In BJT small-signal models there is both re and rπ 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:

rπ = β × re

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

And as definition re = 1/ gm.

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

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

Since there Ic = Ib × β this yields to rπ = β × re

You might think what is my question here. Even though the formulas yield this relation between re and rπ, 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?

  • \$\begingroup\$ Just use KVL. I think you'll see that \$I_B=\frac{V_{TH}-V_{BE}}{R_{TH}+\left(\beta+1\right)R_E}\$. Note here that for base current calcs the emitter resistance is multiplied by \$\beta+1\$. Similarly, \$I_E=\frac{V_{TH}-V_{BE}}{\frac{R_{TH}}{\beta+1}+R_E}\$. And here, the base Thevenin resistance is divided by \$\beta+1\$ to reflect it to the emitter for calcs there. It just depends on what you care about at the time. Of course, \$r_e\$ and \$r_\pi\$ are AC dynamic resistances. But the concept still applies for small signal analysis. \$\endgroup\$
    – jonk
    Commented Apr 9, 2018 at 6:34

3 Answers 3


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).

  • \$\begingroup\$ I'm not into the subject quite new but I think I see what you mean in that sense it is not a resistance. I think some engineers invented the "resistance looking into" thing to make re a resistance. I just saw a user ak-94's answer here: electronics.stackexchange.com/a/157638/161776 I didn't think of it like that before. \$\endgroup\$
    – user1999
    Commented Apr 9, 2018 at 15:38
  • \$\begingroup\$ Yes - looking into a certain node gives the input resistance (which is the resistance between that node and GROUND). But your question was if re is a kind of resistance between E and B (internal emitter resistance or something like that). And the answer is : NO ! A resistor describes the voltage-current ratio between two nodes. However, looking into the E node, tlhis is not the case because the voltage between B and E is Vbe, but the current goes through the node C. \$\endgroup\$
    – LvW
    Commented Apr 9, 2018 at 15:43
  • \$\begingroup\$ I see I mixed up the concepts. \$\endgroup\$
    – user1999
    Commented Apr 9, 2018 at 15:44
  • \$\begingroup\$ Just one more last question: When Vbe changes Ic changes so we can find the transconductance. So Vbe controls Ic in general... But is the opposite correct? In active region, if one causes a change in Ic(maybe by temperature or resistance change) would Vbe change according to the transconductance relation? Is causality valid both way? (I also ask because re is dVbe/dIe sounds like change in Ic causes a change in Vbe) \$\endgroup\$
    – user1999
    Commented Apr 9, 2018 at 15:48
  • \$\begingroup\$ Look at the slope of the emitter-base junction at 1uA or 10uA or 100uA or 1mA or 10mA. That is the gm or 1/gm. \$\endgroup\$ Commented Apr 9, 2018 at 17:05

\$r_\pi\$ is an input resistance looking into the base with emitter terminal at AC ground.

enter image description here

\$\Large r_\pi\ = \frac{d V_{BE}}{dI_B} = \frac{V_T}{I_B} = \frac{\beta}{g_m} = (\beta+1)r_e \$

On the other hand \$r_e\$ is an input resistance looking into the emitter terminal with the base terminal at AC ground.

enter image description here

\$ \Large r_e = \frac{dV_{BE}}{dI_E} = \frac{V_T}{I_E} = \frac{r_\pi}{\beta +1} = \frac{\alpha}{g_m} = \frac{\beta}{g_m (\beta +1)} \$

Sometimes because \$I_E \approx I_C\$ we can we can simplify our life and assumed that \$r_e \approx \frac{1}{g_m}\$



You are confusing two different models.

The hybrid π model includes a rπ element. The T model includes a re element.

The models are similar and the two resister elements represent much the same physical factor but are expressed differently because they sit differently in their respective models.

model circuits Circuits from https://leachlegacy.ece.gatech.edu/ece3050/notes/bjt/BJTBasicsSu10.pdf


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