I'm slowly working my way through understanding negative feedback in simple BJT amplifiers. I'm studying this circuit here:
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).
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?