Output impedance with feedback of a current series feedback amplifier

Question

In a chapter pertaining to feedback and oscillator circuits in the textbook "Electronic Devices and Circuit Theory" by Boylestad and Nashelsky, there is a derivation for the output impedance with feedback of a current series negative feedback amplifier. This is the relevant circuit:

Since we are using a negative feedback circuit, we have \$V_i = V_s - V_f\$. To find the output impedance with feedback, we short the supply voltage, thus \$V_s = 0\$ and \$V_i = -V_f\$. But the book says \$V_i = V_f\$, producing the correct expression, which is \$Z_{of} = Z_o(1 + \beta A)\$. But in the previous section discussing the \$Z_{of}\$ of voltage series feedback, where the feedback signal is similarly connected in series to the input signal, it uses \$V_i = -V_f\$. Why is this so? Also, how to intuit the change in input/output impedance due to feedback in different feedback topologies, respectively?

Answer 1

For your 2nd part -- the intuitive understanding of impedances in feedback.

When the feedback is measuring VOUT, then it will tend to regulate that value -- for a voltage, that is similar to saying it has a low impedance --> the amplifier will reduce the open-loop impedance.

When it is measuring IOUT (as in your diagram), then the feedback will regulate that, and regulating the value of a current means increasing its output impedance.

When the feedback signal is a voltage (it's a from a transresistance here), then it tends to (with a high gain amplifier) drive the amplifier's input voltage to 0, even when a signal is applied -- this means it 'looks' like the input impedance of the amplifier is low.