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This question is connected to the link. Why are base current and emitter current the input current in common emitter and common base configurations respectively?

  1. Is voltage drop across the base-emitter junction in the common base configuration considered the input voltage because this voltage drop mainly triggers electrons/holes to flow from emitter to collector via base?

  2. Why is the voltage drop across the collector-emitter junction in the common emitter junction considered the output voltage but not the voltage drop across the load resistance?

  3. Is the dynamic resistance of the base-emitter junctions considered the input resistance?

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  • \$\begingroup\$ 1) draw the circuit (yeah wel all know what it looks like but we're discussing a circuit so just draw it, a pic form the internet is also fine). 2) Consider that I can either use Vce or V(Rload) as the output voltage. Can we know / determine Vce when we know V(Rload) and the supply voltage ? Is that a simple relation? So does it matter if I use Vce or V(Rload)? \$\endgroup\$ Oct 20, 2021 at 18:42
  • \$\begingroup\$ You were asking why the voltage at the collector is considered as output voltage but not the voltage drop across the load resistance. And my answer was: Both are identical because the load is connected to the collector. \$\endgroup\$
    – LvW
    Oct 20, 2021 at 20:50

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Answer to 3.): The voltage between base and emitter (Dc with superimposed ac) controls the collector current. Therefore, it is considered as "input voltage". However, I would not use the term "voltage drop" - because it is an externally applied voltage.

Answer to 4.): When the load RL is connected to the collector, the voltage at the collector (vs. ground) is the voltage across the parallel combination (r_out||RL). In this expresson, r_out is the output resistance at the collector node.

Answer to 5.): It is the input resistance in C-E configuration only

  • without consideration of the external base resistors
  • and when there is no ac resistance in the emitter path (which would create signal feedback)
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  • \$\begingroup\$ Didn't understand the 4th answer😓 \$\endgroup\$
    – MSKB
    Oct 20, 2021 at 19:07
  • \$\begingroup\$ Sorry - my reply was at the wrong place. Here it is again: You were asking why the voltage at the collector is considered as output voltage but not the voltage drop across the load resistance, right? And my answer was: Both are identical because the load is connected to the collector. \$\endgroup\$
    – LvW
    Oct 21, 2021 at 8:28
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Is voltage drop across the base-emitter junction in the common base configuration considered the input voltage because this voltage drop mainly triggers electrons/holes to flow from emitter to collector via base?

Let's break this up into its two parts:

  • Is voltage drop across the base-emitter junction in the common base configuration considered the input voltage

No, not really. A forward voltage is required in order for the common base configuration to work. That's a given. But the input is taken to be a "small signal" that allows the tangent at the local operating point to be used as a linearization (simplication) for analysis purposes. So the input is usually thought of as the small signal itself, not the DC operating point forward voltage that is also required in order to set up the right situation. Though, I suppose, some might decide to just be argumentative about it.

  • because this voltage drop mainly triggers electrons/holes to flow from emitter to collector via base?

That's an issue for microelectronics engineers and physicists (aka "going down a rabbit hole.") It's just a no-nevermind for the first part of the question. Treat it as an unrelated addition designed to direct your mind away from the issues at hand, and to confuse you instead of enlighten you.

As a final note, I like to think of the common base situation as a place where you use the input signal to "pull" current out (or push it in depending on which bipolar you are considering and how you like to think about things) of the emitter, where this current is then propagated across the BJT and generates matching changes in the collector current. The difference is that the emitter "looks like" a voltage source to the input signal (very low impedance) while the collector "looks like" a current source (very high impedance) to the output signal. This concept makes configurations like the cascode very easy to follow.

(But keep in mind that this isn't actually what is taking place. Active mode BJTs always have a collector current that is related to the base-emitter voltage. It's just that this viewpoint is somewhat less helpful here. You can wrestle with it and in doing so you'll understand even more. But for a first-approximation, you don't need to burrow in like that.)

Why is the voltage drop across the collector-emitter junction in the common emitter junction considered the output voltage

The CE voltage drop is usually not considered the output voltage. It is usually the collector voltage taken with reference to ground that is considered the output voltage here. But it could just as well be the collector voltage taken with reference to the collector supply rail on the other side of the collector resistor, too. There's no difference in meaning between the two. So, technically, it's your call here.

The BJT is a 3-pin device, not a 4-pin device. But signals in and signals out are always 2-wire. So the input and the output must share a common reference, leaving 1 remaining wire for the input and 1 remaining wire for the output (which adds up to 3.)

Or, in different lingo, the input and the output are "single-ended" and not "differential."

It usually is easier to just say that there is a common reference between the input and the output signal that is the ground reference. And so the collector output is assumed to be referenced to ground. However, a voltage source is basically the same thing as a dead short to a signal. So there is no difference between thinking of the ground as the common and thinking that the supply rail is the common. They are equivalent for this purpose.

but not the voltage drop across the load resistance?

As I just wrote, you can take it across the collector load resistance. That's just another way of saying the same thing. So, no problem. Do that if you want. It's fine and it's no different.

Is the dynamic resistance of the base-emitter junctions considered the input resistance?

I'm not sure you are ready for this, as I'm now uncertain you understand the concept of linearizing complex curves in order to simplify analysis. But the dynamic resistance is, once again, just the tangent line to an operating point along an exponential larger-scale curve. It's a simplification. But it only works for one operating point. Different operating points will have different tangent lines and therefore different values.

But yes, it is considered as part of the input resistance when computing it for the purposes of small signal analysis. But it is not considered when trying to work out the DC operating point because it has no relevance there. (Doesn't exist as a concept for DC operating point calculations.)

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