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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 ?

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  • \$\begingroup\$ Possible duplicate of How do you determine the input impedance for an inverting amplifier? \$\endgroup\$ – Nick May 25 '18 at 14:02
  • \$\begingroup\$ @Nick in that question input impedance is trivial. Its just \$R_1\$. Also that question is more about calculating compensating resistance, not input impedance.. \$\endgroup\$ – AgentS May 25 '18 at 14:05
  • \$\begingroup\$ Since the voltage at V- is Vx + Iin*R2. So the V- voltage will depend on RL because Vx voltage depend on RL/ \$\endgroup\$ – G36 May 25 '18 at 16:25
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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ ty :) that formula is from Electronic principles by Malvino. They must have approximated few things to arrive at that simple looking formula.. I'll move on for now, it is okay... this video has the exact same circuit 1www.youtube.com/watch?v=a9kSibPi8yI @Sven B \$\endgroup\$ – AgentS May 25 '18 at 18:08

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