Versione del:mercoledì 13 marzo 2019pessina.mib.infn.it/Corsi_del_IV_anno/Dispense...•...
Transcript of Versione del:mercoledì 13 marzo 2019pessina.mib.infn.it/Corsi_del_IV_anno/Dispense...•...
Amplifiers and feedback gpessina 1
1. INTRODUCTION
PROGRAMMA
• Gli amplificatori ideali e reali, amplificatori operazionali.
• La reazione, o feedback, negli amplificatori: teoria e pratica.
• La risposta in frequenza: stabilità degli amplificatori reazionati,
approssimazione a polo dominante.
• Esempi, l’amplificatore di carica e di tensione.
• La manipolazione di segnali deboli: il rumore. Rumore bianco serie,
rumore bianco parallelo, rumore 1/f, il rumore nei componenti
passivi.
• Introduzione ai componenti attivi (transistori): Transistori
(opzionale) bipolari in Si, SiGe, JFET in Si, HEMT, MOS. Accenno
alle problematiche nella realizzazione di circuiti monolitici.
• Modelli matematici e di piccolo segnale per i componenti attivi:
modelli per il simulatore SPICE.
• Sorgenti di rumore nei componenti attivi.
• Effetti della radiazione sui componenti attivi
• Risoluzione di circuiti in presenza di rumore negli amplificatori e
nei componenti attivi: approssimazione del rumore dominante.
Versione del:mercoledì 13 marzo 2019
Amplifiers and feedback gpessina 2
. ………. introduction 2
The design of amplifiers has changed from the beginning as the technology grew. The starting standard way to design was by connecting transistor by transistor, resistor by resistor, etc.; namely brick by brick.
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.
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 electrical contacts in the circuit;
• The testing of the circuit, done automatically on large series, takes a time duration that is proportional to the measurement nodes.
Amplifiers and feedback gpessina 3
………. introduction 3
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 (micro-controllers, micro-processors, etc.) capable to contain millions of bricks per mm2 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 with respect to present: 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 4
………. introduction 4
At the end of the ‘40 of the last century it started the study of semiconductors, mainly Germanium and Silicon,
The first working transistor, a bipolar transistor, was in Ge and was developed at the Bell Labs.
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.
Since then transistor manufacturing became planar, namely 2D and Si was the standard substrate for it.
Bonding wire
Power transistor ( ∼mm2)
Amplifiers and feedback gpessina 5
………. introduction 5
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 semiconductor-chip inside.
Several are the constraints to be followed: compactness, stress resistance, power dissipation, hence thermal conductance, …
Amplifiers and feedback gpessina 6
………. introduction 6
Having the transistors available, amplifiers can be implemented combining several of them with resistors, capacitors and inductances.
For quite a long time one packages included only one transistor and there was an evolution in packaging from metal can, through-hole (TO) to surface mount devices having much smaller dimension.
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.
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:several transistors are needed to compose an amplifier just as many bricks are needed to construct buildings.
Amplifiers and feedback gpessina 7
………. introduction 7
Through-hole 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 implementing the electrical connections of the pins of the components:
The assembling of SMD circuits is more expensive compared to through-hole circuits and this is a parameter of decision about which technology to use: if space is not an issue through-hole is the choice.
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.
Amplifiers and feedback gpessina 8
………. introduction 8
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” in 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
For the first time resistors, capacitors and transistors share space on the same substrate (Ge).
An oscillatorA flip-flop (memory)
Nobel Prize in 2000
Amplifiers and feedback gpessina
………. introduction 9
9
Since the introduction of the integrated circuits, where integrated is intended with several transistor per chips, namely, the evolution from 3D to 2D dramatically evolved:
ENIAC: il primo PC (1946)
ENIAC: Electronic Numerical Integrator and Computer
Amplifiers and feedback gpessina 10
………. introduction 10
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 to satisfy a very wide range of applications.
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 the volume of asingleatom:
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. Its name is SOT23-5.
Amplifiers and feedback gpessina 11
………. introduction 11
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.
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 of-the-shelf. 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:
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 feed-backed 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 12
………. introduction 12
Then, in the second part of the course, we will enter the black box to try to understand the OA, or Amplifier structure:
Let’s briefly summarize our next steps in our study.
First we will study the use of amplifiers as black-boxes, namely, we assume that the Operational Amplifier, OA, is available and we can manage it by adding some components around:
This is the inside of the black box above.
OA as a black box, a mathematical tool.
Amplifiers and feedback gpessina 13
………. introduction 13
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 for 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 in-between 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 14
2. THE BASIC PASSIVE ELECTRICAL ELEMENTS TO KNOW
RESISTORS:
or
CAPACITOR OR CAPACITANCE:
INDUCTOR OR INDUCTANCE:
L
C
R
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.
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.
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.
v = Ri
v(t) =1C
0
t
)i(t
v(t) = Ldidt
Amplifiers and feedback gpessina 15
………. the basic electrical elements to know 2
NON-IDEAL VOLTAGE SOURCE:
A non-ideal voltage source is modelled with an ideal voltage source having a resistor RG in series.
The resulting signal at the load is then:
To null a voltage source means that the source itself generates a zero-amplitude voltage. A voltage source with zero-amplitude can be modelled with a short circuit.
VG
RG
RL
VO
VG RL
VO
VG=0 V RL
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 zero-series impedance and is ideal: we will show why in a while.
According to the model above, a non-ideal 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.
ZERO-VOLTAGE SOURCE:
≅
VO =RL
RL + RGVG
RG→0VG
(see next page for the evaluation)
= zero V, o GND
Amplifiers and feedback gpessina 16
………. the basic electrical elements to know 3
Let’s evaluate the output voltage:VG
RG
RL
VO
Now VO is the voltage drop across RL:
VO = IGRL
Therefore:
IG
The current IG flows through the series of the 2 resistors:
IG =VG
RG + RL
VO =RL
RG + RLVG
We can arrive at the same result by considering that the voltage VG is linearly shared between both resistors, while the current is the same:
VG: RL + RG = VO: RL
= zero V, o GND
Amplifiers and feedback gpessina 17
………. the basic electrical elements to know 4
This situation is contradictory: an ideal voltage source connected to a short, or 0 Ωresistor.
In practice, the voltage source would be able to deliver a very big energy, since the power dissipation is the product VGIG. What it happens is that the source breaks or the short melts.
The current IG through the short would be:
IG =VG0
= ∞ A
From the mathematical or design point of view the situation is to be avoided for obvious reasons. Said in another way: if we are facing such a situation in our design that means that something went wrong.
VG
IG
Amplifiers and feedback gpessina 18
………. the basic electrical elements to know 5
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 infinite-parallel impedance and is ideal: we will show why in a while.
IG RGRL
IO
IG RL
IO
IG=0 A RL
NON-IDEAL CURRENT GENERATOR
A non-ideal current source is modelled by an ideal current source having a resistor RG in parallel.
The resulting signal at the load is then:
According to the model above, a non-ideal current source tries to be close to an ideal current source as soon as its parallel-impedance tends to infinite, or it is >> larger than the load impedance.
≅
To null a current source means that the source itself generates a zero-amplitude current. A current source with zero-amplitude can be modelled with an open circuit.
ZERO-CURRENT SOURCE:
IO =RG
RL + RGIG
RG→∞IG
(see next page for the evaluation)
Amplifiers and feedback gpessina 19
………. the basic electrical elements to know 6
IG RGRL
IO
The voltage VO is common to both resistors:
Hence:
IG =VORG
+VORL
Let’s evaluate the output current:
VO
VO =RGRL
RG + RLIG
The current through RL is:
IO =VORL
Finally:
IO =RG
RG + RLIG
We can arrive at the same result by considering the current IG is linearly shared between both resistors, or conductances, while the voltage is the same:
IG:1
RG+
1RL
= IO:1
RL
Amplifiers and feedback gpessina 20
………. the basic electrical elements to know 7
IG
IOVO
This situation is another contradiction: an ideal current source connected to an open circuit, or ∞ Ω resistor.
In practice, the current source would be able to deliver a very big energy , since the power dissipation is the product VGIG. What it happens is that the source breaks or the open burns.
The dropout voltage VO would be:
VO = ∞IG = ∞ V
From the mathematical or design point of view the situation is to be avoided for obvious reasons. Said in another way: if we are facing such a situation in our design that means that something went wrong.
Amplifiers and feedback gpessina 21
………. the basic electrical elements to know 8
There is a class of voltage and current generators that are at the basis of the modelling of amplifiers and transistors: the voltage-controlled and current-controlled 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:
+ +V1 V2=-BVN
R1 R2
R3I3
N
As it can be seen voltage generator V2 generates a voltage that is linearly proportional, with coefficient “-B”, to that at node N.
Current-controlled and Voltage-controlled sources are very useful mathematical tools exploited to model amplifiers and transistors.
CURRENT-CONTROLLED/VOLTAGE-CONTROLLED SOURCE:
Amplifiers and feedback gpessina 22
………. the basic electrical elements to know 9
+ +V1 V2=-BVN
R1 R2
R3I3
N
Let’s see the ways the circuit can be solved as they will be exploited soon for the study of the fed-backed OAs.
The direct approach is to solve the network considering the definition of V2.In solving a network we have to write equations equal to the number of unknowns:
V1 − VNR1
=VNR3
+VN − −BVN
R2
V1R1
=1
R1+
1R3
+1 + B
R2VN
VN =1
R1+
1R3
+1 + B
R2
−1 V1R1
Amplifiers and feedback gpessina 23
………. the basic electrical elements to know 10
+ +V1
(V2=-BVN)
R1 R2
R3I3
N
A second way to proceed is to consider V2 independent for a while, solve the network under this assumption, then assign the dependence of V2 at the end:
V1 − VNR1
=VNR3
+VN − V2
R2
V1R1
=1
R1+
1R2
+1
R3VN +
−V2R2
VN −1
R1+
1R2
+1
R3
−1 V2R2
=1
R1+
1R2
+1
R3
−1 V1R1
V2
Now we assign to V2 its dependence:
VN −1
R1+
1R2
+1
R3
−1 −BVNR2
=1
R1+
1R2
+1
R3
−1 V1R1
1R1
+1
R2+
1R3
−1 1R1
+1
R2+
1R3
VN +BVNR2
=1
R1+
1R2
+1
R3
−1 V1R1
VN =1
R1+
1R3
+1 + B
R2
−1 V1R1
Amplifiers and feedback gpessina 24
………. the basic electrical elements to know 11
A last approach for solving the network is to use the superimposition principle, that will be considered in the following pages.
Amplifiers and feedback gpessina 25
3. SUPERIMPOSITION PRINCIPLE
THEO:
ES.
+ +V1 V2
R1 R2
R3I3
V1=6 VV2=-14 VR1=1K ΩR2 =9K ΩR3 =2K Ω
I3=????
+V1
R1 R2
R3IA
VA
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.
CASE a) let’s null V2:
This first result is:IA =
R2R1R2 + R1R3 + R2R3
V1
(see next page for evaluation)
= zero V, o GND
Amplifiers and feedback gpessina 26
………. superimposition principle 2
+V1
R1 R2
R3IA
VA
Let’s write the solving eq for the network:
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 not, then we must proceed to the following node for adding a new equation. This process will end when the number of the written equations equals the number of the unknowns.
For the network above the balance at node VA is:
V1 − VAR1
=VAR3
+VAR2
With only one unknown and:
IA =VAR3
Amplifiers and feedback gpessina 27
………. superimposition principle 3
+V1
R1 R2
R3IA
VA
VB
CONSIDERATION:Let’s see why the voltage, common to all, at potential Can be assumed equal to 0 V.
Let’s call it VB and see what it happens:
V1 + VB − VAR1
=VA − VB
R3+
VA − VBR2
Now, if we call:
VX = VA − VB
We obtain:V1 + VX
R1=
VXR3
+VXR2
Which is the same equation of the previous page.
Amplifiers and feedback gpessina 28
………. superimposition principle 4
+
R1 R2
R3IB
VB
V2
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, currents having opposite direction must be subtracted each other):
CASE b) let’s null V1:
The second result is:
IB =R1
R1R2 + R1R3 + R2R3V2
=R2V1 + R1V2
R1R2 + R1R3 + R2R3
=)9K ⋅ 6 + 1K ⋅ (−14
9K ⋅ 2K + 1K ⋅ 2K + 1K ⋅ 9K= 1.379 mA
I3 = IA + IB
Amplifiers and feedback gpessina 29
………. superimposition principle 5
Now that we know which way the superimposition principle operates, let’s reconsider the previous example with the voltage dependant source to find out the voltage VN.
V1 − VN1R1
=VN1R3
+VN1R2
V1R1
=1
R1+
1R2
+1
R3VN1
+ +V1
(V2=-BVN)
R1 R2
R3I3
N
V2
Step 1: let’ consider present V1 only.
+V1
R1 R2
R3I3
N
VN1 =1
R1+
1R2
+1
R3
−1 V1R1
Amplifiers and feedback gpessina 30
………. superimposition principle 6
+
R1 R2
R3I3
N
Step 2: let’ consider present V2 only, as it were independent.
V2 − VN2R2
=VN2R3
+VN2R1
V2R2
=1
R1+
1R2
+1
R3VN2
V2
VN2 =1
R1+
1R2
+1
R3
−1 V2R2
Amplifiers and feedback gpessina 31
………. superimposition principle 7
Step 3: VN is the sum of VN1 and VN2:
VN = VN1 + VN2 =1
R1+
1R2
+1
R3
−1 V1R1
+1
R1+
1R2
+1
R3
−1 V2R2
+ +V1
(V2=-BVN)
R1 R2
R3I3
N
V2
VN1 =1
R1+
1R2
+1
R3
−1 V1R1
VN2 =1
R1+
1R2
+1
R3
−1 V2R2
Step 4: V2 is given its dependence:
VN =1
R1+
1R2
+1
R3
−1 V1R1
+1
R1+
1R2
+1
R3
−1 −BVNR2
VN +1
R1+
1R2
+1
R3
−1 BVNR2
=1
R1+
1R2
+1
R3
−1 V1R1
1R1
+1
R2+
1R3
−1 1R1
+1
R2+
1R3
VN +BVNR2
=1
R1+
1R2
+1
R3
−1 V1R1
1R1
+1
R3+
1 + BR2
VN =V1R1
VN =1
R1+
1R3
+1 + B
R2
−1 V1R1
Amplifiers and feedback gpessina 32
4. THEVENIN THEOREM
ES. R1
R2
R3
R4
VA+
V1VTH
+
-With respect to the 2 terminals above VTH results in:
R1
R2
R3
R4
To calculate RTH let’s proceed by applying a test voltage, VT, 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 RTH. Before doing this all the independent voltage sources must be nulled:
IT
VT
The result is:
VB
With respect to a pair of its terminals, a netwtork can be modelled by an ideal voltage source, VTH, having in series a resistor, RTH. The voltage VTH is the voltage measured at the terminals when the load impedance is ∞, while RTH is the impedance that can be measured once that all the independent sources of the network are nulled.
V1=0 V
VTH =R2R4
R1 + R2 R3 + R4 + R1R2V1
RTH =VTIT
=R4 R2R3 + R1R3 + R1R2
R4 R1 + R2 + R2R3 + R1R3 + R1R2
(see next page for evaluation)
≈VTH
RTH
Amplifiers and feedback gpessina 33
………. Thevenin theorem 2
R1
R2
R3
R4
VA+
V1VTH
+
-
Network evaluation:
Starting from the left we encounter the first node VA:
V1 − VAR1
=VAR2
+VA − VTH
R3
We see now that the above eq has 2 unknown, VA and VTH, and we need to add an eq.For that we move further on the right:
VA − VTHR3
=VTHR4
This new eq does not add any new unknown and the 2 eqs above allow to solve the network.
VG
RG
RL
VO
CONSIDERATION: Why the voltage VTH is the voltage measured at the given terminals is soon understood if we remember that the output voltage VO of the network below equal VG only and only if RLequals ∞ every time RG has a finite value.
VO =RL
RL + RGVG
RL→∞VG
Amplifiers and feedback gpessina 34
5. NORTON THEOREM
ES.:R1
R2
R3
R4
VA+
V1
R1
R2
R3
R4
VA+
V1
INO
With respect to a pair of its terminals, a netwtork can be modelled by an ideal current source, ITH, having in parallel a resistor, RTH.The current ITH is the current measured at the terminals of interest when the load impedance is 0, a short, while RTH is the impedance that can be measured once that all the independent sources of the network are nulled.
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 INO we consider the following:
From which we obtain:
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:
INO =VAR3
=R2
R1R2 + R1R3 + R2R3V1
VTh = INoRNo, RTh = RNO
INO RNORL
IO
≈
Amplifiers and feedback gpessina 35
………. Norton theorem 2
R1
R2
R3
R4
VA+
V1
INO
Let’ try to understand why the Norton current INO is evaluated this way:
Interpretation: we know that current IO equals INO if and only if RL is zero if RNO has a finite value:INO
RNORL
IO
IO =RNO
RL + RNOIG
RL→0IG
Amplifiers and feedback gpessina
6. A NOTE ON THE TEXT COLOUR OF EQUATIONS
36
In the following a formula coloured like this:
T=−AβRi
Ri +Rs+Roβ
RL RiβA
RL RiβA +R𝑜𝑜
TRool
=−AβRi
Ri +Rs+Roβ
1Ro
1Rof
=1
RL+
1RiβA
+1
Ro1 +
AβRiRi +Rs+Roβ
Represents a standard result of an evaluation.
Equations coloured like this:
represents an intermediate step of an evaluation and can be temporarily skipped.
Equations coloured like this:
Indicates a result that is widely used around.
Amplifiers and feedback gpessina 37
7. AMPLIFIERS: SYMBOLS
VO= AVi
vi vO
The simplest form for representing a voltage amplifier is the following:
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 non-ideal voltage source:
ARS
RLvs
vO
The amplifier, being ideal, has no current that enter its input, so there is no current that flow through RS. The consequence is that the voltage Vi is equal to VS.Being ideal, the output voltage does not depends on the load impedance RL. As a consequence:
vi
VO=AVS
The result is independent from the source resistor RS and load resistor RL. In this latter respect the output of the ideal amplifier behaves as an ideal voltage source.
Ii=0 Ri=∞
Amplifiers and feedback gpessina 38
………. amplifiers: symbols 2
ARS
RLvs
vO
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:
vi
VO=AVS
Moreover, there is another important consequence:The input power, or energy, is negligible, being:
Pi = VSIi~0
The output power is instead:
Po =Vo
2
RL≫ Pi
The amplifier merely takes energy from the supply voltages, bringing it to the load. In this respect the output cannot exceed the supply rail.
ARS
RLvi
VCC
VEEIDC
Amplifiers and feedback gpessina 39
………. amplifiers: symbols 3
RS
RLvs
vo
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 contrast to single ended input amplifiers. OAs have only a disadvantage in some situations concerning noise (we will study this).
viVO=A(V+ − V−)
There is another common, actually the most widely diffused, amplifier configuration: the differential amplifier or Operational Amplifier, OA:
+
-
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 40
………. amplifiers: symbols 4
The model, actual or ideal, is not an abstraction as OAs exist and are widely used. The know-how we will acquire will find practical applications.
+
-
In+In- Out
Vcc
Vee
In+
In-
Vee
Vcc
Out
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 a complicated function of frequency and 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.
In-In+Vee
VccOut
Amplifiers and feedback gpessina 41
8. THE NEGATIVE FEEDBACK CONCEPT
The idea at the basis of the modern electronics circuits is dated 1927: the negative feedback applied to amplifiers to improve or suppress distortion problems.
Amplifiers and feedback gpessina 42
………. the negative feedback concept 2
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:
The idea is to find an amplifier with such a gain off-the-shelf. This is not trivial for 2 reasons: (a) the number of applications is so wide that it would be very difficult for a manufacturer 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).
RS
RLvs
vovi
VO=A(V+ − V−)
+
-
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 107 – 108 V/V of gain, without any requirements concerning precision and stability.
RS
RLvs
vo
viVO=108 (V+ − V−)
+
-
Amplifiers and feedback gpessina 43
………. the negative feedback concept 3
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, or 10-7 V, 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.
RLvs
vovi
VO=108 (V+ − V−)
+
-
RS
Amplifiers and feedback gpessina 44
………. the negative feedback concept 4
Suppose now to be able to take a fraction β of the output signal VOand subtract it from the input, namely 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:
RLvs
vo
viVO=108 (V+ − V−)
+
-
Take a fraction β of VO and subtract it from the input.
From A → ∞ then V+ ≈ V−,
but V+ = VS and V− = 𝛃𝛃𝐕𝐕𝐎𝐎 therefore we obtain: VS ≈ βVO
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 un-fed backed OA.
A number of other useful consequences follows.
We can also be a bit more precise in considering A<∞:
VO = A V+ − V− = AVS − A𝛃𝛃𝐕𝐕𝐎𝐎
VO 1 + Aβ = AVS
𝐕𝐕𝐎𝐎 = VSA
1 + Aβ≈
𝐕𝐕𝐒𝐒𝛃𝛃
RS
Amplifiers and feedback gpessina 45
………. the negative feedback concept 5
Before entering into more details let’s move to consider a more real model of an OA:
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 VA, 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:
+
-
VA=A(v+-v-)
vo
i+
i-
Ri Ro
RS
vs
+
-
VA=108(v+-v-)i+
i-
Ri Ro
Take a fraction β of VO and subtract it from the input.
RL
vo
Amplifiers and feedback gpessina 46
………. the negative feedback concept 6
Being present a finite value of the input impedance we expect that a current would flow through it.
But this current is given by:
RS
vs
+
-
VA=108(v+-v-)i+
i-
Ri Ro
Take a fraction β of VO and subtract it from the input.
RL
vo
I+ = I− =V+ − V−
Ri=
⁄VA ARi
≈0Ri
= 0
The consequence of the large OA gain and the feedback which forces V+ to be close to V- makes negligible the input current. This means that there is no current through RS , too, and, as a consequence, VS equals V+.
And again, given that V+≈V- and also that I+=I-≈0 we can confirm the result of the previous pages:
VO = A V+ − V− = AVS − AβVO
… …
𝐕𝐕𝐎𝐎 = VSA
1 + Aβ ≈𝐕𝐕𝐒𝐒𝛃𝛃
Amplifiers and feedback gpessina 47
………. the negative feedback concept 7
We have another very important result: the gain so far obtained is not, or very weakly, 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.
RS
vs
+
-
VA=108(v+-v-)i+
i-
Ri Ro
Take a fraction β of VO and subtract it from the input.
RL
vo
𝐕𝐕𝐎𝐎 ≈𝐕𝐕𝐒𝐒𝛃𝛃
Amplifiers and feedback gpessina 48
9. THE NEGATIVE FEEDBACK CIRCUIT EXAMPLES
(See also Appendix A)
Let’s apply the rules:
From (2) and (3):
Therefore the currents in R1 and R2 satisfies , with the position (3):
Finally, from (4):
The final result has been obtained very quickly without the need of modelling the feedback by mean of a 2-ports network.
v+ = vs ≈ v−
vo − v−R2
=v−R1
voR2
=R1 + R2
R1R2v−
⇒ vo =R1 + R2
R1v−
β =R1
R1 + R2
1) A = ∞ ⇒ 2 V+ ≈ V−,
3)I+ = I− = 0 4 Vo =1β Vs
+
-
VA=A(v+-v-)
voRi Ro
vs
R1R2
RL
RS
I-=0
and vo =R1 + R2
R1vs
v+
v-
Amplifiers and feedback gpessina 49
………. the negative feedback circuit examples 2
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. This is a good approximation, provided that the gain can be considered large.
Moreover, under the large gain approximation, the result is independent from the source impedance RS and the load impedance RL.
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 correcting terms are added, or summed.
vo =R1 + R2
R1vs hence: β =
R1R1 + R2
+
-
VA=A(v+-v-)
voRi Ro
vs
R1R2
RL
RS
I-=0
Amplifiers and feedback gpessina 50
………. the negative feedback circuit examples 3
Let’s come back to our original result:
𝐕𝐕𝐎𝐎 =A
1 + Aβ VS ≈𝐕𝐕𝐒𝐒𝛃𝛃
And elaborate a bit:
𝐕𝐕𝐎𝐎 =A
1 + Aβ VS =1𝛃𝛃
A𝛃𝛃1 + Aβ VS =
1𝛃𝛃
−T1 − T VS
In this new look the overall gain is put into evidence as a product of the fed-backed gain term, 1/β, with a correcting factor which measures the departure of the gain from ideal, -T/(1-T).
The quantity T=-βA is called loop gain and describes the path that the signal must follows starting from a whatever point in the feedback loop and coming back 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 actually results: T=-Aγ(β), γ(β) being dependent on the elements around (normally |γ(β)|≤1), and it will be easier to determines directly 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 |γ(β)A| is smaller than |A|.
RLvs
vovi
VO=108 (V+ − V−)
+
-
Take a fraction β of VO and subtract it from the input.
RS
Amplifiers and feedback gpessina 51
………. the negative feedback circuit examples 4
𝐕𝐕𝐎𝐎 =1β
−T1 − T VS
The larger is the value of |T|, the closer to ideal is the behaviour. This is what it happens in most of the applications and one could 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 give the recipe for its evaluation. We will see that T will be not equal to βA, as already anticipated, and other correcting factors are present which are dependent on the network around, namely, the network that composes the feedback and the load/source impedances.
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.
+
-
VA=A(v+-v-)vo
Ri Ro
vs
R1R2
RL
RS
I-=0
Amplifiers and feedback gpessina 52
………. the negative feedback circuit examples 5
𝐕𝐕𝐎𝐎 =1β
−T1 − T VS
Now we wish to determine T, in order to estimate better the precision of the result above.
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 or, remembering the superimposition principle, we can consider independent the dependent generator, solve the network, then recover its functionality.
+
-
VA=A(v+-v-)vo
Ri Ro
vs
R1R2
RL
RS
I-=0
The application of the superimposition principle will allow us to set some rules that will make “mechanical” the analysis of a whatever fed-backed OA.
Amplifiers and feedback gpessina 53
………. the negative feedback circuit examples 6
Let’s start… the idea is to study the behaviour due to the source from that which comes from the feedback.
The easiest way to understand what it happens is to superimposethe signals; namely to null one source at a time. The first step is to consider VS=0 and breaking the feedback loop by assuming the voltage controlled voltage source independent and equal to a test voltage VT for a while:
+
-
VA=VT
voRi Ro
Vs=0
R1R2
RL
RS
I-=0
The circuit above is equivalent to this if VS is 0 V:
+
-
VA=VT
voRi Ro
R1R2
RL
RS
I-=0
Amplifiers and feedback gpessina 54
………. the negative feedback circuit examples 7
We can solve the network:VA − VOA
RO=
VOA − V−AR2
+VOARL
VOA − V−AR2
=V−AR1
+V−A − V+A
Ri
V−A − V+ARi
=V+AR𝑆𝑆
The syntax VxA indicates that we are considering the case VS is 0 and VA is the exciting source.
The solution of the network gives:
VOA = αVA
V+A − V−A = γVA
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 expressions on the right will be still maintained with different values of α and γ, in case a different feedback is considered.
We will see soon that we have to determine only γ.
+
-
VA=VT=generic variablevoA
Ri Ro
R1R2
RL
RS
V-A
V+A
Amplifiers and feedback gpessina 55
………. the negative feedback circuit examples 8
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:
We can solve the network:
+
-
vosRi Ro
vs
R1R2
RL
RS
V-s
V+s
VS − V+SRS
=V+S − V−S
RiV+S − V−S
Ri=
V−SR1
+V−S − VOS
R2V−S − VOS
R2=
VOSRO
+VOSRL
Where with VxS is intended voltage nodes evaluated when A=0, namely VA=0.
The result of the system of equations is:
VOS = ADIRVS
V+S − V−S = ξVS
Here we call ADIR the coefficient in the first equation as it represents the direct transmission of the signal through the feedback and take into account that the feedback is not an (ideal) unidirectional network, in general.
We will see soon that we have to determine only ADIR.
Amplifiers and feedback gpessina 56
………. the negative feedback circuit examples 9
Now the OA and the sources are back to their original condition. We remember that we have obtained:
+
-
VA=A(v+-v-)voRi Ro
vs
R1R2
RL
RS
I-=0
VOA = αVA
V+A − V−A = γVA
VOS = ADIRVS
V+S − V−S = ξVS
VO = VOS + VOA=ADIRVS+ αVA
V+ − V− = V+S − V−S + V+A − V−A = ξVS + γVA
Let’s superimpose the above results:
Now we arrived at the point for which it becomes convenient to consider the link we know present between the output and the input through the OA, namely, the voltage generator VA is a controlled source:
VA = A V+ − V− = AξVS + AγVA
Amplifiers and feedback gpessina 57
………. the negative feedback circuit examples 10
From which:
+
-
VA=A(v+-v-)voRi Ro
vs
R1R2
RL
RS
I-=0
We are interested to the voltage VO. To obtain it we have to consider this result in the first eq of the system:
That we can re-write as:
VA = A V+ − V− = AξVS + AγVA
VA = ξA
1 − Aγ VS
VO = ADIRVS+ αVA VO = ADIRVS+ αξA
1 − AγVS
VO = ADIR−αξ𝛾𝛾
−Aγ1 − Aγ VS
Amplifiers and feedback gpessina 58
………. the negative feedback circuit examples 11
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:
+
-
VA=A(v+-v-)vo
Ri Ro
vs
R1R2
RL
RS
I-=0
VO = ADIR−αξ𝛾𝛾
−Aγ1 − Aγ VS
VO =A→∞
1β VS
So that, from above:
ADIR−αξγ =
A→∞
1β
αξ𝛾𝛾 = ADIR −
1β
VO = ADIR−ADIR−Aγ
1 − Aγ+
1β
−Aγ1 − Aγ
VS
VO =ADIR
1 − Aγ +1β
−Aγ1 − Aγ VS
Therefore:
that is:
Since ADIR and γ does not depend on A (ADIR was obtained with A=0, while γ with VA=VT) we have:
limA→∞
ADIR−αξ𝛾𝛾
−Aγ1 − Aγ = ADIR−
αξγ
Amplifiers and feedback gpessina 59
………. the negative feedback circuit examples 12
What does represent V+A-V-A?
V+A-V-A is the signal that is generated by VA which travels the path arriving at the input. Then, if VA is made a voltage dependent again, this signal return to the starting point, once multiplied by the OA gain A.
V-
Indeed, if, from the previous page, we set:
+
-
VA=VT
voRi Ro
R1R2
RL
RS
I-=0
We have:V+A − V−A
VA= γ =
TA
T is the so called loop gain and is evaluated when the input signal is nulled while the controlled source generator is made independent.
It is very easy to show that if the OA is made ideal, namely Ro=0 and Ri=∞, γ reduces to -β
T = Aγ
Amplifiers and feedback gpessina 60
………. the negative feedback circuit examples 13
Considering what we obtained by setting A=0 we have the 3 rulesto apply:
1) A = ∞ ⇒ V+≈ V− and I+ = I− = 0 ⇒ Vo =1β Vs
2) Null the input source and make the input node controlled source (VA, the gain) independent for a while, then, solving the network it results that:
V+A − V−AVA
=TA
3) Null the gain then, solving the network it results that: VOS = ADIRVS
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).
So far to solve our network we have used the superimposition principle, but we do not need to determine all the terms of the resulting system. This because we applied what it happens when A→∞.
Amplifiers and feedback gpessina 61
………. the negative feedback circuit examples 14
vo= 1β
−T1−T vs+
ADIR1−T vs
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 ADIR is already small, in general).
We can then consider 3 levels of accuracy:
1. T is very very large: vo reduces to 1/β times the input;
2. T is large, but we would like to be a bit more accurate: vo is 1/β [–T/(1-T)] times the input;
3. T is large, or even not too large. We can be very accurate and consider both terms in the expression above for vo.
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 estimation of VO, but it is its exact expression.
Amplifiers and feedback gpessina
vo= 1β
−T1−T vs+
ADIR1−T vs
62
10. FEEDBACK PROCEDURE
The only way to obtain the gain 1/β is to consider A=∞, even if it is not true: if A=∞ in the expression for vo all terms hide, except 1/β.
Step #1:
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/(1-T) or not (the term f(β)βA is the upper limit for |T| and gives a first guess for the necessity of calculating the actual value of T).
1) A = ∞ ⇒ 2 V+ ≈ V−,
3)I+ = I− ≈ 0 4 Vo =1β Vs
+
-
VA=∞ (v+-v-)voRi Ro
vs
R1R2
RL
RS
Let’s summarize still another time.
I-=0
Amplifiers and feedback gpessina
vo= 1β
−T1−T vs+
ADIR1−T vs
63
………. feedback procedure 2
To evaluate the loop gain T we null the input source and let vA to be “generic” and not dependent on the input nodes.
Step #2:
Once that we know T we can decide if –T/(1-T) can be considered well approximated with 1 or not.
v+−v− = γvA
+
-
VA=VTvo
Ri Ro
R1R2
RL
RS
T =γA
From the circuit on the left:
then:
Amplifiers and feedback gpessina 64
………. feedback procedure 3
Finally, once we know β and T we can calculate ADIR by nulling the amplifier gain A.
Step #3:
vo= 1β
−T1−T vs+
ADIR1−T vs
Combining all together the non-approximated 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 4-terminals block.
vo=ADIRvs
+
-
voRi Ro
vs
R1R2
RL
RS
Amplifiers and feedback gpessina
11. A FURTHER BIT OF HISTORY
65
The first high gain amplifiers introduced were not monolithic, but based on valves. They were the K2-W and 12AX7 in the ‘40 - ‘50 of the last century.
…and then it came Bob Widlar with the µA700 in 1963:
Amplifiers and feedback gpessina 66
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:
…and calculate:
The above result says that the percentage of change of the final gain, Af, equals the percentage of change of the gain A, but attenuated by the factor 1+γ β A.
Af =vovs
=1β
γ β A1 + γ β A
ΔAf =1β
γ β 1 + γ β A − γ β 2A1 + γ β A 2 ΔA
ΔAf =1β
γ β1 + γ β A 2 ΔA
ΔAf =1β
γ β A1 + γ β A 2
ΔAA
ΔAf =1β
γ β A1 + γ β A
11 + γ β A
ΔAA
ΔAf = Af1
1 + γ β AΔAA
⇒ΔAfAf
=1
1 + γ β AΔAA
Amplifiers and feedback gpessina 67
………. the negative feedback and stability 2
AVCC
VEE
Narrow linear region: the gain is 107 V/VVi
AfVCC
VEE
The gain slope Af is smaller and the input can span a widely region before saturation occurs
Vi
Just for having an idea. If β=0.01, or the gain 1/β is 100, and A=107
V/V, the factor γ β A results in 105, if we assume γ β ≈β,and even a 100% of change of the gain A results in only 10-3% of change in Af.
This effect can be interpreted graphically. The response to the input signal of the open-loop 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:
The gain Af is much smaller than A and the input voltage can spans larger values before the saturation toward the rails occurs:
ΔAfAf
=1
1 + γ β AΔAA
Amplifiers and feedback gpessina
vo − vAR1
−vAR3
=vAR2
vo − vAR1
+ is =vAR2
voR1
= −is +R1 + R2
R1R2vA
voR1
= −is −R1 + R2
R1R2R3is
vo = −R1R2 + R3 R1 + R2
R2is
68
………. the negative feedback and stability 3
Another example:
Again a fraction of the output is taken to the inverting input, where it is subtracted: once again the feedback is negative.
Apply (1) and take care: here v- results 0 just because v+ is connected to ground.
Apply (3)
Again we have obtained the link between the output and input.
v− ≈ v+ = 0
is =v− − vA
R3+
v−RS
=0 − vA
R3+
0RS
= −vAR3
1) A = ∞ ⇒ 2 V+ ≈ V−,
3)I+ = I− ≈ 0 4 Vo =1β Vs
+
- RLR1
R2
R3
vo
is
v-
vA
i-∼0
Rs
Amplifiers and feedback gpessina 69
………. the negative feedback circuit examples 4
Again, (2) is valid, therefore:
Here we have found 2 other properties:
The output is a voltage, while the input is a current: when to disentangle between the input/output variables 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 γ(β)A must be dimensionless.
The answers to both of the above questions will be soon given.
vo = −R1R2 + R3 R1 + R2
R2is
β = −R2
R1R2 + R3 R1 + R2
1) A = ∞ ⇒ 2 V+ ≈ V−,
3)I+ = I− ≈ 0 4 Vo =1β Vs
+
- RLR1
R2
R3
vo
is
v-
vA
i-∼0
Rs
Amplifiers and feedback gpessina 70
………. the negative feedback circuit examples 5
Here we have a first OA A1, for which across its feedback path there is another feed-backed OA, A2.Note that since A2 is inverting the feedback at the input of A1is taken at the non-inverting node, for negative feedback.Again:
From (2)
From (3)
At the input node:
Now from the output, merging the input:
From (4):
We see that input and output have the same dimension, but are currents.
Parameter β is dimensionless.
V1+ ≈ V1− = 0
Is =V1+ − VA
RF+
V1+R𝑆𝑆
= −VARF
V2+ ≈ V2− = 0
Io =V2− − VA
R1= −
VAR1
Io = −−IsRF
R1=
RFR1
Is
β =R1RF
io =vo − v2−
RL=
voRL
1) A = ∞ ⇒ 2 V+ ≈ V−,
3)I+ = I− ≈ 0 4 Vo =1β Vs
-
+
RLR1RF
vo
is
V1+
vA
+
-
V1-
V2+
V2-
io
A1
A2
Rs
Amplifiers and feedback gpessina 71
………. the negative feedback concept 6
Voltage output over voltage input:
Current output over voltage input
vs
βRio
+-
vovi
RL
io
+-vs
vo
βvo
vi
RL
+
+
So far we have became familiar in solving fed-backed 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 signals 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.
The output voltage is read, and a fraction of it is subtracted from the input.
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 72
………. the negative feedback concept 7
Current output over current input
βiio
vo
RL
io
is
Voltage output over current input
βGio
vovi
RLis
The output current is read, and a fraction of it is subtracted from the input.
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 73
………. 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:
+
-
vo=A(v+-v-)
R1R2
vs
vo
RL
I-=0
Rs v+We have found that:
The output voltage is independent from the load impedance RL, under the assumption that the gain is very very large, whereas the output current is:
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.
vo =R1 + R2
R1vs
io =voRL
=R1 + R2
RLR1vs
Amplifiers and feedback gpessina 74
………. the negative feedback concept 9
+
-
vo=A(v+-v-)
R1R2
vs
voRL
I-=0Rs v+
Indeed our result is:
So the final result does not depends on the input source resistor Rs and we can consider the voltage the input variable.
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.
From our approximation we know that the input current is negligible; the consequence is that the voltage drop across Rs is negligible and Vs =V+.
+
-
vo=A(v+-v-)
R1R2
isvo
RL
I-=0Rs
v+
Let’s consider the case the input is a current generator with its source parallel resistor Rs.
We have now that:
Therefore:
vo =R1 + R2
R1vs
v+ = RsIs and v+ = vs
vo =R1 + R2
R1RsIs
Amplifiers and feedback gpessina 75
………. the negative feedback concept 10
We now consider the source resistor Rs in parallel to the input current source. Since v+=0 V and v- ≈ v+, then v-≈0 V and the current through Rs is 0 A: the output signal will not depend on Rs, as it happens, actually.
The output becomes dependent on the source impedance in this case: the amplification is about current.
We have found that the gain is:
Let’s consider the case the input is a voltage generator with its source series resistor Rs.
Let’s consider the second of our examples.
That does not depends on the load RL: the output variable is therefore a voltage.
+
- RLR1
R2
R3
vo
is
v-
vA
i-∼0
Rs
+
- RLR1
R2
R3
vo
vsv-
vA
i-∼0
Rs
is
Now we have that:
So that:
vo = −R1R2 + R3 R1 + R2
R2is
is =vsRs
vo = −R1R2 + R3 R1 + R2
R2
vsRs
Amplifiers and feedback gpessina 76
………. the negative feedback concept 11
We know that V2- is about 0 V so that:
The output voltage depends on the load impedance RL, and the output signal is, therefore, a current.
We have found that the gain is:
and:
Let’s consider the last of our examples.
The input follows the same consideration of the previous case: v1+ is at ground and the current in Rs is nulled.
Io =RFR1
Is
vo = RLIo
vo = RLIo = RLRFR1
Is
-
+
RLR1RF
vo
is
V1+
vA
+
-
V1-
V2+
V2-
Rs
Io
Amplifiers and feedback gpessina 77
………. the negative feedback concept 12
In the circuit on the left one of the inputs, V+, is held at constant potential. As a consequence, the other terminal, V-, is held close at a constant potential, too. This suggests that the input is at low impedance since it does not change by injecting current.
There is another approach for understanding which kind of feedback is working with respect to the input variable.
-
+
RLR1RF
vo
is
V1+
vA
+
-
V1-
V2+
V2-
Rs
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.
+
-
vo=A(v+-v-)
R1R2
vs
vo
RL
I-=0Rs v+
The opposite is true for the voltage at the input: when the 2 inputs are floating and one of the input is connected to the source, or when 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 78
………. the negative feedback concept 13
When the feedback reads the same node of that of the OA output, or one of the nodes of the load impedance RL is held at constant potential, then the voltage is the output variable.
There is another approach for understanding which kind of feedback is working with respect to the output variable.
-+
RLR1RF
vo
is
V1+
vA
+-
V1-
V2+
V2-
Rs
+-
vo=A(v+-v-)
R1R2
vs
vo
RL
I-=0Rs v+
If the load impedance RL has it legs connected to floating nodes, then a current is the output variable.A current is read also in case the feedback is not connected to the load directly. See the example below.
Amplifiers and feedback gpessina 79
13. EXAMPLES OF COMPLETE FEEDBACK EVALUATION
Rs+
-
VA=∞(v+-v-)
R1R2
vs vo
RLi-
Ri Ro
i+
Let’s consider some examples.
Step #1:
If (1) is valid then (2) is valid, too, and v-≈v+=vs therefore:
vsR1
=vo − vs
R2
From which we obtain the same result of the ideal case:
…or:
vo =R1 + R2
R1vs
1β =
R1 + R2R1
1) A = ∞ ⇒ 2 V+ ≈ V−,
3)I+ = I− ≈ 0 4 Vo =1β Vs
Amplifiers and feedback gpessina 80
………. examples of complete feedback evaluation 2
+
-
VA is generic
R1R2
vsvo
RL
i+
i-
Ri RoThe input is nulled
Step #2: v+−v− =TvAA
Let’s solve the network as the feedback is not present (vAconsidered independent) for finding the loop gain T:
vA − voRo
=voRL
+vo − v−
R2
vo − v−R2
=v−R1
+v−
Rs + Ri
v+ − v− =Rs
Rs + Riv− − v− = −
RiRs + Ri
v−
Let’s define: 1R1′
=1
R1+
1Rs + Ri
,
vo − v−R2
=v−R1′
voR2
=1
R1′+
1R2
v− vo =R1′ + R2
R1′v−
v− =R1′
R1′ + R2vo
namely: R1′ = R1‖ Rs + Ri
From the second eq:
Rs
v+ =Rs
Rs + Riv−
Amplifiers and feedback gpessina 81
………. examples of complete feedback evaluation 3
vA − voRo
=voRL
+vo − v−
R2
vo − v−R2
=v−R1
+v−Ri
v+ − v− = −v−
Now the first eq:
v− =R1′
R1′ + R2vo
vARo
=voRo
+voRL
+voR2
−1
R2
R1′R1′ + R2
vo
vARo
=voRo
+voRL
+voR2
1 −R1′
R1′ + R2
vARo
=voRo
+voRL
+1
R1′ + R2vo With the position:
1RL′
=1
RL+
1R1′ + R2
, namely: RL′ = RL‖ R1′ + R2
vARo
=voRo
+voRL′
vA =Ro + RL′
RL′vo vo =
RL′Ro + RL′
v𝐴𝐴
And:
v− =R1′
R1′ + R2
RL′Ro + RL′
v𝐴𝐴
Amplifiers and feedback gpessina 82
………. examples of complete feedback evaluation 4
TvAA = v+ − v− = −
RiRs + Ri
v− =− −Ri
Rs + Ri
R1′R1′ + R2
RL′Ro + RL′
v𝐴𝐴
+
-
VA is generic
R1R2
vsvo
RL
i+
i-
Ri RoThe input is nulled
Remembering that:
We, finally, obtain:
T = −Ri
Rs + Ri
AR1′R1′ + R2
RL′Ro + RL′
Remember that in a negative feedback loop T must be negative (at DC) and dimensionless.
If you do not find such a result two possible drawbacks are: the feedback path was set positive or an error happened during your solution of the system of equations.
Rs
v− =R1′
R1′ + R2
RL′Ro + RL′
v𝐴𝐴
R1′ = R1‖ Rs + Ri
RL′ = RL‖ R1′ + R2
Amplifiers and feedback gpessina 83
………. examples of complete feedback evaluation 5
We null the gain source:
We put:
+
-
VA=0
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
Step #3: vo=ADIRvs
vs − v−Rs + Ri
=v−R1
+v− − vo
R2
v− − voR2
=voR0
+voRL
1Ro′
=1
R0+
1RL
v−R2
=voR2
+voR0′
v− =R2 + R0′
R0′vo vo =
R0′R2 + R0′
v−
From the second:
vsRs + Ri
=v−
Rs + Ri+
v−R1
+v−R2
−1
R2
R0′R2 + R0′
v−
From the first:
vsRs + Ri
=v−
Rs + Ri+
v−R1
+v−R2
1 −R0′
R2 + R0′
Amplifiers and feedback gpessina 84
………. examples of complete feedback evaluation 6
vs − v−Rs + Ri
=v−R1
+v− − vo
R2
v− − voR2
=voR0
+voRL
vo =R0′
R2 + R0′v− =
R0′R2 + R0′
R1′Rs + Ri + R1′
v𝑠𝑠
Again, for convenience, we can put:
vsRs + Ri
=v−
Rs + Ri+
v−R1
+v−R2
1 −R0′
R2 + R0′
vsRs + Ri
=v−
Rs + Ri+
v−R1′
1R1′
=1
R1+
1R2 + R0′
vs =Rs + Ri + R1′
R1′v−
v− =R1′
Rs + Ri + R1′v𝑠𝑠
Finally:
Or:
ADIR =vovs
=R0′
R2 + R0′
R1′Rs + Ri + R1′
Amplifiers and feedback gpessina 85
………. examples of complete feedback evaluation 7
vo =R1 + R2
R1
−T1 − T vs +
R0′R2 + R0′
R1′Rs + Ri + R1′
11 − T vs
Putting all together:
+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
T = −ARi
Rs + Ri
R1′R1′ + R2
RL′Ro + RL′
And:
Amplifiers and feedback gpessina 86
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 a 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:
T = γA
As a consequence:
T = −βA
limRi→∞, Ro→0
γ = −β
In practice, we can consider γ as expressed with the form:
γ = −βf(β)
Namely:
T = −βf(β)A
(f(β) →1 when Ri →∞ and RL →0)
Amplifiers and feedback gpessina 87
…and another consideration
+
-
R1
R2
vs vo
Ri
Avivi Ro
+
VB
+
At first it might seem a complicated procedure to follow the 3 steps for the solution of the network, as we did.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
To come to the final solution several intermediate steps are needed.
At the end the final result does not distinguish the different contributions.
vs − vBRi
+vo − vB
R2=
vBR1
A vs − vB − voRo
=vo − vB
R2vi = vs − vB
vsRi
= vB1
R1+
1Ri
+1
R2−
voR2
vsA
Ro= vB
ARo
−1
R2+ vo
1R2
+1
Ro
⋯
⋯
vo =AR2Ri R1 + R2 + R1R2Ro
R1 + Ri R2 + Ro R2 + A + 1 R1R2Rivs
≈AR2Ri R1 + R2
A + 1 R1R2Rivs ≈
R1 + R2R1
vs cvd!
Amplifiers and feedback gpessina 88
14. INPUT IMPEDANCE OF A FED BACKED OA
+
-
Ri
Avi
Ro+vs
Rs
RL
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 89
………. input impedance of a v-to-v fed backed OA 2
The current is is flowing through Rs and also thought Ri and through this latter satisfies also:
The input impedance of the network is that seen at the input source node, and is given by:
Rif =vsis
is =v+ − v−
Ri
or:
is =v+ − v−
Ri=
vAARi
v-to-v= voltage input and voltage output amplifier
+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+is
Amplifiers and feedback gpessina 90
………. input impedance of a v-to-v fed backed OA 3
Here β’ and ADIR’ refer the fact that node vA is different from node vo, but T is the same, as expected all around the loop.
The expression above is expected to be valid since the evaluation made for node vo was completely general, not based on a particular assumption, so that the same procedure can be applied at every other node of the network.
We now will try to verify if vA can be obtained exploiting the parameters we have just learned to calculate.
Node vA has not special properties with respect node vo or any other node of the network and we can consider it as a fed backed node; therefore we can write:
vA =1β′
−T1 − T vs +
ADIR′1 − T vs
+
-
R1R2
vs voRL
Rs
i-
Ri Ro
i+ vA = 1β′
−T1−T
vs+…vo = 1
β−T
1−Tvs+…
Amplifiers and feedback gpessina 91
………. input impedance of a v-to-v fed backed OA 4
vA − voRo
=voRL
+vo − v−
R2
vARo
=1
Ro+
1RL
1 +R2R1
vs +vsR1
vA = 1 +RoRL
1 +R2R1
+RoR1
vs
or:
namely:
vAvS
=1β′ = 1 +
RoRL
1 +R2R1
+RoR1
+
-
VA=∞(v+-v-)
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
Although not necessary for what is our intended goal, let’s determine β’, just to practice:
=present evaluated feedback node
vA − voRo
=voRL
+vo − vs
R2but v- ≈ vS
Amplifiers and feedback gpessina 92
………. input impedance of a v-to-v fed backed OA 5
Therefore, coming back to the restored amplifier:
Now, if the gain is reduced to 0 we see that node VA cannot change by direct transmission as, being VA an ideal voltage source, it has a zero impedance and ADIR’ results to be 0:
v+ − v− =vAA =
1A
1β′
−T1 − T vs =
1Aβ′
−γA1 − T vs
vA =1β′
−T1 − T vs
v+ − v− =− ⁄γ β′1 − T vs
is =v+ − v−
Ri=
− ⁄γ β′1 − T Ri
vs
+
-
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
T = γA, (γ < 0)
The loop gain T is proportional to A, and we have found to depend on the components of the network. We know it can be put in the form:
=present evaluated feedback node
Amplifiers and feedback gpessina 93
………. input impedance of a v-to-v fed backed OA 6
Finally:
The feedback multiplies the parameter Ri/( ⁄γ β′) by 1-T, 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 Ri/( ⁄γ β′) must become equal to the input impedance that is measured in open loop condition, or when A is set to 0…
is =v+ − v−
Ri=
− ⁄γ β′1 − T Ri
vs
Rif =vsis
=Ri⁄−γ β′ 1 − T
+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
=present evaluated feedback node
Namely:
Riol =Ri⁄−γ β′
Amplifiers and feedback gpessina 94
………. input impedance of a v-to-v fed backed OA 7
Therefore, we can write that:
where Riol (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:
Rif =vsis
= Riol 1 − T
A practical example is given in Appendix A.
+
-
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
Riol = Rs + Ri + R1′
R1′ = R1R2
′ R2′ = R2 + RLRowith: and:
Amplifiers and feedback gpessina 95
………. input impedance of a v-to-v 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 Rif.Namely, in this way the feedback is not perturbed when we apply the A=∞ rule: when A=∞ the current thorough Ridrops to a negligible value and is managed from the feedback.
Moreover, if the testing voltage source is not ideal, namely Rs is not completely negligible, the measured input impedance value is not contributed, or is marginally contributed, by Rs itself.
+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
Amplifiers and feedback gpessina
………. input impedance of a v-to-v fed backed OA 9
96
A consideration (b):
If a current source were instead used for the characterization we would perturb the feedback, acting on the actual value of T.
We see the way this perturbs T:
T=−AβRi
Ri +Rs+Roβ
RLRiβARL+Ro
1RiβA +Ro′
We see that if Rs → ∞ because we use a current source, than T → 0. The direct gain was omitted for brevity:
vo = 1β
−T1 − T vs = 1
β−T
1 − T Rsis
vo =ARi
Ri +Rs+Roβ
RLRiβARL+Ro
1RiβA +Ro′
11 − T Rsis
vo Rs→∞ARi
RLRiβARL+Ro
1RiβA +Ro′
is
Namely, the gain depends on A and not on 1/β only: we are unhappy of this.
+
-
VA=A(v+-v-)
R1R2
Isvo
RL
Rs
i-
Ri Ro
i+
+
-
VA=A(v+-v-)
R1R2
Isvo
RL
Rs
i-
Ri Ro
i+
The conclusion is: an inappropriate instrument or method to determine the input impedance would affect the behaviour of the network.
Amplifiers and feedback gpessina 97
………. input impedance of a v-to-v fed backed OA 10
A consideration (c):
In case a current source were used anyway, it would help to consider a source impedance, Rs, 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.
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 Rs → 0, or take a voltage source if T → 0 when Rs → ∞
+
-
VA=A(v+-v-)
R1R2
Isvo
RL
Rs
i-
Ri Ro
i+
Amplifiers and feedback gpessina 98
………. input impedance of a v-to-v 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:
is =v+Rs
+v+ − v−
Ri
is =v+Rs
+⁄−γ β′
1 − T RiRsis
vA =1β′
−T1 − T Rsis =
1β′
−γA1 − T Rsis
Where, as we have already shown:
is 1 −⁄−γ β′
1 − T RiRs =
v+Rs
And:
Rifp =v+is
= Rs 1 −⁄−γ β′
1 − T RiRs = Rs
2 1Rs
−1
Rif
= Rs2 1
Rs−
1Rs + Rifo
= RsRs + Rifo − Rs
Rs + Rifo=
RsRifoRs + Rifo
And Rif0 is the feedback input impedance measured when Rs is set to 0 in the network, see the next consideration for detail.
+
-
VA=A(v+-v-)
R1R2
Isvo
RL
Rs
i-
Ri Ro
i+
and v+ − v− =vAA(i+=0 se A=∞)
Rifp
Amplifiers and feedback gpessina 99
………. input impedance of a v-to-v fed backed OA 12
A consideration (e):
An useful approach is to disentangle the source impedance, Rs, from the amplifier input impedance.In practice the impedance so far evaluated, Rif, we would like to be express as:
Rif = Rs + Rif0
Starting from the zero gain, or open loop, input impedance, is obvious that:
Riol = Rs + Riol0
We show that it is indeed valid the following expression:
Rif = Rs + Riol0 1 − T0
Where Riol0 and T0 refer to Riol and T evaluated when Rs=0 Ω.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
In the above network we have that:
Riol = Rs + Ri + R1′ so that: Riol0 = Ri + R1′
and: R1′ = ‖R1 R2 + ‖RL RO
Amplifiers and feedback gpessina 100
………. input impedance of a v-to-v fed backed OA 13
In the above circuit we have that:
Riol = Rs + Ri + R1′
+
-
R1R2
vs vo
RL
Rs
i-
Ri Ro
i+
R1’
Where we call R1’ the impedance seen to the right of node V-.
By applying the Thevenin th. to node V- we obtain that the impedance seen is R1’, while αVA will be the voltage, α being the fraction of VA measured across R1 when R1 is unloaded.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
Thev.
Amplifiers and feedback gpessina 101
………. input impedance of a v-to-v fed backed OA 14
In the new equivalent circuit above both Vs and VA’ see now the same impedance and the network assume a symmetry, in someway:
Riol = Rs + Ri + R1′
If now we take Vs=0 we can write, remembering that the dropout across a resistor in a series of resistors is proportional to the resistor value, according to the following proportion:
is+
-
VA’=αVA =αA(v+-v-)
R1’
vs
Rs
Ri
Riol V+ − V− = RiVA′ Riol0 V+ − V− 0= RiVA′
V+ − V−Ri
=VA′Riol
|V+ − V− 0Ri
=VA′
Riol0(RS = 0)
We have:
But from the last 2 eqs:
Ri = RiolV+ − V−
VA′= Riol
V+ − V−αVA
= RiolT
αA
Ri = Riol0|V+ − V− 0
VA′= Riol0
|V+ − V− 0αVA
= Riol0T0αA
Hence:RiolT = Riol0T0
Pay attention, this node is not Vo!
and also:
Amplifiers and feedback gpessina 102
………. input impedance of a v-to-v fed backed OA 15
Now let’s elaborate:
So that: Rif = Riol 1 − T
=Riol-RiolT
=Riol − Riol0T0
=Rs+Riol0 − Riol0T0
=Rs+Riol0 1 − T0
T =Riol0Riol
T0
RiolT = Riol0To
CVD!
T =Riol0
RS + Riol0T0but Riol = RS + Riol0, hence:
Amplifiers and feedback gpessina 103
15. OUTPUT IMPEDANCE OF A V-TO-V FED BACKED OA
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 Vs, provided that the source does not perturb the network, hence the loop gain T.
Again, we expect that:
The output impedance is the impedance seen from the load and is characterized as shown below:
(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)
vo =1β"
−T1 − T
iT +ADIR"1 − T iT
Parameters β’’ and ADIR’’ are only to indicate that vo is excited by a source other than Vs and are different from the respective β and ADIR.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
IT
Amplifiers and feedback gpessina 104
………. output impedance of a v-to-v fed backed OA 2
Let’s begin with considering T=∞:
vo =ADIR"1 − T iT
Therefore vo results 0 and we remain with:
0 = v+ − v− ⇒ is = 0
So that:
Rofp =voiT
=ADIR"1 − T
With T=0 Rofp reduces to the output impedance, Rool (R output, measured in open loop), measured when the gain is zero, hence:
Rofp T=0= Rool = ADIR"
is+
-
VA=A(v+-v-)
R1R2
vsvo
RL
Rs
Ri Ro
ITi2
therefore:
v− = 0, hence I1 =V−R1
= 0 and i2 = 0 ⟹ vo = v− + i2R2 = 0 ∀ IT
vo =1β"
−T1 − T
iT + ⋯
=present evaluated feedback nodei1
Amplifiers and feedback gpessina 105
………. output impedance of a v-to-v fed backed OA 3
So we can write:
Rofp =Rool
1 − T
The feedback works in lowering the output impedance by a factor 1-T, making the output of the network closer to an ideal voltage source.
Of course this behaviour is made more effective if the loop gain T is large.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
IT
=present evaluated feedback node
Amplifiers and feedback gpessina 106
………. output impedance of a v-to-v fed backed OA 4
Just to exercise, now we have that:
iT =voRo
+voRL
+vo − v−
R2
vo − v−R2
=v−R1
+v−Ri
is+
-
R1R2
vs vo
RL
Rs
Ri Ro
IT
Or, simply:vo − v−
R2=
vo
R2 + R1Ri
We have:
iT =1
Ro+
1RL
+1
R2 + R1Rivo
And:
Rool =voiT
=1
Ro+
1RL
+1
R2 + R1Ri
−1
=present evaluated feedback node
Amplifiers and feedback gpessina 107
………. output impedance of a v-to-v fed backed OA 5
A consideration (a):
We expect to evaluate a very small output impedance, then, for not perturbing the loop, it is recommended to use a current source as test signal.
Below, we use a voltage source. See what it happens.
It is easy to imagine that if the voltage source is close to ideal, namely RT close to 0 Ω, the output node is shorted and the loop does not work: the feedback is perturbed.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.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
IT
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
VT
RT
Amplifiers and feedback gpessina 108
………. output impedance of a v-to-v fed backed OA 6
T=−A R1‖ Ri + Rs
R1‖ Ri + Rs + R2
RL′R2′RL′R2′+Ro
TRT→0
0
A consideration (b):
Let’s evaluate the loop gain.For convenience we define:
RL′ = RLRT RT→0RT
Then:
In this case, obviously, again neglecting the direct transmission:
vo = 1β
−T1 − T vs
vo =ARi
Ri +Rs+Roβ
RL′RiβARL′+Ro
1RiβA +Ro′
11 − T vs RT→0
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 RT → 0, or take a voltage source if T → 0 when RT → ∞.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro VT
RT
R2′ = R2 + R1‖ Ri + Rs
Amplifiers and feedback gpessina 109
………. output impedance of a v-to-v fed backed OA 7
Practical explanation about the best way to measure an impedance.An instrument, in doing the measurement, can apply a voltage and measure the driven current or apply a current and measure the voltage. In the first case the instrument behaves as a voltage generator, in the second as a current generator.Let’s study the 2 situations.
RM =VTIT
− RT ⟹∆ITIT
= −∆RT
RT + RM
VT =RMRT
RM + RTIT ⟹
∆VTVT
= −RM
RT + RM
∆RTRT
VT
RT
RM
IT
So the error in the measured currentbecomes negligible if RM >> RT, then a voltage source is a good instrument to measure large value impedances.
IT RTRM
VT
So the error in the measured voltagebecomes negligible if RT >> RM, then a current source is a good instrument to measure small value impedances.
Amplifiers and feedback gpessina 110
………. output impedance of a v-to-v fed backed OA 8
A consideration (c): nevertheless…
The test signal in the position shown here does not affect the loop and can be used for testing.Also in this case we have only direct transmission, given by:
vo =RoR2′
RoR2′ + RovT
vo =RoR2′
RoR2′ + Ro
vT1 − T
=ADIR′′′1 − T
vT
Therefore, finally, when A returns to its actual value:
iT
is+
-
R1R2
vs vo
RL
Rs
Ri Ro
VT
R2′ = R2 + R1‖ Ri + Rs
When A=0
Amplifiers and feedback gpessina 111
………. output impedance of a v-to-v fed backed OA 9
A consideration (d): nevertheless…
Therefore:
iT =vT − vo
RL= 1 −
ADIR′′′1 − T
vTRL
Rofs =vTiT
= RL1
1 − ADIR′′′1 − T
= RL + Rof∞
It can be shown that the result obtained in this measurement configuration for Rof∞ and in the previous configuration with a current source as test signal are the same; where Rof∞ is intended evaluated when RL is considered ∞, see the next pages.
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.
iT
is+
-
R1R2
vs vo
RL
Rs
Ri Ro
VT
vo =RoR2′
RoR2′ + Ro
vT1 − T =
ADIR′′′1 − T vT
th
Amplifiers and feedback gpessina 112
………. output impedance of a v-to-v fed backed OA 10
A last consideration :
An useful modelling is obtained when the load impedance, RL, is disentangled from the amplifier output impedance.In practice the impedance so far evaluated, Rof, we would like to be expressed as:
Rof = RLRof∞
Starting from the zero gain condition, or open loop, output impedance:
Rool = RL RoR2′
We can write, obviously:
Rool = RLRool∞
We will show that it is possible to write:
Rof = RL Rool∞
1 − T∞
Where Rool∞ and T∞ refer to Rool and T evaluated when RL=∞ Ω.
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
IT
R2′ = R2 + R1‖ Ri + Rs
Amplifiers and feedback gpessina 113
………. output impedance of a v-to-v fed backed OA 11
This way the impedance seen from IT and IA are the same and equal to:
is+
-
VA=A(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
IT
Norton
is+
-
IA=VA/Ro= A/Ro(v+-v-)
R1R2
vsvo
RL
Rs
Ri Ro
IT
Rool = RL RoR2′
R2′ = R2 + R1‖ Ri + Rs
IR
Assuming that IT=0 the current IR is a fraction of the total current weighted on the inverse of the impedance on which it flows:
IR⁄1 R2′
=IA
⁄1 Rool
IRRool
=IA
R2′
If we set RL=∞, then: Rool∞ = RoR2′
|IR RL=∞
Rool∞=
IAR2′
But, from above:
Amplifiers and feedback gpessina 114
………. output impedance of a v-to-v fed backed OA 12
is+
-
IA=VA/Ro= A/Ro(v+-v-)
R1R2
vs vo
RL
Rs
Ri Ro
IT
IR
Now it is evident from above that we can write:
IRRool
=IA
R2′
So that from the result of the previous slide:
1Rool∞
|IR RL=∞
IA=
1R2′
We can write:
V+ − V− = ρIR V+ − V− RL=∞= ρ IR RL=∞
So that:
IRIA
=V+ − V−
ρROVA
=TA
ROρ
|IR RL=∞
IA=
|V+ − V− RL=∞
ρROVA
=T∞A
ROρ
1Rool
TA
ROρ =
1Rool∞
T∞A
ROρ
TRool
=T∞
Rool∞
|IR RL=∞
Rool∞=
IAR2′
1Rool
IRIA
=1
R2′
Amplifiers and feedback gpessina 115
………. output impedance of a v-to-v fed backed OA 13
Let’s elaborate:
1Rof
=1 − TRool
Namely:
=1
Rool−
TRool
=1
Rool−
T∞Rool∞
=1
RL+
1Rool∞
−T∞
Rool∞
=1
RL+
1Rof∞
Rof=RLRof∞
TRool
=T∞
Rool∞
1Rool
=1
RL+
1Rool∞
=Rool∞ + RL
RLRool∞
T =Rool
Rool∞T∞ =
RLRL + Rool∞
T∞