NPN BJT transistor: base voltage / emitter current

[1rel]

What happens when the transistor goes into saturation mode? Or closer to saturation, i.e. I_B increases, compared to I_C?

Ib > (Vcc/Rc)/β will saturate our BJT.

Let's say, in the 2nd configuration, top row from first post: R_E = 100 Ω, R_C = 1 kΩ, V_CC = 5 V, β = 100.

Ib > (5 V/1 kΩ)/100 = 10 μA

In our case, like Jony already showed, that is in saturation, and that the base current is much larger than the collector current, so it has to be taken into account:

Ie = 24mA and Ve = 2.4V
Ic = (Vcc - (Vce(sat) + Ve))/Rc = (5V - (0.2+2.4V))/1kΩ = 2.4mA
and Ib = Ie - Ic = 24mA - 2.4mA = 21.6mA

Is that something that is only possible with BJTs? MOSFETs would behave completely differently?

 

[Jony130]

Is that something that is only possible with BJTs? MOSFETs would behave completely differently?

Yes, the MOSFET's will indeed behave completely differently. I know that it may sound strange and confusing but the term "saturation" has a completely different meaning for the FET's in general. In short:

FET's "saturation" = "linear region" or "active region" in BJT's.

BJT's "saturation" = "linear region" or "ohmic region" in FET's.

So the Full ON MOSFET's will i behave just like a very low valuer resistor (lower than 1 ohm's if you use a proper MOSFET with low Vgs(th) voltage). Therefore drain current is equal to the source current and this is equal to Vcc/(Rc+Re) = 4.5mA.

Try read this
http://www.ittc.ku.edu/~jstiles/312/handouts/Applying a Drain Voltage to an NMOS Device.pdf
http://www.ittc.ku.edu/~jstiles/312/handouts/Creating a Channel for Current Flow.pdf

As for the saturation maybe this example explain why saturation occurs:

I assume β = 100

At first we forced Ib current to be equal to 10μA

The resulting collector current will be equal to:
Ic = β * Ib = 1mA and voltage across Rc resistor:

VRc = Ic * Rc = 1V

Vce = Vcc - Vce = 1V (II Kirchhoff law).

Now let see what will happen when we increase the base current to 50μA ?

Well the Ic = Ib*β = 5mA (what a surprise) and corresponding Vce voltage is 5V.

Now it can be seen that if we increase the Ib current (Vbe must also increased) -> the collector current Ic is increased too, and collector voltage (Vce) decreases. As a side note this is why we say that CE amplifier gives 180 degree phase shift.
Now we can wonder, what is the max Ic that can flow in this circuit. We knows Ohms law, so Ic_max = Vcc/Rc=10mA and that give as Ib_max = Ic_max/β = 100uA
But if we force larger base current the BJT will enter the saturation region because there is not enough supply voltage (saturation is reached when increasing the base current no longer makes the transistor conduct any more current).
See another example

As you can see Ic = β*Ib = Ic = 100*1mA = 100mA don't holds any more. I hope you know why.
Because now we have Rc in the circuit and Ohm's, Kirchoff's law must also holds (BJT cannot conduct any more current than Vcc/Rc). BJT tries to create a situation in which the collector current is Ic = β*Ib = 100mA. The mechanism by which it does this is by controlling the "resistance" of the collector-emitter path through the BJT. But this time the BJT no longer has any tools in its bag to increase the collector current up to the point it would like it to be. So the Vce voltage stops dropping and the collector current is less than β*Ib. Our BJT is full ON ( operate in saturation region).

P.S
And here I find a very good summary post by LvW
http://electronics.stackexchange.co...ation-and-active-regions-of-bjt/129423#129423

 

[lychette]

Where is your problem in understanding the load line concept?
* The transitor data sheet contains a set of characteristic curves Ic=f(Vce) with parameter Ib or Vbe (in your case: Ib).
* However, due to the rising voltage drops across Rc and Re (for rising current Ic), the remaining voltage for Vce decreases correspondingly.
That means: You cannot have any combination (Ic,Vce) as suggested by the data sheet curves.
That is the meaning of the load line: It shows (crossing points) which combinations (Ic,Vce)are possible for fixed resistors and a fixed supply voltage.

it is a bit patronizing to ask the OP where his problem is in understanding load lines...If the OP has no problem he would not be here expecting advice

 

[Averagesupernova]

I have to admit that I have skimmed this thread pretty lightly. My take on the current controlled device vs voltage controlled device has always been this: The beta of a transistor in a circuit is always assumed to be at or above a certain number. 100 for example. The Vbe is also assumed to be within a range. With these 2 things known a common emitter amplifier with voltage divider bias on the base the beta can move quite a lot and with proper design it will make little difference. It appears at this point that it is a voltage controlled device. Take the emitter resistor away and this perception goes away. Now the behavior of the amplifier is highly dependent upon beta. If a transistor had a consistent beta, then it is likely that voltage divider bias would have never been conceived. Would some folks still think that BJTs are voltage controlled devices?

 

[LvW]

it is a bit patronizing to ask the OP where his problem is in understanding load lines...If the OP has no problem he would not be here expecting advice

I am sorry if my wording sounds a bit "patronizing". This was not my intention.
Instead of writing "..where is your problem.." I could have written "...to clarify your problem...".
Please note that the main part of my response was an answer to the OP`s problem, OK?

 

[LvW]

...
It appears at this point that it is a voltage controlled device. Take the emitter resistor away and this perception goes away. Now the behavior of the amplifier is highly dependent upon beta. If a transistor had a consistent beta, then it is likely that voltage divider bias would have never been conceived. Would some folks still think that BJTs are voltage controlled devices?

Of course, the BJT is a voltage-controlled device - with or without an emitter resistor RE. The BJT will not alter its properties in dependency on external parts.
However, because we do not know if a base-emitter voltage of 0.65 or 0.68 will result in the desired collector current (and because of the temperature dependence of VBE) we make use of DC feedback (by the way: One should realize that RE provides current-controlled VOLTAGE feedback!).

Apropos "temperature dependence": The well-known VBE tempco of -2mV/K was calculated based on the laws of charged carrier physics!
In words: There are theoretical considerations which proove that and how much the base-emitter voltage VBE must be corrected as a result of a temperature caused variation of Ic. Is there any similar value for the base current?

Regarding your last sentence: Yes, there are some "folks" which can proove that the BJT is voltage-controlled (folks from Berkeley Univ, Stanford Univ., Harvard Univ., Georgia Inst.Tech, MIT and - last not least - the well-known Barrie Gilbert from Analog Dev).
Barrie Gilbert: BJT is a voltage-controlled current-source; the base current is purely incidental (it is best viewed as a „defect“).

 

[Averagesupernova]

If we eliminate the collector resistor in a common emitter amplifier (NPN) and tie the base of a PNP transistor directly to the collector of the CE amp with the appropriate emitter and collector resistors in the PNP, what is happening? ;) Wink, wink.

 

[leright]

The goal is to understand things like current mirrors and differential amplifiers... but I don't even really get how a simple transistor (BJT) works. Oh well.)

Maybe this will help.

First of all, the BJT is very hard to understand and its operation is complicated since two types of carriers and multiple types of currents are at play (drift, diffusion, recombination, and whatnot) but basically, the base emitter junction is forward biased. The base collector junction is reverse biased. In the case of the NPN configuration in your examples electrons are injected into the base (which is purposely made very thin and you'll see why in a second) from the emitter and holes are injected into the emitter from the base. you now have a lot of excess minority carriers in the base. Since the base is short and the electrons can easily diffuse across the base the electrons are sucked up by the reverse biased collector base junction, like a vacuum cleaner (think about the direction of the electric field and you'll see why the electrons are sucked up and pulled across the collector base junction.

http://www.doe.carleton.ca/~tjs/16-bjtop.pdf

The above pdf is a good explanation of BJT behavior from a device perspective. I think it is good to understand the BJT's physics before analyzing circuits with them.

 

[LvW]

If we eliminate the collector resistor in a common emitter amplifier (NPN) and tie the base of a PNP transistor directly to the collector of the CE amp with the appropriate emitter and collector resistors in the PNP, what is happening? ;) Wink, wink.

For my opinion, this is not a good discussion - even worse: It is no discussion at all.
I have asked several questions - no answer.
Instead, new questions from the "other party".
Averagesupernova, my response to your question: I know what you mean, but you could put it simpler: Ask for the working principle of the Darlington pair.
I can assure you - it is possible to explain the working principles of ALL BJT based circuits with voltage-control.
But I can list you many circuits and observable effects which CANNOT be explained with current-control.
But it is not my job here to win over current-control followers and to convince them that the BJT is voltage-controlled.
It was my only intention to explain my view in responding to the question "how a simple transistor (BJT) works" (question from 1rel in post#1).

R. P. Feynman:
“Religion is a culture of faith; science is a culture of doubt.”
"I can live with doubt and uncertainty and not knowing. I think it is much more interesting to live not knowing than to have answers that might be wrong."

 

[Averagesupernova]

For my opinion, this is not a good discussion - even worse: It is no discussion at all.
I have asked several questions - no answer.
Instead, new questions from the "other party".
Averagesupernova, my response to your question: I know what you mean, but you could put it simpler: Ask for the working principle of the Darlington pair.
I can assure you - it is possible to explain the working principles of ALL BJT based circuits with voltage-control.
But I can list you many circuits and observable effects which CANNOT be explained with current-control.
But it is not my job here to win over current-control followers and to convince them that the BJT is voltage-controlled.
It was my only intention to explain my view in responding to the question "how a simple transistor (BJT) works" (question from 1rel in post#1).

R. P. Feynman:
“Religion is a culture of faith; science is a culture of doubt.”
"I can live with doubt and uncertainty and not knowing. I think it is much more interesting to live not knowing than to have answers that might be wrong."

Wow, just wow. Goodbye...

 

[1rel]

This thread exploded, and I'm sorry, for the many questions about the basics without good foundations. - LvW and Jony130, thanks a lot for your great help, we've collected some useful examples on BJTs for beginners like me, that can be used by others too hopefully.

The discussion about the BJT as a voltage-controlled VS current-controlled device should be based on facts. The only thing I know about BJTs for sure now is that there needs to be a base current in order to make them work. - The very basic calculation examples are great for a beginner, but obviously I had already problems with those. As a beginner in analog electronics cannot say more on this topic than: where's a current, there's also a difference in voltage, that controls that current. - Now I'm asking myself if this is only a "semanitc problem" about labels.

 

[LvW]

The only thing I know about BJTs for sure now is that there needs to be a base current in order to make them work.

1rel - I am not sure if you are still interested in my opinion (although I am not the only one - see the knowledge sources I have listed at the end of my post#43).
Nevertheless - one small correction: The base current IB cannot be avoided because IE is always splitted into two parts: IE=IC+IB.
But it is not true that "there needs to be a base current in order to make them work."
Instead, IB is just a kind of "byproduct" (according to Barrie Gilbert it is even a "defect").

 

[1rel]

"there needs to be a base current in order to make them work."
Instead, IB is just a kind of "byproduct" (according to Barrie Gilbert it is even a "defect").

Seeing the current as an unavoidable/necessary defect of the bipolar junction transistor is an interesting perspective! I often read that BJTs are current-controlled devices , and most of the books I've seen are stating it this way as well. Since I'm not able to understand them yet completely at all, I thought I should focus more on simple circuits and analysis/design methods (like the load lines) for now. - But the discussion on the actual semiconductor physics is interesting and worth it, also because it's relevant for actually understanding BJTs instead of "memorizing formulas or oversimplified models".

With my superficial understanding of the topic, I still asking myself: is there a benefit of using BJTs over MOSFETs then? For digital applications, the CMOS/FET technology has the lead, but BJTs are coming up in analog circuits. - But as you see, I cannot really come up with the right question to ask, because I don't understand the basics yet.

Yes, the MOSFET's will indeed behave completely differently. I know that it may sound strange and confusing but the term "saturation" has a completely different meaning for the FET's in general. In short:

FET's "saturation" = "linear region" or "active region" in BJT's.

BJT's "saturation" = "linear region" or "ohmic region" in FET's.

So the Full ON MOSFET's will i behave just like a very low valuer resistor (lower than 1 ohm's if you use a proper MOSFET with low Vgs(th) voltage). Therefore drain current is equal to the source current and this is equal to Vcc/(Rc+Re) = 4.5mA.

Try read this
http://www.ittc.ku.edu/~jstiles/312/handouts/Applying a Drain Voltage to an NMOS Device.pdf
http://www.ittc.ku.edu/~jstiles/312/handouts/Creating a Channel for Current Flow.pdf

As for the saturation maybe this example explain why saturation occurs:

I assume β = 100

At first we forced Ib current to be equal to 10μA
View attachment 103606

The resulting collector current will be equal to:
Ic = β * Ib = 1mA and voltage across Rc resistor:

VRc = Ic * Rc = 1V

Vce = Vcc - Vce = 1V (II Kirchhoff law).

Now let see what will happen when we increase the base current to 50μA ?

Well the Ic = Ib*β = 5mA (what a surprise) and corresponding Vce voltage is 5V.
View attachment 103608
Now it can be seen that if we increase the Ib current (Vbe must also increased) -> the collector current Ic is increased too, and collector voltage (Vce) decreases. As a side note this is why we say that CE amplifier gives 180 degree phase shift.
Now we can wonder, what is the max Ic that can flow in this circuit. We knows Ohms law, so Ic_max = Vcc/Rc=10mA and that give as Ib_max = Ic_max/β = 100uA
But if we force larger base current the BJT will enter the saturation region because there is not enough supply voltage (saturation is reached when increasing the base current no longer makes the transistor conduct any more current).
See another example
View attachment 103612

As you can see Ic = β*Ib = Ic = 100*1mA = 100mA don't holds any more. I hope you know why.
Because now we have Rc in the circuit and Ohm's, Kirchoff's law must also holds (BJT cannot conduct any more current than Vcc/Rc). BJT tries to create a situation in which the collector current is Ic = β*Ib = 100mA. The mechanism by which it does this is by controlling the "resistance" of the collector-emitter path through the BJT. But this time the BJT no longer has any tools in its bag to increase the collector current up to the point it would like it to be. So the Vce voltage stops dropping and the collector current is less than β*Ib. Our BJT is full ON ( operate in saturation region).

P.S
And here I find a very good summary post by LvW
http://electronics.stackexchange.co...ation-and-active-regions-of-bjt/129423#129423

This has been a real eye opener! About a question I've had coming up many times: Why this confusion about the term saturation, used differently for MOSFETs than BJTs. Thanks a lot Jony130!(A practical problem I've had recently, was that I wanted to control the frequency of a (phase shift op amp based) oscillator. I can change it with simple potentiometer, or a digital pot via SPI, but I don't know how to do a similar thing in an analog way (voltage control) with a transistor as a "variable resistor", I don't even know if it is possible (many people are doing it with transconductance amplifiers (lm13700), which are even more mysterious than discrete transistors, or op amps to me). - A MOSFET behaves like a voltage-controlled resistor, for small drain-source voltages... maybe this can be used. Maybe I can build something out BJTs, but ... but I kind of gave up due to missing foundations... -- the digital domain is definitely easier to understand to me)

 

[LvW]

Seeing the current as an unavoidable/necessary defect of the bipolar junction transistor is an interesting perspective! I often read that BJTs are current-controlled devices , and most of the books I've seen are stating it this way as well.

1rel - I know that some books prefer a view which they call "simple view" or "application-oriented view".
But this is simply false. Viewing the BJT as current-controlled does not simplify anything. This subject has really something to do with "religion".
Some guys have learned that the BJT would be current-controlled, they have followed this view for some years - and now they are unable or unwilling to accept that this may be not correct.
I think it makes no sense to stress this subject further, however, without going deep into the physics of the transistor, it is completely sufficient to realize the following simple considerations:
* For a pn-diode, the current follows the applied voltage according to the well-known exponential relation, OK?
* There is no reason that the pn junction between B and E should not follow the same function.
* Therefore, we "open" the B-E junction with a voltage of - let`s say - 0.6V. As a result, there will be an emitter current IE with a certain value (of course, assuming a suitabe collector voltage). Of course, a small part of this current goes into the base node (IB).
* Would you agree that the emitter current will be larger if we increase the voltage up to VBE=0.7V ?
* Hence, the amount of emitter current depends on the applied voltage VBE. That`s logical, is it not?
* Now - do you see any reason why this current - suddenly - should depend on the base current IB and NOT on VBE?
____________________
Regarding your last question (oscillator) I would recommend to open a new question.
This oscillator has many interesting aspects worth to be discussed.

 

[1rel]

1rel - I know that some books prefer a view which they call "simple view" or "application-oriented view".
But this is simply false. Viewing the BJT as current-controlled does not simplify anything. This subject has really something to do with "religion".
Some guys have learned that the BJT would be current-controlled, they have followed this view for some years - and now they are unable or unwilling to accept that this may be not correct.

The true nature of the analog, real world always remains a secret to us, or hard to describe/imaging by us humans I guess. On the other hand, we can perfectly grasp the behavior of the basic elements of digital systems, because they are "symbolic devices", but we can only come up with "humanized" concepts, pictures (sequential stories basically) and abstracted models for the electrical behavior of these semiconductor crystals and all the (even macroscopic) particle behavior on that level. So it is probably fine to argue about it. One side in me says: i only want the most practical simplification to work with those parts. And the other one is saying: no no! You really need to understand the real thing...

it is completely sufficient to realize the following simple considerations:
* For a pn-diode, the current follows the applied voltage according to the well-known exponential relation, OK?
* There is no reason that the pn junction between B and E should not follow the same function.

That's where I have my difficulties. I had often simplifed the diode (pn-junction) to the extreme states of being OPEN or CLOSED. I thought, when the voltage across the two terminals of a forward biased diode is high enough (above the forward threshold voltage), it gets "transformed" into a short circuit with constant voltage drop (or a "constant voltage source"?). - This picture is flawed, I know, because it does not describe the relationship between current and applied voltage, which is non-linear... As you've stated already, the current as a function of voltage is an exponential one (Shockley's equation). - But normally, the simplification often works fine, because the diode is used above its threshold voltage, and the current is only limited (controlled) by a adequate resistance in series.

But ...maybe this should be stated the other way around: the voltage across the diode is controlled by the current that is "allowed" to pass through it, by the resistor. The voltage as a function of the current? The larger the series resistor, the less current can flow, thus the voltage drop across the diode falls below the threshold voltage. It does merely conduct anymore.

(sim link)

Physically this does not make much sense though! The current is the result and the not the cause, of the potential difference required across the diode terminals, to "squeeze" the depletion layer inside the diode together, so that the charge carriers can pass. (?)

However, the diode as a current controlled "voltage-(drop-generator)" or constant voltage source kind of works a concept, and can be practical. - For example, when I want to light up an LED with +5 V, the simplification works. The LED can be seen as a short-circuit, with a constant voltage drop of say 2 V. So the reset of the voltage needs to be dropped across a series resistor, which controls the current that can be set by choice (up to a certain maximum), always assuming that the current is high enough to overcome the "diode barrier". -- Although the voltage across the LED will change a bit depending on resistor size/current (temperature etc.), it can be assumed to be constant (threshold voltage). (?)

* Therefore, we "open" the B-E junction with a voltage of - let`s say - 0.6V. As a result, there will be an emitter current IE with a certain value (of course, assuming a suitabe collector voltage). Of course, a small part of this current goes into the base node (IB).
* Would you agree that the emitter current will be larger if we increase the voltage up to VBE=0.7V ?

That's where my BJT model breaks and is probably totally inaccurate. I think of the base-emitter voltage V_BE as constant, and so I can know where the emitter sits (voltage). That voltage determines V_CE, which decides whether the transistor is in saturation or not (V_CE < 0.2 V). - I have not used/known this before starting this thread, and learned about it from the interwebs and Jony's examples. - It is probably inaccurate and physically wrong (linear(ized 3 state) BJT model).

* Hence, the amount of emitter current depends on the applied voltage VBE. That`s logical, is it not?
* Now - do you see any reason why this current - suddenly - should depend on the base current IB and NOT on VBE?

Physically, it makes absolutely sense! The voltage V_BE controls I_B.

But do I need to take that variable base-emitter V_BE into account, when looking at BJT circuits?

I don't really know what I'm talking about, to be honest. It would be great to have a handy tool (model) that makes building/understanding of active circuits easier. - LvW, What would you suggest instead of that "linear BJT model" posted earlier?

Regarding your last question (oscillator) I would recommend to open a new question.
This oscillator has many interesting aspects worth to be discussed.

As soon as I come back to this, yes.

 

[LvW]

One side in me says: i only want the most practical simplification to work with those parts. And the other one is saying: no no! You really need to understand the real thing...

I think, both views are correct and necessary!
Only if you understand "the real thing" you are able to decide if and under which circumstances you are allowed to make simplifications.
Example: In many cases, opamps may be treated as ideal voltage-controlled voltage sources - but not always!. And it is up to YOU to decide if simplifications for a specific application are possible or not.

That's where I have my difficulties. I had often simplifed the diode (pn-junction) to the extreme states of being OPEN or CLOSED. I thought, when the voltage across the two terminals of a forward biased diode is high enough (above the forward threshold voltage), it gets "transformed" into a short circuit with constant voltage drop (or a "constant voltage source"?). - This picture is flawed, I know, because it does not describe the relationship between current and applied voltage, which is non-linear... As you've stated already, the current as a function of voltage is an exponential one (Shockley's equation). - But normally, the simplification often works fine, because the diode is used above its threshold voltage, and the current is only limited (controlled) by a adequate resistance in series.

As you said: "...the simplification often works fine". Often - but not always. And it is absolutely necessary to know about the real physical behaviour.
In oscillators, we use diodes for dynamic gain regulation - and we need the knowledge about the exponential characteristic for estimating the oscillator amplitudes.

But ...maybe this should be stated the other way around: the voltage across the diode is controlled by the current that is "allowed" to pass through it, by the resistor. The voltage as a function of the current? The larger the series resistor, the less current can flow, thus the voltage drop across the diode falls below the threshold voltage. It does merely conduct anymore.

Yes - this another example for a "simplified view": Current causes voltage drops. But this is not true - in reality it is ALWAYS the voltage with allows a current to exist. But sometimes such a view helps and can simplify calculations. But we must never forget that it is a simplification!

Physically this does not make much sense though! The current is the result and the not the cause, of the potential difference required across the diode terminals, to "squeeze" the depletion layer inside the diode together, so that the charge carriers can pass. (?)

Yes - correct.

That's where my BJT model breaks and is probably totally inaccurate. I think of the base-emitter voltage V_BE as constant, and so I can know where the emitter sits (voltage).

Why do you assume the voltage VBE as constant? This is another good example for my comments above: This simplification of a constant VBE is allowed only if you know under which conditions this is allowed. What are the conditions? Answer: Emitter degeneration ! In this case, the value of VBE (0.6 or 0.7V) plays a minor role only because of all other uncertainties (in particular, parts tolerances). I have tried to explain this using a simple graph (see my post#22 ). In the graph, I have used the name "working line" - it is better to say: "stabilization line".
But realize: The exact value (between 0.6 and 0.7V) does not matter too much - but VBE must exist and must have a value in this region! And the base current IB is only a result of this "opening voltage" - not the cause!.

Physically, it makes absolutely sense! The voltage V_BE controls I_B.

VBE controls IE and - at the same time (because of IE=IB+IC) - also IB and IC. This is the contents of Shockley`s formulas.

But do I need to take that variable base-emitter V_BE into account, when looking at BJT circuits?

Yes - at first, because it is the reason for using an emitter resistor and, secondly, each signal voltage at the base causes a change of VBE which is transferred to a change in IC.
And the amount of IC change can be derived from the transconductance gm which is the SLOPE d(IC)/d(VBE) of the IC=f(VBE) curve. The transconductance gm is the key parameter for calculating the voltage gain of an amplifier stage.

I don't really know what I'm talking about, to be honest. It would be great to have a handy tool (model) that makes building/understanding of active circuits easier. - LvW, What would you suggest instead of that "linear BJT model" posted earlier?

I would propose to use no "model" at all. For designing or analyzing an amplifier you need only four basic rules:
* Ohms law,
* IE=IC+IB
* Transconductance gm=IC/Vt (as a result of Shockley`s equation)
* VBE=(0.6...0.7)V, to be estimated.
Please, have a look again to my post#33. Here I have listed the 5 basic steps for designing a BJT-based amplifier stage. It was not necessary to make use of any model.

(In case you prefer to have a model, I would recommend the small-signal pi-model with voltage-controlled current source ic=gm*vbe. Don`t forget that such a model is small-signal model only which is valied for one single operational point only)

 

[cabraham]

Those who swear by voltage control make assumptions not supported by science. This question extends beyond bjt devices. if a resistor is used as a heater, is it a voltage controlled heat source or current controlled heat source? Does the voltage across R determine current, or vice-versa, or can it be either depending on conditions? PhD profs disagree. You can surf web sites from MIT, Stanford, Caltech, Case Western Reserve, & find published positions supporting either side. Back to bjt, here are some issues with voltage control that science refutes.
1) The Shockley diode equation is often introduced to students as Id = Is*exp((Vd/Vt)-1). This expression emphasizes to beginners that large changes in current take place while the change in voltage is quite small. But when a diode is forward biased, we do not put a CVS (constant voltage source) directly across it, it would thermally run away. Saturation current Is is temperature dependent. As temp increases, Is increases drastically. A CVS would output a current equal to Is*exp( (Vd/Vt)-1), resulting in a temp rise, which raises Is, which raises Id, raising power, raising temp, raising Is, etc. Id inceases until destruction usually. Vt is thermal voltage = nkT/q, which increases with temp, but not enough to offset Is increase, so diode runs away thermally.

But if we bias the diode with CCS (constant current source), it is thermally stable since Vd = Vt*ln ((Id/Is)+1). Increases in temp will increase Is, but Vd goes down since Is is in denominator, hence power goes down & device is thermally stable. Of course power ratings from diode maker must be observed. This why we always drive p-n junctions with current, not voltage, i.e. "current driven". But a CVS plus a resistor is about as good, since current cannot run away. Any increase in current increases the resistor voltage drop, which reduces Vd, so that current cannot run away. There is seldom an argument that diodes, LED, bjt, SCRS, etc., must be "current driven", never voltage driven, because of thermal stability. To be continued.

Claude. :-)

 

[Gandes]

For this time FETs are much reliable like MOSFETs. Fet basics, In this type of transistors, the gate is insulated from the channel with the dielectric layer. The area marked “N+” is heavily doped “N” type semiconductor. In case of E MOS transistors with a voltage of UGS = 0, the channel is blocked (its resistance takes the value of MΩ and the ID current doesn’t flow). By increasing the UGS voltage channel increases its conductivity and after reaching a certain value called UT threshold voltage, the flow of the ID drain current was possible through the channel.

 

[dlgoff]

For this time FETs are much reliable like MOSFETs. Fet basics, In this type of transistors, the gate is insulated from the channel with the dielectric layer. The area marked “N+” is heavily doped “N” type semiconductor. In case of E MOS transistors with a voltage of UGS = 0, the channel is blocked (its resistance takes the value of MΩ and the ID current doesn’t flow). By increasing the UGS voltage channel increases its conductivity and after reaching a certain value called UT threshold voltage, the flow of the ID drain current was possible through the channel.

From the OP's profile:
Last Activity: Oct 19, 2016

 

[cabraham]

1rel - I know that some books prefer a view which they call "simple view" or "application-oriented view".
But this is simply false. Viewing the BJT as current-controlled does not simplify anything. This subject has really something to do with "religion".
Some guys have learned that the BJT would be current-controlled, they have followed this view for some years - and now they are unable or unwilling to accept that this may be not correct.
I think it makes no sense to stress this subject further, however, without going deep into the physics of the transistor, it is completely sufficient to realize the following simple considerations:
* For a pn-diode, the current follows the applied voltage according to the well-known exponential relation, OK?
* There is no reason that the pn junction between B and E should not follow the same function.
* Therefore, we "open" the B-E junction with a voltage of - let`s say - 0.6V. As a result, there will be an emitter current IE with a certain value (of course, assuming a suitabe collector voltage). Of course, a small part of this current goes into the base node (IB).
* Would you agree that the emitter current will be larger if we increase the voltage up to VBE=0.7V ?
* Hence, the amount of emitter current depends on the applied voltage VBE. That`s logical, is it not?
* Now - do you see any reason why this current - suddenly - should depend on the base current IB and NOT on VBE?
____________________
Regarding your last question (oscillator) I would recommend to open a new question.
This oscillator has many interesting aspects worth to be discussed.

You make assumptions refuted by observation. Let us restrict this discussion to a simple 2 terminal diode device, a simple p-n junction.

In order to commence forward current, you need NOT "open up" the junction with 0.65 volts. A diode is connected to a 1.0 kohm resistor & 12 volt battery with a switch in series. The switch is open initially. Diode current Id, & voltage drop Vd, start near zero. The junction barrier voltage at 25 C room temp is around 25.7 mV, the thermal value, Vt.

The switch gets closed. What happens? Not what you think. Positive charges move towards the anode, negative charges towards cathode of diode. Vd is at 25.7 mV. However, charges conduct very well through the diode, which shoulld not surprise us. Positive charges, or *holes* inside the p type anode, move easily through the p type anode, as they are majority carriers. Likewise, when electrons reach the n type cathode, they conduct easily being majority carriers.

The diode voltage Vd is still at 25.7 mV, yet current conducts readily through the diode. No surprise, in the forward direction, p type anode readily conducts holes, & electrons easily move in n type cathode.

After holes travel through anode, they cross junction & enter cathode, then recombine. Holes do not get very far in the cathode, being minority carriers. Likewise electrons in cathode cross junction, enter anode, recombine, being minority carriers. A depletion region is formed by these stored minority charges. A local E field is created, forming a potential barrier. Vd changes as a result of this new current Id. Eventually equilibrium is reached. Shockley studied junction diodes & described the I-V relation as follows:
Id = Is*exp((Vd/Vt)-1). Another form of this equation is as follows:
Vd = Vt*ln((Id/Is)+1). Here is the dispute in a nutshell.

Does the value of Id control Vd? Or vice-versa? Current control deniers, or voltage supremacists if you prefer, insist that Vd determines Id. Simple observation in lab with probes & scopes negate this view.
I have posted sims from Spice in threads showing that current Id changes first, then Vd catches up. A lab test gives the same result.

Of course many will dispute me. But independentt verification is available. Please search using key words *diode reverse forward recovery*.
SE, Shockley equation, holds under steady state, not transient. A reverse biased diode has a Vd negative in value, with SE predicting tiny value of reverse current. However, before Vd goes forward, Id forward current exists. For a moment, current is positive with negative junction voltage. This condition is forward recovery, eventually the junction voltage Vd, changes to 0.65V, & equilibrium is attained.

The voltage across the diode junction is literally determined by forward current. During forward recovery, SE does not hold. Vd does NOT determine Id. It is opposite.

Reverse recovery can be studied using the search engines. My point is that the current control deniers accept as religion that junction forward voltage controls the forward current. This is mere faith. A lab test easily demolishes this.

I will post simulations & reference material Tuesday. Cheers.

Claude Abraham :-)

 

[LvW]

Hello Claude - do you really intend to start the discussion again?
I know your position and you know my position. Still - I am convinced that you are wrong!
It is already the 2nd line of your contribution which is questionable: Why are you using a 1k resistor in your fictitious "test arrangement"?
Why don`t you connect the diode to a test voltage?
I know your answer ("in reality, nobody will put an ideal voltage source across a pn junction"), however, this is a bad argument because we are dicussing physical properties and not practical circuits. In another thread (another forum?) I have asked several questions to you regarding dimensioning of practical circuits (based on, of course, voltage control). But I never got an answer from you. So - let`s stop it at this point.

Regards
LvW

PS: In your contribution (Dec, 28, 2016) you completely misssed the point of discussion. While speaking about physical control mechanisms we should not mix this discussion with practical design aspects (thermal run-away effects). From the physical point of view it is the VOLTAGE across the pn junction that controls the current through it - however, it is a completely other question if it is practical to use an ideal voltage source for this purpose or a voltage source in series with a suitable resistor. Of course, even in this case, we have a control voltage across the junction (principle of volage division).

 

[cabraham]

Law- I simmed a voltage source directly across a pźn junction, & my statements were affirmed.
You are in denial of forward & reverse recovery. Take a simple full wave rectifier network of 4 diodes in a bridge. An a.c. voltage source powers it with caps & load a free the bridge.
Each half cycle the diodes transition, 2 of them from forward to reverse, vice-versa for the other 2. A scope & probes will affirm that a diode in reverse bias has a large negative Vd, with a small negative Id. When polarity of the input source changes, the forward current Id commences pos itively while Vd is still negative.
This is published in the Power Integration company handbook, late 1990s. Forward recovery is the term. If the diode was reverse biased at -10 volts, then polarity changed, with Id transitioning to +1.0 amp, the po wet dissipated is momentarily 10V*1.0A=10 watts. This is very short in duration.
The new value of Id results in Vd transitioning to 0.7 volts, & steady state dissipation becomes 1.0A*0.70V=0.70 watts.
Just search under the key words *diode recovery forward reverse*.
This is well known. In order to change the diode current, you think that the voltage must change first, then current follows. That is wrong. Voltage changes as a result of current. Diode will conduct full forward current with any voltage, +0.7V, +0.1V, 0V, -2.5V, -10V, -50V, etc. But only momentarily.

I have answered every question you ever asked.

Claude

 

[LvW]

In order to change the diode current, you think that the voltage must change first, then current follows. That is wrong. Voltage changes as a result of current. Diode will conduct full forward current with any voltage, +0.7V, +0.1V, 0V, -2.5V, -10V, -50V, etc. But only momentarily.
Claude

Voltage changes as a a result of current? Really surprising!
Just one simple question: Connect a battery voltage of 0.5 volts directly (without any resistor) across a pn diode (forward direction).
Is there a current as a result of the voltage? Yes or no?
Do you really dispute the validity of Shockleys equation Id=f(Vd)?

 

[cabraham]

Voltage changes as a a result of current? Really surprising!
Just one simple question: Connect a battery voltage of 0.5 volts directly (without any resistor) across a pn diode (forward direction).
Is there a current as a result of the voltage? Yes or no?
Do you really dispute the validity of Shockleys equation Id=f(Vd)?

Yes & no. The 0.5 volts is generated via reduction & oxidation in the chemical process, which results in actions (positive) being forced towards the positive terminal, & anions [negative) forced towards negative terminal. The battery voltage comes into existence via internal current. The charges inside the battery move against the E field. Thus battery internal current produces the terminal voltage, since moving charges constitute current.

Of course, any load connected across battery, diode in this case can draw current as a result of the electric field which is related to voltage. But this external current tends to decrease the battery voltage. As electrons exit battery negative terminal, pass through diode, then enter battery positive terminal, voltage would decrease but redox reaction inside battery propels ions against the E field to restore the terminal voltage.

A similar example is to use a charged cap instead of battery. Say a cap is charged to 0.65V, & placed across diode, forward biased. The cap will discharge & diode current results. But the cap loses charge & energy, eventually discharging, then current ceases. Unlike the battery, the cap has no redox going on inside.

Battery generates a voltage via current internal, which is generated via redox. External lies current cam be said to happen due to battery voltage, but the internal current in battery maintains this voltage. If diode current is 1.0 amp, the battery redox reaction must generate 1.0 amp internal to maintain the voltage at the terminals. So both are active, current & voltage.

But remember that the moment the diode is connected the junction barrier voltage is 25.7 mV, thermal value. The current is 0.5V battery value minus 25.7 mV, divides by silicon bulk resistance. As current continues, minority carriers build up the depletion region until the barrier potential settles to a value under 0.5V. The steady state condition results in a steady current equal to 0.5V minus Vd, divided by Rdiode.

Of course a current can happen due to a voltage. But you always omit the fact that that voltage required a current to come into existence. Just as a charged cap can bias a diode into conduction, so can an energized inductor.

Many reading this have seen a diode placed across a solenoid or relay coil. It is reverse biased when power source drives coil. When power is switched off, inductor current continues into the catch diode, & dissipates. The current in the inductor consists of electrons in the conduction band. Upon reaching the diode So material, electrons collide with lattice ions. Electrons fall from conduction band to valence band, a lower energy state. Photons are emitted per Planck law E = if.

Also as the current passes through the diode, voltage builds up in the barrier. Switching power converters display this.

I have stayed forever that a voltage CAN give rise to a current. A current CAN give rise to a voltage. We must examine the conditions for each example.

I have tried posting simulation results before, but my files are already posted in another thread, & a duplicate file won't upload. I'll rerun & post later today.

Yes. Voltage changes due to current, & vice-versa.

Claude

 

[LvW]

Of course a current can happen due to a voltage. But you always omit the fact that that voltage required a current to come into existence. Just as a charged cap can bias a diode into conduction, so can an energized inductor.

In this thread, we are speaking about currents in electronic circuits (driven by an applied constant voltage source) - and, in this context, it is of no importance HOW this voltage is generated.
This brings to an end my participation in is thread. We should agree not to agree.

(If you can't explain it simply, you don't understand it well enough. A. Einstein)

 

[berkeman]

Hopefully the OP's question has been answered. Time to close this thread.