Question about a gain calculation for OPA660

Discussion in 'General Electronics Chat' started by opa627bm, Aug 30, 2012.

  1. opa627bm

    Thread Starter New Member

    Nov 26, 2011
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    Hello everyone,
    I found the introduction of OPA660 on the elektor magazine and I was amazed by the performance of this chip. I never used a OTA before so I want to learn about it. In one of the circuit in the datasheet http://www.ti.com/lit/ds/symlink/opa660.pdf. On page 19, the direct feedback amplifier gain formal has a (R3/2) on top. I don't understand why R3 needs to be divided by 2. I think the G would be G = 1 +R3/R5.
    [​IMG]
    Thanks a lot!
    Lee
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    I found this thread "moderated" (taken out of sight).
    This sometimes happens with new members using links.
    I have approved the thread and made it visible to others.

    Bertus
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The equation give in the datasheet is correct.
    The gain is equal to G = 1+ R3/2R5? The answer is quite simple. This OTA behaves as an ideal transistor. So emitter current must be equal collector current Ie = Ic
    And resistor R3 is connected in parallel to OTA.
    Vin/R5 = Ic + IR3 = Ic + (Vout - Vin)/R3
    This means that current that flows through R3 is equal half of a the emitter current. Or simply half of a Vin/R5 current is provided by the emitter and the second half of a Vin/R5 current is provided by collector.
     
    Last edited: Sep 1, 2012
  4. Harborbreeze

    New Member

    Aug 31, 2012
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    Well this is really interesting question and it also quiet difficult. I need some time to solve this.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, this question is simple if you now how this simply circuit work.
    This "Ideal transistor" is build from connection the diamond buffer and current mirrors.
    [​IMG]

    The current that is flow through R1 is equal Vin/R1 = Ie + Ic
    As you can see the half of a Vin/R1 current comes from diamond buffer (emitter). And the rest of a current comes from current mirrors (collector) and this is why the gain is equal to G = 1+ R2/(2R1).
     
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  6. opa627bm

    Thread Starter New Member

    Nov 26, 2011
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    Thanks for the reply!
    Could you explain more about why the current flows through R3 is 1/2 of Ie ? Thanks!
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    My post 5 is not enough?
     
    opa627bm likes this.
  8. opa627bm

    Thread Starter New Member

    Nov 26, 2011
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    Got it ! Thanks alot!!!!!!!!!:D. I went to Hot Chips conference this week and found the introduction of this chip on a magazine. I asked around (attendees are all electrical engineers and computer architects). To my surprise, none of the people knows about analog anymore :( Is analog a dying art now?
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    :confused:
    1+R2/(2R1) = (2R1+R2)/2R1 ≠ R2/R1

    I agree with the rest of your analysis.
     
    Last edited: Sep 1, 2012
  10. #12

    Expert

    Nov 30, 2010
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    To some degree, yes. As a percentage of all things electronic, analog is becoming a small part. However, analog will never die completely. Interfacing with the real world, like oxygen sensors, Ph probes, strain gauges, air pressure, vacuum measurement, temperature, electric guitars, these things are analog in nature. The art and science of getting analog measurements into digital form will probably never die, and some things are just better in analog form.

    You might say, "The Manifold Absolute Pressure sensor on my car gives a digital output" but somebody had to get it into digital form from its normal analog state.
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Thanks Ron H for point that out. The correct one is given in data sheet.
    And I am a complete idiot not noticing this.
     
  12. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    I dare to say that the whole Nature is analog in nature.
     
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