Differential, Op-amp questions

Discussion in 'Homework Help' started by jegues, Sep 28, 2010.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    See figure for problem statement.

    Assume V+ = V- on both Op-amps, and the current flowing into both Op-amps is 0A.

    I'm confused as to what condition I would need to fufill in order to have a "perfect" differential amplifier?(perfect on paper atleast)
     
  2. Georacer

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    Nov 25, 2009
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    I am not quite sure what the exercise means as a perfect diff. amplifier. I guess it wants the output signal to be an exact analogy of the differense of the two signals i.e. c(V2-V1) and not cV2-dV1. You can calculate R4/R3 by equating the constants in front of V2 and V1 (c=d).
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Can someone verify that the condition for a "perfect" differential is indeed when the two coefficients of V1 and V2 are identical?

    I should be able to handle the rest if that's the case.
     
  4. t_n_k

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    Mar 6, 2009
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    Perfect just means the output is purely a fixed multiple of the difference between V1 and V2 and unaffected by the common mode value of V1 & V2.

    Check this discussion of a practical differential amplifier from National Semiconductor AN 29 "IC Op Amp Beats FETs on Input Current" page 12.
     
  5. Georacer

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    You mean, like I said, c \cdot (V_2 - V_1)?

    And what does the attachment mean my "resistor value matching" and how does that affect the common mode output?

    In our analysis of the OP's circuit, we ended up with a Transfer Function which was unaffected by the Common Mode Signal, were we not?
     
  6. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    So would this be correct?

    See figure attached
     
  7. Georacer

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    That's what I calculated myself too. But since the only guideline is R3=10R4, we could pick any values for those two. Unless t_n_k says different, about mathing resistor values and such.

    To t_n_k: I just noticed that your username is acutally t_n_k and not t n k. The link underline didn't let me see the underscores.
     
  8. t_n_k

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    Hi there,

    Agree with the preceding posts.

    Interesting that the selected circuit values don't give much amplification of the difference signal.

    Don't concern yourself about the tag name Georacer - I'm just jealous I didn't choose something more like yours! Mine's rather 'pedestrian'. Heard on the news that Common Market countries are having issues with workers protesting about the economic strictures being placed on citizens (unwanted wage & pension cuts) - Greece got a mention. Everything OK with you and yours I trust.
     
  9. t_n_k

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    I think you know this but ....

    Resistor value matching applies to actual circuits in which one tries to get the critical values as close as possible.

    In the case of the amplifier in question the relationship for Vo is

    Vo=[1+(R4/R3)]*V2-(R4/R3)*[1+(R2/R1)]*V1

    So if R1=R4 and R2=R3 exactly then Vo is independent of the common mode value.

    For the sake of discussion, let Vcm=common mode voltage.

    Also let V2=Vdiff+Vcm and V1=Vcm. So V2-V1=Vdiff

    Vo=[1+(R4/R3)]*(Vdiff+Vcm)-(R4/R3)*[1+(R2/R1)]*Vcm
    Vo=[1+(R4/R3)]*Vdiff +{[1+(R4/R3)]-(R4/R3)*[1+(R2/R1)]}*Vcm

    So Vo is effected by the common mode voltage to the extent by which the relationship

    {[1+(R4/R3)]-(R4/R3)*[1+(R2/R1)]}≠0

    The requirement being that R1=R4 and R2=R3 for no effect.
     
  10. Georacer

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    Nov 25, 2009
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    Thank you for the concern. It is true that things go from bad to worse in Europe, and I 'm not just talking about the economy. Politics seem to go from right to righter (if you understand what I mean) and many social privileges are being cut in favor of debt payment and introduction to a more free market. Thankfully, since I am a student in university, making money is not my primary concern and my father has a decent enough job to support us all.

    On the matter at hand now: I know what a common mode signal is. But in order to reject it isn't it enought to set \frac{R_4}{R_3}=\frac{R_1}{R_2}? All the math seems to say so...:confused:
     
  11. t_n_k

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    Good to hear life is going OK.

    Re: Common Mode.

    It's one thing to say the R values just have to be 'so', but it's a different story when you actually try to find the exact values in practice - & even with a 0.1% tolerance one ends up with some (albeit pretty small) common mode breakthrough. As the part AP Note I posted in Post #4 points out, 0.1% tolerance on the R values claims a minimum 80dB rejection of the common mode voltage. For what it's worth 80dB is pretty good.

    With a 1% resistor tolerance say, the situation is going to be worse - probably only marginally acceptable. Since a common mode often includes noise, one tends to get more noise at the output, which is the enemy when one is often dealing with very low magnitude difference sensor signals.

    Off-the-shelf true differential amplifiers will have precision laser trimmed gain setting resistors included 'on board'.
     
  12. Georacer

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    So is it ok to say that we don't need exact values for R3 and R4 but an exact match for their ratio? If so, multimeter ftw...
     
  13. The Electrician

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    Oct 9, 2007
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    If the inputs to a differential amplifier are V1 and V2, I would expect the common mode input to be (V2+V1)/2, and the differential input to be (V2-V1).

    Consider your choce of V2=Vdiff+Vcm and V1=Vcm. Then we have Vcommon = (Vdiff+Vcm + Vcm)/2 = Vcm + Vdiff/2, which doesn't seem right. I think it works out better to choose V2 = (Vcm+Vdiff/2) and V1 = (Vcm-Vdiff/2). Then these two expressions:

    Vo=[1+(R4/R3)]*(Vdiff+Vcm)-(R4/R3)*[1+(R2/R1)]*Vcm
    Vo=[1+(R4/R3)]*Vdiff +{[1+(R4/R3)]-(R4/R3)*[1+(R2/R1)]}*Vcm

    are somewhat different.

    Requiring R1=R4 and R2=R3 does give infinite common mode rejection, but that requirement is stricter than needed. As Georacer suggested, it is sufficient to require R1/R2 = R4/R3, and that's easy to do exactly (nominally speaking) without setting R1=R4 and R2=R3; for example, just select another pair of resistors from the same tolerance scheme a decade higher or lower.

    AN-29 shows 0.1% resistors in the schematic, but in the text they say "...a 1 percent mismatch between two resistors lowering the common mode rejection to 80 dB." Perhaps they meant to say "...a 0.1% mismatch..."

    They don't say which two resistors they are referring to; I guess find the worst case pair.

    I don't get 80 dB common mode rejection with a 0.1% mismatch, much less with a 1% mismatch. I get about 60 dB with a .1% mismatch, and about 40 dB with a 1% mismatch. What do you get if you run the numbers?

    Note well, they don't say "using 1% resistors"; they say "a 1% mismatch". If we use 1% resistors, and a pair of them are used in the worst case, we would have one of them 1% low and the other 1% high, for a 2% mismatch.
     
  14. t_n_k

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    Yes - I had a feeling of disquiet about Vcm which you've correctly picked up. I guess the difference depends largely on the magnitude of Vdiff. But you are correct - as I've come to expect from your good self.

    On the CMRR figures, I have done some tinkering overnight. I didn't get the same figure as the NS app note for the stated 0.1% tolerances (?) either. I'm not at home at the moment so I'll check what I did on my Excel spreadsheet when I do get back. Perhaps the app note author(s) include other contributions to the CMRR - say from the amplifier itself - but it's probably to do with the matters of mismatch and tolerance you raised. I'd think the op-amp would have a CMRR much better than 80dB. I think I've read somewhere that even the much berated LM741 has a CMRR~90dB.

    I did comment that a CMRR of 80dB was "pretty good". On reflection it doesn't look so great if the common mode signal is (say) 60dB above the difference signal. Which I guess is possible if one was trying to amplify a difference of 1mV rms sitting on a common mode noise level of 1V rms.

    It would be interesting to get some insights from those guru's who have worked (or work) with this technology. Essentially on what's achievable in a good practical design. And what are the pitfalls.
     
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