Shunt feedback input impedance

Question

I am trying to derive the input impedance for the transresistance amplifier (shunt feedback) on page 119 in the Art of Electronics.

I am considering a voltage \$v_i\$ applied to the input. As described in the book, it is the normal input impedance \$R_i\$ in parallel with something else due to feedback. When \$v_i\$ is applied, it causes a current $$I=v_i(1-A)/(R_f+R_o)$$ to flow from the input (so there is really current flowing into the input with A>1).

So shouldn’t the parallel part of the input impedance be $$v_i/I=(R_f+R_o)/(1-A)$$ rather than what is shown in the image? This is a negative impedance in parallel with \$R_i\$.

Answer 1

Your approach is correct. But what isn’t shown in the figure explicitly, are the non-inverting and the inverting input terminals.

This case, you can safely assume it’s a negative feedback. Hence, the feedback resistor \$R_{f}\$, is connected to the inverting input terminal.

Therefore, the differential input voltage \$V_{+} - V_{-} = -v_{i}\$

After applying the loop equations as you have done, $$ I_{f} = \frac{v_{i} - A(-v_{i})}{R_{f} + R_{o}} $$ $$ I_{i} = \frac{v_{i}}{R_{i}} $$ $$ I = I_{f} + I_{i} $$

and assuming the output resistance \$R_{o}\$ getting getting swamped by negative feedback, you get the same expression for \$Z_{in} = \frac{v_{i}}{I}\$ as given in the book.

Reference: Operational Amplifiers (Page 9)

Answer 2

This is an inverting Transimpedance Amplifier.

The current will see a very small input impedance and here is how to prove the statement: $$I_{in} = \frac{V_{in}}{R_i} + \frac{V_{in}-V_o}{R_f}$$ $$V_o = -AV_{in}$$ $$I_{in} = \frac{V_{in}}{R_i} + \frac{V_{in}+AV_{in}}{R_f}$$ $$\frac{I_{in}}{V_{in}} = \frac{1}{R_i} + \frac{1+A}{R_f}$$ $$Z_{in} = \frac{V_{in}}{I_{in}} = R_i || \frac{R_f}{1+A}$$