Connecting analog power domains in the same chip in ICs (Integrated Circuits)

Discussion in 'The Projects Forum' started by Tako, Nov 3, 2014.

  1. Tako

    Thread Starter New Member

    Oct 21, 2014
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    I have a following issue, which I didn't see in IC books: I would like to know if it is possible to send analog signals (sin waves) between two power domains in an integrated circuit. The issue looks as follows:

    case1.png



    I have never had a problem with sending digital signals between two different power domains as it is done through level shifters / level converters. However, I never thought about sending analog signals between two power domains. Is it done using protection diodes as below?

    case2.png
     
  2. Tako

    Thread Starter New Member

    Oct 21, 2014
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    Of course, signal send from 3.3 V domain is limited to 2.5 V. This sinusoidal 3.3 Vpp emphasizes that signal is send from 3.3 V domain to 2.5 V domain and that maximum signal that can appear on the line is 3.3 Vpp.
     
  3. MikeML

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    Oct 2, 2009
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    The diodes shown in the bottom will clip an analog waveform, causing huge distortion. If you wish to preserve the waveform, just use a two-resistor voltage divider keeping the voltage at the tap within the range allowed by the 2.5V circuit.
     
  4. Tako

    Thread Starter New Member

    Oct 21, 2014
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    Oh my goodness . . . I had an eclipse of the mind at the end of a work day . . . Of course I can connect two power domains using capacitor + voltage divider. See the picture.

    Anyway, @MikeML , reserved biased diodes are going to cause huge distortion?
     
  5. joeyd999

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    Jun 6, 2011
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    This solution will be a) frequency dependent and, b) still clip for high frequencies.
     
  6. Tako

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    Oct 21, 2014
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    Right, but if I'm going to send high frequency signal, that is > 1 MHz then, big resistors are just fine. Power consumption is minimized and high frequencies work well. Right?
     
  7. joeyd999

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    If the frequency is well above the "knee" of the RC filter, then your signal will clip. Do you not understand this?
     
  8. Tako

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    Oct 21, 2014
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    I would like to send a signal from 3.3 V domain that is 2.5 Vpp (0 - 2.5 V). Series capacitor + voltage divider means high-pass filter. Thus my signal should be fine. However, I see that diodes are still necessary for protection. That is caused by the fact that 3.3 Vpp signal can pass as well.

    @joeyd999 I do not understand you. Series capacitor + voltage divider means high pass filter. To be above the knee of RC filter means to be above 3dB frequency of my high pass filter what is what we desire.
     
  9. MrChips

    Moderator

    Oct 2, 2009
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    Big resistors will cause fall off in high frequency response.
    You need a resistive voltage divider:

    [​IMG]
     
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  10. joeyd999

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    That's not what your drawing shows.

    To eliminate any frequency dependent attenuation, the 3db frequency should be a decade lower than your signal frequency. There, your cap will look like a short circuit. Analyze accordingly.
     
  11. MikeML

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    Yes, unless you limit the peak-to-peak voltage of the analog signal to be less than about -0.5V to +3V. That is what the resistive divider is for.

    Do not use a capacitor. First, it will block the dc component of the analog signal, shift it down, and clip the negative peaks. It will also introduce a high-pass or differentiator.
     
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  12. Tako

    Thread Starter New Member

    Oct 21, 2014
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    Ok. Thank you for the discussion. :)
     
  13. Tako

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    Oct 21, 2014
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    I see. But sometimes there is a need for such connection. For example see Fig. 32 and Fig. 49 here: http://www.ti.com/lit/ds/symlink/afe031.pdf . The input signal of the PGA1 is 10Vpp while the PGA2 can support input signal equal to its power domain that is 3.3 V. these two PGAs are connected through low-pass filter LPF and capacitor (C10 in Fig. 49).
     
  14. MikeML

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    You must decide if you need to preserve the DC level inherent in the analog signal or not. If the signal can be capacitively coupled, then use a coupling capacitor. If you do, you may have to use biasing resistors on the subsequent stage, or rely on the IC chip maker having put them there.
     
  15. Tako

    Thread Starter New Member

    Oct 21, 2014
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    Ok, understood. Thanks. :)
     
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