =is generic now Ro It is interesting to verify which values takes T 0 : The effect of on T is marginal when becomes negligible, as = ( )=999 Ω and 0 = ( )=999 Ω. Different is the effect of when it takes very large values: R E = ( )= 196 Ω, while R E = =9999 Ω. The loop gain T, from the value of 661.6, becomes, after that is set to 0 and to : 0 R E T = A = R E R o The factor T/(1T) from the value now takes: T 0 1 T 0 = Amplifiers and feedback gpessina 160

161 . Appendix A (5) Input impedance is determined starting by nulling the gain and measuring the input current consequent to a source excitation: I s v i Ro Input current I s comes from: v = v v v = v v v = R o v = R o I s = v I s = v v = v v = R o R o v = = v v v 1 R o R o R o R o v I s = v I s = v Amplifiers and feedback gpessina 161

162 . Appendix A (6) I T v i Ro = v v v 1 R o R o = R o R o v I s = v = = v R o R o v v R o I s = v v = R o R o = R o R o v I s = v Amplifiers and feedback gpessina 162

163 . Appendix A (7) I T v i Ro v = R o R o = R o R o v I s = v I s = 1 R o 1 R o I s = 1 R o Therefore: And:: r = I s = R o Ω f = R o 1 T Ω 1 T = Ω Amplifiers and feedback gpessina 163

164 . Appendix A (8) The output impedance can be evaluate starting by nulling the input and the gain: v i Ro v T I T i T = v T v T R o v T v i T = v T v T R o v T v v T v = v v v T = v v = ( ) i T = v T v T R o v T v v = v T i T = v T v T R o v T v v v T = v T i T = v T v T R o v T R ool = v T i T = R o Ω v v T = v T and: R of = R ool 1 T = 0.1 Ω Amplifiers and feedback gpessina 164

165 APPENDIX B (1): current amplifier i s v A In the circuit on the left we have: A=1000, R o =100 Ω, =1000 Ω, =4000 Ω, =200 Ω, =10 5 Ω, =10 7 Ω. Determine β, T, A OL, f and R of. The noninverting terminal is at constant potential and we know that the inverting terminal is at the same potential due to the feedback action: therefore we have a current input sensitivity. Assuming A= : v 0, results: = i s β = 1 v A =is generic now v v i Ro Start: v A R o v = v = v v v Now set: = = 990 Ω v A R o v = v = v Amplifiers and feedback gpessina 165

166 . Appendix B (2) v A =is generic now v i Ro v v A R o v = v = v v A = v R o R o = v v R 1 R o v A R o = R o v v = v v A = R o R o v = v Amplifiers and feedback gpessina 166

167 . Appendix B (3) v A =is generic now v i Ro v v A = R o R o v = v v A = R o v = v With: R E = = Ω v A = R 1 R o R E 1 v R 1 = v Amplifiers and feedback gpessina 167

168 . Appendix B (4) v A =is generic now v i Ro v v A = R 1 R o R E 1 v R 1 = v v A = R E R o R E v = v Or: v = R 1 R E R v 1 R E R A o Amplifiers and feedback gpessina 168

169 . Appendix B (5) v A =is generic now v i Ro v v = R 1 R E R v 1 R E R A o Then: T = A R E R E Ro = R E = = ( ) The closed loop gain, or 1/β, does not depends on, but takes part in the loop gain T, lowering its absolute value. In this configuration is not necessary and worse the loop gain and should be avoided. Putting = we obtain: T = A R E R E Ro = Amplifiers and feedback gpessina 169

170 . Appendix B (6) v i Ro v i T i T = v v v v i T = v v v v v = R o v = R o ol = v i T ol = v i T Solving: and: f = ol 1 T = 6.1 Ω Amplifiers and feedback gpessina 170

171 . Appendix B (7) v i Ro v v T I T i T = v T v T R o v T v i T = v T v T R o v T v v T v = v v v v T = v v R ool = v T i T R ool = v T i T Solving: R ool = R o = 65.8 Ω and: R of = R ool 1 T = 0.5 Ω Amplifiers and feedback gpessina 171

172 . Appendix B (8) v i Ro v i T Direct transmission: i T = v v v v i T = v v v v v = R o v = R o i T = v v i T = R o R o v = R o R o v = R o R o i T = R o R o R o v = R o R o = Amplifiers and feedback gpessina 172

173 . Appendix B (9) v i Ro v i T Direct transmission: i T = R o R o R o v = R o R o i T = R o R o v = R o v R o o i T = R 1 R o R v 1 R o o v = R o R o Amplifiers and feedback gpessina 173

174 . Appendix B (10) v i Ro v i T Direct transmission: i T = R o R o v = R o R o Finally: = R o R o i T = R 1 R o R o R i 1 R o R o R T 2 = R o R o R o i T = A DI T = 13 i T Amplifiers and feedback gpessina 174

175 . Appendix B (11) i s v A Summary: = T 1 T i s R o R o R o 1 1 T i s T = A R E R E Ro = Or: = i 13 s i s = 0.992i s 0.099i s The ratio between the gain and the direct transmission is: 0.992i s 0.099i s = Amplifiers and feedback gpessina 175

176 APPENDIX C (1): inverting amplifier V s v A R2 Let s assume A=. Then, v 0, since v =0 and: = = This is called inverting configuration and, if =, it results: v O =. Note that the gain depends on the impedance in series to the source, so the feedback is sensitive to the input current. The loop gain is the same as that of the previous example of Appendix B, but and are now one resistor: T = A R E R E Ro = R E = Output impedance is the same as that of Appendix B, as the output networks are similar. As we saw at the end of chapter 18, the impedance seen from the voltage source satisfies: Being: fs = fp f = lim fp Amplifiers and feedback gpessina 176

177 . Appendix C (2) Onceagain a remark: remember that the impedance seen from the voltage source is not merely divided by (1T). f v A R2 Applying the evaluation done at the note seen at the end of chapter 18, whose result is shown in the pervious page Amplifiers and feedback gpessina 177

178 . Appendix C (3) v A R2 V s It has to be remarked that with this kind of feedback the connection at the input of a voltage source is unnatural and must be considered with care. As it can be seen above if the source has a non negligible series impedance its value enters in the gain since it is in series with : = This means that the source impedance must be known with precision or must be negligible. Normally this setting is used when the stage is driven by a voltage output fed backed stage, with known negligible impedance, as in this example: v A R2 Amplifiers and feedback gpessina 178

179 APPENDIX D (1): consideration about large feedback gain v A R2 Side effect: Let s suppose that is the source impedance, namely a resistor that assumes small values, having a zero value in the ideal case. In this situation the gain is subjected to 2 effects from both β and T: β = s s 0 0, hence 1 β T = A R E 0 0 R E Ro Rs = So: = But we know that T=Af( β)β: T 1 T v 0 s [0] [ 0] 1 = 1 β Af(β)β 1 Af(β)β = Af(β) 1 Af(β)β v β 0 s Af(β)vs = Af β = A T β = = A s s R E 0 v R E Ro A ss R E R E Ro Therefore, if β 0 the closed loop gain can not increase indefinitely since, at the same time, T 0: the resulting gain is limited by the OA gain and it results as the network is not closed loop. Amplifiers and feedback gpessina 179

180 . Appendix D (2) i s v A R2 Another side effect: Let s suppose to increase the closed loop current gain by increasing and increasing. We know that if T= we get: = i s Coming back to the case T is not very very large: β = 1 0 T = A R E 0 R E Ro But: 1 β T 1 T T β = Af β = A R E R E Ro R 2 A R E R E Ro Therefore: = 1 β T 1 T i s = T β 1 T i s = Af(β) 1 T i s Af(β)is = A R E R E Ro i s Again the network behaves as it were operating open loop. To conclude: every time T 0 the circuit behaves as it is in open loop, as it is expected since T is the measure of the loop operation. Amplifiers and feedback gpessina 180

181 . Appendix D (3) A last side effect: Again, if T= we have: = V s If now 0 the close loop gain, but: But: β = 0 0 R E 0 T = A 0 R 1 R E R o T β = Af(β) = A R 1 R E 0 R E A R E Ro R E Ro And so: = 1 β T 1 T V s = T β 1 T V s = Af(β) 1 T V 0 s Af(β)Vs = A Once again the network behaves as it is open loop. R E R E Ro V s To conclude: every time T 0 the circuit behaves as it is in open loop, as it is expected since T is the measure of the loop operation. Amplifiers and feedback gpessina 181

182 APPENDIX E 1: Miller theorem The Miller effect v 1 Z v 2 I We want to show that the simple scheme on this left is equivalent to this: I v 1 v 2 Z 1 Z 2 I The current I must be equal in both branches: I= v 1 = v 2 = v 1 v 2 Z 1 Z 2 Z From which: 1 Z 1 = 1 Z 1 v 2 v 1, Z 1 = Z 1 v 2 v 1 Let s apply the Miller effect to resistor of the example of Appendix B: i s i s Z 1 v v A A 1 ZZ 2 = 1 Z 1 v 1 v 2, Z 2 = Z 1 v 1 v 2 Z 2 By assuming for a while negligible the output impedance of the OA: Hence: v O v A Z 1 = 1 v 1 A = 4 Ω And we get the order of magnitude of the expected results Amplifiers and feedback gpessina 182

183 . Appendix E 2 The result of the previous page has been obtained with some approximation and we can try now to improve the precision: i s Av i v i Ro v We have that: Let s assume that v is about zero:: v O = i S To evaluate the Miller effect we must know the ratio between V O and V. Av v O R O v O v O Av R O = v O R O = v O 1 R O 1 Av R O = v O R O R O v O v = A R O therefore: Z 1 = 1 v R 1 A L R o = = 6.09 Ω With a greater precision. Amplifiers and feedback gpessina 183

184 APPENDIX F (1): unity gain voltage amplifier Let s determine now the input and output impedance of the unity gain buffer amplifier: A=10 4, =10 6, R O =100 Ω, =200 Ω, =50 Ω. V s v i Av i Ro By assuming for a while that A= we know that v v and: =, namely β=1. v A =is generic now v i Ro Let s start with T: V o = R o V A = = 199,96 Ω V = s V therefore: V V = V o Finally: T = A V V = R L V T R A 6666 L R o Amplifiers and feedback gpessina 184

185 . Appendix F (2) Now the input impedance in open loop condition: V T R o From which: f = r 1 T = R o (1 T) = 6,67 GΩ Finally the output impedance: R o I T From which: R or = R o = Ω R of = R o 1 T 10 mω Amplifiers and feedback gpessina 185

186 . Appendix F (3) Finally the direct transmission: V s R o R o = R o = Ω It results: = R o R o = A DIR V s R o And, considering the loop closed: = V s = T 1 T A DIR 1 T = v s = 0, , Amplifiers and feedback gpessina 186

187 APPENDIX G (1): the differential amplifier THE DIFFERENTIAL AMPLIFIER v 1 v 2 R C R D R A R B Let s consider A=, from which we know it is valid that v v, and i i 0, we get the following eq: v = v 2 v R A R D R D R C v 1 = v R B Solving: = R A R B R A R D R C R D v 1 R B R A v 2 We must arrange the coefficients multiplying v 1 and v 2 to be equal for obtaining a true differential amplification., namely: R D = R B R A = R C From which now it is valid that: = R B R A v 1 v 2 Amplifiers and feedback gpessina 187

188 THE DIFFERENTIAL AMPLIFIER. Appendix G (2) v 1 i 2 R A R B The impedances seen from the 2 sources. v 2 R A R B The impedance that the source v 1 sees is always R A R B, whatever is v 2, in the approximation that >>R A, R B. Source v 2 has a different perspective and: v 2 = v 1 : 2 = R A R B, R v 2 = v 1 : 2 = R A R B A R A 2R B v 1 = 0 : 2 = R A This means that if v 1 and v 2 had different source impedances, namely R A is slightly different for the 2 sources, the differential gain would not be symmetrical. If v 1 =0 then v =0 and: i 2 = v 2 R A If v 1 =v 2 then v = If v 1 =v 2 then v = R B v R B and= i 2 = 1 v A R B v A R B A R B R B R A v 2 and: i 2 = 1 R A v 2 R B R B R A v 2 Amplifiers and feedback gpessina 188

189 . Appendix G (3) v 1 i 2 1 R A R B v 2 2 The fact that the impedance seen from the 2 sources are not equal, and, more important, not very large can lead to a not negligible effect that comes from the source resistors 1 and 2 the inputs sources are not ideal. Our original differential gain is modified this way, now: R A R B = R B R A 2 R A 2 R B R A 1 R B v 1 v 2 We can se that if the 2 source resistors are equal the gain look different, but, very important, if the 2 resistors are different the gain is no more differential and a large common mode signal results. The above result can be put in the following form, that highlights the effect: = R B R A 2 v 1 v R A 1 R B v 1 Amplifiers and feedback gpessina 189

190 . Appendix G (4) THE DIFFERENTIAL AMPLIFIER R C R D Let s suppose now that the 4 resistors have not a very precise value. To simplify let s suppose that only inaccurate one is R B R B =R B (1ε). From the general result of the previous page: v 1 v 2 R A R B = R A R B 1 ε R A R D v R C R B 1 ε v D A R B =RD R B = R A R A R B R C R B v 1 εr B R C R B v 1 1 ε v 2 R A =R C R B = v εr B v A R A v 2 εv 2 B If now v 1 =v 2 =v C : In the ideal case of perfect matching: = R B R A v 1 v 2 ε v 2 R B R A R B v 1 R B = ε v R A R c B = A cf v c = R B R A v 1 v 2 = A df v 1 v 2 Therefore we can say that the Common Mode Rejection Ratio, CMRs: CMRR = 20log 10 A df A cf R B 1 R A = 20log 10 R A ε Amplifiers and feedback gpessina 190 R B A df =10 ε=0.1% 81 db

191 . Appendix G (5) THE DIFFERENTIAL AMPLIFIER R A R B Another source of uncertainty is given from the OA itself, since it is not a perfect differential OA. In general it can be written: v c v c R A R B Let s not neglect the common mode gain now: v = R B v R B R c A = A d v v A c v v 2 v c v R A = v v R = R B v B R B R c R A v A R B R o A = A d A c 2 v A d A c 2 v Replacing v and v in the last eq: = A c R B 1 R B R A 1 A d A c 2 v R c A R B R A A c A d R B R A v c Common mode signal is reflected at the input as it is a differential signal in open loop condition.. Closed loop gain R A R B Amplifiers and feedback gpessina 191 v 1 v 1 v com R A v com = R B A c A d v c

192 . Appendix G (6) It is possible to improve the performance of the differential amplifier, at expense of a greater complexity. v 1 1 R A 2 v A R B R B v B R R R R Now the 2 sources see each one a very large impedance, due to the feedback. v 2 Supposing that for the 3 OA is A= : v A v 1 R B = v 1 v 2 R A v 1 v 2 R A = v 2 v B R B v A = R A R B R A v B = R A R B R A v 1 R B R A v 2 v 2 R B R A v 1 The output stage is simply a unity gain differential amplifier: = v B v A = R A 2R B R A v 2 v 1 Amplifiers and feedback gpessina 192

193 . Appendix G (7) Now it is possible to appreciate what happens to the common mode signal: v 1 =v 2 =v c. v c 1 R A 2 v A R B R B v B R R R R If the 2 inputs are equal the current through R A is zero; as a consequence the current will be zero also in the 2 resistors R B s. Therefore v A =v B whatever is the tolerance of R A and R B s and the value of the s. v c If ε is the tolerance of the resistors R of the output OA, their CMRs: CMRR R = 20log 10 2R B R A R A 1 ε If the inputs OAs are very similar, that is the case when they share the same package, then the common mode gain A c is similar, giving similar contribution to the nodes v A and v B. This way the CMRs limited in accuracy by the output OA and: CMRR AMPLI = 20log 10 2R B R A R A A d A c That is to say that the CMRs larger and larger if the gain of the input stage is larger and larger. Amplifiers and feedback gpessina 193

194 . Appendix G (8) Here there is another differential configuration having a large input impedance: v 2 v R C R D Let s apply the superimposition principle and set v 1 =0 first, that implies that v A =0 V and: v A Now suppose that v 2 =0 V, which sets to 0 V the noninverting terminal of the output OA: R A R B v O2 = R A R B R A v 2 = 1 R B R A v 2 v O1 = R B R A R C R D R C v 1 = R B R A 1 R D R C v 1 Therefore: v O = v O1 v O2 = 1 R B R A v 2 R B R A 1 R D R C v 1 = 1 R B R A v 2 R B R A R D R C R B R A v 1 Finally, by setting: R B R A = R C R D we get: v O = 1 R B v v 1 A Amplifiers and feedback gpessina 194

195 . Appendix G (9) Application of the differential configuration. Errors rise when the source and the amplifier are connected to ground references far away: since currents flow around in the environment the 2 references could have slightly different potential. R B v M R A This can be modelled this way: v M R B R A The amplified signal has superimposed the disturbance v M to the signal. If now we introduce our differential amplifier we have 2 reading terminals that can exploited to read the signal across the source only: v M v M R A V A R B R B V B A v M R R v M R R V o vo=a( (VsvM) vm) =Avs This way the disturbance is rejected since the amplifier behaves as it is virtually floating: =A( (V s v M ) v M )=A. But Amplifiers and feedback gpessina 195

196 . Appendix G (10) The way the source is connected to the reading amplifier is important. If in the environment are present sources of EMI disturbances they can generate spurious signals in the connecting loop: v M v EMI Φ t Area EMI R A V A R B R B V B R R R R V o According to the Faraday/Lenz s law when a time varying magnetic field interacts with a coil it generates an electric field, or voltage, in the coil. In our example above this EMI voltage, v EMI, add to the signal voltage. The common cause of magnetic field disturbances are from the main line, 50 (Europe)/ 60 (USA)Hz. Disturbances from the main line are always present and one of the task when we instrument a low noise system is to suppress them with shields and other mechanism that we will try to address in the following. As it can be seen above, the Faraday s law is proportional to the coil area and one way to suppress its effect is by minimizing the sensitive area and or organizing the connections in a series of coils such that in each coil the current is opposed to that of the previous coil, as is the case with the twisted pair connections: v M In this way the signal current induced in one coil cancels in the following coil as the current reverses. R A R B R B V A V B R R R R V o Amplifiers and feedback gpessina 196

197 . Appendix G (11) The better way is to have the source signal differential, then drive the twisted line: v M R A V A R B R B V B In this way the signal current induced in one coil cancels in the following coil as the current reverses. R A V A R B R B V B R R R R V o Now let s consider a practical consideration that very often induces errors. In principle the scheme above is working since there is a true differential signal readout from a true differential input pair. R A V A1 R B R B V B1 V in1 V in2 R A V A R B R B V B R R R R V o v M System on the right and on the left can be floating each other and the 2 grounds could be different in DC voltage: this is not a disturbance, but it depends only on the environment settings. This means that: V A1 = V GND_OFFSET V s V B1 = V GND_OFFSET V s Amplifiers and feedback gpessina 197

198 . Appendix G (12) R A V A1 R B R B V B1 V in1 V in2 R A V A R B R B V B R R R R V o v M System on the right and on the left can be floating each other and the 2 grounds could be different in DC voltage: this is not a disturbance, but it depends only on the environment settings. This means that: V A1 = V GND_OFFSET V s V B1 = V GND_OFFSET V s Where V GND_OFFSET is the voltage difference between the 2 grounds. V in1 V in2 is equal to V s V s, and this can be a small signal. But the absolute value of V in1 and V in2 can be large and even close to one of its rails. This determines a saturation of the input stage and a signal distortion. R A V A1 R B R B V B1 V in1 V in2 R A V A R B R B V B R R R R V o v M R G The workaround of the problem is to set a common reference for both grounds with a feeble connection through which the current cannot flow, to avoid disturbances. A resistor, R G, helps in doing this. Amplifiers and feedback gpessina 198

199 . Appendix G (13) Twisted cables are useful to reject EMI disturbances. But they are not enough and our apparatus, and cables, need to be shielded. We need to shield lectormagnetic filed that consists in electric field and magnetic field. We know that the electric field is shielded by enclosing our circuits in metallic box, exploiting the Faradaycage principle. Metallic shield Incident Electric field Reflected Electric field Coaxial, shielded, cable Electric field is well rejected by shielding both the amplifier and the connection. For better results, the Shield must be connected to a constant potential, ground, to avoid any residual field variation coming, from example, from large currents around. R A V A R B R B V B R R R R V o Amplifiers and feedback gpessina 199

200 . Appendix G (14) Electric field is easily shielded. But magnetic field has the ability to penetrate, especially at small frequency: 50 Hz is very difficult to shield, indeed. The attenuation of the magnetic field at low frequency is given by: A exp t δ with: δ = πfµ r σ r Therefore, at low frequency ff we need High magnetic permeability μμ rr High electrical conductivity σσ rr Which conclusion: to obtain a large attenuation, A, we need a small value of δ. If a small value of δ is not available in the metal shield we are using, then, we need a large thickness of shield, otherwise the disturbances is not shielded, but only slightly attenuated. Amplifiers and feedback gpessina 200

201 . Appendix G (15) A exp t δ with: δ = πfµ r σ r Very large relative permeability, µ r, is found in Iron and some steel, around There are a few special alloy, studied for this purpose: Mumetal, Skudotech, which allow to obtain a value close and larger than This is important as in this way a shield can be obtain with a few mm of material, against cm need for iron. This has impact on the weight of the final apparatus. Amplifiers and feedback gpessina 201

202 APPENDIX H (1): OA specifications An actual OA is not ideal and shows several characteristics that can have affect in the application. Its datasheet list them and one can select the best suited offtheshelf OA for the own application. The characteristics are listed in the datasheet and, for a good OA, several are listed for a good detail. Generally one focalizes the parameters important for the own application and do not consider the others. Here some examples: Amplifiers and feedback gpessina 202

203 . Appendix H (2) Other important parameters are the noise; The frequency response and the ability / inability to drive low value impedances and capacitive impedances; The value of the output impedance; The upper and lower limits of the output signals with respect to the supply voltages; Power dissipation; The maximum output current and short circuit current; Etc. Amplifiers and feedback gpessina 203

204 . Appendix H (3) An actual OA is not ideal and shows several characteristics that can have affect in the application. Its datasheet list them and one can select the best suited offtheshelf OA for the own application. The characteristics are listed in the datasheet and, for a good OA, several are listed for a good detail. Generally one focalizes the parameters important for the own application and do not consider the others. Here some examples: Amplifiers and feedback gpessina 204

205 . Appendix H (4) Other important parameters are the noise; The frequency response and the ability / inability to drive low value impedances and capacitive impedances; The value of the output impedance; The upper and lower limits of the output signals with respect to the supply voltages; Power dissipation; The maximum output current and short circuit current; Etc. Amplifiers and feedback gpessina 205