inversion of coil secundary capacitive picked up signal

Discussion in 'Analog & Mixed-Signal Design' started by patpin, Jun 27, 2016.

  1. patpin

    Thread Starter Member

    Sep 15, 2012
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    Hello,Can anyone help me on the way... I would like to invert an analog signal received from the secundary of a car ignition coil, which I picked up with a capacitive probe 1000:1 described here ( http://www.lotuselan.net/forums/download/file.php?id=9946) .
    The resulting signal has peaks up to 40V (40 KV on secundary/1000), but the wave shape should be some thing like below: "oscillo secundary normal".
    In fact (with the probes described above) I get the mirrored image (on a horizontal axe). This is rather difficult to interprete since textbooks on the subject always display the signal with the largest peak goiing upward; I was thinking of a 1:1 transformer. Any one has some experience with those things?
     
  2. crutschow

    Expert

    Mar 14, 2008
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    What do you intend to do with the signal?
    Why does it need to be positive?
     
  3. patpin

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    Sep 15, 2012
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    I just want to study the wave shapes on my bitscope ( for car diagnostics). Positive only because easier and better to compare with existing examples.
     
  4. crutschow

    Expert

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    Doesn't the bitscope have a polarity inversion option for the input signal?
     
  5. patpin

    Thread Starter Member

    Sep 15, 2012
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    Haven't found that on it. http://bitscope.com/product/BS05/?p=specs

    BTW I forgot to mention that I use a scope Probe 10:1 after the 1000:1 capacitive pickup probe described in #1, since the Bitscope inputs can only be +10V->-7 or so.
    I v got an old modem transfo (probably 1:1; ohmic 156 ohm both windings). Could I do it with that? And how to connect so that the wave shape doesnt alter (beside it's mirrored).
     
  6. crutschow

    Expert

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    Depending upon the frequency components of the signal, I think the transformer will likely significantly distort the signal wave shape.
    Your best bet might just be a high frequency op amp configured for a inverting gain of -1.
    What is the horizontal sweep speed of the oscilloscope display in the first post?
     
  7. patpin

    Thread Starter Member

    Sep 15, 2012
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    Not my own capture but my best guess is 8ms/div.
     
  8. patpin

    Thread Starter Member

    Sep 15, 2012
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    Which one would you use. I have a LM358 here. Is the input impedance high enough to 'NOT distort' the signal and how would you dimension it? Is the negative goiing part of the signal a problem? I dont want to loose signal quality.
     
  9. crutschow

    Expert

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    The LM358 has a typical gain-bandwidth around 1MHz so the frequency response at a gain of -1 would be about half that, which probably will be sufficient if the sweep speed of that posted display is only 8ms/div (which is an odd number since usually oscilloscope sweep speeds are in a 1, 2, 5, 10 sequence).

    Actually, considering your use of a 10:1 probe, you could use a 10 MegΩ input resistor and a 1 MegΩ feedback resistor to run the op amp at an inverting gain of -0.1, giving you the same output signal level as with the probe.
    That will also increase the bandwidth to near 1 MHz.

    If you want both positive and negative signal transitions then you either need to capacitively couple the input and output, and bias the op amp positive input at 1/2 the supply voltage with a resistive divider, or use a dual (plus and minus) supply voltage.
     
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  10. patpin

    Thread Starter Member

    Sep 15, 2012
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    Thanks for yr answer.
    It (time/div) was an estimation. To be more accurate: the horizontal part after the large positive peak (spark line ) is typically 1.3 msec before the oscillations start.
    As for the diagram for the opamp: do U mean some thing like this http://www.analog.com/library/analogDialogue/archives/35-02/avoiding/index.html . Which one of the described diagrams would U prefer. I would consider diagram 3? But how to reduce the gain to -0.1. I would try to sim it with LTspice
     
    Last edited: Jun 28, 2016
  11. crutschow

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    Yes, Figure 3 should work.
    As I stated, you can make R1 = 10 MegΩ and R2 = 1 MegΩ to give a high input impedance with a gain of -0.1.
    1μF for C2, and Cout should be sufficient.
    You don't need C1 since your signal is already capacitively coupled.
     
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  12. patpin

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    Sep 15, 2012
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    Many thanks for info. Can C2 and Cout be polarized?
     
  13. crutschow

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    Yes.
    C2 positive side to the junction of Ra and Rb, and Cout positive to the output of the op amp.

    Note that you will need a single supply voltage of at least 10V to avoid clipping the signal.
     
  14. patpin

    Thread Starter Member

    Sep 15, 2012
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    OK i ll try to sim that. cU later!!
     
  15. crutschow

    Expert

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    Actually the supply should probably be as least 12V
     
  16. patpin

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    Sep 15, 2012
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    Please consider the sim. The mirroring is far from OK. Any idea? HV input protection met TVS en opampLM358 capacitive02 zonder TVS.jpg
     
  17. patpin

    Thread Starter Member

    Sep 15, 2012
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    after changing voltage to 12v and after changing the PWL in order to check if I still have the oscillations at the end of the signal... please have a look. I miss the oscill. in the output..; HV input protection met TVS en opampLM358 capacitive03 zonder TVS.jpg
     
  18. patpin

    Thread Starter Member

    Sep 15, 2012
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    a more realistic PWL gives following output. I miss the small oscillations and the voltage sinking slowly after 4.0ms (this line should rise on the mirroring output). I think the capacitive probe is the problem and off coarse the decoupling on the output too. Is there a solution? HV input protection met TVS en opampLM358 capacitive04 zonder TVS ander PWL.jpg
     
    Last edited: Jun 28, 2016
  19. crutschow

    Expert

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    What is the purpose of the 1 meg resistor to ground at the input?
    You shouldn't need that with the 10 meg input of the op amp.

    Just noticed a newbie error that is common when first using Spice.
    You have R5 as 10m which is interpreted as 10 milliohm by Spice.
    It should be 10meg.
     
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  20. patpin

    Thread Starter Member

    Sep 15, 2012
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    Indeed I have very little experience with LTSPICE. Thanks for mentioning the error. The output is now perfect !!!!! Many thanks for yr help!!!
    in the article ref.d in #1 on page 5 they explain the R as "

    A second capacitor of about l000pF should be connected as shown. The capacitive divider thus formed divides the
    input signal by about 1000:1 thus reducing the input signal to a workable 3 - 20 volts. A 1M resistor should be

    connected across the l000pF capacitor to provide a dc load. "
     
    Last edited: Jun 29, 2016
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