Calculating input impedance of ICIS opamp circuit

Question

My textbook just gives the formulas w/o derivations and I've been trying to prove them when possible. But I feel stuck while calculating the input impedance of below ICIS amplifier.

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My work :

Since almost all the input current goes through \$R_2\$ , voltage at the meeting point of the 3 resistors is given by \$-i_{in}R_2\$ .

Then the output current is given by $$ - i_{out} = i_{in} + \dfrac{i_{in} R_2}{R_1} $$

So the output voltage equals $$V_{out} = i_{out}R_L - i_{in}R_2$$

Also we have \$V_{out} = -A_{VOL}V_{-}\$

With some algebra to above 3 equation we get $$\dfrac{V_-}{i_{in}} = \dfrac{\left(1+\dfrac{R_2}{R_1}\right)R_L + R_2}{A_{VOL}} = \dfrac{\dfrac{\color{red}{R_L}}{B} + R_2}{A_{VOL}} $$

This is no where close to the given formula and I don't seem to simplify any further. Also I feel my work is wrong because \$\color{red}{R_L}\$ should not be there... Any help ?

Answer 1

I believe you have accidentally assumed that \$V_- = 0V\$ when you said

Since almost all the input current goes through \$R_2\$, voltage at the meeting point of the 3 resistors is given by \$−i_{in}R_2\$.

While it actually should be the voltage drop:

$$V_- - V_x = i_{in}R_2$$

You can still keep the equation

$$V_{out} = -A_{VOL}V_-$$

But the KCL at the crossing point (\$V_x\$) becomes:

$$i_{in} + \frac{V_{out} - V_x}{R_L} = \frac{V_x}{R_1}$$

This results in the expression

$$Z_{in} = \frac{V_-}{i_{in}} = \frac{(R_1 + R_2)R_L + R_1R_2}{R_L + (A_{VOL} + 1)R_1}$$

You can verify this formula by simulating the schematic below. I tried some combinations and I'm fairly certain that the input impedance is correct.

I haven't been able to get to the equation in your book though. I may need more information about their assumptions.

^{simulate this circuit – Schematic created using CircuitLab}