Op amps

Discussion in 'Homework Help' started by joka2188, Nov 20, 2006.

  1. joka2188

    Thread Starter New Member

    Nov 10, 2006
    9
    0
    Iv just had a go at three questions, which are in the attachment, can anyone please tell me if my answers are correct so i could go back and figure out where to go from there. Thanks for all your help, much apprieciated.
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,271
    1,065
    You might want to show your work. It will help the members provide the proper guidance.
     
  3. joka2188

    Thread Starter New Member

    Nov 10, 2006
    9
    0
    I done most of it on scrap, so ima have to get the equations again and do them and then hopefully if i get the same answers il post them up if not ima b scratchin my head in stress lol thnks anyway
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Can you scan the work on the scrap paper? You will then be able to upload the scanned image as a Jpeg.

    Dave
     
  5. richbrune

    Senior Member

    Oct 28, 2005
    106
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    The LED current is not correct, here's a clue:

    The op-amp configuration shown is called a comparator. Whenever the positive input (pin 1) is at a higher voltage then the negative input (pin 2), the op amp output will swing to theoretical 12v (actual about 11.5v). When the voltage at the positive input (pin 3) is at a LOWER, voltage then the negative input, the op amp output will swing to theoretical 0 volts.
    Calculate the voltage at the inputs as follows:
    Vpositive input = (R4/R3+R4)*12
    Vnegative input = (R2/R2+R2)*12
    Next you must do the following calculation for current, if the op amp is "on":
    (12v-1.2v)/1000 (where 1.2 volts is the forward voltage drop of the LED)
    or this, if the op amp is "off" (or any voltage less than the 1.2v of the LED turn on voltage):
    0v) 0/1000=0 (no current)
    Please note that the bottom example question is also a "trick" question:
    U1's input has positive feedback. Any positive signal applied to the negative input will cause the output to increase, which will in turn feed back into the positive input, causing DEVICE DESTRUCTION.
     
  6. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    The last question has been dealt with in another thread.

    It looks like the first circuit has been as carelessly put together as is the last one. The circuit is of a comparitor, but no value is given for R2 and no details are given of the op-amp output characteristics.
     
  7. richbrune

    Senior Member

    Oct 28, 2005
    106
    0
    The value of R2 is imbedded in the drawing it's hard to see, but I think it's 1Kohm.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Right you are richbrune R2 is 1kOhm, if you zoom the page in Word it becomes moderately clearer.

    Dave
     
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