Textbook Active Hi-Pass Filter Not working... what am I missing?

Discussion in 'Wireless & RF Design' started by dr.evil, Nov 20, 2017.

  1. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
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    Gents,

    Please have a look at my awesome schematics:
    Screenshot from 2017-11-20 19-29-16.png
    And this is the signal entering the non-inv input: y:2v/div x: 500ns/div
    Screenshot from 2017-11-20 19-22-35.png
    I know that since I have single supply, the negative spikes will be lost, and this is desired. But this is the output:

    .... absolutely NOTHING, a complete flatliner 0V with some negligible crud on top.

    Now I know the op-amp works, because if remove R3, I have a square signal of 4V p-p and it gets amplified to rail of 5.5V p-p

    So what am I doing wrong here?

    Evilly yours
    Dr. Evil
     
  2. cmartinez

    AAC Fanatic!

    Jan 17, 2007
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    This is an interesting problem whose answer I'd like to know too... I'll take the liberty of paging @OBW0549, if he doesn't mind...
     
  3. MrChips

    Moderator

    Oct 2, 2009
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    If it is a high-pass filter, what is the expected corner frequency?
    Set your generator to a low-frequency sine wave and monitor the output amplitude as the frequency is increased.
     
  4. OBW0549

    Distinguished Member

    Mar 2, 2015
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    I don't see anything inherently wrong with the circuit, other than the fact that the C1-R3 is only 5 nanoseconds and that may be too short a pulse for even a fast op amp like the AD8066 to process. But I'd expect to see something, at least, on its output anyway. If you're not seeing anything at all, I'd say check your connections.
     
  5. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    MrChips,

    Is "corner frequency" the same as "cut-off frequency" ?

    In any case I don't know, I just messed a bit with the R3 value until it produced the desired output signal. I have a measurable signal, so how would an expected corner frequency value benefit? (asking out of pure ignorance)
     
  6. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
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    Mr. OBW0549

    Yes, but if I change these values I won't get the desired spikes. But I do see that 5nS translates to 400V/uS and the datasheet says typical slew rate is 160V/uS, could that be it? If yes, what does a man have to do to get a 5nS pulse?

    If I remove R3, I have a square signal of 4V p-p and it gets amplified to rail of 5.5V p-p, so everything is a-ok.
     
  7. crutschow

    Expert

    Mar 14, 2008
    22,050
    6,375
    Here's the LTspice simulation of your circuit.
    It shows an output, so I don't understand why you are seeing nothing. :confused:
    Interestingly, it shows a positive output for both edges of the input pulse, which I'm not sure is real.
    (If this is being done on a plug-in breadboard, the high stray capacitances of that will seriously change the response).

    upload_2017-11-20_12-40-34.png

    Note that R1 and R2 are a little high for a high frequency signal due to stray capacitance at the (-) input node.
    Better to change them to about 2kΩ or so.
    The simulation below shows an improved response with that change.

    upload_2017-11-20_12-44-36.png
     
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  8. MrChips

    Moderator

    Oct 2, 2009
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    Yes, corner frequency is the same as cut-off frequency.
    This the frequency below which a high-pass filter will be attenuating sine waves.
     
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  9. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
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    @crutschow, thank you very much for your response.

    Well, a simulation is a simulation i.e. theory. And you are right, the pulse on the falling edge is false. Anyway, it's not done on a breadboard, it's old-school tight soldering ;-)

    I followed your advice and changed to 2K2, and you were right it is a capacitance issue, now at least something is getting through the op-amp, please see below, RED being the input pulse: Screenshot from 2017-11-21 08-28-45.png
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Why are the two positive input pulses different amplitudes?
    And why does the lower input amplitude have a higher output amplitude? :confused:
     
  11. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    I'm asking myself the same thing, I suspect it's some overtones (if I can call it that), probably need some more decoupling and grounding somewhere. I will be investigating the issue today and let you know.

    Here's the full picture:

    Screenshot from 2017-11-21 10-02-23.png
     
  12. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    I did some debuggin': The physical distance from square gen output to op-amp non-inv input is about 20mm (3/4 inch), so it's quite tight for a prototype. I then disconnected the op-amp non-inv input, and measured before and after C1 (see below scope dump). The square is rock steady, but after C1 all the hubbub begins and I have no idea why. I even switched to battery power to eliminate any potential noise from the power supply, but the problem persists.

    Any ideas?

    RED is after C1, and 0.4V/div
    Screenshot from 2017-11-21 16-17-30.png
     
  13. MrChips

    Moderator

    Oct 2, 2009
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    With
    C = 22pF
    R = 2200Ω
    the cut-off frequency = 1/(2 x pi x R x C) = 3.3MHz

    Your input square wave at 1MHz is below the cut-off frequency. Since this is a high-pass filter, 1MHz is in the stop-band of the filter.
    What you are seeing is the derivative of the square wave, i.e. spikes on the positive and negative transitions.

    Use a sine wave input and measure the attenuation Vout/Vin as a function of frequency.
     
  14. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    Thank you for your suggestion, but the resistor is only 220Ω, so the cut-off freq is about 330KHz which is way below the 1MHz I would say.
     
  15. crutschow

    Expert

    Mar 14, 2008
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    Your calculator needs batteries.
    The cutoff frequency is 32.9MHz.
     
  16. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    Let me try again, I never claimed meth err i mean math, was my core competency:

    1/(2 x pi x R x C)

    1/(2 * π * 220Ω * 2,2×10⁻¹¹F) = 32.899.931,56 Hz

    OK, you are right, the result is 32.9MHz, but it wasn't a battery problem ;-)
     
  17. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    OK, now C3 is 1nF which is 723KHz, and the scope looks like this:

    RED: op-amp +IN BLUE: op-amp output
    Screenshot from 2017-11-21 19-16-07.png

    So how do I make it more spiky/pointy like it was before? Increase R3?
     
  18. MrChips

    Moderator

    Oct 2, 2009
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    What exactly are you trying to do?
     
  19. dr.evil

    Thread Starter Active Member

    Aug 18, 2010
    80
    1
    Turn a transistor on (and off) with the pulse.
     
  20. crutschow

    Expert

    Mar 14, 2008
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    No.
    You reduce R3 or C1.
    Then why are you differentiating the pulse?
    Do you want the transistor to be on only for a short time at the positive pulse edge?
     
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