How measure hie of a bipolar transistor?

[audioguru]

Hi Audioguru, don't you think the above formulation (...."internal emitter resistance") is a bit unclear?
I think, a formula can avoid confusions. Thus: "internal emitter resistance"=1/g (see the formula in post#17 ?

The internal emitter resistance that is used to calculate voltage gain and input impedance of a transistor is simply 26mV/Ic.

 

[Winterstone]

The internal emitter resistance that is used to calculate voltage gain and input impedance of a transistor is simply 26mV/Ic.

Yes, no doubt about it (see my formula in post#17).
My comment was related to the verbal description ("internal...) only. For my opinion, this wording is a bit uncommon and, therefore, it is - possibly - not understood by everybody.

 

[The Electrician]

A web search turns up 138 hits for the phrase "intrinsic emitter resistance" and 3510 hits for the phrase "internal emitter resistance", both of which I've seen used in connection with bipolar transistors.

 

[Winterstone]

OK, may be that my assumption („uncommon term“) was a bit to optimistic.
On the other hand, I think it shouldn't be the task of this forum just to quote Internet contributions, but primarily to discuss different approaches to explain physical and technical phenomena.
And that's what we are doing now.
Therefore, I still dislike this term (intrincic or internal emitter resistance) because it can lead to a lack of understanding of the BJT operation principle.
And this is an important aspect – in particular for newcomers.
I will try to explain my sight in short:

A resistor is defined as the property of a current-conducting path between two points - in our case: Between the outer emitter node and ..... ? I am not able to define such an internal point.
Perhaps, an answer can be found by measurement. Can this resistor be measured?
Yes, it can. It is the input resistance in common-base configuration. Thus it is the resistor that can be measured between the outer emitter node and ground.
But this involves the whole npn device because the applied emitter voltage causes a current that is split in two parts: Base current and collector current.
Therefore my question: Is it correct to allocate this resistor solely to the emitter part of thje BJT?
Such a sight leads to a totally incorrect „explanation“ that can be found, for example, in the Internet (www.allaboutcircuits.com) :

Quote: However, bipolar transistors have a small internal resistance built into their Emitter region called Re .
The transistors semiconductor material offers an internal resistance to the flow of current through it and is generally
represented by a small resistor symbol shown inside the main transistor symbol.

I am afraid, many newcomers will think this resistor is something like a “path-resistor” caused by the material only – and that obviously would be false.
Why not simply state: Common base input resistance is equal to (1/transconductance g) ?
In this case, everybody can immediately recognize that the input resisteance directly depends on Ic.
More than that, this "resistor" cannot be found in the small-signal equivalent BJT circuit diagrams. This may also confuse some beginners.
Was it simply neglected? No - it is implicitely contained in the output current source.

Sum total (recommendation): Not every technical information that can be found in the internet is correct by 100%.
As another example, forum members may ask themselves: Is the BJT a voltage or current-controlled device? My first impression: Most answers from the Internet are wrong.
Comments are welcome.
W.