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Simple op-amp circuit help.

  1. Jul 15, 2009 #1
    I have a simple op-amp circuit that I need help understanding:
    http://img200.imageshack.us/img200/3939/opamp.jpg [Broken]

    I know that the positive input is noninverting and the negative input is inverting.

    I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 15, 2009 #2
    Have you considered that the output might saturate?
     
  4. Jul 15, 2009 #3

    berkeman

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    Staff: Mentor

    Opamps are not used open-loop. You need to add feedback to have an opamp "circuit".
     
    Last edited by a moderator: May 4, 2017
  5. Jul 15, 2009 #4
    You made a slight mistake in your formula. It's a common mistake and I've noticed experienced engineers in this forum making the same mistake. The formula should be (2*V2 - V1) * Aol though opamps are rarely used open loop. This formula holds even for a standard inverting amplifier with feedback. For instance in your circuit, if you put a 1k resistor between V1 and the inverting input and another 1 k resistor between the output and the inverting input the amplifier becomes an amplifier with a gain of -1. If we put 5V at V2 and 4 V at V1, what is the output voltage? Is it 5V - 4V = 1V or is it 2*5V - 4V = 6V? Try it.

    In your example, because the gains are so high your outputs were always at the + or - rail except once, when you got -200mV. Lets plug your input voltages into the formula and calculate the result. (2*1uV - 3uV) * 200K = (2uV - 3uV) * 200k = -200mV. Okay?
     
    Last edited by a moderator: May 4, 2017
  6. Jul 15, 2009 #5
    I am still having problems with both formulas. For example lets do the first two. First with my formula and then with yours.

    V1=4mv V2=1mv

    my way- (1mv-4mv)* 200k= -600 volts
    your way - (2x1mv-4mv)* 200k = -400 volts

    So why is the answer 9 volts? Does it have something to do with the sources?



    V1=2 micro v V2=1 micro volt

    my way- (1 micro volt- 2 micro volts)*200k = -200mv
    This actually is the answer that is written down why?

    Your way (2X 1 micro volt- 2 micro volts)*200k = -400mv
    Which is not the answer written down.


    Can someone try to explain to me what I am not getting?
     
  7. Jul 15, 2009 #6
    Yes, this is just academic to see how one works.
     
  8. Jul 15, 2009 #7

    berkeman

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    Staff: Mentor

    Yep. How would the output of the opamp swing to hundreds of volts, using only +/-10V power supplies?
     
  9. Jul 15, 2009 #8

    The Electrician

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    Gold Member

    It's probably because the opamp isn't a so-called rail-to-rail output opamp. This means that when the opamp output is saturated, the outputs can't swing all the way to the rail (supply) voltages. They can't make it all the way to 10 volts; only 9 volts.
     
  10. Jul 15, 2009 #9
    Oh, I get it guys. Thanks a lot.
     
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