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Described Problem

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?

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  • \$\begingroup\$ How would you define the input resistance of a BJT seen from the base into transistor than? electronics.stackexchange.com/questions/407868/… \$\endgroup\$
    – G36
    Apr 3, 2020 at 8:33
  • \$\begingroup\$ 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. \$\endgroup\$
    – LvW
    Apr 3, 2020 at 9:01

1 Answer 1

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This is only Ohm‘s law. Vb is the voltage loss from base input to ground. Taking the current passing this line and using Ohms law, you get the resulting resistance. This resistance is parallel to 100k.enter image description here

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