bjt - Confusion about the meaning of re and rπ

In BJT small-signal models there is both r_e and r_π parameters. They both represent the dynamic resistor between the base and the emitter terminals.

But I read that they are different by a factor of β as:

r_π = β × r_e

I know the concept of transconductance g_m. It is the slope of the Ic Vbe plot at a fixed bias collector current i.e: g_m=∂Ic/∂Vbe.

And as definition r_e = 1/ g_m.

So what I understand is that r_e is the change in Vbe with respect to a change in Ic.

Secondly r_π is the change in Vbe with respect to a change in Ib.

Since there Ic = Ib × β this yields to r_π = β × r_e

You might think what is my question here. Even though the formulas yield this relation between r_e and r_π, I don't quite understand how they represent the same thing aka the "dynamic resistance between the base and emitter terminals".

I mean their definitions are same but yet they are different things. They are both the dynamic resistance between the base and emitter terminals in my mind. But they differ by a factor of β. I'm really confused about the approach. Is it about where we look at the base from?

Answer

I don't quite understand how they represent the same thing aka the "dynamic resistance between the base and emitter terminals".

You are completely right. The term re does NOT represent any resistance between the nodes E and B. My suggestion: Forget the term re and always use 1/gm instead of re.

Some people say (and think... because they have read this somewhere) that the term re is something like a dynamic emitter-base resistance. But this is wrong!!

The quantity 1/gm has "ohms" as unit (because it is the inverse of a conductance) but, in fact, it is not a resistive element at all.

The quantity gm is a "transconductance" and it does NOT describe the current-to-voltage relation between two nodes - and the same applies, of course , to the inverse 1/gm. Hence, it is not a resistive element at all.

There is only one case, where it makes sense to write re=1/gm ....in the common base stage the input resistance at the emitter node is re=1/gm. But still - it is the input resistance (betwen E and common ground) and NOT the resistance between E and B.

Comment: A detailed analysis shows that the rather small input resistance at the E node can be explained as a result of negative feedback internal to the BJT (the relatively large E-B-resistance is reduced drastically to 1/gm due to negative feedback).