Differential Input Voltage zero in Opamp

Discussion in 'General Electronics Chat' started by ankurtech, Jan 26, 2012.

  1. ankurtech

    Thread Starter New Member

    Jan 26, 2012
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    I am Grad student. I was reading basics of opamps & read that diff voltage is ideally zero. So that voltage at inverting terminal(v1) is equal to non-inverting terminal(v2) ideally. (mostly in all derivation this has been done). i.e v1=v2. But if I apply volatge at V1 only & ground v2 to use as comparator. Then how does it works if volatge is equal. I found this confusing.
    ->. In case of opamp as comparator,(besides it drawbacks), wll give high/low if v1 not equal to v2. So what about vid now.
    As I have read vid = Vout/A. A being very large, vid ~=0. Isn't it contradiction b/w two points.


    ->. Even in differential opamp with feedback,
    Vo= (-Rf/R1)(v1-v2) //terms carry their usual meaning
    if v1=v2, then above vout gets zero.(Obviously this don't happen, just some poor assumptions of mine, as I didn't get that concept till now).
    Kindly help me to understand this.
     
  2. ankurtech

    Thread Starter New Member

    Jan 26, 2012
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    any help :)
     
  3. Audioguru

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    The voltages at the inputs of an opamp are almost the same when the opamp has negative feedback and the output is not saturated, because its open-loop voltage gain is extremely high (200,000 or 1 million).
    A comparator has no negative feedback and its output is always saturated so the input voltages can be different.
     
  4. ankurtech

    Thread Starter New Member

    Jan 26, 2012
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    In case of opamp as comparator,(besides it drawbacks), where v1 not equal to v2. So vid is not zero. It may go upto (V1-v2) or 0.7V if have internal clamp diode But I have read vid = Vout/A. A being very large, vid ~=0. This two equation does not fit together. How to explain this.
     
  5. Audioguru

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    The output of an opamp is limited by the voltage of its power supply.
    If the opamp has a voltage gain of 1 million then with an input of +15mV its output will try to go to 15 thousand volts which is impossible. Its output simply clips at a voltage that is its saturation voltage.
     
    Last edited: Jan 29, 2012
  6. ankurtech

    Thread Starter New Member

    Jan 26, 2012
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    Got ur point. But how to explain this in equivalent circuit diagram of opamp.How to explain it there.
     
  7. Audioguru

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    The equivalent circuit of an opamp has a voltage gain of infinity (1 million is almost infinity).
    The input voltages of an opamp operate properly only within its allowed common-mode voltage range.
    Its output has a saturatrion voltage loss.
     
  8. ankurtech

    Thread Starter New Member

    Jan 26, 2012
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    In ideal case, we take V1=V2 =0V, no current flows into terminals of opamp since infinite input resistance, we also take these while deriving equations of practical opamp.

    If there no voltage at pin, no current at pins that means no signal, that what does opamp amplify or work on?
    (Some poor understanding of topic of mine, any help will be appreciated)
     
  9. Jony130

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  10. Audioguru

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    The DC input voltage has nothing to do with the DC input current. An NPN transistor will have a positive input current and a PNP transistor will have a negative input current.

    Then there is no input DC voltage but there is a signal input voltage that the opamp amplifies.

    There are many opamps with Jfet transistors or Mosfet transistors at the input that use no input bias current. Their input resistance is so high it is almost unmeasurable.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    The voltage across the input terminals of a real op amp is not completely zero if the op amp is operating as a linear amplifier with negative feedback. The input differential voltage (ignoring any offset) is equal to the output voltage divided by the open-loop gain (1V divided by 1 million gives 1μV, for example). In doing the calculations for an ideal closed-loop op amp circuit you just take V1-V2 = 0V since it simplifies the calculations and the error in doing this is normally small as compared to a real op amp.
     
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