Op-Amp circuit with a gain of 4

Discussion in 'Homework Help' started by TL314, Apr 7, 2015.

  1. TL314

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    Mar 29, 2015
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    Last edited: Apr 8, 2015
  2. Jony130

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    But haven't you notice that Rf resistor has already 1kΩ resistance. And capacitor together with inductance will form a resonant tank ?
    So you do not need any formula for parallel network. Simply ask yourself one simple question. What is the network impedance at parallel resonance?
    http://www.electronics-tutorials.ws/accircuits/parallel-resonance.html
     
  3. TL314

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    Mar 29, 2015
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    So would the value of capacitor not even matter since the current would be zero in this resonant tank?
     
  4. Jony130

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    No wrong, it will matter, the circuit must be at resonance. But at which frequency ??
     
  5. TL314

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    Mar 29, 2015
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    Resonance the parallel LC tank circuit acts like an open circuit with the circuit current being determined by the resistor, R only. So the total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and Z = R as shown.
    Frequency 40kHz?
     
  6. WBahn

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    The LC tank is at parallel resonance at a particular frequency. That frequency may or may not be 40kHz, but at whatever frequency it is it has an infinite impedance at that frequency.

    In order to get a gain of -2 your circuit, because of the two R values in that circuit, whatever is in parallel with the feedback resistance has to have infinite impedance at the frequency you want it to have a gain of -2 at.

    What does that tell you about what the resonant frequency of the LC tank needs to be for this circuit?
     
  7. TL314

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    Mar 29, 2015
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    that the C value has to make the LC tank have infinite impedance?
     
    Last edited: Apr 7, 2015
  8. WBahn

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    But at what frequency does it have to make the LC tank have infinite impedance? I can choose a value of C that makes it have infinite impedance at 1MHz. Or I can choose a value of C that makes it have infinite impedance at 1kHz. Get in the habit of stating your conditions in terms that you could turn the problem over to someone that is a bit behind you in your studies and that they could then do it.
     
  9. TL314

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    since it tells us a gain of -2 at 40kHz wouldnt I need to find the value of C that creates the LC tank to have infinite impedance? like C=1.58314e^-8
     
  10. WBahn

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    I'm not sure that you are still grasping things (maybe you are and I'm just not getting that warm fuzzy).

    If it asked for a gain of -1.5 at 40kHz, you would NOT be looking for the value of C that causes the LC tank to have infinite impedance at 40kHz just because it tells use to find a certain gain at 40kHz.

    You want to find the value of C the causes the LC tank circuit to have infinite impedance at 40kHz because you need the LC tank circuit to have infinite impedance at the frequency in order for the gain of the circuit to be -2 at that frequency.
     
  11. TL314

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    im not grasping it, so the 40kHz is irrelevant in looking for a value of C that will cause the LC tank to have infinite impedance?
     
  12. TL314

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    the resonant frequency = 1/(2 pi sqrt(LC))
    so any value of C gives a different resonant frequency when compared to L=.001H
    so what resonant frequency would I be looking for to find the value of C that corresponds with it I guess would be my question


    earlier you said "In order to get a gain of -2 your circuit, because of the two R values in that circuit, whatever is in parallel with the feedback resistance has to have infinite impedance at the frequency you want it to have a gain of -2 at."

    the question asked "what capacitor will this circuit have a gain of -2 at 40 kHz. "

    so 40kHz is the frequency I need, so my C value would be 1.58314e^-8
    that would give me a resonant frequency of 40kHz

    right?
     
    Last edited: Apr 7, 2015
  13. TL314

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    oh but then you stated "If it asked for a gain of -1.5 at 40kHz, you would NOT be looking for the value of C that causes the LC tank to have infinite impedance at 40kHz just because it tells use to find a certain gain at 40kHz."

    so those seem to be conflicting statements so im confused again :/
     
  14. WBahn

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    This is right, except that C can't have a value of 1.58314e^-8 because that is just a number, not a capacitance. That's like saying that my height is both 72 and 6. That's meaningless. A physical quantity is inherently tied to the units with which it is measured. I am 72 inches tall and I am 6 feet tall. (72 inches) is equal to (6 feet), but 72 is NOT equal to 6.

    So you got a value for C that is 1.58314e-8 F (or farads).

    Next, you are not justified in expressing this to six sig figs. In fact, since the inductance is only given to one sig fig you technically aren't justified to more than one or two sig figs. In engineering, the convention is to use three sig figs unless the situation clearly justifies more. The convention is also to use engineering prefixes, so you would have C=15.8 nF (or even just 15 nF).

    Now, putting all of that aside, look at the reasoning presented here. We are finding the value of C that results in resonance at 40kHz not because 40kHz happens to be the frequency we are operating at. That is the point that I'm not sure that you are catching.
     
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  15. WBahn

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    Okay, so let's change the problem slightly and make that feedback resistor 2000Ω instead of 1000Ω. Everything else is the same, namely you are trying to find the value of C that makes the gain of the circuit -2 at a frequency of 40kHz.

    How would you proceed?
     
  16. TL314

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    Mar 29, 2015
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    oh yes im sorry, I wouldnt be given full credit without the units thanks you.

    I am kind of understanding it but what if the frequency and value of C wasnt given, how would I find the resonance and value for C? or in your case your question. if my fb resistor was 2000 then my gain would be -4 if I kept everything the same right?
     
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