Question dealing with gain bandwidth of inverting op-amp

Discussion in 'Homework Help' started by uofmx12, Aug 31, 2011.

  1. uofmx12

    Thread Starter Member

    Mar 8, 2011
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    In this circuit:
    http://i812.photobucket.com/albums/zz41/uofmx12/E31ckt.jpg

    A) what two estimates of the gain bandwidth product of your circuit can you make
    B) Needed to make a circuit with bandwidth of 50kHz with gain of 1000V/V, would this circuit be chosen?
    -No?

    C) What if requirements were 10k Hz and 100V/V, would this circuit be chosen?
    -Yes?
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    Is there any more information provided? For exampe, did they mention which OPAMP it is, or give the gain-bandwidth product for the OPAMP?
     
  3. uofmx12

    Thread Starter Member

    Mar 8, 2011
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    inverting op-amp circuit and 100Hz frequency. That was provided.
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    OK, that helps. Do you understand what that means? In other words do you know the meaning of gain bandwidth product? We need to have a baseline of whether your trouble is that you don't understand the definition, or if you just don't know how to apply the definition to that particular problem.

    Your above information is still a little vague though. So is the 100 Hz frequency referenced to a particular inverting opamp, or to the one in your circuit? Or, is it referenced to the OPAMP itself, without any components (open loop)?
     
    Last edited: Aug 31, 2011
  5. Audioguru

    New Member

    Dec 20, 2007
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    The input resistors attenuate the input 101 times then the feedback resistors cause the opamp to amplify 1001 times.
    The result is an amplification of only about 9.9 times. The output will probably be very noisy (hiss).

    An audio opamp (something better and newer than a lousy old 741 opamp) that is open loop has a gain of 33,000 to about one million at 100Hz.
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    Isn't it 1000 times for the 1M and 1K feedback portion, and net gain of -9.8 due to voltage attenuation and source resistance of the input voltage divider?
     
    Last edited: Aug 31, 2011
  7. Audioguru

    New Member

    Dec 20, 2007
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    The input resistors attenuate 101 times and the feedback causes 1001 times gain.
    So the result is a gain of about only -9.9 times.
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    How are you defining your feedback portion? If it's the two resistors 1M and 1K, then this is an inverting amplifier configuration and the gain of that portion is -R2/R1=-1000.

    Looking at the whole circuit, you can make a Thevenin equivalent source of Vi*10/1010 and a source resistance of 10 in parallel with 1000, which is about 9.9 Ohms. Or Vi/101 and 9.9 ohm source resistance. Now the effective input resistance on the inverting amplifier (looking in from the ideal Thevenin voltage of Vi/101) is 1009.9 ohms due to the source resistance, and the net gain of the inverting Opamp is -1000000/1009.9=-990. Now combine this with the attenuated voltage and the net gain is -990/101 which is about equal to 9.8. Isn't it?

    Alternatively, you can analyze the full circuit with equations and you come up with Av=-R3*R2/(R4*R1+R3*(R4+R1)) where R2 and R1 are the OPAMP feedback resistors 1M and 1K, respectively; and R4 and R3 are the voltage divider resistors 1000 and 10, respectively. Again, the gain is about -9.8.
     
    Last edited: Sep 1, 2011
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Another way of calculating the signal gain is to note that Ra, Rb and R1 form a T network. Use the delta-wye transformation to convert it into a pi network. Then the shunt resistors of the pi network can be ignored because the one across the input voltage has no effect if the input voltage source has zero output impedance, and the shunt across the - input of the opamp has no effect because that terminal is a virtual ground.

    We are left with the series element of the pi network, which is 102000 ohms. Then the signal gain is 1000000/102000 = 500/51 = 9.80392
     
  10. steveb

    Senior Member

    Jul 3, 2008
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    Very elegant method !

    I'm still a little confused on the OPs question. I have the gist of what he's asking, but the precise question and the precise information he was given is not fully clear to me.

    For example, "What are the two estimates one can make?". I'm not sure what this means. It seems we need to have the OPAMPs gain-bandwidth product to make any estimates, and once we have that, shouldn't we just have one estimate for the entire circuit GB product?

    Maybe I'm missing something?
     
  11. The Electrician

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    Oct 9, 2007
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  12. uofmx12

    Thread Starter Member

    Mar 8, 2011
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    That is all that was asked. I left off no other information. Not an important question now.
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Thought I'd simulate the results for a couple of different op-amps ...
     
  14. ftsolutions

    Active Member

    Nov 21, 2009
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    sometimes I think that professors need to actually look at a databook once in a while so that their questions are not quite so open to variable interpretation (or generate more questions). Maybe that was the intent?
     
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