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Operational amplifier exam questions

  1. Jun 22, 2011 #1
    1. The problem statement, all variables and given/known data

    See attached file for relevant questions. I'm finding the Op-amp questions in section A very difficult. Any help with these questions would be much apprietiated.


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 22, 2011 #2

    uart

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    I think you should choose one particular question (part) to ask about. You can always ask another question later.
     
  4. Jun 22, 2011 #3
    Okay, that seems more sensible.
    I have attemped Question 1. Am I correct in saying that you would draw the Inverting Op-amp circuit diagram, remove the op-amp component since the inputs have a very large input impedance (and thus no current flows into them), and label the virtual earth point between the feedback and input resistors?
    Also, the gain would be given by -Rf/Rin, therefore, -1M/100K = -10
     
    Last edited: Jun 22, 2011
  5. Jun 22, 2011 #4

    uart

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    Yes that is correct, however this question is more about frequency response and gain bandwidth product.

    From the question see if you can figure out the following.

    - What is the opamp DC gain (without feedback)
    - What is the opamp 3dB bandwidth (dominant pole)
    - What is the opamp gain bandwidth product.

    * BTW. There is a bit of a trick with this question in that the relevant closed loop gain (for calculating gain-BW product) for the inverting opamp is actually 1+|A_cl|.
     
  6. Jun 23, 2011 #5
    Thanks for your reply. To caluate the DC gain (without feedback) would I use:
    A_d = 2x10^5 / (1+jf/7) and set f=0 since it is DC. Therefore, I get A_d = 2x10^5. In dB, this is a gain of 106.02 dB. To find the -3dB point, do 106.02 - 3 = 103.02dB and convert back to magnitude using, 10^(103.02/20) = 141579.
    Am I on the right path to the solution?
     
  7. Jun 23, 2011 #6

    uart

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    Yep A_0 = 2x10^5 is good. :)

    The -3dB point does occur at approx 141400 (gain) but that's a really long winded way to go about finding it. The -3dB points are defined as having a gain of 1/sqrt(2) times the maximum, so you could just divide 2x10^5 by sqrt(2) to get there more easily.

    More importantly however you don't actually need to find that -3dB gain (141400) in order to find the -3dB frequency. Just look at the transfer function A_d = 2x10^5 / (1 + j (f/7) ) and tell me what value of "f" you need to make the magnitude of the denominator equal to sqrt(2).

    Hint : What is the magnitude of (1+j) ?
     
  8. Jun 23, 2011 #7
    The magnitude of (1+j) is sqrt(2) or 1.414. So therefore f would be 7Hz at the -3dB point.
     
  9. Jun 23, 2011 #8

    uart

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    Yep that's correct. So therefore the gain bandwidth product is 7x2x10^5 = 1.4x10^6 Hz.
     
  10. Jun 23, 2011 #9
    Okay. So Gain-bandwidth product = Open-loop gain X cut-off frequency. Is this the answer to the small signal bandwidth or is there a lot more computation to do? (I'm guessing there is more to do)
     
    Last edited: Jun 23, 2011
  11. Jun 23, 2011 #10

    uart

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    No there's not much more to it, that's the simplicity and beauty of the "gain bandwidth product" method.

    The open loop gain times the open loop -3dB freq is approx equal to the closed loop gain times the closed loop frequency. That is, the gain bandwidth product is approximately a constant.

    BTW. The gain bandwidth product is most accurate for the non-inverting configuration, so make sure you review my hint in reply #4 above.
     
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