Abnormal gain in op amp inverter configuration

Discussion in 'The Projects Forum' started by arenwi, Jul 6, 2015.

  1. arenwi

    Thread Starter Member

    Dec 29, 2014
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    Hi every body,

    I was making test on an inverter configuration op amp (picture atached) but mi gain calculations are not ok under 500Hz the gain start to go down.

    IMG_6185.jpg
    As I think I make the calculation of the condenser to have an hight pass filter with cut frecuenci on 159 Hz but the gain until 400 or 500 hz are smaller than the teorical gain.
    I was testing different op amps: TLV2264; MCP6022
    Any body know witch could be the reason.????

    I would like to have a linear amplifier betwing 160 Hz and 20 Kz.
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    @arenwi
    Your rail-to rail output voltage is 5 volts since your supply is only 5 volts. I did not check the Datasheets, those op amps could be less than rail-to-rail. Now, I am assuming that your input signal is more than 0.5 volts peak to peak.

    If the input is that large, you are saturating your output and you see no gain. Look at your output with an o-scope.
    At low frequencies, you are seeing gain because you are attenuating your input with a low-pass filter (starts cutting off at 175 Hz) and signal is low enough to stop saturating somewhere below 159 Hz. The high-pass filter is your 0.1uF cap with the 10k and 100k in parallel (for 9090ohm effective resistance).

    Now, to prove this, use an input voltage of 0.2 volts or less so your output is 2 volts or less (Peak-to-peak) or less than 0.07v RMS input to get 0.7V RMS output.

    You should see the 10gain at lower input voltages.
     
  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    You don't say what the actual gain of your circuit is, only what you calculated. How far off is the measured performance?

    Your corner frequency calculation is correct. However, the gain of a single pole highpass filter is down 3 dB at the corner frequency for each stage, or -6 dB for the combined circuit. Also, the frequency response is not perfectly flat down to the corner frequency.

    Calculate the capacitive reactance at 500 Hz, 320 Hz, and 160 Hz. At each frequency, add the capacitor's impedance to the 10K input resistor for the true input series impedance, and re-calculate the gain. Do this for both stages, multiply the gains, and you will see that your circuit is behaving correctly for its design.

    To have 12 dB /octave attenuation that is only -3 dB down at the corner frequency and with less gain rolloff in the passband, change to a true 2-pole lowpass filter rather than two one-pole filters in series.

    At 20 kHz the 2264 has an open loop gain of only 35 dB (datasheet page 29), which means that each stage of your circuit has less than 20 dB of negative feedback. This will affect the gain and frequency response of your circuit. 40 dB of gain at 20 kHz is 2 MHz gain-bandwidth; that's a lot. Most general purpose opamps won't work in this application, even with two amps cascaded as in your circuit, which is why there are specialty audio opamps. They usually require higher operating voltages and draw more power. If you can provide +12 V or +/-5 V, consider a 5532 dual audio opamp, with the first stage as a 2-pole highpass filter with a gain of 2, and the 2nd stage with a flat gain of 50.

    A 2-pole, gain-of-2, active filter (highpass or lowpass) is one of my favorites because the component values work out nicely:
    http://www.calculatoredge.com/electronics/equal comp high pass.htm

    ak
     
  4. OBW0549

    Well-Known Member

    Mar 2, 2015
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    Setting your high-pass rolloff frequency to 159 Hz is NOT going to give you a flat response down to 160 Hz. 159 Hz is the -3 db frequency (-6 db for two cascaded stages) point, but gain will begin rolling off well above that frequency. I would suggest changing your 0.1 uF capacitors C1 and C2 to at least 1 uF, preferably more.
     
  5. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Nope. The inverting input is a virtual ground, isolating the input from the feedback impedance (as long as the loop is closed). The corner frequency is correct in his drawing.

    ak
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    What is your application for this circuit? How accurate does the filter corner frequency have to be? There are many active filter circuit designs, and all of them have the same problem when it comes to component values and tolerances.: 1% resistors are cheap. 1% capacitors are not. So the first thing to do is determine what filter characteristics you really need (corner frequency accuracy, passband flatness, etc.) as opposed to ones you would like to have. For example, do you really need 12 dB/octave attenuation, or do you just need a specific amount of attenuation at a specific frequency?

    ak
     
  7. arenwi

    Thread Starter Member

    Dec 29, 2014
    37
    0
    HI to every body and many thanks to your answers.

    The filter with cut frecuency at 159 Hz and 6 dB / octabe is ok for my aplication.

    In the paper I think that this desing must be ok but in practice:

    With the right imput and no saturating, The circuit give me an output with gain of 10 each stage, a total of 100= 40dB. That is right but is only in frecuencies biger than 400hz (and lower than 300KHz), when you start to go down with the frecuency generator under 400 Hz the gain start to go down. Acording to mi calculations the gain must be linear on 10 by stage until 159Hz (when the hight pass filter starts to work).
    Y had changed the condenser C1 of a midle value (50nf is 75hz hight cut frecuenci) but the inestability of the gain are exactly the same. Under 400 Hz the gain starts to go down.

    Many thanks.

    I would like to know witch could be the reason of this bandwith.
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Low freq roll-off as a function of CV, where CV steps from 0.05uF to 1.6uF, doubling each step.

    89a.gif
     
  9. arenwi

    Thread Starter Member

    Dec 29, 2014
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    0
    Hi MikeML I did´nt understand your answer. This teorical calculates is what I sow in the first post.

    The problem is becouse the real conduct as i told one post before is not like the therical.

    With the right imput and no saturating, The circuit give me an output with gain of 10 each stage, a total of 100= 40dB. That is right but is only in frecuencies biger than 400hz (and lower than 300KHz), when you start to go down with the frecuency generator under 400 Hz the gain start to go down. Acording to mi calculations the gain must be linear on 10 by stage until 159Hz (when the hight pass filter starts to work).
    Y had changed the condenser C1 of a midle value (50nf is 75hz hight cut frecuenci) but the inestability of the gain are exactly the same. Under 400 Hz the gain starts to go down.

    Many thanks.

    I would like to know witch could be the reason of this diferences betwing teorical and real bandwith.
     
  10. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    0db on the plot I show is relative to a gain of 100. As the value of the coupling capacitors increase (there are two of them), the low freq rolloff begins at progressively lower frequencies.

    Red: CV=0.05uF
    Org: CV=0.1uF
    Yel: CV=0.2uF
    Grn: CV=0.4uF
    Blu: CV=0.8uF
    Vio: CV=1.6uF

    As you have already been told (several times), the reason that your measured response is different than your calculated one is that you forgot that there are two stages (not one).
     
    Last edited: Jul 6, 2015
  11. arenwi

    Thread Starter Member

    Dec 29, 2014
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    MIchael,

    Measuring only one stage I had the same limitation.

    I´m using 0.1uf capacitor and 10K resistor Fcut=1/(2*pi*10K*0.1uf)= 159 Hz if you use two or tree similar stages this frecuency cut will be always the same it only change the angle of the atenuation 3db/octave by stage. 3 stages = 3*3db/oct. Remember that stages are identical.

    But It did´nt solve mi question about this strange gain in frecuencies lower than 400Hz.
     
  12. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    ""if you use two or tree similar stages this frecuency cut will be always the same it only change the angle of the atenuation 3db/octave by stage. ""

    No, it won't. As explained above in post #3, stacking up several 1-pole stages is not the same as a single multi-pole stage. The frequency responses below, at, and above the corner frequency change with each additional stage.

    ""3 stages = 3*3db/oct.""

    No, it doesn't. The voltage response attenuation slope is 6 dB per octave per pole, not 3 dB.

    ""But It did´nt solve mi question about this strange gain in frecuencies lower than 400Hz.""

    Again, even though a Butterworth filter is called "maximally flat", it is not perfectly flat. Period. The frequency response is down 1 dB one octave above the cutoff, and 3 dB at the cutoff. If you calculate the capacitor's impedance at various frequencies as I explained above, you will see this for yourself.

    Also, the attenuation slope is not truly 6 dB per octave until 2 octaves away from the cutoff. In terms of the difference between theory and the real workd, filters are the worst.

    ak
     
    arenwi likes this.
  13. arenwi

    Thread Starter Member

    Dec 29, 2014
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    Hi AnalogKid many thank for your answer, I´m starting to know your answer, "In terms of the difference between theory and the real workd, filters are the worst."

    But I had asembled on board the circuit with different capacitor values and always the anileality starts in 400 Hz also with a capacitor of 0.1nf. I can see the diferent incline of the filter but before it it is also atenuation. Always starting in 400Hz.

    I was looking about butterwoth schematics but there is a lot of diferent circuits, somebody know a good butterworth filter schematic??

    Many thanks
     
  14. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Post #3.

    ak
     
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