Would you explain why op-amp behave like that, Please?

Discussion in 'General Electronics Chat' started by Michael George, Feb 7, 2016.

  1. Michael George

    Thread Starter Member

    Feb 8, 2015
    52
    2
    Hello,
    I build a low pass filter on a breadboard using op-amp as a non-inverting amplifier.
    This is the schematic of the circuit:
    download.png


    The op amp was in the saturation mode and it needs two biasing resistors. The waveform was this:

    Untitled.jpg
    The input is the green waveform. The output is the red waveform.


    Then I added a coupling capacitor at the input.

    download cap.png


    Then, The op-amp is biased without putting a biasing resistors. The waveform was like this:

    Untitled.jpg

    How is one capacitor able to bias the op-amp? If this cap blocks DC, The op amp would still saturated. How the capacitor biased the op-amp without using two biasing resistors?
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    It is not saturation. You can't expect an opamp with a single supply voltage to have the output swing below ground -- or can you?
    BTW -- what is the purpose of C2?

    Hint: Bias the opamp at 4V = Vcc/2, then apply the AC input and feed back to the negative terminal. Did you dream up this configuration or did you copy it from some source? If so I think you need another source.
     
    Last edited: Feb 7, 2016
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  3. Michael George

    Thread Starter Member

    Feb 8, 2015
    52
    2
    Hello, @Papabravo
    Thank you for your answer,

    Of course I don't expect an opamp with a single supply voltage to have the output swing below ground.
    What I think is:
    The input voltage is 1.5v peak to peak.
    Before adding the capacitor, the input voltage swings from -0.75V to +0.75.
    The opamp output responds to the positive half cycle only. So, the output wave form is due to the range of input voltage from 0 to +0.75.

    I wanted to bias the opamp at Vcc/2. So, I first inserted the capacitor to block any DC signal that might come form the input source. Then, I surprised that the output wave form has changed and the opamp responded to both positive and negative half cycles. I think that the input signal is affected by the capacitor and the input swings from 0 to 1.5 volts instead of from -0.75 to +0.75. But I don't know how could this be? That's why I'm asking here.

    I did not dream up this configuration :) I found the circuit here:
    http://www.electronics-tutorials.ws/filter/filter_5.html

    But in this website the opamp is not biased and it does not work for the negative half cycle. That's why I started adding a capacitor and biasing resistors. But I'm surprised that the op amp works for the entire cycle after adding the capacitor only!!
     
  4. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,221
    In real life it wouldn't.
    Either the op amp model doesn't simulate the op amp input bias current, or you didn't run the simulation long enough for the bias current to saturate the op amp.
     
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  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,418
    488
    Hi,

    Spice sometimes assumes that you dont do anything that is too far "off the wall", which means too far out of the ordinary. Not using a negative supply or not biasing the input to 1/2 of Vcc would be too far from ordinary here.

    To see more realistic effects, try using a clamping diode from the non inverting input to ground. That will ensure that the input is not driven to a non spec'd value which would actually blow out a real chip.

    It could be that the cap powers the circuit from the non inverting input or something strange like that. We actually see this happen in real life with some chips but i dont think this chip will allow that i think it would blow out in real life.

    So the best bet is to bias the input at 1/2 Vcc with two resistors and go from there. Let us know how you make out if you can.

    Another simpler example of this shows up in some diode models, where the reverse voltage breakdown is sometimes set wayyy higher than the actual spec on the data sheet. For example, 500v for a 40v part. In simulation the diode would work up to at least 450v, but in real life it would soon blow out.

    So when you see strange phenomena like this always check your assumed limit parameters.
     
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  6. RichardO

    Well-Known Member

    May 4, 2013
    1,228
    382
    Actually it might -- sort of. The coupling cap might DC restore against an intrinsic / protection diode. My battery is failing. More details later.
     
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  7. hp1729

    Well-Known Member

    Nov 23, 2015
    1,938
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    Is this a better design?
    I don't think it is mandatory that R2 = R3. We can adjust the operating point by adjusting this????
     
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  8. Michael George

    Thread Starter Member

    Feb 8, 2015
    52
    2
    Hello, @MrAl
    Actually, it is not spice. I build the circuit on a real breadboard. The oscilloscope is PC based Soundcard Oscilloscope.

    I did biased the input at 1/2 Vcc with two resistors and it works very well. The biased circuit was similar to the circuit that is provided by @hp1729

    It is a very very good idea to use a clamping diode. I will try it as soon as possible. Thank you very for the information you gave me.
     
  9. Michael George

    Thread Starter Member

    Feb 8, 2015
    52
    2
    Yes, The design works very well. Thank you very much for your demonstration by sending me the schematic.
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
    2,418
    488
    Hello again,


    Ohhhh, so it's a sound card scope :)
    Does that have DC coupling or just AC coupling? With just AC coupling it will always look like a zero based sine signal even when there is a large DC bias.
     
  11. Michael George

    Thread Starter Member

    Feb 8, 2015
    52
    2
    @MrAl It is AC coupling. I don't think it has the option of DC coupling. When I apply a pure DC, The wave form is zero voltage.
    I think the issue of my circuit is the capacitor was previously charged before inserting it in the circuit. So, when I inserted it in the circuit, it somehow provided some bias (Without using biasing resistors).
    Now, I tried the circuit again. I discovered that: when I apply and then remove the biasing resistors, the wave form is still biased and I have to wait for about half a minute so that the capacitor discharges and then the wave form will be changed.
    That's why I thought that the capacitor was able to bias the circuit without the two biasing resistors :)
     
  12. atferrari

    AAC Fanatic!

    Jan 6, 2004
    2,644
    759
    Hola Miguel,

    The necessity of bringing the non-inverting input to Vcc/2 has been covered already.

    I downloaded and saved this useful image created by Audioguru, a member of this forum. It covers the possible conditions you could encounter. Take your time to understand it. It is worth the effort.

    upload_2016-2-8_10-56-36.gif
     
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  13. atferrari

    AAC Fanatic!

    Jan 6, 2004
    2,644
    759
    I recall using an opamp with the Vref to the non-inverting input quite closer to one of the rails to accommodate a signal. Worked flawless. Sure, it does not seem a common case.
     
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  14. Michael George

    Thread Starter Member

    Feb 8, 2015
    52
    2
    Thank you very much :)
     
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