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Inverting op amp with positive input NOT grounded

  1. Feb 9, 2010 #1
    For an inverting op amp with the positive input grounded, the equation for V(out) is (Vin)*Rf/Rin. Thats fine

    But what if the positive input is not grounded? What if there is a voltage applied to that input as well? How would I find V(out) when the positive input voltage is larger than the negative input? How would I find V(out) when the positive input voltage is smaller than Vin?
     
  2. jcsd
  3. Feb 9, 2010 #2

    berkeman

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    Staff: Mentor

    When you close negative feedback around an opamp from the output to the - terminal (through whatever network), that feedback will try to do what to the difference in voltage between the + and - inputs? Why does that happen? How can you use this to answer your question?
     
  4. Feb 9, 2010 #3
    So lets say I have a 5V DC inputted to the - terminal ( V(in) ) and Rf/Rin = 10. Now, instead of the + terminal being grounded I have instead a 3V DC inputted (V2). What would be V(out)?

    This is a completely made up question, but if I apply the equation my teacher has in the class power point slides, an equation I cannot find ANYWHERE else on the internet because all examples of an inverting amp have + grounded, I get the correct answer. The equation is:

    V(out) = (-V(in)*Rf/Rin) + V2 + Rf/Rin

    Why does this equation work? Is this the correct equation to use in this situation? I mean, it works and all but why can I not find this equation anywhere else??

    I know that the feedback is trying to keep voltage at the - input and + input equal, and in the case where + is grounded this would make the - input 0 as well because of the virtual ground. But if the + input is 3 will the - input be 3 too? What if the - input is 2? What if its 5? And if it is then why cant I just use the normal equation then?
     
  5. Feb 9, 2010 #4

    berkeman

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    Staff: Mentor

    It's kind of hard to talk about this wiithout pictures. And you don't "input" a voltage into the - terminal. You have a feedback network that connects the output to the - input, and optionally the input voltage and ground and other stuff.

    And forget about blindly writing equations for this. It's much more important for you to start by understanding what the high gain of the opamp and closing negative feedback do to the difference in voltage between the + and - inputs of the opamp....
     
  6. Feb 9, 2010 #5
    berkeman is right, it is important to understand what is happening when you use negative feedback with a high gain op amp. There are many applications for op amp circuits that extend beyond resistive feedback networks.

    I think a good place for you to start to figure this out is to remember the summing point constraint. Remember that the differential input voltage (the voltage between the negative and positive terminal of the op amp) is driven to 0V with negative feedback. Also remember that in an ideal op amp the input impedance is infinite, which means no current flows into the op amp. Try applying these two constraints on the circuit and see if you can figure it out with some KCL or KVL equations.

    Once you learn how to calculate the gain this way you will easily be able to know (not memorize) the gain for any circuit configuration so long as negative feedback is being applied to a high impedance amplifier.
     
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