Trouble Understanding Single Supply Op-Amp Setup

Discussion in 'General Electronics Chat' started by madscientistdan, Mar 6, 2012.

  1. madscientistdan

    Thread Starter Member

    Mar 6, 2012
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    Does anyone know how to explain non-inverting amps?
     
    Last edited: Aug 20, 2012
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I redraw your diagram

    [​IMG]

    And as you can see this diagram is almost a classic diagram.

    R2, R4 , R4 provide DC bias voltage. Because every "active device" to work properly as an amplifier supplied from a single source need proper bias circuit. When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region". In case of single supply op amp you have to do the same think. We need to bias the op amp somewhere in the middle of his "linear region".

    C2 ensure that the DC voltage gain is equal to 1.

    But AC gain is equal to
    Av = 1+ R6/R5 = 101

    As for this 0.584V it is a measurement error.
    Becaues R4 has 100M and you use a multimetr which has 10M internal resistance.
    So you form a voltage divider, solve for Va in this circuit

    [​IMG]

    Va ≈ 6V * 10M/(10M + 100M) ≈ 6* 10/110 ≈ 0.5454V
     
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    Last edited: Mar 6, 2012
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  3. madscientistdan

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    Mar 6, 2012
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    Hi Jony130

    Wow thank you so much this makes way more sense now! I just realized I should have drawn a better schematic like you did, sorry about that. I didn't realize the voltage reading error was due to the internal resistance of 10 MegaOhms from the multimeter.



    Is the D3 diode for reverse polarity protection, or does it also serve another purpose?
    I'm guessing the C3 capacitor is for filtering?
     
    Last edited: Mar 10, 2012
  4. Jony130

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    This resistor determined the input resistance for AC signals
    Rin = R2||R3 + R4 ≈ R4


    Only reverse polarity protection.

    C2 and C3 determine the amplifier frequency response.
    See the example for the ideal op amp case

    [​IMG]

    But for your op amp OPA637 the gain bandwidth product is equal to 80MHz
    So the gain will start to drop at 235KHz at rate 6dB/octave and at Fc = 80MHz/101 = 790KHz gain start drop even faster at rate 12dB/octave. Google bode plot.

    So C3 prevent circuit form osculation. To be sure the full bode plot is needed.
     
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  5. madscientistdan

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    Mar 6, 2012
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    Hi Jony130,

    Thanks so much for the response!
     
    Last edited: Aug 20, 2012
  6. Jony130

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    R1 and R2 are simply input/output protection resistor.

    R7 is pull down resistor. Prevents output from floating.
     
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  7. Audioguru

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    Dec 20, 2007
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    Because the piezo transducer must feed into a 100M extremely high resistance. Therefore R4 is 100M. It separates the input of the opamp and also the piezo from the 50k total resistance of the R2 and R3 voltage divider.

    Because your multimeter loads down the 100M resistance which causes the voltage measurement to be too low.

    Because the R2 and R3 voltage divider causes the input of the opamp to be +6V and since the opamp has a DC voltage gain of 1 then its output is also +6V.

    I repeat, your meter is loading down the input DC voltage. The DC input is +6V. Therefore the output DC voltage is also +6V. C2 causes the DC gain to be 1.
     
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  8. madscientistdan

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    Mar 6, 2012
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    Thanks for the explanations.
     
    Last edited: Aug 20, 2012
  9. Jony130

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    As for C3 I made a stupid mistake.
    F3 is not equal to 23.5MHz It should be
    F3 = 0.16/ ( C3* R6 ) = 235KHz
     
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  10. jimkeith

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    Oct 26, 2011
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    Check out page 13 of the datasheet http://www.ti.com/lit/ds/symlink/opa637.pdf

    They indicate layout as being a factor (stray capacitance), but the real hooker is the low, but non-specified input capacitance that adds phase shift to the feedback.
     
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  11. madscientistdan

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    Mar 6, 2012
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    Thanks jimkeith for your response. Its still not easy for me to understand how that capacitor interacts with the non-specified input capacitance but I'm guessing its hard to without looking at a bode plot.
     
  12. madscientistdan

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    Mar 6, 2012
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    Thanks Jony130.

    Is there an easy analogy or explanation for computing the F1, F2, F3, and F4 cutoff frequencies of the amplifier response?

    I cannot find any of those computations in the design references that I have for op amps. For instance, why is F2 computed without R6 and C3 in the equation? I don't understand why F1 doesn't use C3, and why F3 doesn't use R5 and C2, and why F4 doesn't use C2 in the equations.

    Thanks.
     
  13. Jony130

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    Read this form page 13
    http://www.analogzone.com/acqt0704.pdf
    And you will see how C3 helps.
     
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  14. Jony130

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    You need to find the transfer function.
    And by knowing transfer function you can find pols and zeros.
    And since C3<<C2 we can analyze the circuit much easier.
    Simply if we analyze C2 influence we remove C3 from the circuit, and vice versa, we remove C2 when we analyze C3 influence.
    http://www.electro-tech-online.com/general-electronics-chat/120949-cutoff-frequency.html#post997029
     
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  15. madscientistdan

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    Mar 6, 2012
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    Thanks again Jony130. I will read the material in the links you attached and try and learn about both the op amp frequency response and stability.
     
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