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 VBE 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 VBE controls IB.
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 VBE 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,
* 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)