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New to Op-Amps

  1. Mar 1, 2007 #1
    I'm new to op-amps, and would like some guidance. I have a quick question on this HW problem. I'm not asking you to do it for me, but I want to see if my hunch is correct:

    [​IMG]

    Now, using ideal op-amp model... I'm going to assume:

    V+ = V1

    V- = V1 because of neg. feedback?

    Then apply KCL at the node where the R's meet?
    So,

    Vx / 1k + (Vx-V0) / 2K + (Vx-V2)/2k = 0
     
    Last edited: Mar 1, 2007
  2. jcsd
  3. Mar 1, 2007 #2

    AlephZero

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    When you say "I'm going to call it V1" I hope you meant "the terminals of an ideal op amp are at the same voltage, so V+ = V- = V1"

    If you didn't mean that, calling two different voltages V1 was not a good plan.

    When you applied KCL, you forgot the current flowing in the other 2K resistor (between V2 and the opamp).
     
  4. Mar 1, 2007 #3

    Yes, that's what I meant.. sorry for the confusion. I've updated my KCL eq. in the first post -- does it look good?
     
  5. Mar 1, 2007 #4

    AlephZero

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    Yes, it looks OK now.

    The equations are right, but it is not because of "the neg. feedback", it's because an ideal op-amp always forces V+ = V- (whatever the rest of the circuit consists of).
     
  6. Mar 1, 2007 #5

    berkeman

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    No, it is indeed because of the negative feedback. There is no internal mechanism for an opamp to hold its inputs together. It's the high internal gain and the external negative feedback that holds V- = V+. That's important to understand.
     
  7. Mar 1, 2007 #6
    Thanks guys...

    I simplified it to 4Vx - V0 - V2 = 0...so,

    V0 = 4Vx - V2, where Vx = V- = V+ = 7V? and V2 = 7V (because of the balance since it is an ideal op-amp)? I ended up gettin 4*7 - 7 = 21.0 V for V0.. it says Im incorrect.
     
    Last edited: Mar 1, 2007
  8. Mar 1, 2007 #7

    AlephZero

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    OK I understand that is true for real op amps and for designing practical circuits.

    But isn't it true that for analysing a circuit with the ideal op amp model, then the assumptions are always i+ = i- = 0 and v+ = v- ?
     
  9. Mar 1, 2007 #8

    AlephZero

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    Correct.

    No. V1 is the same as V+ because they are joined by a wire link.
    V1 = V+ = V- = 4V.

    V2 is not the same as V-, because they are joined by the 2k resistor which has current flowing through it.
     
  10. Mar 1, 2007 #9
    I see.. Since Vx is in the same node (shares the same wire) as the V-, Vx = V- = 4V.

    So

    (4 * 4) - 7 = 9 Volts, thanks guys!!!
     
  11. Mar 1, 2007 #10

    berkeman

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    Fair enough, but as soon as you are asked to do the calcs for transfer functions of real opamps, you have to take things like the finite (but high) gain and non-zero input currents (and offsets) and non-zero input voltage offsets into account. Also, if you connect up an opamp as a comparator with positive feedback to set the hysteresis, you will definitely not get V- = V+. I just wanted to be sure the OP kept that straight for future applications. o:)
     
  12. Mar 1, 2007 #11

    rbj

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    no. if you do not have negative feedback, your op-amp becomes a comparator and saturates to very nearly the positive power supply voltage if v+ > v- and saturates to very nearly the negative power supply voltage if v+ < v- . since they are extremely high impedance inputs, v+ and v- are determined by analyzing the rest of the circuit surrounding those nodes.

    actually the op-amp does not know and does not care if there is negative feedback or not, but, because of the extremely high gain, if it is in a negative feedback configuration, the output voltage will be forced to adjust to what it has to in order for v- to very nearly be equal to v+.
     
    Last edited: Mar 1, 2007
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