new to op amps: question

Discussion in 'Homework Help' started by Vanush, May 25, 2008.

  1. Vanush

    Thread Starter Active Member

    Apr 19, 2008
    hi. im having trouble figuring out this op amp and there are no examples in the book to follow apart from a 'voltage follower' configuration.


    how do i figure out the output voltage vo? the answer says it is 3 V - but how is that possible? i gather its a non-inverting configuration, and thus the gain can never be less than 1?
  2. Caveman

    Active Member

    Apr 15, 2008
    Typically, you analyze opamp circuits by using the ideal op amp model.
    In this model, the + and - inputs have the same voltage, and they draw no current.

    So in this circuit, you can see that no current can flow through the 3k resistor because it would have nowhere to go. Since the two inputs are the same, the negative is equal to the positive voltage. I think you can probably carry it from there.
  3. hgmjr


    Jan 28, 2005
    You can take a look at the information on the topic of opamps contained in the AAC ebook. This will provide you with an introduction that may help clear up some of the issues you have with your opamp analysis.

  4. circuitashes

    Active Member

    May 13, 2008
    As i see it there are two ways to solve/view this problem. However, in either case you would have to assume what caveman called 'the ideal op-amp model' ie. that for a differential op-amp using negative feedback to maintain 'equilibrium' the differential input must be 0 volts. In other words:
    (+ve input) - (-ve input) = 0 volts. Or (+ve input) = (-ve input)

    Method 1:
    Since +ve input = 4v, then -ve input = 4v ( you can visualize this by inserting a 4v bat. in the feedback loop between the 1v bat. and the -ve input of the op-amp) Additionally, it is assumed that ideal op-amps have infinite input impedances, and so do not draw any current from the input sources. If this is so, then there is no current flowing through the feedback loop or indeed the 3k resistor. This means that the pd. across that resistor is 0v. This would only happen if the emf on the left side of the resistor is equal to the emf on the right side. Emf on the left
    = 4v-1v = 3v. Therefore emf on the right = 3v = Vo.
    NB: The emf at the point of intersection b/t feedback loop and output(or upper part of the 1K resistor, or the right side of the 3K resistor) is Vo

    Method 2:
    Lets assume we do not know that the op-amp draws no input current.
    Using KVL on feedback loop: (4V - 1V) + I(3K) = Vo
    I = (pd. across resistor)/Resistance = ((4 - 1) - Vo)/3K amps
    Therefore, (4 - 1) + [((4 - 1) - Vo)/3K]*(3K) = Vo
    3 + 3 - Vo = Vo
    2Vo = 6; Vo = 3V
    Now to verify that no current flows, substitute Vo = 3 into eqn. for I and you get I = 0 amps

    I hope this is helpful. If anyone out there spots any mistakes i have made, kindly let me know. This is my first time replying to a post.