OPAMP question

Discussion in 'Homework Help' started by Schoppy, Jul 9, 2014.

  1. Schoppy

    Thread Starter New Member

    Jul 7, 2014
    Ok so not directly an OPAMP question but rather a very simple one and I just can't seem to understand it. For V1=-40E3(i2) where does the -40E3 come from? It's been two semesters since I took my first Networks class and I'm embarrassed to say I can't recall how to arrive there, Voltage Division maybe? That doesn't seem to work though. Thanks guys.
  2. Schoppy

    Thread Starter New Member

    Jul 7, 2014
    Got it, it's because i2 = -v1/R because of it being an inverting amp, makes sense now
  3. bertus


    Apr 5, 2008

    I have turned the picture over and changed the perspective.
    This way the others can have a better look at the picture.


    Last edited: Jul 9, 2014
  4. Schoppy

    Thread Starter New Member

    Jul 7, 2014
    Thank you, and I was wrong with my original assumption, it looks like it equals that because let's call the note next to the 2k resistor Vs. Then i2 comes from (Vs-Vi)/40k.
    But why does Vs = 0? That's the only thing that seems to make sense but I am having trouble seeing why.
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
    Because of Virtual Ground, note the use of capital letters.

    The none inverting input of op amp is connected to ground. There is 0 volts at none inverting input.

    The inverting input of the op amp MUST be equal to the none inverting input. So, V inverting = V none inverting, therefore V inverting = 0 volts because V none inverting is connected to ground which means it is 0 volts.

    If you want more background, google: op amp virtual ground.

    That is also how they found i3.
    V none inverting = 0 volts.
    V inverting = V none inverting = 0 volts.
    i3=(V source - V inverting)/2k=(150 mV - 0)/2k = 75 microA
    Your basic nodal analysis.