Where does the input resistance formula of a BJT amplifier come from?

I just solved the (a) question of this problem from one of my past exams. And I am a bit confused on to why the solution is as follows:

Rin = 100kΩ || Rib

Rib = (rπ + (1 + β)Re)

And the way Rib is derived is using this equation:

Vb = rπib + (ib + βib)Re

Rib = Vb / ib = [rπib + (ib + βib)Re] / ib = (rπ + (1 + β)Re)

Now, what I don't understand: Why is the input resistance Rib = Vb / ib? Where did that come from? Is that derived from the Thevenin equivalent of the BJT?

• How would you define the input resistance of a BJT seen from the base into transistor than? electronics.stackexchange.com/questions/407868/…
– G36
Commented Apr 3, 2020 at 8:33
• To complete the picture (verification based on system theory with beta=hfe and rpi=hie)): With (1+hfe)=ge (transconductance ge=d(Ie)/d(Vbe)) we arrive at r,ib=hie(1+geRe). What we can see now is how signal feedback (caused by Re) increases the input resistance of the B-E path (hie) by the factor (1+loop gain). This is in full accordance with sysytem theory.
– LvW
Commented Apr 3, 2020 at 9:01