Basic transistor bias q... sorry

Thread Starter

clam

Joined Apr 23, 2007
6
Using electron flow... in a transistor, once the initial bias is set and ac current is moving, how do the electrons get from the emitter into the collector if the base to collector is reverse biased? I'm sorry for such a basic question. It seems as if all the electrons, as they enter into the base region and fill up holes and need to move out, would run head on into the reverse bias condtion and not be able to move any farther in spite of the power supply's attraction. This has always confused me. They flow easily from emitter to base because of the forward bias. Why then do they flow from the base to the collector with a reverse bias? It seems like they would hit a "wall" and not be able to move.
 

Distort10n

Joined Dec 25, 2006
429
Someone with more transistor knowledge will be able to add more meat (Haha..get it? Thanksgiving!) to my answer, but here is how I understand it.

Yes, the base-collector junction is reversed biased; however, since the base is lightly doped with holes the excess electrons are pulled across the BC junction because of the attraction of the collector supply.

It would be interesting to see more of a physics answer to this question.
 

Thread Starter

clam

Joined Apr 23, 2007
6
Thank you both. I'll look more deeply. My question, in a nutshell I guess, is: why doesn't the base to collector diode act as a typical rectifier at ac? Why isn't the upswing cut off at roughly .7 volts at a forward bias below the conduction band and the downswing shut off completely at a reverse bias condition?
 

thingmaker3

Joined May 16, 2005
5,083
My question, in a nutshell I guess, is: why doesn't the base to collector diode act as a typical rectifier at ac?
Its not a diode.:) Its a PN junction (same as one might find in a diode) but there's another PN junction within a critically close physical distance. So its not a diode and therefore doesn't quite act like one.
 

Thread Starter

clam

Joined Apr 23, 2007
6
Thank you guys very much for responding.

From the Wikipedia article it says...

"The collector–base junction is reverse-biased, so little electron injection occurs from the collector to the base, but electrons that diffuse through the base towards the collector are swept into the collector by the electric field in the depletion region of the collector–base junction".

This is about as far as I get usually and get stumped. Collector to base movement would be leakage current. Base to collector movement would be a typical charge movement moving toward the attracting potentional of the power supply. Because the base is P doped with few negative carriers and the voltage bias opposes the typical carrier flow no negative base charges are able to get from the base to the power supply through the collector. Why do the same reverse bias forces which oppose an electron movement from base to collector suddenly allow for mass movement from the emitter to the collector. In any other condition, say rectification, wouldn't you see this as the diode reaching it's breakdown point? Is it something about the word "diffuse" which I'm not understanding.
 

tidmoh

Joined Jan 12, 2007
3
:rolleyes:
Thank you guys very much for responding.

From the Wikipedia article it says...

"The collector–base junction is reverse-biased, so little electron injection occurs from the collector to the base, but electrons that diffuse through the base towards the collector are swept into the collector by the electric field in the depletion region of the collector–base junction".

This is about as far as I get usually and get stumped. Collector to base movement would be leakage current. Base to collector movement would be a typical charge movement moving toward the attracting potentional of the power supply. Because the base is P doped with few negative carriers and the voltage bias opposes the typical carrier flow no negative base charges are able to get from the base to the power supply through the collector. Why do the same reverse bias forces which oppose an electron movement from base to collector suddenly allow for mass movement from the emitter to the collector. In any other condition, say rectification, wouldn't you see this as the diode reaching it's breakdown point? Is it something about the word "diffuse" which I'm not understanding.
 

Thread Starter

clam

Joined Apr 23, 2007
6
I'm sorry you see my post as worthy of "eye rolling" tidmoh. I'm not intending to try your patience. Your response is exactly the reason I hesitated in even posting my question though. I see this attitude way too often and don't normally ask things because of responses like yours. Nonetheless the question is asked in good faith. I don't understand what is happening here, that's all.
 

tidmoh

Joined Jan 12, 2007
3
hello,
Just think in terms of majority and minority carriers in each region of the BJT transdistor. In the ussumed case of reverse biased collector-base junction NPN transistor the collector is connected to the positive terminal of the power supply. This would attract any negative charge carrier from the depletion region. These negative charge carriers are still coming from the emmitter after crossing the (thin) base region.There would be no incoming negative charge carriers to the collector if there was no negative charge carrier removal from the base (or addition of positive charge carriers). This latter phnomenon is called saturation. Thus small base current would help in the existance of large collector current and this is the essence of current amplification.
 

GS3

Joined Sep 21, 2007
408
clam, there's no need to get defensive. Your question is a good one and even if many may think otherwise it's a good one for you and that's all that counts. It is best to try to ignore responses which you do not find helpful and try to concentrate on those which you do find helpful. Otherwise you will find yourself mired in arguments with those you find unhelpful and those who could be helpful will probably just go away, not wanting to participate in an argument.

Having said that, here's a couple more tidbits which may or may not be helpful.

Think that the question you are asking is a pretty common one and is explained in many places. If you have read many places and still do not understand it the reason is probably not because all those are bad explanations and we can give you a better one which you will understand at once. Rather, understanding complex things takes time and study and going over the same things over and over many times. One day you realize you now understand things which seemed impossibly complex before. It is just that all that study has settled into something which now makes sense. It is like studying a language or learning the layout of a new building. It is not enough for someone to tell you once what the words are in French or what the layout of the building is. You need to practice and get accutomed to that new environment. I have studied quite a few things in my lifetime which seemed just impossible to understand at first but after many, many hours of thinking about it in my head I have understood them easily. Then, I try to explain it to someone and I realise he is just as lost as I was and he is not understanding a word I am saying. He needs to spend time, just like I did.

You do not need to understand everything in great depth unless you especiafically DO need it. Unless you are going to be designing the actual transistors you do not need to understand in depth how they work to be able to make good use of them. It is much better to have a very good knowledge of what they do and their limitations than to have a very good knowledge of the principles on which they work. Especially at first. Later you can learn more and refine your model but if you try to ingest too much at first you will just get mental indigestion.

The basic idea of a bipolar transistor is that the B-E junction behaves like a diode and that by something magic whatever current you put into the base is multiplied at the collector and with a higher impedance. If you have that simple mathematical model in mind then you can do and understand a lot. Later on you can refine it with factors which add limitations.

You really do not need to understand the mechanics of the internal workings of a transistor but imagine the base as a very thin separation between emitter and collector. Now imagine the charges shoot from the emitter into the base and what happens is that most of them have shot right through the base (because it is so thin) and into the collector before they can even stop to think.

Maybe some of this will help but, as I say, your best bet is to print out the relevant parts of several articles or books and keep going over them until... you give up and realise you can work with transistors without knowing exactly how or whay they work.
 

Thread Starter

clam

Joined Apr 23, 2007
6
Ok, thank you very much. I think I get it. I got to that second part just now Ron H. Thank you.

Is it safe to say then that modelling the PN junctions as diodes applies only in reference to how the base relates to the other terminals and not to how the emitter relates to the collector? Is this what thingmaker3 is alluding to and what Mr. Beaty refers to as an "air gap"? The emitter doesn't see the collector as a glass-like barrier like it would normally if it were a real diode arrangement between the two; needing to meet a threshold voltage before conducting? I think this might be the crux of my misundertanding.
 
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