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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 enter image description here

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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

  • \$\begingroup\$ It's not clear what your second question actually is -- did you not finish that thought? \$\endgroup\$
    – TimWescott
    Jun 20, 2019 at 18:54
  • \$\begingroup\$ @sunita: "op-amp" is a contraction of "operational amplifier". It is not an initialisation so it doesn't get capital letters. \$\endgroup\$
    – Transistor
    Jun 20, 2019 at 20:43

3 Answers 3


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.

  • \$\begingroup\$ regarding your first question basic amplifier is OPAMP along with Loading by feedback i.e R1 between inverting terminal and ground and R2 between output terminal and ground ( I am not sure what exactly will be loading ) Regarding second query feedback impedance meant input impedance after feedback and how do I get that expression R/1+AB as A tends to infinity we can say it comes out to be (R/AB) after it got struck so I wrote simply R/L where l was loop gain (sorry I just typed feedback gain in original) I don't know how to proceed further. \$\endgroup\$ Jun 20, 2019 at 19:23

When you do the feedback calculations, you need to do the calculation on the amplifier with feedback. This means that the input impedance you use is the input impedance of the amplifier with the feedback network added. So the raw amplifier has infinite input impedance and zero output impedance, but as it's used in circuit, the amplifier has an input gain of R2, because there's a path from the input pin to the output.

Then the input impedance of the amplifier + feedback is \$\lim_{a \to \infty} \frac{R2}{a}\$, and it all makes sense.

  • \$\begingroup\$ what is input gain and output gain I am only familiar with one gain \$\endgroup\$ Jun 20, 2019 at 19:02
  • \$\begingroup\$ I meant to say 'impedance'. I've edited my answer -- sorry for any confusion. \$\endgroup\$
    – TimWescott
    Jun 20, 2019 at 19:07
  • \$\begingroup\$ I am not able to correlate how this has to justify that we can use feedback method to verify or find input impedance of this circuit. \$\endgroup\$ Jun 20, 2019 at 19:25
  • \$\begingroup\$ You use the feedback method to find that the open-loop impedance is R2. Then you use the feedback method to find that the closed-loop impedance is \$ R2 / \infty \$. Your confusion confuses me -- could you point to a reference on the method you are trying to use? \$\endgroup\$
    – TimWescott
    Jun 20, 2019 at 19:31
  • \$\begingroup\$ "You use the feedback method to find that the open-loop impedance is R2" I though open loop impedance of OPAMP is infinity for analysis purpose I have taken it as R where R tends to infinity and then by formula for ideal OPAMP shunt feedback Rfeedback should be R/L where L is loop gain and as r and l both are tending to infinity so I am struck after that. \$\endgroup\$ Jun 20, 2019 at 19:42

Another logical perspective is when the inverting output applies an opposite voltage such that the differential voltage is always zero, (regardless of voltage gain).

You can see Vin+=0V so Vin/Rin=Zin you input impedance. Even if there was a level shift on input to Vcc/2 and same with signal! The Zin is the same the “virtual ground “ on Vin.

This only applies to linear operation and the assumptions breakdown when the output is saturated.


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