Operational Amplifier Circuit

Discussion in 'Homework Help' started by anglia2011, May 26, 2011.

  1. anglia2011

    Thread Starter New Member

    Mar 21, 2011
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    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.
     
  2. #12

    Expert

    Nov 30, 2010
    16,862
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    Looks like a multiple choice quiz. Only one answer is correct.
     
  3. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    What is the voltage at the inverting terminal of the opamp when in input resistance is low?
     
  4. anglia2011

    Thread Starter New Member

    Mar 21, 2011
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    I think the voltage will be high when the input resistance is low.
     
  5. Georacer

    Moderator

    Nov 25, 2009
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    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?
     
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  6. jegues

    Well-Known Member

    Sep 13, 2010
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    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.
     
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  7. Georacer

    Moderator

    Nov 25, 2009
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    Nice! The time has come when you would teach me something.

    Thanks for the tip!
     
  8. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    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.
     
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  9. anglia2011

    Thread Starter New Member

    Mar 21, 2011
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    Thanks guys for your contributions. So, what is the conclusion?
     
  10. Georacer

    Moderator

    Nov 25, 2009
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    @ 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.
     
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  11. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    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.

    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.
     
  12. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Are you seeking to understand something, or just looking for an answer for an assessment question? We are not supposed to supply the latter.
     
  13. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    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.
     
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