I'm slowly working my way through understanding negative feedback in simple BJT amplifiers. I'm studying this circuit here:
[![enter image description here](https://i.stack.imgur.com/ksLpJ.png)](https://i.stack.imgur.com/ksLpJ.png)

I read that one of the advantages to negative feedback in this topology is that it raises the input impedance by a factor of \$1+KA\_V\$ (where \$K\$ is the feedback factor and \$A\_V\$ is the open-loop gain of the internal amplifier). I did a simple analysis of the input resistance \$R\_{in}\$ and found that

\$R\_{in}=R\_B||R\_{ib}\$

\$R\_{in}=[(85k||106k)||(\beta(R\_E+r\_e)]\$

\$R\_{in}\approx 45k\Omega\$

I was able to verify that this is the correct input impedance. By simulating a test signal \$v\_s=20mV\$, I measured the current entering the circuit at \$i\_{in}=444nA\$, which checks out.

Here's the part that doesn't make so much sense to me. If the negative feedback increases the input impedance by a factor of one plus the loop gain:

\$R\_{in}=R\_{in'}(1+KA\_V)\$

What *was* the internal amplifier's input resistance before the negative feedback was added? From my previous work with this circuit, I know that ...

\$K=(\frac{1}{A\_F})-(\frac{1}{A\_V})\approx 1\$

\$A\_V\approx g\_m(R\_E) = 0.12 \* (4000) = 480\$

\$A\_F\approx 1\$

Plugging all this in, I should be able to find the internal amplifier's nominal input resistance \$R{in'}\$before the feedback network was added.

\$R\_{in}=R\_{in'}(1+KA\_V)\$

\$45k\Omega=R\_{in'}(1+(1\*480))\$

\$R\_{in'}\approx 93.6\Omega\$

For the life of me, I can't see how the internal amplifier could be analyzed to have an input resistance of 93.6. I'm fairly certain my loop gain is correct as all those calculations check out in a previous question ([here](https://electronics.stackexchange.com/questions/536905/voltage-voltage-feedback-and-common-collector-analysis?noredirect=1#comment1390088_536905)).

As an aside, I did try comparing this circuit to one with \$R\_E\$ removed to see if that helped any (it didn't, I don't think). I reduced \$V\_{cc}\$ to a low enough level that my \$I\_{CQ}\$ remained at \$3mA\$. My input resistance fell as expected to \$R\_{in}\approx \beta\*r\_e \approx 2080\Omega\$. So, again, my analysis of input resistance proved out OK and checked out in simulation. But adding the feedback resistor \$R\_E\$ only increased the input resistance by a factor of around 21, not the expected 1 + loop gain factor.  

**Can anyone explain what expression for an internal amplifier's input resistance will work with the \$R\_{IN}=R\_{}in'(1+KA\_V)\$ equation?**