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Op-Amp circuit with applied external compensation

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    FIGURE 3(b) shows the THS4021 in a practical op-amp circuit with applied external compensation. Determine an expression for the low frequency gain of the circuit.

    2. Relevant equations


    3. The attempt at a solution
    I really don't know where to start.
    Is there a virtual earth at the inverting input?
     

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  3. Sep 16, 2016 #2

    rude man

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    Big hint: " ... low-frequency gain ...".
     
  4. Sep 16, 2016 #3
    Thanks Rude Man
    I did see that in the question.
    [tex] f_c=\frac{1}{2πRC}[/tex]
    This is the only equation I can find for low pass RC circuit.
     
  5. Sep 16, 2016 #4

    gneill

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    I think you've missed @rude man 's hint. What's the lowest frequency you can think of?
     
  6. Sep 19, 2016 #5
    Fractions above zero?
     
  7. Sep 19, 2016 #6

    gneill

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    What's the lower limit?
     
  8. Sep 19, 2016 #7
    0Hz

    Here is some more info on the op-amp.
     

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  9. Sep 19, 2016 #8

    gneill

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    Yup. So What's the 0 Hz gain of the circuit?
     
  10. Sep 20, 2016 #9
    From the op-amps characteristic graph we can see 90 dB?
    so how do I "Determine an expression for the low frequency gain of the circuit"
     
  11. Sep 20, 2016 #10

    gneill

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    The op-amp is surrounded by a feedback network. Start with the ideal op-amp model and find the gain. If the result is much smaller than 90 dB then you can take the result as accurate enough. Otherwise, if what you get is a large fraction of 90 dB you'll have to introduce a more realistic model for the op-amp and go through the work of analyzing it. But that's unlikely to happen here, since I can see by inspection of the circuit that the gain won't be large...
     
  12. Sep 20, 2016 #11

    rude man

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    Hint: #1: you don't need any "op amp characteristics" here.
    Hint #2: what can you say about a capacitor at dc after a "long time" has passed?
     
  13. Sep 20, 2016 #12
    Ah so when the cap has charged current through it stops.

    and the general op-amp inverting input gain equation is [tex]{V_{out}}=-\frac{R_f}{R_1}V_{in}[/tex].
    or [tex]A_v=\frac{V_{out}}{V_{in}}=\frac{Rf}{R_1}[/tex]
     
    Last edited by a moderator: Sep 21, 2016
  14. Sep 20, 2016 #13

    rude man

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    OK, what would that be using the nomenclature in your figure?
     
  15. Sep 20, 2016 #14
    So the gain Av
    [tex]A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}[/tex]
     
  16. Sep 20, 2016 #15

    rude man

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    Ooh, almost. Sign?
     
  17. Sep 21, 2016 #16
    I put it in the original equation then missed it out on the second.
    Because it is inverting
    [tex]A_v=\frac{V_{out}}{V_{in}}=-\frac{R_2}{R_1}[/tex]
     
  18. Sep 21, 2016 #17

    rude man

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    Right.
     
  19. Sep 21, 2016 #18
    Thanks rude man and gneill for all your help.
     
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