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Offline SredniTopic starter

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A simple transistor question
« on: May 25, 2022, 09:17:50 pm »
The purpose of this post is twofold.
Bipolar transistors are arguably the hardest transistor type a beginner has to study; with FETs and MOSFETs the carrier transport is unipolar and the 'pinched hose' model provides a reasonably intuitive simple model to explain how they can provide amplification. But BJTs are a little more complicated due to the interplay between drift and diffusion currents and the role played by both carrier types.

Is it possible to use the doubts a student is exposing in a seemingly naive question to teach them how a bipolar junction transistor actually works as an amplifier without delving too much (or at all, if possible) into semiconductor theory?
Here is the question:

Quote
In the following PNP transistor circuit, we know that due to forward biasing the resistance is low in the p−n part and high in n−p part due to reverse biasing. And I learnt that voltage drop in p−n part is low due to the low resistance and voltage drop in n−p part is high due to the high resistance.


https://i.postimg.cc/qvK6ZPq9/screenshot.png
(my note: a better looking picture can be found e.g. on Sedra Smith, or Millman Halkias, when they first introduce BJT)

But in the first p−n junction, the current IE is a high current and the voltage, let's say Rpn is low, so by ohm's law, voltage difference between E and B is V1=VE−VB = IE Rpn.
Similarly V2 = VE−VB = IC Rnp.

Now, how do we conclude V2 > V1?

Even though Rpn < Rnp, we have IE > IC. So there is the possibility of being V1 and V2 equal even.
Then how is it plausible to deduce V2>V1 and not the other way around?
Sorry for any misconceptions.

My question to the users of this forum are two:

1) First, can you understand what the student's doubt(s) is/are?
2) Can you answer this question in a simple but meaningful manner  so that the student will learn how a transistor in that configuration can be used to amplify power and/or adapt impedances?


I already have an answer in mind, but I will add it later in order not to put boundaries on your creativity and imagination.
Let's say this is a small didactic experiment.
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Offline SredniTopic starter

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Re: A simple transistor question
« Reply #1 on: May 25, 2022, 09:19:25 pm »
Analysis of the question

First, there is an error in one of the formulas (maybe introduced with the formatting of the original question), but I see that everybody here has automatically intepreted what the student meant, nevertheless (I wrote this part before Kleinstein spotted it). He wrote "V2 = VE - VB" instead of VC - VB and there is also a question of signs (VC - VB or VB - VC?) that might be discussed but from the actual question asked it seems clear that the student is talking about absolute values.

I have to disagree with the conclusion reached by rfeecs that we cannot say V2 > V1, because the drawing, as ugly as it is, shows that the battery that is reverse biasing the collector junction is bigger than the one forward biasing the emitter junction. This is the classical circuit configuration used in several textbooks to explain how transistor works and it is implied that the transistor is in active mode. If VBE is at least 0.6V, then |VBC| is at least 0.6V. Now, this might seem to void the question of any meaning, since if we already know (from the batteries sizes) that the output voltage is in absolute value greater than the input voltage the question seems already answered. But it's not. What the student is grappling with is how is it possible that in the input mesh we have a current IE that gives rise to a voltage V1 and in the output mesh we have a smaller current IC that (never mind the sign for the moment) gives rise to a voltage V2 that is bigger than V1.

And this brings us to the interesting point of the 'resistances' Rpn and Rnp. A diode is indeed a resistor, albeit a nonlinear one as correctly noted by TeslaCoil (one reference to rule them all: Chua, Desoer, Kuh, "Linear and Nonlinear Circuits"). We are used to deal with the dynamic - differential - resistance of the device because our way to analyse nonlinear circuits is by linearizing them in the neighbour of their operating points, but this is not the case: here we are considering a large signal behaviour and the resistance (the value of the ratio V/I) happens to be a nonlinear function of the voltage (or the current).
The coexistence of these two similar but completely different concepts of resistance is nothing new: we do this for hFE and hfe when the relation between IB and IC is not linear, and in physics for phase and group velocity when the dispersion relation is not linear.


https://i.postimg.cc/9MBpqFn2/Nonlinear-relations.jpg
fig. nonlinear relationships and global and differential quantities

So, the resistances the student is referring to are the values of the ratio V/I for a given voltage (or current). From the point of view of the input mesh we see a small voltage V1 associated with a current IE, while in the output mesh we see another voltage V2 associated with a current IC. And the student is puzzled: if IE>IC how can we be sure that V2 > V1?

In my view, there are several opportunities here: explain why IE > IC is actually IE ≈ IC, and then what is the mechanism that allows basically the same current to be associated with different voltages. I would even go further and add a source resistor to 'program' IE and a load resistor to 'pour the programmed current into' and show that I can get power amplification (in this configuration, it's voltage amplification).

Many interesting answers so far, and I am grateful to all participants. Some of the answers are quite similar to the one I had already written. free_electron makes a valid point about the doping asymmetry of the BJT but I feel that is a bit premature to discuss at this student's level. That could be the subject of a follow-up question by the student.

 I will post my answer shortly (it contains parts that have already been put forward), after producing a couple of pictures. And then I will explain - in a separate post so that it can be moved if necessary - the second purpose of my question here (even if I have just been threatened in PM to "mind my own business" if I am "not a teacher". LOL



So, here's my idea of answer. I modified my initial stance adding some more detail of the inner working, but I am not sure it is a good idea for a student at this level. Excessive detail and still to many simplifications (like my completely ignoring the electron current going from base to emitter) might lead to confusion.

My take on the question

First: in this configuration and with the base junction directly biased and the collector junction reverse-biased, the transistor is in active mode. This means that the collector current is alpha times the emitter current, with alpha slightly less than one (typical values can be 0.95 - 0.99 and higher) and that means that even if the collector current is technically smaller than the emitter current, we can consider IC ≈ IE.
From this, the reasoning of comparing Rpn IE with Rnp IC, when IE ≈ IC and Rnp >> Rpn resolves in favour of Rnp IC > Rpn IE. (It can be shown how we can add external resistors to the input and output mesh to exploit the fact that the same current can go through different resistances to provide amplification.)

Here is the simulation of a fairly aggressive biasing using the simple default model of BJT in LTSpice. As you can see, even if  they are mathematically different due to the relation IE = IB + IC, the input current IE and the output current IC are nearly the same - the differ by roughly 1%. For this example we can infer a value of alpha = IC/IE = 59.7/60.3 = 0.99.


PNP BJT with ideal generators

The above explanation, though, is little more than a math trick because it does not actually explain why we can have basically the same current  in both the input and output meshes when the resistances encountered can be so different.

The 'coupling-decoupling' mechanism
So, why is IC nearly the same as IE? And why, what basically is the same current can have two different voltages associated with it?
It has to do with the dual mechanism of transport (drift and diffusion) inside a semiconducting material. Inside the copper conductors, the current is exclusively a drift current, where electrons are pushed along by the tiny (typically a few microvolt/m) electric field in copper. In the middle of the transistor, that is in the base, the current becomes a diffusion current that depends on the gradient in the concentration of carriers and not on the electric field.

If we focus our attention of the main contributors to the currents in a PNP transistor which are holes (see footnote), the oversimplified nitty-gritty details are as follows:

Inside the transistor, the current in the emitter is largely a hole drift current, acted upon by the tiny electric field in the bulk of the emitter region; the large number of holes that overcome the weakened voltage of the first depletion region cross the EB junction and, once in the thin base, they become a diffusion current that depends on the concentration of carriers. In the base, these holes are minority carriers that diffuse independently of the electric field and almost all of them - due to the geometry and concentration profiles - reach the base-collector depletion region, ending up in the collector bulk as if they were a reverse saturation current for the reverse-biased BC junction. We can call 'alpha' the fraction of carriers that from the emitter cross the thin base without recombining and without 'exiting' the base terminal.
In the bulk of the collector, the current becomes mostly a drift current again, this time subject to another electric field - the one associated with the small (we're talking mV, here) collector bulk voltage. The now majority carriers end up being collected by the negative collector terminal thanks to the electrons that are pushed there by the external circuit.

The large collector current is essentially an 'artificially inflated' saturation current for the reverse biased BC diode (in the VI characteristics, it is the nearly flat horizontal part of the exponential curve that sits right under the voltage axis) and, by being essentially independent on the base-collector voltage, it translate to the nearly flat horizontal output characteristics of the transistor.
As a matter of fact, a BJT transistor can be seen as a reverse biased diode (the BC junction) whose reverse saturation current is dictated by the carriers injected by the emitter.


Here are the input and output characteristics of the common base transistor, along with the vertical characteristics of the generators, seen as one-ports attached to the BJT terminals:


Input and output chars with ideal voltage generators drive

How is this 'coupling-decoupling' modeled?
The circuit under the characteristics is a large signal model for the PNP transistor drawn as a two-port: the input port sees a forward biased diode (the PN emitter junction), while the output port - with its horizontal characteristics at constant current (representing the reverse biased NP collector junction) - is modeled by a current controlled current generator.

To reiterate: the fundamental mechanism that makes it possible to have a 'transistor action' is the existence of a 'bridge phenomenon' (diffusion) that is not ordinary conduction and that allows us to disjoint the effects of conduction current in the input mesh from the effects of (roughly the same) conduction current in the output mesh.

The need for diffusion as the main transport phenomena in the middle of the transistor is also the reason why we cannot build a transistor by putting two discrete diodes back to back, but that's another story...
Similar 'bridge phenomena' can be identified in the workings of FETs and MOSFETs. In all cases it is this 'decoupling' that allows us to model a transistor (in this example for large signals, but a similar reasoning applies to the linearized small signal circuits) with controlled generators - usually in a T topology.


Large and small signal equivalent circuits of the transistor

We can turn this three terminal device into a two-port by putting one terminal in common, the base in this particular example.
And it is again the 'decoupling' between the input and output currents that allows us to either have amplification (in a CB configuration it is voltage amplification) or match impedances (the current that is paired with a large voltage in the output mesh is seen as a current paired with a small voltage in the input mesh).

Practical Application of transistor action
When the transistor is in active mode, the input mesh can sustain a maximum voltage,  across the transistor BE terminals, of less than 1V (typically, it's 0.7V) and the input current is strictly linked to that voltage. The output mesh, on the other hand, can sustain  the much higher voltages of a reverse biased junction, with a current that is essentially independent of that voltage.. Therefore, with an adequate supply, the output port can reach higher voltages than those provided in the input circuit. Since Ie = Ic + Ib, the current in the load will be roughly the same as that flowing in the input circuit. This simultaneous link between input and output current (they are basically the same) and decoupling of the voltages associated with such currents in the input and output meshes, is what makes amplification and impedance matching possible.

As a practical application of transistor action we can use an emitter resistor RE to 'program' the current in the input mesh, and then letting the transistor transfer it to the output where it can flow in another resistor, the load resistor RL.
How the voltage in the output mesh redistributes itself between CB junction and load resistor depends on how much voltage the current set by the input will develop across the load.

Two extreme examples are the following: in the first case we program a current of 3A in the input circuit by choosing RE = 0.78 ohm. The nearly 3A flowing in the load will result in almost all 24V of Vcc to drop across RL, and nearly nothing across the CB junction.


circuit 1: A load with big current and big voltage

Actual values for this circuit are

    VEB = 0.98 V
    VCB = - 0.45 V      IE = 2.973 A       Vload = 23.55 V

In this other circuit, where we program 1mA in the input circuit, only 8 mV will develop across the load resistor, and basically all of VCC will show up across the reverse biased CB junction.


circuit2: A load with small current and small voltage

Actual values for this circuit are

    VEB = 0.773 V
    VCB = - 23.992 V      IE = 0.972 mA       Vload = 0.008 V @ 0.962 mA

The independence of the output current from the collector-base voltage can be easily shown by changing the value of the load resistor, without changing the 'programming' in the input mesh. For example, if in the above circuit we set RL = 10 kohm, we get nearly 10V across the load, 14V across the reverse biased collector junction,

    VEB = 0.773 V
    VCB = - 14.378 V      IE = 0.972 mA       Vload = 9.62 V @ 0.962 mA

and (with the idealized BJT model of LTSpice) the current did not even flinch.

Footnote:
What I am neglecting completely in the typical already simplified analysis of BJT currents that appear on introductory books, is the electron current that from the base goes into the emitter
This current can be made small by strongly doping the emitter and lightly doping the base. It's not exactly negligible but its inclusion would have complicated the exposition. Moreover, the "holes 'exiting' the base" are actually electrons entering the base and recombining with that 1% of the unlucky holes that do not make to the collector. In the charge controlled model of the transistor in a common emitter configuration, they are the reason a bigger current flows from emitter to collector.

...
I said that the purpose of my question was twofold, so I am going to state the second purpose in a separate post that happens to be post # 23.
« Last Edit: June 02, 2022, 09:20:21 pm by Sredni »
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Offline TimFox

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Re: A simple transistor question
« Reply #2 on: May 25, 2022, 09:28:01 pm »
Perhaps the common-base configuration (sketched in the question) is the easiest to explain in the hand-wavy department.
You inject current into the emitter from an outside source.
Giving the charge carriers free will, once in the base region they see a higher field attracting them to the collector region, and most decide to flow there instead of into the base contact.
The ratio IC/IE = "alpha" is slightly less than 1.  The usual current gain "beta" = IC/IB = [alpha]/(1 - [alpha]) and is reasonably large.
Transistors are actually voltage activated, but when forward-biasing the base-emitter junction by an appropriate voltage (like 0.6 to 0.7 V for silicon), the ratio of collector-lead current to base-lead current is high and roughly constant.
Although transistors are often treated as current amplifiers, the physical reality is that they are voltage-operated devices with substantial input (base) current.
 

Offline rfeecs

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Re: A simple transistor question
« Reply #3 on: May 25, 2022, 09:59:00 pm »
For a PNP transistor like you have drawn:

The first PN junction (emitter to base) is a diode, not a resistor.  Ohms law doesn't apply.

In a diode, when the P region is higher voltage than the N region, holes flow from P to N and electrons flow from N to P.  The transistor is made so there are a high concentration of holes in the P region and a low concentration of electrons in the N region.  So there are much more holes flowing from P to N than electrons flowing from N to P.  This is why you have current gain.

The small current of electrons flow in from the base, resulting in a high current of holes flowing in the emitter.  When the holes enter the base, they are swept up into the collector because the collector is negative relative to the base.

V1 is the forward bias voltage of the diode and is always around .6V for a silicon diode.  V2 is a reverse bias and can be anything from zero volts to a high voltage as long as it doesn't break down the junction.

So actually we can't conclude that V2 > V1.
 
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Offline TimFox

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Re: A simple transistor question
« Reply #4 on: May 25, 2022, 10:21:48 pm »
However, if we have the same current flowing into a higher (externally-applied) voltage, that is a power gain.
 

Offline T3sl4co1l

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Re: A simple transistor question
« Reply #5 on: May 26, 2022, 12:08:50 am »
The arguments concerning resistance suggest a too tight reliance on linear theory; which, a transistor most certainly is not [linear].

Perhaps it would be illustrative to introduce Thevenin equivalents, and then nonlinear elements (e.g. diodes), and compare average (Rav = V/I) and incremental (Rinc = dV/dI) resistances.  This contradicts the prior case, which itself (the contradiction) perhaps needs to be explained away, that earlier teachings are intentionally oversimplified, and can't be taken as complete and whole truths, and few teachers take the opportunity to note this about their material.  (Perhaps some other examples might be given, like how you can't take sqrt(-1)... except when you can.)

And then to make it clear that, collector current is still irrelevant based on just this; i.e. clearly the collector incremental resistance (h_oe to be technical) is high regardless of Ic (as long as Vce > Vce(sat)) so Ic cannot possibly be due to this either.  So we need still more advanced theory to explain this.

And this should maybe be taken as par for the course, at this point; because, again, nonlinear systems can be nonlinear in so many interesting ways.  Ways that, often, our linear toolkit isn't wholly inapplicable to -- but they must be applied more narrowly and carefully.

If, having exhausted all these more conventional explanations, further explanation is still left wanting -- perhaps a minor introduction to semiconductor physics would be in order?  For example, the fact that a diode has reverse leakage current is pretty ordinary; whatever mechanistic explanation might be offered to motivate that, is probably fine.  Well, we can simply say that, by gluing two diodes together (very closely), it just so happens that the B-E junction causes the B-C junction to leak, indeed with a carefully constructed device, it causes such great leakage that Ic dominates over Ib.  And this is a similar effect to the leakage depending on applied light (photodiode / PV) (which however does generate a small amount of power, as seen by nonzero current flowing at zero voltage; we might argue that the B-E voltage drop accounts for this in the BJT, hence we don't see negative Vce(sat) in that case).

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Offline free_electron

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Re: A simple transistor question
« Reply #6 on: May 26, 2022, 12:13:20 am »
let me answer with another question :
flip that emitter and collector around. see what happens .
who says emitter and collector are where they are drawn ?

if you can answer that , you have the answer you seek.
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Offline aneevuser

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Re: A simple transistor question
« Reply #7 on: May 26, 2022, 06:31:36 pm »
I'm not sure exactly what the question is that you're asking. However:

1) the student seems to believe that the BJT behaves as if it has a couple of resistors inside it, if I understand what they have written correctly. You need to explain that this isn't the case, else they won't get far.

2) I think that the key to understanding BJT operation is:

a) that of understanding that the reversed biased base-collector junction has a constant-current property - if the emitter injects I A/s of charge carriers into the base, then (almost) I A/s flows into the collector, and hence flows out of the collector.

This is true regardless of the emitter-collector voltage (over some range) - although the rate of drift of charge carriers in the collector varies with the electric field therein, that doesn't change the current that passes through it, which is determined by how much charge the emitter injects (the electric field in the collector will however determine the charge density in the collector, which will be larger when the collector drift speed is smaller - intuitively, when drift speed is lower due to smaller electric field, the charge carriers will be bunched closer together, but we will still have the same number entering and leaving the collector each second)

b) to understand that that the current injected into the base by the emitter is determined by the base-emitter voltage V_{BE}, and is exponential in V_{BE} (this is physically due to the fact that the charge carriers on either side of the junction follow (to a good approximation) the Boltzmann distribution - somewhat strangely, given that they're fermions...)

c) to understand that the base-current-to-collector-current relationship/beta is more like a correlation than a causation - collector current is "caused" by the V_{BE}, but correlates well with the base current.

The extent to which that can be put across to any given student will vary, of course.

[prediction: this thread will gradually devolve into the standard heated argument as to whether BJTs are current-controlled or voltage-controlled]
 

Offline rfeecs

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Re: A simple transistor question
« Reply #8 on: May 26, 2022, 07:03:25 pm »

c) to understand that the base-current-to-collector-current relationship/beta is more like a correlation than a causation - collector current is "caused" by the V_{BE}, but correlates well with the base current.

The extent to which that can be put across to any given student will vary, of course.

[prediction: this thread will gradually devolve into the standard heated argument as to whether BJTs are current-controlled or voltage-controlled]

You cannot show that voltage causes current or that current causes voltage.  You have a relationship, not a cause/effect.  So better not to bring it up.  :)
 

Offline TimFox

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Re: A simple transistor question
« Reply #9 on: May 26, 2022, 07:08:37 pm »
For devices with a monotonic relationship between voltage and current, it is often useful to label them "voltage-activated" or "current-activated" depending on the curvature.
For linear resistors, this label is not useful.
For a PN diode, it is easier to consider the voltage to be a function of the current, since the current as a function of voltage is a very steep function, and subject to small shifts in the voltage parameters, especially those to do with temperature.
 

Online Terry Bites

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Re: A simple transistor question
« Reply #10 on: May 26, 2022, 08:19:57 pm »
The behaviour of the bipolar transistor viewed as semiconductor physics is a little complicated. It takes the magic out of it for me, I can tell you.

Picture that the small current from the base bias potential is making the base more transparent to the C to E current flow. You only a little current need to keep the feed enough carriers into depletion regions and push the energy barriers down. 

Base bias is used to overcome the energy barriers created by the depletion regions. Tons of pictures and animations out there.
The depleted regions lack charge carriers and so no current can flow there. Extra charge flowing into the base undoes the depletion, it reduces the width of depletion regions allowing charge carriers in the collector and emitter to be in much closer proximity and do a bit of mingling (and other stuff for exams). This allows current to flow from E to C in an NPN or C to E in a PNP.

Theres always more to it but WTF, solder like a boss!







 

Offline TimFox

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Re: A simple transistor question
« Reply #11 on: May 26, 2022, 08:25:52 pm »
In my first electronics class, the teacher started with the basic Shockley equation for a P-N junction current as a function of voltage.
Adding the B-C junction on top of the base region, and applying an appropriate voltage on the collector pin attracts most of that current to the collector pin.
That governs the collector current as a function of VBE.
From that, one calculates the transconductance dIC/dVBE
Making a less accurate assumption that the fraction of the emitter current that flows in the base lead instead of the collector lead gives the usual parameter (beta), which is not truly constant, and an estimate of the base current.
From that, one gets an estimate of the input resistance of the common-emitter circuit  dVBE/dIB.
We were taught how to reduce the dependence of our circuit on that base current, due to its wide variation.
You can get a lot of low-frequency bipolar transistor behavior from those simple assumptions.
« Last Edit: May 26, 2022, 11:09:22 pm by TimFox »
 

Online Kleinstein

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Re: A simple transistor question
« Reply #12 on: May 26, 2022, 09:11:47 pm »
The argument with V2, which is supposed to be the base to collector voltage (in the text it is written as emitter to base like V1) is not really helpfull. The point is, that the base to collector voltage can even change the sign.

Using the base configuration and thus alpha is a good way, though one may argue that a NPN transistor may be a little easier as electrons carry most of the current.

The understanding of the transistor depends a lot on the picture one has of the diode. The usual picture is balance of the dirft and diffusion currents, which can be quite confusing. Another picture is the depletion zone as a high resistance zone and the bottelneck in forward direction, which is a bit misleading.

A 3rd less common picture, but closer to the quantitative modeling is realizing that the recombination process as the one to limits the current flow in forward direction. So in a forward biased PN junction holes from the P side pass through the depletion zone (with essentially no obstacle) and than cause a kind of congestion of holes in the N zone waiting for recombination. The rising concentration of holes in the N zone is what limits the current through the depletion zone, not the low number of carriers in the junction region.  The voltage at the PN junction directly translates to the ratio in the hole / electron concentrations on both sides. In forward direction this can increase the minonrity carrier concentration a lot (exponential), but there is not much room to decrease it as it can't get negative and there are only a few to start with. So the generation current (reverse direction) is limited and kind of saturates while the forward current can increase exponentially with the voltage.

In a PNP transistor the holes in the base than have the choice of waiting for recombination and flow to the base terminal or passing all the way through the base and going to the collector without a recombination. With a emitter - collector voltage large enough for the holes to be wellcome in the collector (e.g. > some 50 mV or about 2*kT/e) a lot of the holes choose the collector instead of the base.  The holes in base still see mainly the other holes and this way the emitter potential and not really the base potential.

For the start one can keep the approximation of current control, assuming a constant gain.
 

Offline SredniTopic starter

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Re: A simple transistor question
« Reply #13 on: May 26, 2022, 11:07:18 pm »
The argument with V2, which is supposed to be the base to collector voltage (in the text it is written as emitter to base like V1)

Well spotted  :-+. Yes, that should be the collector voltage and then there is the question of the sign that should be consistent with the BJT being is active mode. This is a beginner question and that configuration is that used to illustrate the working of a transistor in active mode.

Personally I believe a PNP is better to explain a common base configuration because the outside current and the dominant current inside the device (hole current) have the same sign. But this is a matter of tastes.
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Offline rfeecs

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Re: A simple transistor question
« Reply #14 on: May 27, 2022, 12:14:44 am »

Analysis of the question

First, there is an error in one of the formulas (maybe introduced with the formatting of the original question), but I see that everybody here has automatically intepreted what the student meant, nevertheless (I wrote this part before Kleinstein spotted it). He wrote "V2 = VE - VB" instead of VC - VB and there is also a question of signs (VC - VB or VB - VC?) that might be discussed but from the actual question asked it seems clear that the student is talking about absolute values.

I have to disagree with the conclusion reached by rfeecs that we cannot say V2 > V1, because the drawing, as ugly as it is, shows that the battery that is reverse biasing the collector junction is bigger than the one forward biasing the emitter junction.

I didn't notice that.  The level of sophistication of the drawing went way over my head.  ::)

It might be worth considering the importance of notation.  I think most textbooks have some convention of using capital letters for DC values, like bias points, and lowercase letters for AC values.  This simple habit helps to keep things straight in your mind.

Keeping track of positive and negative is also essential.  Holes, electrons, PNP, NPN are easy to get mixed up.  Conventional current, electron current, hole current, total current ...
 

Offline bostonman

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Re: A simple transistor question
« Reply #15 on: May 27, 2022, 03:21:44 am »
NPN transistors were always easier for me to understand, but, when it came time for PNP, I always get confused.

Something that never quite made sense to me. BJT's are "current" gain transistors and MOSFET's are "voltage" gain.

A BJT will not turn on without the base being approximately 0.7v higher than the emitter. Although the beta of the transistor times the base current equals the collector current, if used in a switch configuration, it seems like the math revolves around the voltage.

Is a BJT considered a current gain because beta times the base current equals the collector current?

In the case of a MOSFET, the drain will have more current than the gate, so that seems like it's also a current gain.
 

Offline SredniTopic starter

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Re: A simple transistor question
« Reply #16 on: May 27, 2022, 03:26:30 am »

I have to disagree with the conclusion reached by rfeecs that we cannot say V2 > V1, because the drawing, as ugly as it is, shows that the battery that is reverse biasing the collector junction is bigger than the one forward biasing the emitter junction.

I didn't notice that.  The level of sophistication of the drawing went way over my head.  ::)


Well, I am on my way to open an art gallery in London. But let's get that Nobel prize, first. :-)
Seriously, there is a reason I made that picture as ugly as it is, without copying the original one. I have to say that the original picture had the arrows showing the current directions, which I forgot to add.

Quote
It might be worth considering the importance of notation.  I think most textbooks have some convention of using capital letters for DC values, like bias points, and lowercase letters for AC values.  This simple habit helps to keep things straight in your mind.

Keeping track of positive and negative is also essential.  Holes, electrons, PNP, NPN are easy to get mixed up.  Conventional current, electron current, hole current, total current ...

I agree, but we must keep in mind that this is the first encounter with a transistor and that at this stage of the study there is only one kind of voltage and one kind of current : the small signal model with its artificial mathematical decomposition of iB = IB + ib (or, as I prefer in ASCII Ib = IBQ + ib) is a few lessons away. There is no bias + signal, just current (or voltage).
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Offline TimFox

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Re: A simple transistor question
« Reply #17 on: May 27, 2022, 03:28:58 am »
They are both voltage-actuated.  The difference is that the DC current in the base is substantial, and the DC current in the gate is not.
"Beta" is called the current gain, but it can be considered as the reciprocal of the fraction of the emitter current that flows in the base lead.
In the linear approximation, both devices have a "transconductance", the derivative of the output current with respect to the input voltage for small variations around a quiescent (bias) point.
For a silicon bipolar transistor, gm is approximately the emitter current divided by 26 mV.
 

Offline bdunham7

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Re: A simple transistor question
« Reply #18 on: May 27, 2022, 03:33:59 am »
They are both voltage-actuated.  The difference is that the DC current in the base is substantial, and the DC current in the gate is not.

So a traditional bipolar transistor is still a 'field effect' transistor in way?
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline bostonman

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Re: A simple transistor question
« Reply #19 on: May 27, 2022, 03:53:19 am »
My first education in transistors was given the beta (i.e. 100), and calculating the BJT circuit. When I got my first job, I remember looking at schematics trying to find where the greek symbol beta was located to show the beta value.

When I returned to school, I learned more about beta and transistor circuits, however, I've never needed to dedicate myself to the finite calculations. Plus, my education was a technical degree rather than strictly EE.

In any case, this was one reason why I asked why BJT's are considered current gain. Also, as for beta, I use to get confused on whether I needed to calculate the base current first, and then estimate where the beta is based on the range so I can calculate the collector current, or get the beta based on the graph.

Then I'd get caught in a whirl of basically thermal runaway where the BJT gets warmer, so beta changes, that changes the collector current, etc...

Maybe it's my slight lack of understanding BJTs that confuse me about PNP's, but they just don't flow correctly in my head when I see one used in a circuit.

 

Offline T3sl4co1l

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Re: A simple transistor question
« Reply #20 on: May 27, 2022, 05:42:04 am »
They are both voltage-actuated.  The difference is that the DC current in the base is substantial, and the DC current in the gate is not.

So a traditional bipolar transistor is still a 'field effect' transistor in way?

Even better, there's a SS equivalent to the vacuum tube triode, the Static Induction Transistor.  It's made something like a FET but with a grid, which like the triode, allows some electric field through -- rather than the strong shielding effect of a fully-pinched-off channel -- thus greatly increasing g_os (or however you want to label output conductance).

The fields in a BJT become clearest when using bandgap diagrams, but that is quite fully into semiconductor physics so probably isn't helpful/illustrative for the OP's purposes; but that is indeed where the voltage dependence comes from.

Incidentally, tubes have the same exponential cutoff, for the same reason (Maxwell-Boltzmann statistics of the charge carriers), though it happens to be quite far down the transfer curve (Child-Langmuir law takes over at useful signal currents), and easily masked by leakage currents, so it's not very consequential, and as far as I'm aware, no textbooks bother to discuss it.  I've measured it for a 6AL5 for example; the modeled emission factor of 3.15 presumably implies about as much higher an absolute temperature, which... it's not quite that high, as it happens, but that's plausibly part of the explanation.  Probably tubes have a "non-ideality factor" to them as well, so that the temperature seems a bit higher than otherwise.  (This is the same "n" that commonly appears in the Schockley equation, typically around 2 for Si diodes, which are decided not simmering at 586 K in typical use. :) )


Maybe it's my slight lack of understanding BJTs that confuse me about PNP's, but they just don't flow correctly in my head when I see one used in a circuit.

Fully grasping all the symmetries, once you've already spent so much time learning just the basics, how to use one side of them -- seems confusing, daunting, like relearning all of it; once you realize it's just turning your head sideways, it's quite a lot easier.  But it takes that insight to get there, and it can feel really weird / uncertain while you're still getting used to it.

Even myself, it was relatively recently I pushed through to fully understand/use the series-parallel transformation.  Enjoy a fruit of this labor, the inverted Cockroft-Walton current multiplier (drew this ~2018):



This is one transformation that you tend not to use very much, so I suppose it's no accident I didn't learn it sooner (and even then, it was out of curiosity; one can go a professional career without needing it).

Tim
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Electronic design, from concept to prototype.
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Offline aneevuser

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Re: A simple transistor question
« Reply #21 on: May 27, 2022, 07:25:59 am »
They are both voltage-actuated.  The difference is that the DC current in the base is substantial, and the DC current in the gate is not.

So a traditional bipolar transistor is still a 'field effect' transistor in way?
Yes, very much so. The field in question is the electric field across the BE junction. Feynman discusses the behaviour of PN junctions quite nicely in terms of the E field in sections 14.4 and 14.5 here:

https://www.feynmanlectures.caltech.edu/III_14.html

If you are not familiar with the Boltzmann distribution (which is a idea from statistical mechanics), then you'll have to take the exponential voltage-current relationship on trust though, as he doesn't derive that in this section.
 

Offline aneevuser

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Re: A simple transistor question
« Reply #22 on: May 27, 2022, 08:01:57 am »
Maybe it's my slight lack of understanding BJTs that confuse me about PNP's, but they just don't flow correctly in my head when I see one used in a circuit.
I find that it's easiest to keep the NPN-vs-PNP voltage arrangements clear in my mind by thinking about a BJT push-pull circuit. There, you can imagine a mirror image line though the middle of the circuit, where all the voltages are reflected, NPN voltages above, PNP voltages below.

So, for example, if the NPN device has its collector 10 V above its base, then the PNP device will have its collector 10 V below its base.

And you can imagine a sine wave, with +ve and -ve half-cycles, driving the base - in the +ve half cycle, the NPN turns on when its base is 0.6 V more +ve than the emitter - mirror imaged, this means that the PNP turns on when its base is 0.6 V more -ve than its emitter, which happens in the -ve half-cycle.
 

Offline SredniTopic starter

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Re: A simple transistor question
« Reply #23 on: May 30, 2022, 02:09:13 am »
Serving a youthful porpoise

What was the second purpose of my question here? To verify that the quoted question was not an unanswerable, not focused or ill posed question. And that it could be easily interpreted (despite an ugly picture and one typo) and answered in a way that could help other beginners with similar doubts.
I would go even further to say that the doubt expressed by the student is fairly reasonable for a beginner and provides a couple of hooks that help driving home the concepts of transistor action and how the currents in the input and output meshes are simultaneously linked (IE approx IC) and independent (they show up in two one-ports with a completely different voltage).

Some of the gentlem-- gentlepersons who donated their time to answer, might wonder "why did you need to verify that?". The least I can do is to offer an explanation of where I found the original question and why I did bother with all this. Consider the following a general observation on what should constitute good didactics and also a reflection on the quality of another EE-related website, like I have seen many in the chat section (on distributors, customer services, developers, designers). And I would have posted this in the other thread I had opened in the chat section, if it hadn't been deleted already.

The original quoted question was asked (not by me) on EE Stack Exchange and nearly immediately closed and downvoted so that it could be deleted after a few days. The stated reason was that it was "not focused". I asked on the corresponding Meta the reasons for closing it (and deleting it) but i received no real explanations  for the closure except a tautology ("it was closed because three users with x points decided to close it") and generic bogus reasons ("it was low quality", "you can google it" - yeah like every other question) that were not actually explained. Neither by those who voted to close it, nor by the other high rep users and moderators that follows that Meta.
I needed a reality check to see if I was the only one on this planet to see the question as understandable and answerable (if possible even in a way that could address the doubts expressed by the OP, and also being useful to others.)

On a more general note (and this is why I called this an EE didactic experiment and wasn't sure whether to post it here or in the chat section), what happened shows how a legitimate (and all in all well formulated, considering it exposed the student's reasoning and gave a few key points to work on) student question can be shut down by a lazy or incompetent teacher... just because.
The real problem, though, is when the other 'teachers' band together to defend an opaque decision and cannot even explain why such decision was taken (especially in the case of such a simple and short question). I have seen more talkative witnesses at a mafia trial.

One more thing.
I left a code at the end of my query on EESE Meta. I did this after modifying it in its longest form that asked for specific reasons why the OP question had been closed - therefore before I got downvoted from 0 to -2 and I received further comments that were not specific and gave generic 'low quality' reasons (someone even suggested I was affiliated with the pic hosting site and could be making money out of their banners - which by the way I never saw)
Anyway, it's this code here:
3d8098c310eb3583a51c95f338b340c5

What is it? Well, it is the md5sum of the following txt file that completes my question and should be read after my last line "I hope it is clear what I am asking":

It's about the "we don't know, we cannot know" part. But let's see if my crystal ball is working: I forecast downvotes that will have the effect to hide this Meta question and no real explanation about why the OP question was closed. Except mabye a vague and unsubstantiated "it's low quality" 'reason'. I would not be surprised to see some excuse not to answer the questions posed.

And exactly yesterday (well, when I wrote this it was yesterday) my question went from -2 to -3, and still no actual answers.
Dang, that gipsy was right. I made a bargain!

64c9c56dc1e08ebe882cd07ce33559ec     >:D
All instruments lie. Usually on the bench.
 

Offline SredniTopic starter

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Re: A simple transistor question
« Reply #24 on: May 30, 2022, 02:18:49 am »
They are both voltage-actuated. 

I have to disagree with this. Voltage and current, like electric and magnetic field are concomitant manifestations of a single entity. There is not a cause and an effect (like in an EM wave it's not the change in magnetic field that causes the change in electric field and so on).
One can choose to see a BJT as voltage controlled or as current controlled, or even as charge controlled.

The 'bias' toward the voltage is due to the fact that in solid state physics we deal with energy and potentials, so everything seems 'caused' by a potential.
Then there is the fact that - as you said in another message - it might be easier to see the transistor as current controlled because the voltage would be too sensitive to fiddle with, but fundamentally, one can explain how a BJT works by injecting current - see my answer here https://electronics.stackexchange.com/questions/547625/working-of-bipolar-junction-transistor-with-electron-flow
that has a bit of references at the end, including the original paper by Ebers and Moll who had no problem in modeling the BJT as current controlled.
All instruments lie. Usually on the bench.
 


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