Electronic – How to find input impedance of this OPAMP circuit

analogfeedbackimpedanceinput-impedanceoperational-amplifier

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Assume OPAMP to be ideal with infinite gain (a) and infinite input impedance
I am clearly able to see that the input impedance of this circuit is R1 (correct me if I am wrong) but I am unable too solve using feedback analysis on OPAMP
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I have two queries regarding this circuit

Q1. As the OPAMP is taken to be ideal thus the input impedance of basic amplifier becomes infinite.

Thus feedback impedance should be infinity/loop gain which comes out to be infinite

Q2. If I am given the second diagram that is the circuit diagram with current source at input to OPAMP shown in above figure then can I still find input impedance without using source conversion and converting non ideal source to non ideal voltage source

If so then as the voltage across the resistance R1 in circuit with current source is zero thus resistance turns to be zeroenter image description here

Best Answer

As the OPAMP is taken to be ideal thus the input impedance of basic amplifier becomes infinite.

I am unsure what you mean by "basic amplifier" Maybe you mean: the opamp has inputs with an infinite input impedance. It simply means that no current flows into the opamp's inputs. This does not say anything about the circuit which is opamp + feedback network.

Thus feedback impedance should be infinity * feedback gain which comes out to be infinite

What do you mean by "feedback impedance" ? I also do not understand why you're multiplying the input impedance of the opamp with the "feedback gain". Maybe you're confused with the relations regarding the output impedance of an opamp in a feedback configuration?

The fact that the opamp is considered ideal make this easy to understand. As the output will have a certain voltage, the voltage difference at the input of the opamp will be

\$V_{in} = V_{out} / A = V_{out} / \infty= 0 \$

As the + input of the opamp is grounded this means that the - input will also be at ground level voltage. That then means that the input impedance is equal to \$R_1\$

Transforming \$V_1\$ and \$R_1\$ into \$I_1\$ and \$R_1\$ is simply done by applying Thevenin.

But doing that changes the input circuit such that you cannot consider \$V_N\$ to behave in the same way!

The left circuit needs a voltage source to drive it (and \$Z_{in} = R_1\$) while the circuit on the right needs a current source to drive it and the input impedance (as seen by the current source) is \$Z_{in} = 0\$

=> The two circuits do not have the same input impedances.