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Unknown voltage of a Op amp

Discussion in 'Electronics Homework Help' started by Offshore again, Jun 24, 2016.

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  1. Offshore again

    Offshore again

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    Jun 24, 2016
    Hi,
    Hoping for a prompt in the right direction regarding a unkown voltage on a Non Inverting Op Amp.
    Reading through my notes i understand how the resistors R3 and R2 set up a potential divider which supply the Inverting input on the Amp.
    And the fact that neither input is connected to Earth means the Virtual earth principle cannot be used.

    My course notes show that:
    V−=V+ because the op-amp is ideal and V+=Vin

    So is this as simple as as saying V- =100mV ???
    upload_2016-6-24_13-12-42.png

    Any help would be greatly appreciated
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
  3. Offshore again

    Offshore again

    11
    1
    Jun 24, 2016
    Thanks Harald,

    I assume its the same course that you mention (its a distance learning course) but not sure how many or who the other people are.

    The question threw me a bit as it seems to easy....so thanks for the re-assurance
     
  4. LvW

    LvW

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    Apr 12, 2014
    The principle behind this simple answer is as follows:
    As long as the opamp is operated in its linear region (and negative feedback, as in your circuit, enlarges this region considerably) the input differential voltage of a real opamp is in the µV range (because of the very large open-loop gain Aol).
    During calculation of voltages and currents within such a circuit this small differential voltage can be neglected if compared with all other voltages. That means: We can treat the opamp as an IDEAL unit with infinite gain and (consequerntly) zero input voltage.Hence V+ = V- (V+ - V-=0) and V+=0 for V-=0 (virtual ground principle).
     
    Offshore again likes this.
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    one should also point out the voltage drop across R1 for a perfect op-amp would be...?
     
  6. Offshore again

    Offshore again

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    Jun 24, 2016
    In an ideal op amp if R1 is amp impedance then it should be near infinite resistance so therefore the full voltage would be dropped.....is that correct?
     
  7. LvW

    LvW

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    Apr 12, 2014
    In your circuit, R1 is an external ohmic resistor of - perhaps - som kOhms.
     
  8. Offshore again

    Offshore again

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    Jun 24, 2016
    So the voltage drop across the resistor would be proportional to its ohmic value.
    If it were a variable resistor (POT) then it would adjust the gain......
    Am i on the right lines or going down the wrong track??
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What is the input current for a theoretically perfect op-amp?
     
  10. LvW

    LvW

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    Apr 12, 2014
    Yes - of course. Ohms law must be applied.
    However - what about the current ?
     
  11. Offshore again

    Offshore again

    11
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    Jun 24, 2016
    Current should be zero in a ideal op amp due to the impedance being infinite.
     
    (*steve*) likes this.
  12. LvW

    LvW

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    Apr 12, 2014
    So you can find the answer to your questions (post#6 and 8) by yourself?
     
    (*steve*) likes this.
  13. Offshore again

    Offshore again

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    Jun 24, 2016
    So because there is no current (ideal OP-Amp) then the full 100mV is dropped across the resistor.
    And with V- = V+ this is why the Unknown voltage V (in original post) is 100mV
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Reconsider this in light of V = IR
     
  15. Offshore again

    Offshore again

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    Jun 24, 2016
    V=IR
    V=0xR
    V=0

    Seem to be confusing myself now.
     
  16. LvW

    LvW

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    Apr 12, 2014
    Yes.
     
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Possibly, but what does the equation say the voltage drop across the resistor will be?
     
  18. Offshore again

    Offshore again

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    Jun 24, 2016
    If there is no current then the Voltage drop must be zero across the resistor
    V=IR If I is zero then the V must be zero
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Be careful with what you say. "Voltage" and "Voltage drop" are two different things.
     
  20. Offshore again

    Offshore again

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    1
    Jun 24, 2016
    I have done a bit more digging and think I see where we are going.
    Using Kirchoffs Voltage laws
    Vin (100mv)-V1 (volt drop across resistor)= V+

    So therefore volt drop across resistor is 100mV and V+ = 0V
     
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