Versione del: lunedì 9 marzo 2020

Amplifiers and feedback gpessina 1

1. NOTE INIZIALI ORGANIZZATIVE

L’inizio delle lezioni viene confermato per lunedì 9 marzo (una settimana successiva alla data ufficiale);

Molto probabilmente inizieremo con lezioni video-registrate come questa prova. Questa è quindi una occasione per potere verificare e segnalare tutti i possibili problemi tecnici che potete riscontrare nella fruizione del video.

Naturalmente è possibile porre domande/commenti. Ad ogni lezione sarà associato un feedback, completamente anonimo, presente subito dopo al link al video. La risposta alle domande/commenti sarà fornita a tutti.

Nel caso si facesse lezione in aula i giorni dedicati sono il lunedì dalle 13:30 alle 15:30 ed il mercoledì dalle 14:30 alle 16:30. Questi orari non sono vincolanti e potremmo spostarli per cercare di combinare le esigenze di tutti voi.

Tutto il materiale didattico è disponibile in formato pdf su questa pagina web o quella dedicata, http:/pessina.mib.infn.it/Corsi_del_IV_anno/Indice.htm

La modalità d’esame prevede uno scritto ed un colloquio. Buona parte dei temi d’esame passati si trovano nella pagina dedicata: http:/pessina.mib.infn.it/Corsi_del_IV_anno/Indice.htm.

Dovreste votare se vi va bene che la prima lezione venga caricata entro questo giovedì.

Versione del:lunedì 9 marzo 2020

http://pessina.mib.infn.it/Corsi_del_IV_anno/Indice.htm

http://pessina.mib.infn.it/Corsi_del_IV_anno/Indice.htm


………… note iniziali organizzative

Una piccola introduzione al contenuto del corso.

Il corso non è per interessati all’indirizzo Elettronico, ma per tutti quelli che si vogliono occupare di Fisica sperimentale ed anche di analisi dati.

Gli argomenti trattati vogliono indicare le conoscenze di base che portano alla comprensione di tutte le problematiche e le ottimizzazioni necessarie ad un tipico sistema di rivelazione di un esperimento di Fisica. In particolare alla comprensione di cosa si occupa la parte del sistema che sta tra il rivelatore ed il sistema di digitalizzazione/produzione dei dati: il sistema di amplificazione.

Tuttavia, questo aspetto non è limitativo per chi si vuole avvantaggiare delle conoscenze più prettamente limitate all’elettronica giacché un sistema di acquisizione di un rivelatore è un esempio abbastanza generico rispetto alle applicazioni che si possono trovare nella vita reale.

Per essere un poco più oggettivi, ci occuperemo di capire come funziona un amplificatore reazionato, sia a bassa che ad alta frequenza, gli aspetti riguardanti la stabilità, ed il rumore. Vedremo molti esempi applicativi.

Una nota per gli interessati all’Elettronica: per un buon approfondimento sarebbe opportuno completare lo studio con un lavoro di tesi sull’argomento.

Per essere ancora più espliciti….


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:lunedì 9 marzo 2020


. ………. 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.


………. 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.



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)



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, …



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.



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.



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


11

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



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 a single atom:

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.



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.

+ + …



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.



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).


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


………. 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



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



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 =VG

0= ∞ 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



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)



IG RGRL

IO

The voltage VO is common to both resistors:

Hence:

IG =VO

RG+

VO

RL

Let’s evaluate the output current:

VO

VO =RGRL

RG + RLIG

The current through RL is:

IO =VO

RL

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



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.



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:



+ +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 − VN

R1=

VN

R3+

VN − −BVN

R2

V1

R1=

1R1

+1

R3+

1 + BR2

VN

VN =1

R1+

1R3

+1 + B

R2

−1 V1

R1



+ +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 − VN

R1=

VN

R3+

VN − V2

R2

V1

R1=

1R1

+1

R2+

1R3

VN +−V2

R2

VN −1

R1+

1R2

+1

R3

−1 V2

R2=

1R1

+1

R2+

1R3

−1 V1

R1

V2

Now we assign to V2 its dependence:

VN −1

R1+

1R2

+1

R3

−1 −BVN

R2=

1R1

+1

R2+

1R3

−1 V1

R1

1R1

+1

R2+

1R3

−1 1R1

+1

R2+

1R3

VN +BVN

R2=

1R1

+1

R2+

1R3

−1 V1

R1

VN =1

R1+

1R3

+1 + B

R2

−1 V1

R1



A last approach for solving the network is to use the superimposition principle, that will be considered in the following pages.


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 others 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 (we will see).

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 =

R2

R1R2 + R1R3 + R2R3V1

(see next page for evaluation)

= zero V, o GND

V2=0


………. 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 − VA

R1=

VA

R3+

VA

R2

With only one unknown and:

IA =VA

R3

=0 V



+V1

R1 R2

R3IA

VA

VC

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 + VC − VA

R1=

VA − VC

R3+

VA − VC

R2

Now, if we call:

VX = VA − VC

We obtain:V1 + VX

R1=

VX

R3+

VX

R2

Which is the same equation of the previous page.



+

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

V1=0

V2 − VB

R2=

VB

R3+

VB

R1

IB =VB

R3



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 − VN1

R1=

VN1

R3+

VN1

R2

V1

R1=

1R1

+1

R2+

1R3

VN1

+ +V1

(V2=-BVN)

R1 R2

R3I3

N

V2

Step 1: let’ consider present V1 only (setting V2=0).

+V1

R1 R2

R3I3

N1

VN1 =1

R1+

1R2

+1

R3

−1 V1

R1



+

R1 R2

R3I3

N2

Step 2: let’ consider present V2 only, as it were independent.

V2 − VN2

R2=

VN2

R3+

VN2

R1

V2

R2=

1R1

+1

R2+

1R3

VN2

V2

VN2 =1

R1+

1R2

+1

R3

−1 V2

R2



Step 3: VN is the sum of VN1 and VN2:

VN = VN1 + VN2 =1

R1+

1R2

+1

R3

−1 V1

R1+

1R1

+1

R2+

1R3

−1 V2

R2

+ +V1

(V2=-BVN)

R1 R2

R3I3

N

V2

VN1 =1

R1+

1R2

+1

R3

−1 V1

R1VN2 =

1R1

+1

R2+

1R3

−1 V2

R2

Step 4: V2 is given its dependence:

VN =1

R1+

1R2

+1

R3

−1 V1

R1+

1R1

+1

R2+

1R3

−1 −BVN

R2

VN +1

R1+

1R2

+1

R3

−1 BVN

R2=

1R1

+1

R2+

1R3

−1 V1

R1

1R1

+1

R2+

1R3

−1 1R1

+1

R2+

1R3

+B

R2VN =

1R1

+1

R2+

1R3

−1 V1

R1

1R1

+1

R3+

1 + BR2

VN =V1

R1

VN =1

R1+

1R3

+1 + B

R2

−1 V1

R1


4. THEVENIN THEOREM

ES. R1

R2

R3

R4

VA+

V1VTH

+

-

With respect to the 2 terminals above VTH results in:

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.

VTH =R2R4

R1 + R2 R3 + R4 + R1R2V1

(see next page for evaluation)

≈VTH

RTH

VTH

RTHSetting the voltage generator on the left with the same value VTH just found above we would measure the same voltage of the circuit above with the circuit on the left, independently from RTH as long as the impedance of the measuring instrument has the value of ∞.


………. Thevenin theorem 2

R1

R2

R3

R4

VA+

V1VTH

+

-

Network evaluation:

Starting from the left we encounter the first node VA:

V1 − VA

R1=

VA

R2+

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 − VTH

R3=

VTH

R4

This new eq does not add any new unknown and the 2 eqsabove 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 RL equals ∞ every time RG has a finite value.

VO =RL

RL + RGVG

RL→∞VG



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 value of the impedance RTH. Before doing this all the independent voltage sources must be nulled:

IT

VT

The result is:

VB

V1=0 V

RTH =VTIT

=R4 R2R3 + R1R3 + R1R2

R4 R1 + R2 + R2R3 + R1R3 + R1R2

VTH

RTH

We cane verify the equivalent in our model:

…where we null the voltage generator:

VTH=0

RTH

Obtaining the same result.

(See next page for the solution)



R1

R2

R3

R4

IT

VT

Let’ start from the right:

VB

V1=0 V

IT =VT

R4+

VT − VB

R3

VT − VB

R3=

VB

R2+

VB

R1

From the system above we can evaluate IT as a function of VT, then the expression of the previous page.


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 (see next page):

IMPORTANT: if we apply the Norton Theo to a network to which we have previously applied the Thevenin Theo, or vice versa, it can be shown that:

INO =VA

R3=

R2

R1R2 + R1R3 + R2R3V1

VTh = INoRNo, RTh = RNO

INO RNO

IO

≈


………. 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 IOequals INO if and only if RL is zero if RNOhas a finite value:INO

RNORL

IO

IO =RNO

RL + RNOIG

RL→0IG

Let’s start from the left:

V1 − VA

R1=

VA

R2+

VA

R3

INO =VA

R3

and:


6. A NOTE ON THE TEXT’S COLOUR OF EQUATIONS

40

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.


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: 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



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.



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


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.


………. 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−)

+

-



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 consider 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 VCC

VEE

Is≈0



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 when the OA is un-fed backed.

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≈

𝐕𝐕𝐒𝐒

𝛃𝛃

RS

Is≈0



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


RL

vo



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


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 ≈𝐕𝐕𝐒𝐒

𝛃𝛃



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


RL

vo

𝐕𝐕𝐎𝐎 ≈𝐕𝐕𝐒𝐒

𝛃𝛃


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

vo

R2=

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-


………. the negative feedback circuit examples 2

In obtaining the above result we did not apply any considerations about the input/output impedances of the fed backed OA, nor the direct/indirect transmissions 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 soon see 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 =

R1

R1 + R2

+

-

VA=A(v+-v-)

voRi Ro

vs

R1R2

RL

RS

I-=0



Let’s come back to our original result:

𝐕𝐕𝐎𝐎 =A

1 + Aβ VS ≈𝐕𝐕𝐒𝐒

𝛃𝛃And 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−)

+

-


RS



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



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.



Let’s start… the idea is to disentangle the effect introduced from the input 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



We can solve the network:VA − VOA

RO=

VOA − V−A

R2+

VOA

RL

VOA − V−A

R2=

V−A

R1+

V−A − V+A

Ri

V−A − V+A

Ri=

V+A

R𝑆𝑆

The syntax VxA indicates that we are considering the case VS is 0 and VA is the exciting source.

From the circuit theory we know that the above system is solvable, and the solution will give the unknown variables VOA, V-A and V+A proportional to VA, with coefficients that will depend on the value of the components present in the network.

+

-

VA=VT=generic variablevoA

Ri Ro

R1R2

RL

RS

V-A

V+A



VA − VOA

RO=

VOA − V−A

R2+

VOA

RL

VOA − V−A

R2=

V−A

R1+

V−A − V+A

Ri

V−A − V+A

Ri=

V+A

R𝑆𝑆

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 will be interested in only the value of γ.

+

-

VA=VT=generic variablevoA

Ri Ro

R1R2

RL

RS

V-A

V+A

Namely, we can write:



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+S

RS=

V+S − V−S

RiV+S − V−S

Ri=

V−S

R1+

V−S − VOS

R2V−S − VOS

R2=

VOS

RO+

VOS

RL

Where with VxS is intended voltage nodes evaluated when A=0, namely VA=0.

According to what considered before, the result of the system of equations can be put in the form:

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 interested in only ADIR.



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



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



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, α and γ does not depend on A (ADIR was obtained with A=0, while α and γ with VA=VT) we have:

limA→∞

ADIR−αξ𝛾𝛾

−Aγ1 − Aγ = ADIR−

αξγ



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, as we considered at the beginning of this discussion.

T = Aγ



Considering what we obtained also when we set A=0, we have the 3 following rules to 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−A

VA=

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 we know to happen when A→∞.



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.


vo= 1β

−T1−T vs+

ADIR1−T vs

67

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 of vo all terms mask, 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 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


vo= 1β

−T1−T vs+

ADIR1−T vs

68

………. 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=VT

voRi Ro

R1R2

RL

RS

T =γA

From the circuit on the left:

then:


………. 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.

Note that ADIR represents the direct path of the signal through the network and/or the amplifier. This comes from the fact that both the feedback and the amplifier are not perfect.

vo=ADIRvs

+

-

voRi Ro

vs

R1R2

RL

RS


11. A FURTHER BIT OF HISTORY

70

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:


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 =vo

vs=

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

⇒ΔAf

Af=

11 + γ β A

ΔAA


………. 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. So, 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:

ΔAf

Af=

11 + γ β A

ΔAA


vo − vA

R1−

vA

R3=

vA

R2

vo − vA

R1+ is =

vA

R2

vo

R1= −is +

R1 + R2

R1R2vA

vo

R1= −is −

R1 + R2

R1R2R3is

vo = −R1R2 + R3 R1 + R2

R2is

73

………. 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

= −vA

R3

1) A = ∞ ⇒ 2 V+ ≈ V−,

3)I+ = I− ≈ 0 4 Vo =1β Vs

+

- RLR1

R2

R3

vo

is

v-

vA

i-∼0

Rs



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



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= −

VA

RF

V2+ ≈ V2− = 0

Io =V2− − VA

R1= −

VA

R1

Io = −−IsRF

R1=

RF

R1Is

β =R1

RF

io =vo − v2−

RL=

vo

RL

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



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.



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.



Now the question is: how to figure out the model even from a simple circuit like this one?

Why to try to fit a model in a complicated way, just for verifying which kind of feedback is active?

+

-

vo=A(v+-v-)

R1R2

vs

vo

RL

I-=0

Rs v+

+-vs

vo

βvo

vi

RL

Rβ

?



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 to be as independent as possible from the source and load impedances, thus 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 completely depends on the load impedance: the output variable in this configuration is, therefore, the voltage, or the reading of the output current makes no sense as it depends on the load.

vo =R1 + R2

R1vs

io =vo

RL=

R1 + R2

RLR1vs



+

-

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



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 =vs

Rs

vo = −R1R2 + R3 R1 + R2

R2

vs

Rs



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 =RF

R1Is

vo = RLIo

vo = RLIo = RLRF

R1Is

-

+

RLR1RF

vo

is

V1+

vA

+

-

V1-

V2+

V2-

Rs

Io



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.



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 both 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.

So, when both nodes or no-nodes of the load are floating the current is read from the feedback.


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:

vs

R1=

vo − vs

R2

From which we obtain the same result of the ideal case:

…or:

vo =R1 + R2

R1vs

1β =

R1 + R2

R1

1) A = ∞ ⇒ 2 V+ ≈ V−,

3)I+ = I− ≈ 0 4 Vo =1β Vs


………. 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 with the input source nulled and vAconsidered independent, for finding the loop gain T:

vA − vo

Ro=

vo

RL+

vo − v−

R2

vo − v−

R2=

v−

R1+

v−

Rs + Ri

v+ − v− =Rs

Rs + Riv− − v− = −

Ri

Rs + Riv−

Let’s define: 1R1′

=1

R1+

1Rs + Ri

,

vo − v−

R2=

v−

R1′

vo

R2=

1R1′

+1

R2v− vo =

R1′ + R2

R1′v−

v− =R1′

R1′ + R2vo

namely: R1′ = R1‖ Rs + Ri

From the second eq:

Rs

𝐯𝐯+ =𝐑𝐑𝐬𝐬

𝐑𝐑𝐬𝐬 + 𝐑𝐑𝐢𝐢𝐯𝐯−



vA − vo

Ro=

vo

RL+

vo − v−

R2

vo − v−

R2=

v−

R1+

v−

Ri

v+ − v− = −v−

Now the first eq:

v− =R1′

R1′ + R2vo

vA

Ro=

vo

Ro+

vo

RL+

vo

R2−

1R2

R1′

R1′ + R2vo

vA

Ro=

vo

Ro+

vo

RL+

vo

R21 −

R1′

R1′ + R2

vA

Ro=

vo

Ro+

vo

RL+

1R1′ + R2

vo With the position:

1RL′

=1

RL+

1R1′ + R2

, namely: RL′ = RL‖ R1′ + R2

vA

Ro=

vo

Ro+

vo

RL′vA =

Ro + RL′

RL′vo vo =

RL′

Ro + RL′v𝐴𝐴

And:

v− =R1′

R1′ + R2

RL′

Ro + RL′v𝐴𝐴



TvA

A = v+ − v− = −Ri

Rs + Riv− = −

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



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− − vo

R2=

vo

R0+

vo

RL

1Ro′

=1

R0+

1RL

v−

R2=

vo

R2+

vo

R0′v− =

R2 + R0′

R0′vo vo =

R0′

R2 + R0′v−

From the second:

vs

Rs + Ri=

v−

Rs + Ri+

v−

R1+

v−

R2−

1R2

R0′

R2 + R0′v−

From the first:

vs

Rs + Ri=

v−

Rs + Ri+

v−

R1+

v−

R21 −

R0′

R2 + R0′



vs − v−

Rs + Ri=

v−

R1+

v− − vo

R2

v− − vo

R2=

vo

R0+

vo

RL

vo =R0′

R2 + R0′v− =

R0′

R2 + R0′

R1′

Rs + Ri + R1′v𝑠𝑠

Again, for convenience, we can put:

vs

Rs + Ri=

v−

Rs + Ri+

v−

R1+

v−

R21 −

R0′

R2 + R0′

vs

Rs + 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 =vo

vs=

R0′

R2 + R0′

R1′

Rs + Ri + R1′



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:


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 for the fact that the fed backed network include non ideal terms and the OA itself is not ideal, having 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)


…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 − vB

Ri+

vo − vB

R2=

vB

R1

A vs − vB − vo

Ro=

vo − vB

R2vi = vs − vB

vs

Ri= vB

1R1

+1Ri

+1

R2−

vo

R2

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 + R2

R1vs cvd!


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 the 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.


………. 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 =vs

is

is =v+ − v−

Ri

or:

is =v+ − v−

Ri=

vA

ARi

v-to-v= voltage input and voltage output amplifier

+

-

VA=A(v+-v-)

R1R2

vs vo

RL

Rs

i-

Ri Ro

i+is



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 to node vo or any other node of the network and we can consider it as a fed backed node, just for a while; 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+…



vA − vo

Ro=

vo

RL+

vo − v−

R2

vA

Ro=

1Ro

+1

RL1 +

R2

R1vs +

vs

R1

vA = 1 +Ro

RL1 +

R2

R1+

Ro

R1vs

or:

namely:

vA

vS=

1β′ = 1 +

Ro

RL1 +

R2

R1+

Ro

R1

+

-

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 − vo

Ro=

vo

RL+

vo − vs

R2but v- ≈ vS



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− =vA

A =1A

1β′

−T1 − T

vs =1

Aβ′−γA

1 − Tvs

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:




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 =vs

is=

Ri⁄−γ β′ 1 − T

+

-

VA=A(v+-v-)

R1R2

vs vo

RL

Rs

i-

Ri Ro

i+


Namely:

Riol =Ri⁄−γ β′



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 =vs

is= 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:



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+



102

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:

We see that if Rs → ∞ because we use a current source, than T → 0 (R1’ and RL’ do not include Rs). The direct gain was omitted for brevity:

vo = 1β

−T1 − T vs = 1

β−T

1 − T Rsis

vo = 1β

RiRs + Ri

AR1′R1′ + R2

RL′Ro + RL′

11 − T Rsis

vo Rs→∞

1β

AR1′

R1′ + R2

RL′

Ro + RL′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.

T = −Ri

Rs + Ri

AR1′

R1′ + R2

RL′

Ro + RL′



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+



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=

RsRifo

Rs + 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− =vA

A(i+=0 se A=∞)

Rifp



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



In the above circuit we have that:


+

-

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.

αVA

R1’



In the new equivalent circuit above both Vs and VA’ see now the same impedance and the network assume a symmetry, in someway:


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− 0

Ri=

VA′

Riol0(RS = 0)

We have:

But from the last 2 eqs:

Ri = RiolV+ − V−

VA′= Riol

V+ − V−

αVA= Riol

TαA

Ri = Riol0|V+ − V− 0

VA′= Riol0

|V+ − V− 0

αVA= Riol0

T0

αA

Hence:RiolT = Riol0T0

Pay attention, this node is not Vo!

and also:



Now let’s elaborate:

So that: Rif = Riol 1 − T

=Riol-RiolT

=Riol − Riol0T0

=Rs+Riol0 − Riol0T0

=Rs+Riol0 1 − T0

T =Riol0

RiolT0

RiolT = Riol0To

CVD!

T =Riol0

RS + Riol0T0, but Riol = RS + Riol0, hence:


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


………. 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 (is=(v+-v-)/Ri)

So that:

Rofp =vo

iT=

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:

RofpT=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



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 effective if the loop gain T is large.

is+

-

VA=A(v+-v-)

R1R2

vs vo

RL

Rs

Ri Ro

IT




Just to exercise, now we have that:

iT =vo

Ro+

vo

RL+

vo − v−

R2

vo − v−

R2=

v−

R1+

v−

Ri + Rs

is+

-

R1R2

vs vo

RL

Rs

Ri Ro

IT

Or, simply:vo − v−

R2=

vo

R2 + R1‖ Ri + Rs

We have:

iT =1

Ro+

1RL

+1

R2 + R1‖ Ri + Rsvo

And:

Rool =vo

iT=

1Ro

+1

RL+

1R2 + R1‖ Ri + Rs

−1




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



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 = 1β

A R1‖ Ri + RsR1‖ Ri + Rs + R2

RL′R2′RL′R2′+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



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 + RT =𝑉𝑉𝑇𝑇

IT⟹

∆IT

IT= −

RT

RT + RM

∆RT

RT

RMRT

RM + RT=

VT

𝐼𝐼𝑇𝑇⟹

∆VT

VT= −

RM

RT + RM

∆RT

RT

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.



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

vT

1 − T=

ADIR′′′

1 − TvT

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



A consideration (d): nevertheless…

Therefore:

iT =vT − vo

RL= 1 −

ADIR′′′

1 − TvT

RL

Rofs =vT

iT= RL

1

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

vT

1 − T =ADIR′′′

1 − T vT

th



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



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

IR

Rool=

IA

R2′

If we set RL=∞, then: Rool∞ = RoR2′

|IR RL=∞

Rool∞=

IA

R2′

But, from above:



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:

IR

Rool=

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 IRRL=∞

So that:

IR

IA=

V+ − V−

ρRO

VA=

TA

RO

ρ

|IR RL=∞

IA=

|V+ − V− RL=∞

ρRO

VA=

T∞

ARO

ρ

1Rool

TA

RO

ρ =1

Rool∞

T∞

ARO

ρT

Rool=

T∞

Rool∞

|IR RL=∞

Rool∞=

IA

R2′

1Rool

IR

IA=

1R2′



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∞ =

RL

RL + Rool∞T∞

Versione del: lunedì 9 marzo 2020

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Transcript of Versione del: lunedì 9 marzo 2020