Semi conductors(especially transistors).

[Urmi Roy]

Hi everyone,
I've been doing semiconductors lately,and believe me,I'm going absolutely nuts over it!
This is especially when it comes to transistors.

I have worked pretty hard,and I need to clear up some details before I can be confident in this topic.

I'll put forward two questions that have been driving me crazy...

1. In common collector configuration (CC)of NPN transistor,there is a battery (Vce) connecting the collector (C) and emitter (E) which makes the C positive and the E negative due to the polarity applied...and there is another battery (Vbe)connecting the base(B) and E...which forward biases the B-E junction (am I okay till now?)
Now,this is exactly the same as the CE configuration..so the electrons must flow in the same way in this configuration (CC)as in the CE ,and the magnitude of the current must vary in the same way on varying Vec and Vbe as in CE...so where's the difference?

Also,I have noticed that in all three possible configurations,the B-E junction is forward biased and the C-B junction is reverse biased...then what differentiates them?

2. When we say that the 'input voltage' is through the base and the 'out put voltage' is through the Collector in the CE...and this results to voltage amplification...what exactly do we mean?
(As I said,in all configurations,in all three possible configurations,the B-E junction is forward biased and the C-B junction is reverse biased,so what allows to get different output voltage amplifications and different output characteristics in the three configurations?)

These are very basic questions,I know,but I really need to get them cleared!

 

[Urmi Roy]

Well,it's almost been two days since I made this post,and no one's replied yet.Please realize how desperate I am to get this doubt resolved...please please please help!

 

[DrDu]

I also do know little about transistors, but keep in mind that the emitter and the collector are usually very different in size, with the collector being much larger I think. Hence e.g. the resistivity of the E-B junction is much larger than the C-B junction. Also the capacities are dissimilar.

 

[sophiecentaur]

It's good to find out these fundamentals - very useful for the future.
Remember, the transistor doesn't 'know' what circuit it's connected to - it just has volts and currents supplied to it and behaves according to its characteristics. The different circuit configurations are just exploiting that to get the desired result but it is arranged for the BE junction to be forward biased (for 'Class A' operation, at least)

In a CE amplifier, the voltage you 'see' on the collector is due to the current which flows through the load resistor 'at the top'. That current is set by the current in the base times the Beta of the device when the emitter is grounded. As the Base volts go up, the base and collector currents go up. Hence, the PD across the collector load goes up and the value of Vc goes down further relative to the positive supply - it's an inverting amp.
In the CC configuration, there is no voltage gain because there is 100% voltage feedback. AS you increase Vb, you get more current through the device and through the emitter resistor - which increases Ve, which is very useful. When I learned about these things, the circuit was called an emitter follower because the emitter voltage 'follows' the base voltage faithfully. (It's a non-inverting amp which can supply much more current than is supplied to the base, whereas the CE configuration gives an inverting amp - that is also very relevant, at times). The current through an emitter follower depends upon the PD between base and ground (minus 0.7V) and the value of the emitter resistor.
It is common to have resistors in both collector and emitter circuits. The emitter resistor provides voltage feedback (a simple CE amp is very non linear, despite a huge current gain) which reduces the gain because Ve follows Vb and defines the current through the transistor. In this case, for 'ideal' transistors with very high beta values, the small signal voltage gain is given by minus R(collector)/R(emitter). What could be simpler to remember?

BTW, it is best not to talk of "volts into". They are always applied 'across' two terminals or, possibly 'at' or 'on' a point. It's the current that flows "into" places. This is a bit of a hobby horse of mine but it's worth bearing my fad in mind!

 

[Urmi Roy]

Thankyou sooooooo much sophiecentaur for your reply,after waiting eagerly for so long!Please hang on with me till I get my fundamentals clear...even my teacher doesn't have her fundamentals clear!

I read through your post...and there are some places I don't understand..please clarify these before I proceed any further...

Remember, the transistor doesn't 'know' what circuit it's connected to - it just has volts and currents supplied to it and behaves according to its characteristics.

See,I don't understand,in the first place,how we feed a current into the transistor...for example,in the CE configuration,the base current is the 'input current'...but it's flowing out of the transistor...so how can we input the current through the base?

Also,what do we mean by the output of the transistor...is it a voltage?...how do we use this output?

In a CE amplifier, the voltage you 'see' on the collector is due to the current which flows through the load resistor 'at the top'. ...and the value of the emitter resistor.

Now,apparently the output is the 'voltage' across the collector resistance...but I thought that for a fixed value of the collector load resistance,the output voltage should increase whenever we increase the collector current (like when we increase Vbe)..as per Ohm's law.

It is common to have resistors in both collector and emitter circuits. The emitter resistor ...What could be simpler to remember?

NOw,somehow,I have a feeling that the 'grounding' of the common terminal in a particular configuration is very important,in the sense that grounding doesn't allow the potential of the grounded terminal to change (from zero potential)...for example,in the CC configuration (NPN transistor),there is a battery between the base and collector,with its -ve on the collector and its +ve on the base...but the base-collector is supposed to be reverse biased...so I suppose that since the collector is grounded,the negative potential applied to it by the battery doesn't really affect the reverse bias...instead it just has the effect of incresing the +ve potential of the base...is this right?

 

[sophiecentaur]

I sympathise with all of this; I was there too (about a hundred years ago, it feels)!

"a transistor doesn't 'know'..."
What I mean is that is it just like any other component (i.e. v. dumb). A resistor will let a certain amount of current through it according to the voltage and its value of 'resistance', as we call it. Whatever circuit you put it in, it will behave in the same way. The circuit makes use of these properties.

imho, the best way to start to look at transistors is in the simplest way you can get away with. In about 1961, someone gave me a diagram of a transistor equivalent circuit and I really couldn't make any sense of it at all. Once I was given some very simple working rules - like the ones above - I was able to build simple circuits which actually did what I expected (result!).

The easiest way to think of the operation of a transistor (NPN, in this case) is that, once you put the collector volts a bit higher than the emitter volts and you forward bias the base / emitter junction, the current that will flow into the collector will be Beta times the current into the base. This collector current is more or less independent of the actual voltage that's applied across CE; we call the collector a current source, for that reason. That's all you need to say about the transistor itself. (At the simplest possible level).
To make an actual amplifier, you need some resistors in the circuit. These will set up the conditions for the transistor to work in.

"Now,apparently the output is the 'voltage' across the collector resistance...but I thought that for a fixed value of the collector load resistance,the output voltage should increase whenever we increase the collector current (like when we increase Vbe)..as per Ohm's law."

The collector voltage is set by the current through the collector load resistor. Remember the collector will let a certain amount of current through it, irrespective of the actual potential of the collector. You could say that a varying current produces a varying voltage across the resistor. If you are using a simple oscilloscope, it is the actual voltage that you will see and that's what you would call the output. Of course, because the voltage across the resistor increases with increasing collector current, the actual Volts (relative to Earth) on that point will go Down - hence the idea of the inverting amplifier.

",there is a battery between the base and collector,with its -ve on the collector and its +ve on the base...but the base-collector is supposed to be reverse biased"
No, that's the wrong way round - the collector is always kept positive wrt the base; your battery should be the other way round. But you wouldn't connect a battery directly that way, in any case - you would use a resistor, possibly, to help define the base current.
Look at the diagrams here
http://en.wikipedia.org/wiki/Common_emitter
The 'complete amplifier diagram shows you how you bias the base, how the volts on the emitter vary and how the resulting current affects the collector current and, hence, the voltage which turns up at the collector. You see it depends turns out as V_e R_c/R_e
and V_e follows V_b
This amplifier is described as a Voltage Amplifier.

 

[Urmi Roy]

I'll probably take some time to absorb all this...please bear with me...
by the way,it would be really nice if you said something about my first doubt...without clarifying it,it's difficult for me to imagine things...
"I don't understand,in the first place,how we feed a current into the transistor...for example,in the CE configuration,the base current is the 'input current'...but it's flowing out of the transistor...so how can we input the current through the base?"

Also,is my idea about the 'grounding' business completely wrong?

Thanks,again.

 

[sophiecentaur]

Current flows from the positive bias on the base through to the emitter, which is 0.7V lower in volts.
A transistor doesn't know, explicitly about "ground". How can it? You can connect just one of its terminals to a fixed voltage - it could be Earth, +6V, -9V or anything, To make it operate you need to ensure that Vc is positive wrt Ve and to make Vb appropriately higher by appropriate resistors etc.. Under those conditions it will behave like a transistor - letting current go through (c to e), according to the base current value and its 'current gain'..
What was it you wanted me to "clarify"? I'll try.

 

[Urmi Roy]

Hi sophiecentaur,I'm reading through your posts over and over again...I think I'm slowly getting it...however,in the mean time,if youcould just elaborate on certain points...(sorry to bother you like this..I'll try to get through this as quickly as possible)...

1.The basic biasing of the different junctions (forward bias on the emitter-base and reverse bias on the collector base junction) seem to be the same in all configurations,but,as you said,we obtain the required results by just changing the locations of the resistors,and the batteries...now what I don't understand is as to why we have to change the locations of the batteries (I mean connecting a battery betweenC -E in one config.,and then connecting a battery between C-B in another) (since the batteries are only responsible for the biasing,which is the same for all the congigurations)when changing configuration...since if we only change the locations of the resistors(which are what give us the output),we would get what we want.

2.When we input the current through different terminals in the different configurations,though the biasing is essentially the same in all configurations,the input current seems to undergo different situations in each config.,leading to different output characteristics...how do we explain this?

3. In a CE amplifier, the voltage we 'see' on the collector is due to the current which flows through the load resistor and as the Base volts go up, the base and collector currents go up. Hence, the PD across the collector load goes up and the value of Vc goes down further relative to the positive supply - it's an inverting amp...I've understood this.

What I need to clarify is,even if the net potential of the battery Vce goes down,w.r.t. ground,does it still have the same effect on the biasing of the C-B junction?

Also,in this case,the net potential drop across the collector resistor is reduced...how does this help us in practical purposes?

Please don't get fed up with me! I promise I'm trying hard to understand this..I'll do it as quickly as possible.

 

[sophiecentaur]

Hi
1. You don't have to connect a battery directly to a transistor anywhere - see the complete amplifier diagram in that Wiki link.
2. The input is made to vary in all configurations and will always produce the same current flow (assuming Vce is sufficient).
3. You don't "bias the cb junction". You bias the be junction. If the battery volts go down then you have just changed the operating conditions so I'm not sure what you mean. IF the battery volts drop as result of taking more current (a bad battery), this is the equivalent of having a higher value of Rc - so the overall (small signal) voltage gain will increase. If you just use a lower voltage battery, the small signal gain will stay the same BUT the available range of possible voltages that the collector can attain (voltage swing between transistor 'off' and transistor 'hard on' conditions) will reduce. This is very relevant at times - if you need the circuit to handle 'big signals'. (I have hearing aids with lots of gain but working off a single cell and it's output power is V. low "What? speak up I'm not deaf young man!" howl squeak) Power Amplifiers use higher voltage supplies for this reason. If you try to design an amplifier to work at 12V (automotive applications), you need to have transistors which will handle large currents to make up for it. But that is another story.
I do believe we're getting somewhere.

 

[Urmi Roy]

You won't believe it...I think I'm getting things straight!

Please confirm if my conception is correct in the following statements..(it'll do if you just say 'yes' wherever you think I'm okay,and if you just tell me where I'm wrong in the rest.)
1. Due to grounding, the current in the common terminal is taken to the earth,so we don't need to consider that current (like in CB,the base current isn't analysed,since it is taken to the ground).

2.The net potential across the output (load) resistor is what determines the output voltage.
(in CB,the net potential across the collector decreases from the original value of Vbc applied to it ,also the output current through the collector is less than the input current through the emmitter…whereas in CE,the output current through the collector is larger than the input,and the net potential across the collector decreases from the original value of Vce applied to it)

3.As I said before,the net voltage across the output/load resistor is what we call as the output voltage...in order to use this net potential,do we connect a resistor in parallel to the load resitor,so that we get the same potential difference across this parallel...which we could perhaps use for some practical application.

4.In CC, as we increase Vcb,the input current (Ib) has to decrease due to increase in the thickness of the depletion layer in the C-B junction.(input characteristics of the CC)

5.In CC, the output current Ie attains a constant value after reaching a certain Vce (for fixed Ib),as the forward bias across the B-E junction cannot be increased any further by increasing Vce.

Just three more questions to get things clearer..

1. Why,in CC configuration,is the voltage gain is less than one (it says so in my book...is this wrong?)?

2.You said in your previous post that 'You don't "bias the cb junction". You bias the be junction.'..but I thought we change the reverse bias of the C-B by changing the Vcb (like when obtaining the input characteristics of the CC configuration)

3.In CC or CE,increasing the Vce increases the forward bias across E-B,so automatically,the Ib increases...but then the input current itself changes! Are we really supposed to allow the input current to change?

 

[sophiecentaur]

Hi
I have a feeling that you are browsing through a number of sources of info about transistors and picking up on several statements which are getting you worried. Once you have picked up the basic principles of operation to a good enough level then the next layer of complexity will still be there for you to tackle. I am just trying to cut through the hard stuff and give you a bluffer's guide to making transistors work for you. In 1968, Vic Devereux (a really clever bloke who knew more than I could ever hope to know) gave me these basics one afternoon - scribbled on the back of an envelope. They have worked for me 99% of the time. Operating at high powers, high frequencies, high linearity requirements and with very low level signals, things get more complicated and the simple rules aren't good enough.
Your questions:
btw, the CB configuration is the hardest one to consider, I reckon, and is probably better to tackle once you have the two others sorted in your head. Also, there are more circuits which involve NOT connecting a terminal to any supply voltage or ground. CB,CC,CE are just a start.
1. How can it not matter? Its value is always relevant to the whole operation of the device. It could matter a lot if it was a high value and you needed to factor that into your power supply choice. The base current in CB stages is not "not considered" so much as 'neglected' because it is so small. If there were none then the transistor would not be operating, would it? So it must count to some extent. Electronic Engineering is full of approximations and this is just one example - like the assumption the Vbe is 0.65V all the time. It can't be exactly constant, can it?, because of the diode characteristic, but we assume it in most cases.

2 and 3.(CE config.) It is true that, if you want to pass the output signal on to anything other than an 'ideal' measuring device, which takes no power, then the resistance of what follows affects the voltage gain. One way of looking at your "output signal' problem for a CE amp could be to look upon the transistor as a variable resistor (The name transistor comes from 'transfer resistor', I believe). Along with the collector load resistor (and any other resistors what may be there) you have,effectively, a potential divider with a variable resistor at the bottom and with a value which varies instantaneously with the input signal. The collector volts will be set by the ratio of the resistors. There may, in fact, be no collector load, explicitly; the collector being connected directly to the input of another transistor stage.

4. Is this relevant in a first-level appreciation of what goes on? Look at the data sheets of some transistors (the net is full of them). Whatever happens to the depletion layer, there is a lot of feedback at work - enough to bring the voltage gain to <1, even.

5. Again, is this relevant when you realize that Ie will follows a value which brings Ve just less than Vb. The higher the gain of the device, the nearer it follows - that's how feedback works.

1a. The gain of a CC stage will be slightly less than unity because the effective base junction input resistance is finite and Vbe varies a bit with the base current.

2a. If you really want to insist that the cb junction is reverse biased the I can't argue with you; you are correct, in principle but we don't set up the collector base bias in our designs. We set up the base emitter bias because that's what makes the device operate in a way we want. A transistor can operate pretty much the same, in an amplifier, over a huge range of Vce values.

3a An easy one! A transistor is, inherently non-linear and effects like this will affect it characteristic. We either accept the non-linearity or include feedback to reduce the effects to an acceptable degree. This can involve using extra stages of gain, in order to get the required overall gain an yet include feedback.

That's a lot to take on board - good luck with it!

 

[Urmi Roy]

Hi,

I've been reading through your posts and I sort of think I'm getting it. Yesterday,I tried to think for myself,according to the basic principles that you mentioned,what exactly goes on inside each of the configurations,and tried to predict what the input and output characterisitics could be...I suppose was right in atleast some cases.

The thing is,even though my first priority is in understanding my basics,I've got an exam on Monday,and I've got to understand the other stuff for it too.

Anyway,coming back to the main issue,

I wasn't really saying that Ib doesn't matter in CB,actually I was asking that since it is not used in either the output or the input,it is taken to the ground...ofcourse,I realize that Ib has to be there,else the transistor wouldn' run!I hope that's okay.

Also,please tell me if I'm right in saying that the net potential across the load resistance is what we consider as output voltage.
I was thinking,that to put this voltage into use,we would connect something in parallel to it,and we would thus get the same potential drop across this parallel device.

Personally,I found the CB configuration easier to understand,since initially I was confused about how we could connect a battery directly between C and E in the CE and CC configurations...but that is not there in the CB...am I wrong somewhere?



Again,as I said,since,I tried to figure out what is happening in each of the configurations on my own,I analysed and came up with the following.-
1.in CB,the net potential across the collector decreases from the original value of Vbc applied to it ,also the output current through the collector is less than the input current through the emmitter

2.whereas in CE,the output current through the collector is larger than the input,and the net potential across the collector decreases from the original value of Vce applied to it)

3. In CC,the net potential across the emitter load is also less than the applied Vec...so the out put voltage will again be less than the applied voltage to the emitter.

Do you think I have improved?

 

[sophiecentaur]

A lot of what you say is 'a matter of how you look at it". Are you getting your cause and effect mixed up?
When you say that Ib "isn't used" in the CB configuration, I don't know what you mean. It is a factor in the operation of the device and that is all. Kirchhoff tells you that Ib +Ie = Ic, so, although it is only a small fraction, Ib is still relevant. In CB amplifiers, the 'input current' is, unusually, almost the same as the 'output current' - there is (virtually) no current amplification at all - the difference is all in the 'voltage swings' at the c and e. The input resistance is very low (low voltage swing for a given input current) and the output impedance is high (higher voltage swing for the same current). Whatever current goes into the emitter, the collector 'forces' the same amount of current through a high resistance load, by the fact that the Vc ends up at a suitable value - producing power amplification.
Also, when you say that the "net potential across the collector" is the output, that is very loose terminology. C is one point so you must mean the 'potential at C(?) - relative to ground (implied).
I would argue that you don't "apply" a potential across ce. The potential there depends upon the current that the transistor is allowing to pass through it. The collector is, essentially, a current source and the voltage that appears at c depends upon that current and the resistance of the network connected to it. That is not the same as what you are implying.
3. This doesn't make sense at all. The Potential across the emtter load PLUS the Vce equal the supply volts. Vce can be anything form near zero to near supply volts - depending whether the transistor is passing its maximum current or zero current. (Kirchhoff 2 and potential dividers).
"Personally,I found the CB configuration easier to understand,since initially I was confused about how we could connect a battery directly between C and E in the CE and CC configurations...but that is not there in the CB...am I wrong somewhere?
"
Here you go again - you don't normally "apply a potential" between c and e. That potential appears as a result of the currents flowing in the associated resistors.
I do agree that you could connect a transistor between +12V and Earth and vary the current through it by controlling the base current but there would be no "output voltage signal" because all the current would be flowing from supply directly to earth. That is not the case in either CE or CC amplifiers.
It is possible to connect a transformer as a load - in either emitter or collector -, of course, in which case, the DC conditions and the AC conditions are different and the analysis just gets harder. We don't need to go there yet, though! And, at a pinch, I suppose you could control the DC through a solenoid in series with c or e but that doesn't really help with the basic problem of understanding simple, basic circuits, does it?
Keep going!

 

[Urmi Roy]

...the difference is all in the 'voltage swings' at the c and e. The input resistance is very low (low voltage swing for a given input current) and the output impedance is high (higher voltage swing for the same current).

I didn't really understand about the 'voltage swings.'Does the operator make the input resistance very low and output resistance very high on purpose?

The collector voltage is set by the current through the collector load resistor.

(your second post)
I looked up the diagram on wikipedia,and I noticed that Vout was at a single point at one end of the collector resistance...does that mean that whatever be the potential at that point,will be used as output?
Now,again at the root level (since I think I have some major misconception about this 'output voltage' thing),...when we say that the Vout is an amplified voltage...amplified w.r.t to what?
What role does the battery (or simply positive potential applied)...V+ in the wiki article...connected to the collector play?Does it have no bearing on the output?

Whatever current goes into the emitter, the collector 'forces' the same amount of current through a high resistance load, by the fact that the Vc ends up at a suitable value - producing power amplification.
I would argue that you don't "apply" a potential across ce. The potential there depends upon the current that the transistor is allowing to pass through it. The collector is, essentially, a current source and the voltage that appears at c depends upon that current and the resistance of the network connected to it.

Well,what could be getting me confused is that in my book,it keeps on saying that we adjust the output characteristics of CE by holding Vbe constant,and by varying Vce (thus getting different Ic)...to me,it seems that varying Vce means varying the potential of the battery(or positive potential V+ in the wiki article you suggested)put across the emitter and collector...perhaps this is the source of confusion?

Sorry if I'm messing up a bit...I'm working hard,I promise!

 

[sophiecentaur]

1. If you take resistance as being the ratio of V to I, then it is the transistor itself that is exhibiting a low resistance at the input of a CB amp and high resistance at the output. The operator will 'see' these resistances by the voltages and currents measured when the device is in the circuit.
2. Inputs and Outputs: These are only 'names' and just refer to the signals (volts or currents) measured at some chosen points on the circuit. When you 'impress' a voltage, say, at a certain point, you call this an input and when you measure (or pass on) a voltage at another point, this would be called an output. The same goes when your 'signals' happen to be currents. In a CE amp, we 'say' that the output is the potential at the collector but, if you were to put a resistor in the emitter, as well, then this voltage could also be regarded as an output. In fact, with equal values of Rc and Re, the two output voltage swings are identical but in antiphase (Ve goes down when Vc goes up)
3. The voltage SWING - the AC bit is amplified wrt the swing of the base current - which will be affected by the voltage swing, normally at the other end of a resistor in series with the base (if we are making a voltage amplifier).
4.V+ is there to make current flow through the circuit. Nothing at all would happen without it. the transistor, by control of its base current, allows varying current to flow into the collector and this changes the collector volts (as I said before - it's part of a potential divider).
5. I think I have identified another of our problems here. You seem to be referring here to the situation when you are measuring the characteristics of a transistor - not using it in an amplifying circuit. (See my previous comments about "holding Vce at a certain value and how this is not what you normally do in an amplifier) Yes, if you want to produce a set of characteristic curves then you would do just what you are saying - the device is just 'being itself' in that situation, just as it is when used as a normal amplifier. I think you need to regard the two situations differently. Transistors were only developed as much as they have been in order to use them as amplifiers - not just to measure their characteristics. The characteristics are relevant, of course, in the design of amplifiers. If you look further into how these curves are used in detailed amplifier design you will see that you can use these 'families' of curves to determining how output signals will result from input signals - but you don't necessarily 'stick' to one curve when operating in a real circuit. My point has been that many / most transistors these days have such good 'performance' that their detailed characteristics are not particularly relevant at a first level of use. Unless you can understand where the transistor 'fits' in a circuit and what it is doing, in essence, then I don't think you will appreciate the consequences of its detailed performance.
Playing a musical instrument is a different matter from designing and building it. If you have no idea about what is required by the player then there is little chance of understanding the nuances of bridge height, string weight and fingerboard clearance etc.. It could be said that I'm looking at the problem from the player's standpoint.
Don't appologise for trying to get something right!

 

[Urmi Roy]

1. If you take resistance as being the ratio of V to I, then it is the transistor itself that is exhibiting a low resistance ...

Actually,in my book,it specifies the input and output resistances used in the different configurations,so I was wondering if the user always has to stick to these values.

It seems that the V+ in the wiki article is for the biasing of the CB junction (while measuring the characterisitics of the transistor)...but you were saying something about the voltage at the collector being determined by the Ic only...so can I consider the net output at the collector as the algebraic sum of the V+ and the potential drop across the collector resistor due to Ic?

3. The voltage SWING - the AC bit is amplified wrt the swing of the base current - which will be affected by the voltage swing, normally at the other end of a resistor in series with the base (if we are making a voltage amplifier).

Perhaps it'd be better if I stick to d.c input/output right now...especially since that's what I have in the syllabus,and I find it easier to undersatnd d.c.

Again, when we say the output voltage across the collector is amplified,what is it amplified with respect to?
(As you pointed out,I am referring to measuring the characteristics of transistor..since that's what we have in our syllabus at present.)

 

[sophiecentaur]

The fact that most of your comments refer to measuring characteristics makes things a lot clearer now.
1. "Output resistances used" Are you referring to the resistors which you put into the circuit or are you referring to the effective resistance 'looking into' the transistor? You obviously have a choice what you connect externally (depending on what AC and DC levels you need and the signal gain you require but the transistor does what it wants to, according to the conditions you give it so its effective resistance depends on these conditions.

2. If you are considering DC, the concept of 'gain' can only be applied to the current gain. That has probably been a source of confusion! No wonder you ask the question "Voltage gain wrt what?" For DC, all you can say about the voltages would be to plot a graph of volts applied to the input series resistor (for instance) and the voltage measured at the collector. The slope of this graph would then give you the AC voltage gain - 'swing out' vs 'swing in'.
In the Wiki article they refer to "delta V", afair, this refers to signal variations about the static condition.

I still have a bit of a problem with knowing what you (or your course) are after. The transistor characteristics can be measured in a circuit but I can't see why the measurement circuit would be classified in terms of an amplifier format (CC,CE,CB). Hence my wondering at the idea of connecting a battery across ce - something you don't do for an amplifier because there's always a load of some sort for an amplifier to drive.
It makes me wonder whether your tutor has ever designed or built a transistor amplifier (?). Could be an embarrassing question I feel that it would have been useful if the transistor had been put in the context of its actual use. That Wiki article attempts to do this.

 

[Urmi Roy]

1. "Output resistances used" ...effective resistance depends on these conditions.

Actually,we have a table at the end of the chapter,comparing CB,CE,CC...and it compares the 'input' and 'output' resistances used in each,as a property,as if the values in the table are staple for the configs.

These are only 'names' and just refer to the signals (volts or currents) measured at some chosen points on the circuit.

So that means because we use Vbe to change the value of Ib in CE config,we could call that input voltage?

2. If you are considering DC, ...

Does that mean what I understood about the role of Vce in adjusting the output characteristics was okay?

However,the fact remains that the output of dc in a transistor is taken in terms of voltage,so I guess the concept of 'net' (of Vce and the potential drop due to Ic)potential across the collector resistance remais valid?

I still have a bit of a problem with knowing what you (or your course) are after...

Actually,we're at a very preliminary level...we're just trying to understand what goes on inside a transistor...and so we mainly have to understand and know about the input and output characteristics of a transistors in all the configurations...all in dc.

 

[sophiecentaur]

You still haven't made it clear. Are these 'resistances' that have been put into the circuit or are they the measured (internal) resistances that the device displays? If they are resistances "used" in / soldered into the circuit then they are hardly properties of the device. Do you see the distinction?

You seem to be centering on the significance of the terms 'input' and 'output'. The only 'output' that you can ascribe to a transistor (on its own) will be the current which it allows to pass through it due to the current which you put into its base and the overall biasing from the supplies. Without some external components there can be no 'output voltage' because you are defining its terminal voltages by connecting to 'voltage sources' i.e. constant voltages.
The actual parameter which you control in a transistor is the current it passes.

If your course is simply dealing with how the conditions presented at the terminals affect its currents then it really is not valid to talk of inputs and outputs. The potentials on each of the three terminals are all 'inputs' and the currents that flow are all 'outputs' until you start to include the device in a real circuit and select which of the currents and voltages associated with it are 'cause and effect'.
As you clearly want to get to grips with this topic you will need to make a distinction between characteristics and uses of transistors. It seems that, at this stage, all you really need to do is to look at the characteristic curves of the relationships between the various parameters and relate them to whatever theory /maths that your course provides you with.
When you come to consider the transistor as used in an amplifier then these relationships start to have some practical relevance and 'inputs and outputs' become meaningful concepts - else I should just stick to using terms like Vc, Ie etc.

"However,the fact remains that the output of dc in a transistor is taken in terms of voltage,"
I would argue strongly with that assumption. A transistor is, essentially, a current operating device. It can still work as a transistor between a constant value of Vce - as has been said many times before but, according to that statement, it would have no 'Voltage output', in as far as it couldn't alter the applied voltage (the power supply would win). The only reason of an 'output' to be presented as a varying voltage will be when a load of some resistance value is added. Then the 'output voltage' is not from the transistor but from the amplifier as a whole and is there because of the current (Ic) that is passing through the resistor.

The wording in most textbooks is fairly precise. I suggest you look in detail at the terminology used and I think you will get what I mean.

 

[Urmi Roy]

You still haven't made it clear. Are these 'resistances' that have been put into the circuit or are they the measured (internal) resistances that the device displays?

The thing is,I don't really know, myself! As I said,its just a table,with no explanations,and even in the book,the authors have not stated what the resistances are.

But I guess the resistance of the transistor,meaning the resistance of its terminals wouldn't really be considered...after all, in an ideal transistor,the resistance of the B-E junction is supposed to be zero,and that of the C-B is supposed to be infinite...and since we're just at the basics,everything is treated as 'ideal'...right?

The actual parameter which you control in a transistor is the current it passes.

I get that now(especially since its all dc.).

Also,could you look through the following description of 'what goes on in a CC configuration',as far as I've figured it out...as before,I'm dealing with dc. and a NPN transistor...I'm basically trying to explain the input and output characteristics of the CC.

Here goes...
Input Characteristics:
(Vce constant,Ib varied by varying Vcb)
For Vec=0 and Vbc=0,no current flows,due to the absence of any biasing.

For Vce=0,and gradually increasing Vcb,the reverse bias across the C-B is increased,whereas the B-E junction is not biased at all (Vce=0),so no current should flow for any value of Vbc applied.(because if B-E is not forward biased,no current can flow at all)

For Vce >0 (at a certain fixed value),increasing Vbc, current has started to flow,since the B-E is now forward biased,,but increasing Vbc brings in the early effect, Ib reduces in value.

For greater values of Vce,the corresponding values of Ib shoud be greater.(because greater fixed vale of Vce means a greater forward bias across the B-E,so emitter current increases and Ib correspondingly increases)

Output Characteristics:
(Vbc is kept constant,Vce is varied,)

For Vbc=0 and Vce =0,again, no current flows due to absence of biasing.

For Vbc=0,and increasing Vce, the forward biasing of the E-B is increased (due to increase of Vce),whereas the C-B junction is not biased (Vbc=0)...so current Ie flows similar to that in a forward biased diode.

For Vbc>0,(at a certain fixed value),Increasing Vce increases the current Ie,due to increase in forward bias of E-B junction,the early effect however increases Ic,and reduces Ib.

 

[sophiecentaur]

Yes. mostly. but
"For greater values of Vce,the corresponding values of Ib shoud be greater.(because greater fixed vale of Vce means a greater forward bias across the B-E,so emitter current increases and Ib correspondingly increases)"
That depends upon what and how you are biasing the base (your actual test circuit). At this point the transistor is starting to behave as a transistor and you would need to specify what base bias you were applying - it is an 'input' of some sort so Ib would be an independent variable - definitely not dependent on emitter current - the other way round, surely?
I realize that you have been told to understand this but, if you look at the curves of a 'real' transistor, rather than what the Wikkers article shows (in order to show the Early Voltage), the effect of Vce on Ic is very small (the slope of the curves) the effect of the base current is the dominating one (which is why the devices are described as having 'gain', in fact). The lack of Early effect is very marked for low values of base current. Note that the curves are normally shown for pre-determined (impressed) values of Ib so Ib is independent of the transistor. That's my reason for objecting to your statement at the top.
Look at the current gain of the 'typical' transistor in this link: http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/active/BiPolar/bpcur.html
see what I mean? Early is not very relevant in practice because you are usually operating above the knee. (rude engineering jokes would normally come in here)
But I sympathise; someone has told you to learn this so you've got to. Just remember to distinguish between your independent variables and dependent variables (better terms to use here than input and output, I could suggest).

 

[Urmi Roy]

At this point the transistor is starting to behave as a transistor and you would need to specify what base bias you were applying - it is an 'input' of some sort so Ib would be an independent variable - definitely not dependent on emitter current - the other way round, surely?

I'm not sure what you mean by 'specifying the bias',since Vce acts to forward bias the B-E junction always.

Also,by saying that Ib would be an independant variable...do you mean that we should study the variation of Ie w.r.t Ib and not the other way around?

I realize that you have been told to understand this but, ...transistor. That's my reason for objecting to your statement at the top.

Thanks for pointing this out, I should really keep this in mind.

By the way,I was looking through my explanation in the morning,and comparing with the graphs in my book...there's something wrong with-

1.For Vbc>0,(at a certain fixed value),Increasing Vce increases the current Ie,due to increase in forward bias of E-B junction.

(I found that in my book,the value of Ie,with increasing Vce is a flat line,parallel to the horizontal),so it doesn't change at all with Vce...what's wrong with my explanation here?

2.Also, in regard to "For Vce >0 (at a certain fixed value),increasing Vbc, current has started to flow,since the B-E is now forward biased,,but increasing Vbc brings in the early effect, Ib reduces in value"

If,as you said the influence of the early effect is negligible,how do I explain the decrease of Ib with increasing Vbc (for a fixed Vce)?

(in my book,Ib decreases along a straight line with increase in Vbc.

3.In CC,looking at it in a general view,we could change either Vce or Vcb to change Ib...since the Vce influences the emitter current,and hence the base current...at the same time,Vcb changes the reverse bias of C-B,thereby decreasing Ib.

Is that right?

 

[sophiecentaur]

Hi
Do you know, it keeps looking as if that you are getting confused by the apparent contradictions between the transistor characteristics and their measurement, on the one hand, and the operation of a transistor amplifier, on the other. Once you've spotted which is which, I think you won't need to ask a lot of these questions.
1. You would need to specify the circuit used for the measurement, in particular what is in the base circuit. That could make more difference to Ib than the Early effect could do. I don't know much about this - forgot most of it as soon as I learned it, unlike most of the useful stuff which will be with me till the old folks' home and I'll bore all the nurses with it. In this question are you just considering what goes on below the knee of the Vce Ic graph?.

"in my book,the value of Ie,with increasing Vce is a flat line,parallel to the horizontal)" That would be describing an 'ideal current source, which is what we start off by assuming and which is what I have been batting on about. In that case, the family of curves, for different Ib values would ba a set of horizontal lines and the current gain would be Ic/Ib always.

Incidentally, have you seen any graphs like Fig 2 in here?
http://users.monash.edu.au/~kaminski/ele2262/lab21/lab21.htm
Note that, in the diagram, the transistor is operated away from the knee(saturation) - which would introduce hideous distortion.
Beware of the typo in the equation just below the diagram. It has missed out the R_L in the Kirchoff 2 calculation!
You really can't believe anything you read on the Internet, can you?
It's the basics of designing a CE amplifier. It shows a 'load line', which corresponds to the volts at the collector which result from Ic passing through the load resistor. Ohm's law on R means the voltage dropped across R will increase proportionally with the Ic and the remaning volt (Vc) follow a diagonal straight line. Once you have chosen your value for R, the line is defined and the output voltage signal (of the amplifier!) follows that line as the base current is varied..
2. Again, what is the measurement circuit?
3. Are you sure that you have been told that this particular aspect of a transistor is really crucial? In a real circuit, the current flowing Into the base from a preceding circuit is set by that circuit - either a high value of resistor oar a constant current source, like the collector of the preceding transistor. I can't recall ever seeing a curve of how Ic or Ie affect Ib. It's always been the other way round - reverse cause and effect.

If you already had a perfect grasp of the more relevant stuff about transistors and only this point was proving difficult then I could understand following it to the end of the road but you don't appear to be at that stage yet. Ask your teacher just how important this all is before you worry any more. Is she sure about this? You said,earlier, that she was not too sure of the basics.

 

[Urmi Roy]

Hi
1. You would need to specify the circuit used for the measurement, in particular what is in the base circuit. That could make more difference to Ib than the Early effect could do.

I didn't know that were so many complications! Its just the simplest circuit for measurement possible...no capacitors or anything... there's only a battery Vbc connected between the base and collector (+ve on base),one battery Vce between the emitter and collector (-ve on emitter),and as always,the variable resistor on the base circuit,to regulate the inpit,and the output resistance on the emitter circuit.

It's just that everything in my explanations seem to fit in,but the variation of Ie w.r.t increasing Vce,according to me should be increasing,but apparently,on the graph,it's not.

Yes,my graphs look like those on
http://users.monash.edu.au/~kaminski/ele2262/lab21/lab21.htm

3. Are you sure that you have been told that this particular aspect of a transistor is really crucial?

This is just my own analysis...I thought it would be interesting to try and predict how we could change Ib.

 

[sophiecentaur]

That description doesn't actually make sense. You say you have a battery to establish Vce but you also have an emitter resistor? So the transistor and resistor 'share' those battery volts, according to Ie.
Then there's a battery to establish Vcb (forward biasing the cb junction; do you really mean that?) - and again there's a resistor in the base; its value is relevant.
I'm very confused. Can you do a picture? Or do you mean it's the same as in the 'monash' link and you have described it oddly because it's a CE amplifier ?

Are you actually quoting your own measured results? If the Vce /Ic lines are horizontal it's because the transistor is working 'well' with the collector behaving as a current source. Look at some published transistor characteristic curves, too.

I still think you are confusing causes with effects.

 

[Urmi Roy]

I've been trying hard to send you a picture,but the drawing,in 'paint' is too large to send on physicsforums(that's what it keeps saying..my file is 585Kb,whereas the maximum size is 300Kb)...what should I do?

 

[sophiecentaur]

Send the picture as a JPEG of the appropriate size. There'll be no problem with resolution.

 

[Urmi Roy]

I have no idea as to how to do that yet...I'll try,though...in the mean time,I feel a better description of my circuit could be given as:
a battery Vbb with positive terminal at B,and the other end grounded(since the collector is grounded),a battery Vee with -ve on E and the other end grounded...there is a variable resistor between Vbb and base,and one between Emitter and Vee.
(Vbb provides the potential difference between Band C(Vbc) and Vee provides potential difference Vce between base and emitter.)
I got this idea from a website,so please forgive me if this description is even worse!

 

[sophiecentaur]

You can do a screen shot in Windows can't you? Then crop it and send it.
I can't make any sense of your verbal description. Is this transmitter NPN or PNP?
I do wonder whether you are making the best use of your time with this when you could be discussing and learning something more relevant to transistors.
Give us a link to that site, perhaps.

 

[HP.builder]

@sophiecentaur
thats all about the flow of electron isn't it, either before it enter the Emitter (CC) or after leaving Collector (CE). BTW, how to intuitively thinking about input / output impedance on both CE and CC configuration.

http://www.photoshop.com/accounts/edd7dd06eb124776b1d8f64eb6b36a8a/px-assets/fe07bff9eaa3449c90bbf3c36024ab73

 

[vk6kro]

I've been trying hard to send you a picture,but the drawing,in 'paint' is too large to send on physicsforums(that's what it keeps saying..my file is 585Kb,whereas the maximum size is 300Kb)...what should I do?

When you save your picture go to "save as" then click the little arrow next to where it says "BMP" and choose "PNG".

Then save your file. It will be a lot smaller than a "BMP" file.

You should be able to attach it after that.

 

[sophiecentaur]

"intuitively thinking" can very easily lead to wrong conclusions unless intuition is based on a very firm base.
I recommend using first principles and calculations until you have more experience and can rely on your intuition.
I think we need a diagram and a link before we can procede.

 

[Urmi Roy]

I did it!...here's the picture...

 

[sophiecentaur]

That diagram doesn't really make sense, does it? I assume that the isolated battery / resistor on the left are actually connected to the base - but that is not the problem. Why not draw the diagram the 'right way up' like the circuit diagrams you can see everywhere else? Are the circles meant to be Voltmeters? The standard is to put a V inside the circles. Are there any ammeters in circuit?
You are wiring the device in a very strange way - the actual values of your resistors are very relevant. You appear to be connecting the base to a positive voltage and the emitter to a negative voltage. In this arrangement it is possible to connect positive supply directly to negative supply via the be junction - owch! - they should never be more than 0.7V. The controls of your parameters are not independent but will affect each other. The bc junction seems to be forward biased (?) or where is 'Earth'?
Don't forget that the transistor was developed to be used in a certain way to act as an amplifier. Is there any point in trying to use it in another way, unless you really have a clue about what's goig on? It's just a good way of getting null results or cooking transistors.
I suggest that you look at websites which can advise you on how to measure transistor characteristics (Google "Measure transistor characteristics", for example) and get all that sorted out before you embark on your own investigation. Random measurements like that are very unlikely to be very fruitful or advance your understanding.
You still haven't given the link to this website you say you have looked at. If you are following what it says then I suggest it may be doing you more harm than good.
This thread is really going nowhere now, is it? I thought, originally, that you really wanted to further your knowledge and understanding of transistors. I have my doubts, now.