Opamp Differental signal?

Discussion in 'General Electronics Chat' started by Vindhyachal Takniki, Feb 11, 2015.

  1. Vindhyachal Takniki

    Thread Starter Member

    Nov 3, 2014
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    1. Can opamp amplify a differential signal which is not ground reference to its won ground.
    2. I have attached a circuit.
    3. If yes, then how to calculate if I am in Vcm range of device in this case.
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The opamp in your circuit will just go to Vout max (~10V) or Vout min (~2V). The input offset voltage is about the same size as your 1mV input, so the output voltage is the open loop gain times the input offset voltage (which is unknown).

    You can never run an opamp open loop, except possibly while using it as a comparator.
     
  3. WBahn

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    Mar 31, 2012
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    1. Yes.
    2. But not like that.
    3. N/A.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    To amplify a differential signal with an op amp you can use a differential configuration such as this.

    For better performance (in IC form it has better common-mode rejection, adjustable gain, etc.) you can use an instrumentation amplifier.
     
  5. Vindhyachal Takniki

    Thread Starter Member

    Nov 3, 2014
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    The circuit I had attached is for refernce only. So any input voltage across it.

    @crutschow , in the circuit you had attached, do V1 & V2 are referenced to ground of opamp used.
    Or they can be from different source whose ground is different from opamp gorund
     
  6. crutschow

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    The V1 and V2 voltages have to be referenced to the op amp power supply ground and must stay withing the maximum common-mode voltage range of the op amp.

    If you want complete isolation then you need to have an isolation circuit that uses optical, capacitive, or magnetic means to allow completely separate grounds between input and output.
     
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  7. WBahn

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    That makes it pretty hard to explain how a hand-held voltmeter works.

    Analyze the differential configuration you showed if V1 and V2 are supplied simply by putting a battery between them. You'll see that it is very well behaved. Failing that, build it and see.
     
  8. tindel

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    Sep 16, 2012
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    Here's how you can do it.
    Screenshot (2).png
     
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  9. crutschow

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    Not at all.
    The ground of the meter input is common with the circuit ground and the output is optically isolated (between the display and your eyes). :rolleyes:
     
  10. tindel

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    A better solution maybe? Uses a instrumentation amplifier. The part is more expensive and I couldn't find a part that is rail-to-rail in and out so it requires a dual supply. The setup only requires two parts though - so less real estate is used. As crutchow points out - you do need one of your inputs referenced to ground somehow... it can be indirectly referenced as the last example shows. The last example was reference to ground, but indirectly and also loads the circuit that you're testing... this one will not load your circuit. Screenshot (4).png
     
  11. WBahn

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    Many do, but not all. There's still the example of the circuit you posted, in which V1 and V2 do not have to be ground referred. Also, what about scopes and other instruments that have true differential inputs?
     
  12. WBahn

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    Have you tried running the sim without tying one of the inputs to ground?
     
  13. crutschow

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    The common mode voltage needs to be within the limits of the amplifiers in these instruments. So it may not be "ground referred" but the two ground voltages need to be within these limits.
    For cases (perhaps not common) where the two grounds can be hundreds of volts different you need some sort of galvanic isolation.
     
  14. WBahn

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    If an input is not ground referred, then what does it mean for ground voltages to be within limits? What's the "ground voltages" in that case?

    Take the circuit you posted as a convenient reference for discussion. The input source (if just a battery between V1 and V2 such that Vbat = (V1-V2)) establishes the differential mode input. The common mode input voltage is established internally (what I call "self-referred", but I don't know what, if any, standard terminology applies), which is not to say that it is independent of Vbat (it would be nice if it were, but that particular topology doesn't achieve that). In this case, Vcm = -[Rg/(R1+R2)]Vbat. The main thing to note, and it was makes it "self-referred", is that if you connect two ground-referred sources V1 and V2, such that (V1-V2)=Vbat still, you get a very different expression for the common mode voltage, namely Vcm = [Rg/(Rg+R2)]V2. Now not only is the common mode signal independent of the differential mode signal, but it is even the opposite polarity.
     
  15. BobTPH

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    Jun 5, 2013
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    Doesn't the statement " the two grounds can be hundreds of volts different" imply that they are not isolated? How can you determine a voltage difference between isolated points?

    If the circuits are isolated, won't they be no longer isolated when you make a measurement as in post #8?

    Bob
     
  16. WBahn

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    "Electrically isolated" and "ground referred" are two different, but often interrelated, things. As soon as you connect any point in one circuit to any point in another circuit, they are no longer isolated and you only get to pick one point, somewhere, to call "ground". But that doesn't necessarily make your input signal "ground referred". In a non-ground-referred diffamp, the input signal from one circuit is only establishing a voltage difference between two particular nodes in the other circuit. That other circuit is going to respond, relative to its ground, to that voltage difference. If you were to establish a specific ground reference between the two circuits, that response will almost certainly change, perhaps considerably.
     
  17. crutschow

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    You can make a measurement without breaking isolation.
    For example, a capacitive voltmeter can detect the voltage difference between two points without making contact.
     
  18. Vindhyachal Takniki

    Thread Starter Member

    Nov 3, 2014
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    @crutschow ,
    What I get differential input applied is ground referred one way or another.
    Case may be directly ground referred or via any other node in the circuit. But it can't be completely isolated. Right? unless we use isolation amps.

    What i get is
    1. Vcm = (V1+v2)/2
    2. Vicr - < VCM < Vicr+

    Now one of opamp I am using has Vicr+ = 8.5(min) & 9.3 (typ)
    Vicr- = -10 (min) & -10.3(typ)

    1. S0 I should take minimum values while designing right?

    2. Along with Vcm ratio, there would be max input voltage applied at V1 or v2 should be there. What is that parameter called? is it Input voltage range?

    3. http://www.ti.com/lit/ds/symlink/lm741.pdf
    In this datasheet I didn't find Vicr parameters ?

    4. I am using ADS1248 , 24 bit adc with PGA, it defines its Vcm range as. Its VCM changes with gain
    AVSS + 0.1 + [(PGAGAIN)(|VINdifferential|)/2] < VCM < AVDD - 0.1 - [(PGAGAIN)(|VINdifferential|)/2]

    5. Do same applies to instrumentation amplifier whose gain is set by single external resistor. Do they have any standard equation?
     
  19. tindel

    Active Member

    Sep 16, 2012
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    Screenshot (5).png
    Yes - here is what it shows... it doesn't know what to do. The simulation could be lying here, but I bet it's not. Maybe I'll request a sample part or two from the manufacture and test it out.
     
  20. crutschow

    Expert

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    Maximum amplifier input voltages are given in the Absolute Maximum Ratings Table as Input Voltage.
    Never use typical or nominal values when designing a circuit.
    For the 741 the common-mode info is listed under Common Mode Rejection Ratio. (Don't use a 741 for a new design if you have any need for good DC accuracy or low noise).
    The instrumentation amplifier values are given in its data sheet, similar to an op amp. There is no standard equation.
     
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