transitor: the base pin.

[T3sl4co1l]

i am not arguing with kirchoff. kirchoff is right , but only for static systems. you cannot use kirchoff to solve transient effects.

Uh, I just said you can.

take three capacitors , connect them in star. two go through a resistor to ground. apply a [ed: something?] between the third input of this network and ground. good luck with kirchoff..
accoording to kirchoff no current flows as capacitors block dc. yet electrons have moved !

The current flowing through a capacitance is called displacement current.

Tim

[c4757p]

You can't seriously be claiming that Kirchoff can't solve this...

You're screwing with us, right?

[photon]

You push current into one terminal, the same total amount necessarily must flow out the others, period, no ifs, ands or buts.

But I thought we were talking about pushing electrons? Displacement current does not consist in moving electrons and only plays a role in EM waves, not in transistor theory, as far as I know.

[T3sl4co1l]

What do you think makes electrons "push"? 

As one of the four of Maxwell's equations (in the standard Heaviside format), displacement current is an inseparable aspect of E&M theory on any level.  Whether it's pushing electrons in a quantum mechanical solid, or waves propagating in space (displacement current exists whether or not there is matter present to be polarized by the field).

Ed: slightly disingenuous because electrons also obey the Pauli Exclusion Principle.  That's more of a statistics thing than a pushing thing, but I guess they're equivalent since energy is energy.

Tim

[photon]

What do you think makes electrons "push"?

The E field. The magnetic field has no effect on an electron.

[LvW]

Wow - up to now more than 100 contributions triggered by my claim (replies#2 and#5) regarding the control mechanism of BJTs.
I didn`t expect that - on the other hand it reflects the (surprising, funny, unbelievable) situation that many decades after the BJT was invented there still exist (in books, lecture notes, articles, papers) two different explanations why/how the BJT works.   

I must admit being not a physicist I cannot contribute too much to the discussions/explanations on charged carrier level.
I rather feel as an engineer who is able only to observe and to interpret measurements.
More than that, I am trying to bring these results in relation to commonly accepted rules and principles.
And there are some phenomena which can be explained - as far as I know - based on the voltage control principle only (Shockley`s exponential relation Ic=f(Vbe)).

Some of them:
* Voltage feedback principle (emitter resistance) with increasing input resistance at the base;
* Early effect - caused by (electrical) field increase;
* Transfer characteristic for diff. amplifier (tanh function);
* Working principle of a voltage source based on the „Vbe multiplier“ concept.
* Principle of translinear circuits (introduced by Barry Gilbert).
 
Up to now, nobody was in the position to show how these effects/principles can be explained using the current-control approach.
More than that, I didn`t hear about one single circuit principle/effect which can be explained on the current-control principle only.
(To avoid misunderstandings: Of course, I do not deny the fact that there is a base current Ib=Ic/hfe, which must be taken intoaccount during the design process; however, we are talking here about the control mechanism only).

Thus, I see no reason to treat the BJT as a device having a collector current that is controlled by Ib.
In contrary, tests/measurements have convinced me that both currents (Ic and Ib) are controlled and determined by Vbe only (Shockley`s equation).

[free_electron]

You can't seriously be claiming that Kirchoff can't solve this...

You're screwing with us, right?

That schematic is right , but , i did not say apply a pulse. I said apply a step.
Meaning flick the switch on that power supply so its out put goes from zero to let's say 5 volt and stays at 5 volts forever.

All your resistors are 1k. caps are 1 microfarad , 2 microfarad and 3 microfarad. You can assume ideal parts, no leakage etc.
Please, humor me. Apply kirchoff. All your currents will be zero , yet that ampere needle of that power supply did deviate the moment i threw the switch.

I agree you can do kirchoff for an ac signal, as you can calculate the impdeance of the capacitor at that frequency. But not for a transient.

That is what i am trying to explain when italk about whathappens when a junction goes from non co duction to co duction and back (including the voltage across the junction going to zero).
Anyway, i've had enough of this topic. We are talking two different things here : what happens in the box and how you use the box.

I just saw a little light go off when LvW wrote 'schockleys equation' ic = f(be). 
He is talking the input characteristic of the transistor. And indeed that curve is used to make amplifiers and figure out a circuit. You can make a diagram with 4 curves , this one sitting top right.

So yes, for all means and purposes : when designing with the box : that is function you need.

When designing the box itself that function is a result of doping and the physical structure.
I've been inside the box too long...

But, as a closing thought, using that curve : draw a line that shows ic versus ib and look where they intersect.

[dannyf]

Please, humor me.

Wow!

Q: how many eevblog experts does it take to understand Kirchoff?
A: as many electrons as in the Dead Sea.

[free_electron]

Please, humor me.

Wow!

Q: how many eevblog experts does it take to understand Kirchoff?
A: as many electrons as in the Dead Sea.

Well i would like to see an answer, because i don't know how to do it.
I can do kirchoff for an ac signal. But not for a step. Kirchoff will tell you that the currents in that thing are zero. Yet there was clearly an inrush current. You may be able to figure out how high that inrush current was (pretty easy actually as the caps are essentially a short at that point). And you may even be able to figure out what the current is if the caps are at 1/5 th of their max charge (although i threw a spanner in that one by making them different so they will reach their 1/5 th point at a different point in time )
The math gets very complex, very quickly. Same reason simulators have tremendous problems with capacitors and cant solve the matrix.  Put two caps in series in a sim and see what willl happen. It'll throw a fit.

Anyway all that stuff is moot. We're arguiing the wrong thing. There is nothing wrong with kirchoff , no laws of physics are harmed i. The i ternal function of transistors, there is only temporary charge storage.

If you want the raw maths : i posted a link to that document that shows how a diode goes into conduction and out. Have a ball.

[T3sl4co1l]

You can't seriously be claiming that Kirchoff can't solve this...

You're screwing with us, right?

That schematic is right , but , i did not say apply a pulse. I said apply a step.
Meaning flick the switch on that power supply so its out put goes from zero to let's say 5 volt and stays at 5 volts forever.

All your resistors are 1k. caps are 1 microfarad , 2 microfarad and 3 microfarad. You can assume ideal parts, no leakage etc.
Please, humor me. Apply kirchoff. All your currents will be zero , yet that ampere needle of that power supply did deviate the moment i threw the switch.

Does this meet with your request and expectations?
http://seventransistorlabs.com/Images/Kirchoff_for_FE.png

Note: the blue trace is colored to show that it is drawn completely over the red trace.  Of course, since they match up exactly, I can't show both the red and blue traces, in the same place, at the same time...

I agree you can do kirchoff for an ac signal, as you can calculate the impdeance of the capacitor at that frequency. But not for a transient.

What?

Then how do you think transient simulation, or the real world itself, ever functions?

In the pure time domain, capacitors do not have reactance, they have capacitance.  The fundamental capacitance equation is:

I = C * dV/dt

You perform an integration over time to solve the circuit for its transient response.

In the SPICE case, it's done in small, approximate steps, a necessary solution given the messy things SPICE normally has to deal with.  This linear circuit can be solved symbolically though, for any arbitrary input which satisfies the rules of electrical signals (i.e., excluding pathological functions from the darker side of analytical calculus).

That is what i am trying to explain when italk about whathappens when a junction goes from non co duction to co duction and back (including the voltage across the junction going to zero).

How does a conducting junction become non-conducting?  That's the definition of a junction, that it is always conductive, and therefore connects the component pins which share that junction.  Just because no average current is flowing through it in some instantaneous moment, doesn't make it not a conductor.  Again, if you'd like to bring that all the way down to the quantum level, it remains true.  (In fact, quantum mechanics does an excellent job proving this, since, classical measurement requires that you have current flow to measure the resistance: if R = V/I has exactly no current, R is undefined (but certainly not non-conducting).  QM fluctuations are always present, and depend on the medium; the fluctuations in a conductor are characteristic, as are the fluctuations in 'empty' space -- the vacuum state.)

Tim

[T3sl4co1l]

What do you think makes electrons "push"?

The E field. The magnetic field has no effect on an electron.

I expect a vast number of experiments would contradict that second part...  But, at least within present technology semiconductors, it is the case that magnetic field and spin have no impact on the current flow.

Well... no, I can't even allow for that.  Because Hall effect sensors are very much a real, useful application of magnetic fields upon electrons within a semiconductor.

It would be more accurate to say, general purpose semiconducting devices (those not intended for interacting with magnetic fields) are generally insensitive to magnetic fields.  Your average BJT probably doesn't do anything different until a tesla or a few, at which point, weird things start happening (I would guess, reduced hFE, nonuniform current density leading to early failure of power transistors... um... probably not much else though?).

As for the E-field, precisely: change in the E-field does not occur for free, because the field stores energy.  And a change in that energy is driven by -- guess what -- displacement current.  (Regarding free EM waves: that current, in turn, necessitates a magnetic field, which arises from the changing electric field -- the four states, from the four equations, for the four quadrature phases of a complete circle (relating directly to the fact that d^4/dt^4(sin t) = sin t), and hence the propagation into infinity of an electromagnetic wave.)

Tim

[Dave]

All your resistors are 1k. caps are 1 microfarad , 2 microfarad and 3 microfarad. You can assume ideal parts, no leakage etc.
Please, humor me. Apply kirchoff.

Sure, no problem.

All your currents will be zero , yet that ampere needle of that power supply did deviate the moment i threw the switch.

In the steady state, yes, all currents will be zero. However the magic of the laws is that they always apply. You can quite easily analyse the behavior of the circuit when a voltage step is applied to it.

Here is the circuit with everything marked, to avoid potential confusion:


KVL1: V_g = V_C1 + V_C2 + V_R1
KVL2: V_g = V_C1 + V_C3 + V_R2
KCL: i = i_C2 + i_C3

[photon]

What do you think makes electrons "push"?

The E field. The magnetic field has no effect on an electron.

I expect a vast number of experiments would contradict that second part...  But, at least within present technology semiconductors, it is the case that magnetic field and spin have no impact on the current flow.

I was referring to the force law F = q(E + v x B) in the Maxwell model. F gives the force on an electron in terms of the E and B field. If the electron is moving, i.e. a current, then both the E and B field contribute to the force on the electron. If the electron is not moving, i.e. a charge, then only the E field contributes. I was attempting to add some carefulness in the use of the words "electron" and "current" in this discussion.

[T3sl4co1l]

Oh, as in a static electron.

Of course, electrons are constantly milling about thermally -- but we can refine that even more pedantically by saying, not just an electron, but a population of them whose net velocity is zero, so it still works in that case.

Tim

[dannyf]

Wow, this has really turned into a stump-the-champ exhibition for our experts.

I will provide a different way of thinking, using Kirchoff, .

The two legs are identical. So they can be viewed as one branch, with 2C and 1/2R in serial, which itself is in serial with a C on the upper body.

You can then combine the 2C from the lower leg with the C from the upper body, yielding a 2/3C.

Ie., this circuit, viewed from the source, is identical to a RC circuit with 2/3C and 1/2R.

The peak current is then V/(1/2R), declining from there exponentially to zero.

You can further conclude that the current in the lower two legs are 1/2 of the current going through the upper body.

Yes, Kirchoff holds.

[Dave]

All your resistors are 1k. caps are 1 microfarad , 2 microfarad and 3 microfarad.

The two legs are identical.

Nope.

Yes, Kirchoff holds.

It certainly does.

[dannyf]

Nope.

The same process follows in the analysis.

[Dave]

The same process follows in the analysis.

Sorry, nope.

As soon as you have two different capacitors in the bottom branches, voltage and current waveforms no longer follow a simple exponential rise or fall. You need a different approach to analyzing the behavior of the circuit.
One of the ways to do it would be to write out some differential equations:
v_g = v_C1 + v_C2 + R₁*C₂*dv_C2/dt
v_g = v_C1 + v_C3 + R₂*C₃*dv_C3/dt
C₁*dv_C1/dt = C₂*dv_C2/dt + C₃*dv_C3/dt

These are actually the same three equations I wrote in my post above, just written with differential equations.

KVL1: V_g = V_C1 + V_C2 + V_R1
KVL2: V_g = V_C1 + V_C3 + V_R2
KCL: i = i_C2 + i_C3

You would then have to mash the three equations together and solve for v_C2, for example. But solving for a Heaviside (step) input voltage function would be quite difficult in the time-domain, so a better solution would be to switch into the s-domain (Laplace transform).
The same three equations then become:
v_g = v_C1 + v_C2 + R₁*v_C2*C₂*s
v_g = v_C1 + v_C3 + R₂*v_C3*C₃*s
v_C1*C₁*s = v_C2*C₂*s + v_C3*C₃*s
The Heaviside function is quite simple to write in the s-domain:
v_g = 1/s

Solve for desired quantity and perform an inverse Laplace transform, so you get the output function in the time domain.

This is how one would solve the circuit analytically. The process is rather tricky and time consuming, but (contrary to what some in this thread claim) it can be done. Kirchoff's laws apply. Always.
It is a good idea to help yourself solve problems like these with programs like Matlab.

______________________________________________

We could simplify our life and approach the problem by slapping everything into SPICE and running the simulation. This is what you get, if you choose C₁=2µF, C₂=3µF and C₃=1µF:


You can see something interesting happening, which may seem somewhat counterintuitive at first glance. The current through C₃ goes negative at one point (V(n3) is voltage across R₂, therefore directly proportional to current in that branch).
Just after switching, the capacitors are all at 0V and the currents are only dictated by the resistors. Both resistor values are equal and therefore the currents are equal. The function of voltage on a capacitor is 1/C * integral(i_C dt). Because C₃ is only a third of C₂, the voltage on the former starts increasing three times as fast as on the latter. However, at the end of the transient, they both need to settle at the same voltage (as they are connected in parallel), so the current in C₃ has to go negative in order for the voltage to drop and match the other capacitor. Interesting, huh?

[LvW]

Gentlemen - I am afraid, some of you don`t like to continue the discussion (disput?) about the control mechanism of the BJT (major subject of this thread).
As we all know, we didn`t arrive at a - more or less - common conclusion.
Nevertheless, let me try a kind of summary:
Roughly, I have counted five contributors supporting the voltage-control approach and also five forum members heavily defending the current-control alternative (sorry to say - but in many cases with polemic attacks rather than technical arguments).

So I am still waiting for answers to two simple technical questions (contained in my reply#61) as well as some comments to examples I have listed in my reply#105.
Is there really no member of the „current-control“ group who is able or willing to answer?   

I have to correct myself a little. I got two answers. Here are they:
(1) "any attempt at answering your question is irrelevant to the discussion. It clarifies nothing other than your stubbornness "
(2) "you insist on asking people to answer questions that have nothing to do with this discussion,"

Lack of technical arguments? I`ve got the impression that - for some of you - the question under discussion seems to be a „religious“ one (just a matter of faith).

Here is another comment I got: 

"The fact that some authorities somewhere else had cited it has no bearing whatsoever in a "technical" discussion."

In principle, I agree to this.
However, I think - from case to case - it might be interesting to know the name of the „authority“. 
As an example, here is the answer to a corresponding question from one of the leading developers in the world of electronics: Barrie Gilbert (Analogue Devices):

The old current-in, current-out seems view simple at first, but that's about as far as it goes.
We clearly agree that the BJT should be seen in the same way as an MOS device, explaining that the DC base current of the BJT is actually due to a defect (of sorts) and only a nuisance.
At Analog Devices we have made BJTs (under special conditions) having a DC beta of over 25,000.


Regards
LvW

[T3sl4co1l]

Well if you insist;

1.) First statement: The input resistance of an amplifier goes up if a voltage is fed back to the controlling inverting node and it goes down if a current is fed back. (These are basic rules from feedback theory).

This is ambiguous, because both voltage and current can be fed back in series or parallel.

Indeed, if they are fed back in combination, the amplifier impedances converge (with rising Av) to a constant ratio V/I = resistance.

There is no necessity that the input or output ports have the same or different impedances, because the feedback can be wired to account for either purpose.

2,) Observation: The input  resistance of a BJT/FET goes up in case we apply negative feedback using an emitter/source resitance.

I kindly ask you to answer two short questions:
a) Are both points 1) and 2) above, correct or not?
b) Can we derive from the observation in 2) any information about the question if a voltage or a current is fed back to the inverting node?

This also makes the question much more complicated than necessary, because you are talking about the impedance of a good capacitor: the FET case.  The impedance is already infinite (nearly) at DC, but capacitive at high frequencies, neither of which is very easy to measure.  What frequency should it be measured at?  How are we to determine whether the feedback has an increasing or decreasing effect on it?

Tim

[dannyf]

let me try a kind of summary

You don't seem to be able to understand it on a device level, nor a circuit level - both of which we have tried on you.

Here is another attempt - which I indicated quite a while for you.

The following is a simulation of two ideal devices, G1 on the left is a voltage controlled current devices and F1 on the right is a current controlled current devices. Both are configured into a gain stage.

The graph shows a dc sweep where Vin goes from 0v to 100mv, and Vout1 / Vout2 goes from 5v to 4v, as expected - Vout2 was shifted upwards by 0.1 so it is not right on top of Vout1.

Now, F1 and G1 are completely different devices in that one is voltage-controlled and the other is current controlled.

Question for you, which of them is a bjt and which is a mosfet?

:0

[G0HZU]

In all my 25 years as an RF designer I've never ever heard any of my colleagues argue/bitch over the labelling of how a BJT is 'controlled' in terms of voltage or current. However, they do often argue over which 'model technology' is most appropriate for a particular design or study task but this is usually an argument for/against using small signal s parameters instead of trying to build a decent SPICE model.

So I don't really have a horse in this (pointless?) labelling race but I do have a suggestion. Try analysing the BJT in its old school configuration. i.e. in common base. This is how I was taught to understand the basic physics of an NPN BJT at college many years ago. It makes the device much easier for a beginner to understand because you can see that the electrons flowing in at the emitter (input) are all meant to transfer across to the collector (output) in an ideal BJT. i.e. current gain = 1.

In reality some of them fail to transfer and these are seen as wasted electrons that uselessly flow to the base. But the idea is that they nearly all transfer to the collector load resistor. There's lots of interpretations of what TRANSISTOR means but I was taught that for the simple common base configuration:

transistor = 'transfer input current >> collector resistor'

So you can get power gain by choosing a suitable collector resistance.

This very simple 'transfer' configuration shows that the common base BJT will follow the Shockley equation for device current in that the device current will be a function of Vbe. So it's a bit like a basic diode except that the electrons transfer to the collector rather than flow to the base. It's difficult to argue that the BJT performance is being 'controlled' here by how many failed* electrons arrive at the base.

* Failed as in failed to transfer to the collector.

Some may argue that current out is a function of current in for the above case or other people may prefer to simply argue for a current control label because Ic = Ib * beta when in the common emitter configuration but then I'm not really bothered about attaching a voltage or current control label to the device. I don't think it adds much value.

So I'm not trying to pick a winner here.

[jaxbird]

Oh, I am intimately familiar with the base, she always claim she needs noting, but we all know that current is what she craves, just a little bit she says, but we all know it ends up being much more than that, often way beyond our budget, but what can we do, she needs the current now to perform what we expect of her.

[dannyf]

Nicely done, jaxbird.

[jaxbird]

Nicely done, jaxbird.

Thanks, I find a different perspective can help general understanding