Faithful and unfaithful amplification

[sudar_dhoni]

I have somehow managed to learn unfaithful amplification from this book.

http://books.google.co.in/books?id=CKTS6BzQqm8C&pg=PA226&dq=transistor+amplifier&lr=&ei=uBUtS_L2N6K-lASzxqmtAw&cd=6#v=onepage&q=transistor%20amplifier&f=false

The book has very beautifully explained the amplification in terms of electron movement after which i was amazed having read it. But i can't visualize in a similar manner how faithful amplification takes place in terms of electron movement.
Could anyone explain how faithful amplification takes place i.e how does it occur in terms of electron movement.

 

[Adjuster]

The book you refer to is intended to introduce young people to basic notions of science and technology. It contains humorous references to mammoths and cave-men, to help attract a young person's attention to matter which might otherwise seem boring. It would seem unreasonable to expect to develop a deep understanding of the subjects described just by reading this book.

If you want to learn more about electronics, you will need to get hold of some more serious material, but it will need to be of a kind appropriate to your present level of knowledge. Maybe you are still a student, in which case perhaps a member of staff at your school or college can advise you. It may help people on this forum to advise you if you can give a rough indication of your level of studies.

I don't know what you really mean by unfaithful amplification, although it is usual to talk about high-fidelity amplification, where the output is indeed a faithful copy of the input. Are you really trying to say that you find it possible to understand distorted amplification, but not undistorted, in terms of electron flow? What do you see as the distinction between them?

 

[sudar_dhoni]

The book you refer to is intended to introduce young people to basic notions of science and technology. It contains humorous references to mammoths and cave-men, to help attract a young person's attention to matter which might otherwise seem boring. It would seem unreasonable to expect to develop a deep understanding of the subjects described just by reading this book.

If you want to learn more about electronics, you will need to get hold of some more serious material, but it will need to be of a kind appropriate to your present level of knowledge. Maybe you are still a student, in which case perhaps a member of staff at your school or college can advise you. It may help people on this forum to advise you if you can give a rough indication of your level of studies.

I don't know what you really mean by unfaithful amplification, although it is usual to talk about high-fidelity amplification, where the output is indeed a faithful copy of the input. Are you really trying to say that you find it possible to understand distorted amplification, but not undistorted, in terms of electron flow? What do you see as the distinction between them?

FAITHFUL AMPLIFICATION-The process of raising the strength of a weak signal without any change in its general shape i.e amplifying both the half cycles of a signal is called Faithful amplification.

UNFAITHFUL AMPLIFICATION- The process in which we get an amplified output of the signal with its negative half cycles completely cut off is called unfaithful amplification

Why do you worry about the humorous referrences given in the book.
You just see how it has explained the amplification involved using electron movement which no book has explained . But it explains only unfaithful amplification and not faithful amplification.

 

[sophiecentaur]

The pictures are entertaining and engage ones attention but, as with most Science, the explanation is not really an explanation. The operation of any semiconductor device is based on very sophisticated concepts and, by trying to explain it all in one go, it falls a bit flat.

sudar_dhoni
The operation of the transistor is very non-linear ("faithless" or "low fidelity"). A very simple answer to your initial question is to introduce the idea of 'biasing'. This involves adding an small amount of DC current to the input signal (into the base). When you do this, the transistor is always passing some Collector Current and is never 'off'. The variations of the input (microphone) signal (positive and negative) vary the base current about a mean value and the collector current also varies about a mean value - giving you 'both halves' of the cycle - but not about Zero, of course. This is referred to as Class A operation and involves a lot of wasted power, when dealing with high power amplifiers - it doesn't matter, of course, in low power circuits.

Even when biased, all Junction transistors are, inherently, very non-linear ('faithless') but they have very high gain. This allows negative feedback to be used (in practical circuits) which will eliminate most of the distortions and provide more and more 'faithful' reproduction until you end up with nearly linear amplification: "Hi Fi = High Fidelity = High Faithfullness"

A word of caution. Books like the one you quote are good fun but the ideas they present must be taken with a pinch of salt. Predictions made on such simple models may be very suspect. Move on to something more 'meaty' next.

 

[sudar_dhoni]

1 ) In the faithful amplification (please don't mind the term i am using it) the output is alternating only in magnitude but still flowing in the same direction.
But in the working of a loud speaker, it has been mentioned that the alternating current flows into the coil inside the loudspeaker and the polarity of the coil keeps on changing and as a result the coil is attracted and repelled.
But how can the output collector current be alternating i.e changing its direction .Thats where i am getting stuck and unable to progress.

The operation of the transistor is very non-linear ("faithless" or "low fidelity"). A very simple answer to your initial question is to introduce the idea of 'biasing'. This involves adding an small amount of DC current to the input signal (into the base). When you do this, the transistor is always passing some Collector Current and is never 'off'. The variations of the input (microphone) signal (positive and negative) vary the base current about a mean value and the collector current also varies about a mean value - giving you 'both halves' of the cycle - but not about Zero, of course. This is referred to as Class A operation and involves a lot of wasted power, when dealing with high power amplifiers - it doesn't matter, of course, in low power circuits.

2) I think what you are conveying is that due the action of signal + supply connected to emitter and base the net voltage applied to the EB keeps varying as a result the base current electrons coming out of the base keeps varying which in turn varies the collector electron current coming out of the collector.
i.e in the illustration given in the book where initially during the negative cycle electrons from signal first occupy the holes in base and thereby making the base negatively charged(additional electrons than the neutral level) and preventing electrons from emitter to reach the collector, and during the positive half cycle electrons are withdrawn from the base by the signal and creating holes and thereby causing electrons to move from emitter to collector.

in case when an additional supply is connected what will happen is that the resultant voltage still exists on EB and it remains forward biased . As a result the holes in base diffuse into the emitter and electrons from the valence band of the emitter fill up the holes in the base and these electrons are drawn by the net voltage applied on the EB and when these electrons from base reach the positive terminal of the resultant voltage and equal number of electrons start from the negative terminal and reach the emitter and fill up the holes in the emitter which actually diffused in the emitter from the base. And simultaineously electrons from Vcc would just pass by the emitter and reach the collector since the electrons which had earlier filled the holes had made the base negatively charged and blocking the way for electrons from collector.Now in the base since the blocking electrons have been removed , new holes are created tempts the electrons from Vcc to cross the base and reach collector and go back to the Vcc.
What My doubt is that
if we increase the voltage on EB will the electrons which are present in the valence band of the base also be attracted along with those who filled the holes due to diffusion,
and thereby making the base acquire a net positive charge( note that eventhough base has positive holes when the transistor is unbiased, it is electrically neutral).WHat i am asking is that will additional electrons also be removed from the valence band of the base other than those electrons which filled the holes and thereby making the base deficient of electrons i.e lesser than the neutral level.
If this happens will the depletion region of EB be further reduced and therby decreasing Vbe.
CAN THIS HAPPEN?


3) Also another important doubt is
WHAT IS MEANT BY VCE
Vbe is the barrier voltage and is equal to .6V
I can't understand how will there be a voltage drop for the electrons which travel from emitter to collector via base.Infact they will be accelerated by the electric field caused due to reverse bias . But in my referrence books in many problems Vce is about 10 V when Vcc is 20 V. This is quite silly that how can there be a voltage drop there?

 

[sophiecentaur]

1. It doesn't matter that a continuous DC current happens to be flowing through the loudspeaker. The position of the cone will be offset a bit from its normal rest position but will still vary as the collector current varies and produce both positive and negative air pressure variations according to the audio signal.. (A real design wouldn't do it that way but that is irrelevant to the argument).

2. Except when used as a 'switch' (as in a logic gate, for instance) a transistor is usually biased 'on' all the time. (Not true for some power amplifiers but, again, this is not relevant to normal small-signal circuits).
When a few electrons are taken from the base terminal, the avaiability of carrier electrons in the base (which is very thin) allows many more electrons to flow directly from the emitter to the collector.
I would strongly advise you, if you really want to find about this, to read something which is not quite so superficial as the link you gave. I, frankly, gave up on the pages because I, personally, would rather get the facts, unadorned rather than with distracting 'jolly' pictures of irrelevant stuff. I may be biased (as is the base terminal) because I learned about transistors the conventional way and I realize that they are just not that simple.
otoh, my Dad explained the operation of a thermionic triode valve (when I was about ten), in a way which was along the same lines and which turned out, later, to have been spot on. But the behaviour of free electrons is a lot simpler than the way they behave in the solid state.

It would probably be better to get familiar with the way a transistor operates in an ideal circuit before including the complexities of microphone and loudspeaker operation, imho.

For instance, 3. the 'barrier voltage' doesn't stay exactly at 0.6V and it is not at all obvious, from the article,as you say, why there should be a voltage drop. The fact is that only a certain number of conducting electrons are actually produced 'in' the base region. This means that over a large range of supply voltages, you still get the same current. The collector is referred to as a Constant Current Source, for this reason. If there were, say 10mA flowing and a 10V supply between c and e, the transistor would be just like a 1kohm resistor. Reducing the base current so that there were 5mA would make the transistor behave as if it were a 2kohm resistor. Hence the name TRANsfer resISTOR - how about that?
btw Vce is the voltage between c and e but Vcc refers to supply voltage to the whole circuit - including other components in series with the transistor. In your example, there could be 10V dropped across a resistor load connected between collector and +supply. (Not so silly!)

 

[Adjuster]

Hello again sudar_dhoni,

You raise a fair few objections to other people's ideas of how amplifiers operate. Nevertheless, real amplifiers designed on these principles do work, and no amount of protest will change that. Furthermore, it does not seem reasonable to say:

...in many problems Vce is about 10 V when Vcc is 20 V. This is quite silly...

Until you have more experience of these matters, and perhaps some formal training, it would be better to avoid saying things like this. The principle of defining a dc operating or bias point for an amplifying device, on which the ac signals are superimposed, is a very old and established one. In fact it dates back to the dawn of analogue electronics, somewhere around a hundred years ago.

For small-signal amplification, it is quite usual to bias a device about half-way up the available range of current or voltage. I didn't think this was silly when my lecturer explained it to me, some time in the 1970s, and I don't now.

 

[sudar_dhoni]

1. It doesn't matter that a continuous DC current happens to be flowing through the loudspeaker. The position of the cone will be offset a bit from its normal rest position but will still vary as the collector current varies and produce both positive and negative air pressure variations according to the audio signal.. (A real design wouldn't do it that way but that is irrelevant to the argument).

Could you explain how a loud speaker works with varying dc pulse traveling in the same direction with varying magnitude??


2) Since you are telling that you are strong in vacuum tube triode , Could you explain how the two types of amplification (faithful and unfaithful) takes place in the triode??

Also could you suggest a very good referrence book which explains my doubts i.e it should deal with the electrons movement happening inside the solid state device.Is there any book which explains all these?

btw Vce is the voltage between c and e but Vcc refers to supply voltage to the whole circuit - including other components in series with the transistor. In your example, there could be 10V dropped across a resistor load connected between collector and +supply. (Not so silly!)

Yes 10V also drops in the resistor RL also 10 volts drops as Vce.
VCC = Vce + Ic Rl

whats my doubt is that,
How can there be a voltage drop Vce.
Vbe is a sure event because of barrier voltage but how does Vce drop occur??


[/QUOTE]

 

[sophiecentaur]

Most of these question are arising because you seem determined to use this one, rather bizarre, source for your information. Just look at what Wikipedia has to say about 1. Loudspeakers and 2. Transistors. 3. Elementary Circuit Theory. You need not buy a book to get most of what you need.
You are cherry picking with your questions and you will not gain understanding unless you start somewhere near the beginning of all this. You seem to be trying to pick a fight with something with which you are not familiar enough.

 

[sudar_dhoni]

Most of these question are arising because you seem determined to use this one, rather bizarre, source for your information. Just look at what Wikipedia has to say about 1. Loudspeakers and 2. Transistors. 3. Elementary Circuit Theory. You need not buy a book to get most of what you need.
You are cherry picking with your questions and you will not gain understanding unless you start somewhere near the beginning of all this. You seem to be trying to pick a fight with something with which you are not familiar enough.

Ok
DO you know the answer for my question or not.
If so please type your answer for my question on what is Vce (physical meaning behind it , don't say its the voltage between collector and base as i know that but i don't know how can it be a dropped voltage )??
This is a very direct question, atleast answer this one

 

[sophiecentaur]

Don't get shirty dear boy.
Vce is the voltage that you measure between the c and the e (not surprisingly).
As I said. The transistor let's through a certain amount of current, independent of the Vce and governed by the base current. This current, when passing through the load resistor will cause a voltage drop across this resistor. The remaining voltage will appear across the transistor. Take it or leave it.
A bit of thought plus getting familiar with elementary circuit theory (Kirchoff's laws etc.) would make this understandable. There is no point in having each of your questions answered one at a time so that you can learn the whole of electronics. That job is not mine nor anyone else on a forum.

 

[sudar_dhoni]

Don't get shirty dear boy.
Vce is the voltage that you measure between the c and the e (not surprisingly).
As I said. The transistor let's through a certain amount of current, independent of the Vce and governed by the base current. This current, when passing through the load resistor will cause a voltage drop across this resistor. The remaining voltage will appear across the transistor. Take it or leave it.
A bit of thought plus getting familiar with elementary circuit theory (Kirchoff's laws etc.) would make this understandable. There is no point in having each of your questions answered one at a time so that you can learn the whole of electronics. That job is not mine nor anyone else on a forum.

I just asked in a casual way.Why did you take it up seriously and so intense.Asking doubts is not wrong and moreover i asked only a small thing and not the entire electronics.

What you are saying is that the 20V gets dropped to 10v in a resistor above the collector and the current entering the collector has 10V so the remaining voltage will appear across the transistor.But who will drop the remaning 10V.
If current has to go back to the negative terminal it still has 10v left with it. WHo will drop that 10 V?

 

[sophiecentaur]

I don't know what circuit you refer to but I must assume that there is a load resistor connected between collector and Vcc.
If you are still referring to the 'toy' diagram then count me out. It is a waste of time.

 

[sudar_dhoni]

I don't know what circuit you refer to but I must assume that there is a load resistor connected between collector and Vcc.
If you are still referring to the 'toy' diagram then count me out. It is a waste of time.

I AM not referring to the toy diagram i am referring to this one

What you are saying is that the 20V gets dropped to 10v in a resistor above the collector and the current entering the collector has 10V so the remaining voltage will appear across the transistor.But who will drop the remaning 10V.
If current has to go back to the negative terminal it still has 10v left with it. WHo will drop that 10 V?

 

[sophiecentaur]

On the thumbnail diagram there would be 10V dropped across the RL. That is because of the current which the transistor let's through. (Ohms Law applies for all resistors) the remaining Potential Difference appears across the transistor. IF there were different current flowing then the 20V would be shared differently. As I wrote before, the transistor behaves like a resistor which varies according to the base current. There is no other answer which can satisfy your demand for a 'reason'.
LikecI also said. Learn some basic circuit theory and don't assume that your elementary ideas related to the original link can explain everything to do with transistors. More groundwork is needed.

 

[sudar_dhoni]

the transistor behaves like a resistor which varies according to the base current.

I think you are telling that there is a resistor inside the transistor .
I can understand how the base current varies the collector current.
But with that concept how can there be a resistor inside the transistor.
The base current controls the collector current. I have already explained that in my previous thread

I think what you are conveying is that due the action of signal + supply connected to emitter and base the net voltage applied to the EB keeps varying as a result the base current electrons coming out of the base keeps varying which in turn varies the collector electron current coming out of the collector.
i.e in the illustration given in the book where initially during the negative cycle electrons from signal first occupy the holes in base and thereby making the base negatively charged(additional electrons than the neutral level) and preventing electrons from emitter to reach the collector, and during the positive half cycle electrons are withdrawn from the base by the signal and creating holes and thereby causing electrons to move from emitter to collector.

in case when an additional supply is connected what will happen is that the resultant voltage still exists on EB and it remains forward biased . As a result the holes in base diffuse into the emitter and electrons from the valence band of the emitter fill up the holes in the base and these electrons are drawn by the net voltage applied on the EB and when these electrons from base reach the positive terminal of the resultant voltage and equal number of electrons start from the negative terminal and reach the emitter and fill up the holes in the emitter which actually diffused in the emitter from the base. And simultaineously electrons from Vcc would just pass by the emitter and reach the collector since the electrons which had earlier filled the holes had made the base negatively charged and blocking the way for electrons from collector.Now in the base since the blocking electrons have been removed , new holes are created tempts the electrons from Vcc to cross the base and reach collector and go back to the Vcc.
What My doubt is that
if we increase the voltage on EB will the electrons which are present in the valence band of the base also be attracted along with those who filled the holes due to diffusion,
and thereby making the base acquire a net positive charge( note that eventhough base has positive holes when the transistor is unbiased, it is electrically neutral).WHat i am asking is that will additional electrons also be removed from the valence band of the base other than those electrons which filled the holes and thereby making the base deficient of electrons i.e lesser than the neutral level.
If this happens will the depletion region of EB be further reduced and therby decreasing Vbe.
CAN THIS HAPPEN?

You said that there is a resistor like inside the transistor. Which is the resistor there.
Please don't get angry with me. I have this doubt for a long time

 

[f95toli]

You said that there is a resistor like inside the transistor. Which is the resistor there.
Please don't get angry with me. I have this doubt for a long time

No, he is saying that there is a voltage drop across the transistor, this means that it sort of ACTS as a resistor; but it doesn't mean that there is a "real" resistor inside. You seem to think that we can only have voltage drops across resistors, but that is definitely not the case.

The problem here is that semiconductor components like transistor are quite complicated and in order to understand how they actually work you need to know quite a lot about physics. There is no "easy" way to explain this, in order to understand where to voltage drops REALLy comes from you need to start looking at band diagrams etc and in order to understand what they really represent you need to know quite a bit about solid state physics.

A good (and easier to understand) example of a component which has similar properties to that of a transistor is the diode (which is sort "half" a transistor). Across a typical silicon diode you will always have a forward voltage drop of about 0.7V regardless of how much current you pass through it (whereas in a resistor the voltage drop is obviously proportional to the current). Moreover, the diode will only conduct current in one direction (which is what makes them useful), so the effective "resistance" in the back direction is in theory infinite.

So my suggestion is that you look up the wiki (or some other resource) on diodes, make sure you understand that before you worry about transistors.

 

[sophiecentaur]

That's a lot of writing.
I shall try to address some of the points.
Holes only exist in the p type material. Of course, the charge 'really' flows in the form of electrons - no positive particles actually move. Under forward bias, electrons flow into the p material from the n material in the emitter and electrons flow fom the n type collector into the base. The cb junction is reverse biased but no depletion layer forms because the are many electrons available in the very thin base layer. Vcb is arranged to be high enough to keep these electrons moving up towards the collector. The electrical potential across the transistor is 'used up' in moving these electrons. That is where your 'hidden resistance' occurs. Power is dissipated in the transistor whenever it is conducting. When used as a switch, Vce is made as small as possible by passing a high (relatively) base current.
Vc is always kept higher than Vb so a current will always flow from c to e.
It is possible (I do not know) that the effect you propose can occur if the geometry and doping levels are not optimal. It took some years to get transistor design right!

 

[sudar_dhoni]

That's a lot of writing.
The electrical potential across the transistor is 'used up' in moving these electrons. That is where your 'hidden resistance' occurs. Power is dissipated in the transistor whenever it is conducting.

Thats what my doubt is.You have said there is a hidden resistance which dissipates power from the electrons.What is that hidden resistance .Please explain in the same way in terms of electron movement.
I think now you would have understood my doubt.
The electrons flow from emitter to base and are swept by the large electric field in collector.And during this flow of electrons Power is dissipated in the transistor whenever it is conducting due to hidden resistance.WHAT IS THAT HIDDEN STUFF.It can't be the depletion region as the electrons in base are now like minority carriers and experience no barrier due to the barrier potential. In fact they are swept easily and readily.Then from where did that hidden resistance come which caused a voltage drop called Vce. I think now you would have understood my question.

 

[f95toli]

Thats what my doubt is.You have said there is a hidden resistance which dissipates power from the electrons.What is that hidden resistance .Please explain in the same way in terms of electron movement.

But again, do you understand how a diode works (i.e. where the 0.7V voltage drop comes from)? If not there is no chance it the world that you will understand transistors.
Also, you can't really think about this in terms of "electron movement", that is classical physics and does not work very well for semiconductors. In general we use semi-classical models (which includes elements of quantum mechanics) to describe transistors etc, but then we are normally working in momentum(k) space and not real space.

 

[sophiecentaur]

sudar_dhoni
Why do you assume that no energy is involved in getting the electrons through the transistor?
If energy is lost then you have the equivalent of a resistor - i.e. a finite current for a given Potential Difference.
If you have electrons in a thermionic triode which have been produced by the filament in the cathode the current which hits the anode is far more dependent upon the grid volts than upon the anode volts (over a wide range). In that case there is a PD and there is a Current. The ratio of those two has the same units as Resistance - but there's no actual piece of resistive material there. The power that is dissipated is largely when the fast electrons hit the anode (it will glow red!) but you can measure an actual resistance by relating V and I.
It so happens that, of a metal, the ratio of V to I is constant under constant temperature conditions. You feel familiar with a metal resistor but it is no more fundamental 'resistance' than other conducting devices.
I agree with f95toli. You either have to accept that there is 'an equivalent resistance' of you have to learn some difficult solid state Physics to explain the energy loss mechanism. There ain't a noddy answer to that one.

 

[sudar_dhoni]

sudar_dhoni
Why do you assume that no energy is involved in getting the electrons through the transistor?
If energy is lost then you have the equivalent of a resistor - i.e. a finite current for a given Potential Difference.
If you have electrons in a thermionic triode which have been produced by the filament in the cathode the current which hits the anode is far more dependent upon the grid volts than upon the anode volts (over a wide range). In that case there is a PD and there is a Current. The ratio of those two has the same units as Resistance - but there's no actual piece of resistive material there. The power that is dissipated is largely when the fast electrons hit the anode (it will glow red!) but you can measure an actual resistance by relating V and I.
It so happens that, of a metal, the ratio of V to I is constant under constant temperature conditions. You feel familiar with a metal resistor but it is no more fundamental 'resistance' than other conducting devices.
I agree with f95toli. You either have to accept that there is 'an equivalent resistance' of you have to learn some difficult solid state Physics to explain the energy loss mechanism. There ain't a noddy answer to that one.

OK you are saying that when the electrons flow from emitter to the collector, eventhough it flows easily (i.e attracted by collector) yet it suffers a voltage drop. That voltage drop only is Vce.OK done .But how is that its values are always are quite high like 10V. The electrons actually are attracted then will there be such a high voltage drop when it flows from emitter to collector.

And
Mr f95toli
i know the working on pn diode,that its depletion region is responsible for .7V.
But here i am talking about when the electrons flow from base to collector , they are like minority carriers and they actually suffer no barrier.
The forward biased pn diode which you are talking is no way related to this one.

I don't know why you all are getting angry for this small doubt.
If you feel that i am boring please inform me, i will stop asking more on this topic

 

[sophiecentaur]

Actually, this IS getting boring because you are insisting on an explanation which is prcisely on your own terms. You have no read the posts which were written or you wouldn't have written your last post. Unless you accept that Ohm's law and Kirchoff's laws apply universally in electronics, there is no hope for you. I shall not bother to post any more on this thread.

 

[Adjuster]

Well sudhar_dhoni, here is my parting shot:

I have been watching this thread for a little while since my earlier posts. For a while I assumed you would eventually accept what you were being told. Would it not become obvious that different people were saying essentially the same thing? How could it be credible that people from different backgrounds all shared common delusions about BJT bias and dissipation?

Do you think we all belong to some kind of secret society, perhaps the "CCC" or Cosy Collector Cult? Are we behind an international conspiracy to raise the price of aluminium by making the world buy tonnes of unnecessary heat-sinks? When some of us who actually practice electronics think we have burnt our fingers on an overheating transistor, is this really the result of some kind of mass hypnosis?

 

[Averagesupernova]

When some of us who actually practice electronics think we have burnt our fingers on an overheating transistor, is this really the result of some kind of mass hypnosis?

Of course it is. The cut I got on my hand from jerking it away from the hot transistor quickly and scraping the case of the product I was working on was imaginary too.