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Gain in inverting and noninverting opamps

  1. Sep 14, 2009 #1
    Can you explain the yellow highlighted portions (the derivation part)? How do we get to know which voltages to subtract? Is there also a good link which provides a good explanation of negative feedback mechanism?
    Correction: (V2 in fig 18.28 is actually Vin)

    http://img89.imageshack.us/img89/7360/new1ed.jpg [Broken]
    http://img199.imageshack.us/img199/5502/new2ob.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2009 #2

    berkeman

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    Not sure I understand the question. It's a Differential amplifier. What voltages do you think should be subtracted?
     
  4. Sep 14, 2009 #3

    chroot

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    Do you understand the virtual short and virtual open principles of op-amps? The currents and voltages are easy to derive once you understand those two principles.

    - Warren
     
  5. Sep 14, 2009 #4
    I'm talking about the calculation of currents through resistors in op amp (inverting and non-inverting) which is given in the images I posted. I need help in those (highlighted part),, how did the author do it?

    Nope.
     
  6. Sep 14, 2009 #5

    chroot

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    An ideal op-amp behaves as a "virtual short," because it attempts to force its inputs to have equal voltage on them.

    In your Figure 18.28, for example, one of the inputs is tied directly to ground, or zero volts. The other input will then be driven to the same voltage. Since resistor R1 has Vin on one side and ground on the other, the current through it is Vin/R1.

    - Warren
     
  7. Sep 14, 2009 #6
  8. Sep 14, 2009 #7
    Exactly, but why confuse the reader by writing Vin-V-/R1 and then equalizing it to Vin/R1? Why the subtraction? And similarly for I2 and the currents in the 2nd figure of noninverting amplifier..
    Are we trying to find net voltage or something?

    I don't understand how you got that value for V- in the 3rd step.
     
  9. Sep 14, 2009 #8

    chroot

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    They're not confusing the reader. They're being explicit, which is usually the opposite of confusing.

    To find the current through a resistor, you take the difference in voltage across it, and divide that by the resistance. To find a difference, you use subtraction.

    - Warren
     
  10. Sep 14, 2009 #9
    okaaay.......well that was quite simple...I guess all the new stuff is interfering with my old concepts
    Thanks!
     
  11. Sep 14, 2009 #10
    From Bob S
    Redbelly98's illustration in post #2 is an excellent example of a voltage follower with gain.
    The equations are:

    1) Vout = A(V+ - V-) [ where opamp gain A is arbitrarily large]
    2) Vin = V+
    3) V- = R1*Vout/(R1 + R2)
    4) Vout/A = Vin - R1*Vout/(R1+R2)
    5) Vin = Vout(1/A + R1/(R1+R2))
    Let A ==> infinity
    6) Vout = [(R1+R2)/R1]*Vin


    The equation relating Vout and V- in line 3 is just a voltage divider circuit, since the ideal negative opamp input has infinite impedance and has no input bias current.
    Bob S
     
  12. Sep 14, 2009 #11
    Dear uzair ha91
    The virtual short, virtual open and virtual ground concepts are explained well in the following two reference books (these concepts are explained in the chapter devoted for opamp) :

    1) Thomas Floyd, Electronic Devices [Exellent book for those who like practical work]

    2) Adel Sedra, Microelectronic Circuits [A universal reference book for Electronic and Microelectronic]

    You can download these books together with lots of other books from:

    Admin: link deleted
     
    Last edited by a moderator: Sep 15, 2009
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