|Feb7-10, 01:21 PM||#1|
Semi conductors(especially transistors).
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!
|Feb9-10, 05:54 AM||#2|
Well,it's almost been two days since I made this post,and no one's replied yet.Please realise how desperate I am to get this doubt resolved....please please please help!!
|Feb9-10, 07:23 AM||#3|
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.
|Feb9-10, 10:25 AM||#4|
Semi conductors(especially transistors).
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!
|Feb10-10, 07:07 AM||#5|
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...
Also,what do we mean by the output of the transistor...is it a voltage?....how do we use this output?
|Feb10-10, 08:56 AM||#6|
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,apparantly the output is the 'voltage' accross 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
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 Ve Rc/Re
and Ve follows Vb
This amplifier is described as a Voltage Amplifier.
|Feb10-10, 12:14 PM||#7|
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 someting 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?
|Feb10-10, 12:40 PM||#8|
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.
|Feb11-10, 06:50 AM||#9|
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 accross 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.
|Feb11-10, 07:35 AM||#10|
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.
|Feb12-10, 08:32 AM||#11|
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 accross 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 accross 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 accross 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 accross 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?
|Feb12-10, 10:20 AM||#12|
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.
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 realise 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!
|Feb12-10, 09:35 PM||#13|
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 realise 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 accross 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 accross 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 accross 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?
|Feb13-10, 05:07 PM||#14|
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?
|Feb13-10, 09:23 PM||#15|
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?
Sorry if I'm messing up a bit...I'm working hard,I promise!
|Feb14-10, 05:37 AM||#16|
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!
|Feb14-10, 07:05 AM||#17|
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 accross the collector resistor due to Ic?
Again, when we say the output voltage accross 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.)
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