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Voltage Divider feeding Op-Amp Confusion.

Discussion in 'General Electronics Discussion' started by Balrock, Jun 15, 2012.

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  1. Balrock

    Balrock

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    0
    Mar 1, 2012
    Hello everyone. I have been able to get back to my project again the last week. I have been struggling with the same problem for many days and I have finally run out of things to try.

    I am trying to feed my Op-Amp Voltage Inverter with a Potential divider but with no accuracy.

    The Potential Divider. - With a supply of 5 volts Point A gives me a ratio of 1:4 and correctly gives me 1.25volts.

    The Op-Amp Inverter - If I supply say + 2 volts directly at point X from a power supply rail point Z correctly gives me - 2 volts. This is due to the fact that

    Z (volts out) = - R4 / R3 x X (volts in)
    so the real numbers are
    - 1 / 1 x 2 = - 2

    So far so good.

    If I connect point A to Point X I do not see - 1.25v at point Z. I see -0.63 volts

    If I remove R3 and replace R4 with a 3.9k resistor so that the Op-Amp maths looks correct I still do not see 1.25v at point Z. I see -5.02 volts

    I have come back to this several times but I am stuck. I know that I am obviously missing fundamental knowledge here. Any and all help is very welcome.


    Thanks kindly Paul.
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,280
    2,583
    Nov 17, 2011
    Easy: Pin 2 of the OpAmp is effectively at ground (0 V).
    Therefore R3 is effectively parallel to R2.
    Do you see what follows and why?

    There's no need to have R3/R4 at such a low resistance. Use 100 kOhm instead and the effect becomes almost negligible.

    Harald
     
  3. Balrock

    Balrock

    39
    0
    Mar 1, 2012
    thanks for the prompt replay Harald :)

    I tried simulating it and it was failing past 20k. In the real world though 100k was fine, after I replaced a faulty Op-Amp that is!

    Thank you for the sweet relief ! I will come back to this again later after a rest and try to work out why it works this way.
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    11,280
    2,583
    Nov 17, 2011
    A hint for understanding the operation of the circuit: The OpAmp will produce an output that forces V(+)-V(-) = 0 V (an approximation good enough for many purposes).

    To be exact, it produces Vout = A*(V(+)-V(-)) where A is the open loop gain of the OpAmp. Since for a typical OpAmp A is on the order of >10^5 and Vout is limited to Vcc, you get V(+)-V(-) = Vcc/A ~ 0V.

    Harald
     
  5. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Another way of looking at it is to consider the Thevenin equivalent of the R1/R2 divider. It is equivalent to a 1.25V source and a 975R series resistor. So that resistance is in series with the 1K R3 and therefore the gain of the amp is - 1K / 1.975K or 0.506. Multiply this by the 1.25V input and you get -0.63V.

    Bob
     
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