Because of the holidays, TI E2E™ design support forum responses will be delayed from Dec. 25 through Jan. 2. Thank you for your patience.

  • State TI Thinks Resolved
  • Locked Locked
  • Replies 1 reply
  • Answers 1 answer
  • Subscribers 46 subscribers
  • Views 888 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Qustion for Absolute Value Circuit

Eric dang
Prodigy 10 points

Hi All,

 Please reference the following link:

http://www.analog.com/library/analogdialogue/archives/44-04/absolute.html

 In the Improved Absolute Value Circuit page, there have some describe for difference input signal. 

" the input signal is positive, A1 acts as a voltage follower. Both inputs of A2 are at the same potential as the input signal, so A2 simply passes the positive signal to the output."

 Is anybody help me how to understand "A2 simply passes the positive signal to the output."? 

  Any suggestion are well-come! 

Regard,

Eric


  • 0
    TI__Mastermind 22825 points

    Hi Eric,

    The image from ADI's application note that you sent shows that when VIN is above ground, A1 is just a voltage follower and thus will have an output which is the same as VIN. In that case, there is no current in A2 inverting input resistor R (and also none in its feedback resistor) and A2 output will be the same as VIN. Remember no current through A2 feedback resistor means whatever voltage is on its input(s) will appear at its output.

    The Absolute Value function comes about when VIN goes below ground (remember that A1 and A2 are operating with a single positive supply and cannot swing below ground) and A1 output is stuck at 0V. Then, A2 will act as a gain of -1 amplifier with its non-inverting input grounded, so that whatever is at VIN, will appear inverted at A2 Vout.

    Hope this answers your question.

     

    Regards,

    Hooman