Question:Why is this circuit not suitable for amplifying a signal from a high impedance source?

From my research, I found the following.

1) Because of the loading Ri at the input, there is a low impedance at the input and therefore the circuit is not suitable for amplifying a signal from high input impedance source.

2) Because of infinite input impedance and hence zero current and zero voltage at the inverting input.

However, I am not sure the above points are correct because it seems as if they are contradicting!

House please I need your opinions.

Thanks.

 

Looks like a multiple choice quiz. Only one answer is correct.

 

What is the voltage at the inverting terminal of the opamp when in input resistance is low?

 

What is the voltage at the inverting terminal of the opamp when in input resistance is low?

I think the voltage will be high when the input resistance is low.

 

I don't see a problem with any output impedance of its source, provided that Ri is large enough. The OpAmp can provide good isolation if Rf and Ri are adequately large.

Can somebody elaborate on this?

 

In the case of the inverting op-amp, you'd like to have a large Ri but if the required gain is also large then R2 could become impractically large.(i.e. >2MΩ)

The inverting configuration suffers from low input resistance.

A solution to this is the inverting op-amp with a T network configuration for it's feedback resistors.

 

Nice! The time has come when you would teach me something.

Thanks for the tip!

 

The voltage at the inverting terminal will always be nearly GND since the noninverting terminal is at GND. The input resistance value does not have a large effect on that.

In the case of an ideal opamp, the voltage at the inverting terminal is always equal to the voltage at the noninverting terminal. And, the current into the inverting and noninverting terminals are zero.

 

Thanks guys for your contributions. So, what is the conclusion?

 

@ StayatHomeElectronics

Yes but the issue was that with a given (high) input resistance we couldn't achieve a large enough gain with the standard inverting OpAmp configuration. Without inserting an extra chip, jegues gave a solution.

 

1) Because of the loading Ri at the input, there is a low impedance at the input and therefore the circuit is not suitable for amplifying a signal from high input impedance source.

That circuit is the classic inverting amplifier. The gain is Rf/Ri, and the input impedance is Ri.

For any op amp the two input terminals are roughly equal in voltage. Grounding the + input makes the - input also at ground potential. Thus any voltage at the input is also across Ri, and the input current is Vin/Ri

And that is why the input impedance is the same as Ri.

2) Because of infinite input impedance and hence zero current and zero voltage at the inverting input.

Well, the op amp itself has a very high (not infinite but pretty close) input impedance, not the circuit using the op amp. I think you are getting the two confused:

Op amp: very high input impedance
Inverting amplifier (using an op amp): input impedance = Ri

Everything you said is true, just true at different places.

 

Thanks guys for your contributions. So, what is the conclusion?

Are you seeking to understand something, or just looking for an answer for an assessment question? We are not supposed to supply the latter.

 

@ StayatHomeElectronics

Yes but the issue was that with a given (high) input resistance we couldn't achieve a large enough gain with the standard inverting OpAmp configuration. Without inserting an extra chip, jegues gave a solution.

The solution may have been given but the OP doesn't necessarily fully understand the solution... Just trying to help explain the operation of the circuit.