Integrated Circuit 741

Discussion in 'Homework Help' started by chrischristian, Feb 24, 2008.

  1. chrischristian

    Thread Starter Active Member

    Feb 22, 2008
    Plese, can someone help me to understand from the internal strcture of 741 opamp that how possitive input give positive output and how negative input give inverted output, especislly after the differential amplifier section. The internal structure is on!! and plese try to make it discriptive than mathamatical , I know this is too much but plese help me !!!!!!!!
  2. Papabravo


    Feb 24, 2006
    Your description is slightly off the mark. The inputs do not affect the output independently as you have implied. The absolute magnitude of the inputs is irrelevant as long as they remain within the "Common Mode Range". This parameter is given on datasheets and typically is contained to a range which is smaller than the power supply rails. There are amplifiers whose common mode range includes one or both of the supply rails.

    The critcal point, about all the opamps you will ever see, besides all the ones that you won't, is that the output depends ONLY on the difference between the two inputs.

    Example #1
    Av = 100,000, V+=12 V, V- = -12V, CMR = {-10.5,..10.5}
    IN+ = 8.000001 VDC
    IN- = 8.000000 VDC
    Vo = (8.000001 - 8.0)*100,000 = 100 mV

    Example #2
    Same as Example #1
    IN+ = 6.000002
    IN- = 5.999998
    Vo = (6.000002 - 5.999998)*100,000 = 400 mV

    Example #3
    Same as Example #1
    IN+ = 8.000000
    IN- = 8.000001
    Vo = (8.000000 - 8.000001) * 100,000 = -100 mV

    Example #4
    Same as Example#1
    IN+ = -8.000000
    IN- = -8.000001
    Vo = (-8.000000 - (-8.000001)) * 100,000 = +100 mV !!

    It should be clear from these examples that the differential input stage acts only on the difference between the two inputs and not their absolute magnitude.
  3. chrischristian

    Thread Starter Active Member

    Feb 22, 2008
    Thank you for your time , but my question was different, please read the attachment !!!
  4. Papabravo


    Feb 24, 2006
    You're right, I don't know how to explain it to you. Until you stop considering the inputs one at a time I don't think anyone will be able to get through to you. That's just your tough luck I guess.
  5. spar59

    Well-Known Member

    Aug 4, 2007
    I think I am right in this - op amp internal circuitry is often extremely complex since transistors are easy to create and other componenets more awkward - hence a glut of transistors.

    T8 (in conjunction with other components) forms a constant current source.
    Output to the second stage is taken from the collector of T6.

    A positive input at the inverting input increases current through T2 & T4 raising the voltage at T6 collector. Since this current flows through T8 (constant current source) the current through T1 & T3 decreases (not that we are bothered at this stage).

    A positive input at the non-inverting input increases current in T1 & T3 and due to the constant current source (T8) the curent in T2 & T4 decreases lowering the voltage at T6 collector !

  6. SgtWookie


    Jul 17, 2007
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    Does this help?
  8. chrischristian

    Thread Starter Active Member

    Feb 22, 2008
    Hey! Every one out there ! thank you very much for your time and effort now I feel relieved.