My understanding of a power line carrier circuit

Discussion in 'Wireless & RF Design' started by benvolz, Nov 23, 2017.

  1. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    hi,
    would you be able to help me to understand a part of this circuit circulated in red please. my understanding of this circuit so far is that the
    Mov = protects against voltage spikes
    C1 and L2 = some sort of filter (high ?)
    L1 and C2 = some sort of filter (low ?)
    C3 = to remove any DC voltages
    C4 = to remove any DC voltages
    D1 = over voltage protection

    any math would also help

    thank you in advance


    upload_2017-11-23_20-37-35.png
     
  2. DickCappels

    Moderator

    Aug 21, 2008
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    Both the C1,L1 and C2,L2 circuits are 110 kHz bandpass filters.

    C3 - your are correct, it prevents the TX_OUT from seeing a DC short to common.

    C4 -similarly allows the internal circuitry to set the DC bias to the RX_IN circuitry independent of the voltage on the TX_OUT pin.

    D1 - you understand correctly, to protect the TX_OUT and RX_IN circuits.

    To find the frequency at which an inductor and capacitor are resonant:

    upload_2017-11-23_19-2-3.png
     
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  3. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    thank you for you help ! it good to know im on the wright path.
    the other part of this drawing i cant get my head round is the linking of the 230VAC to 0VDC surely when the sine go's up it going to go bang ?
     
  4. DickCappels

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    Aug 21, 2008
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    The ground symbol being used is the one for chassis ground, not earth ground, so it can, in reality be connected to line or neutral without hurting anything except perhaps some poor technician who goes to troubleshoot the circuit without knowing that the chassis might be hot.
    upload_2017-11-29_17-55-57.png
     
  5. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    ok that does not seam very good. is this common to find ?
     
  6. DickCappels

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    Aug 21, 2008
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    Yes
     
  7. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    oh this is really confusing. how would this not damage the micro ?! as it would see 230vac on the ground line. wouldn't it stop current flow at some point of the sine wave due to no potential difference ? and current flow from 230vac to 5vdc as they try and equal out ?
     
  8. DickCappels

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    Aug 21, 2008
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    It is a lot like a bird standing on a power line. The controller is supposed to be referenced to neutral but even if the connection is switched (and sometimes it is!) there will not be much interaction with earth -only a small capacitive coupling, so no significant current flows. The microcontroller is powered from the Line but the voltage is reduced by the 7V5 Zener and the components that precede it so the 5V regulator and the other low voltage circuitry can operate.
     
  9. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    i dont understand how this cant cause a problem would you mind explaining some more. surly if i have 0vdc and 24vac joined current will flow ?
    thanks for your help
     
  10. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    i have tried this on my PCB and ended up with burnt traces.
     
  11. DickCappels

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    What is it that you see causing a problem? Where is the current path?
     
  12. Sensacell

    Senior Member

    Jun 19, 2012
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    This circuit represents the classic problem where the critical details of the ground connection path are not shown, it's just assumed you understand the danger of mains connected circuits. The source of the +5 is floating, so it "rides" on the 220 V mains voltage!

    "Ground" of this circuit is local and arbitrary, it is not at a safe 0 volts level, it's just the reference point for the circuit itself.

    If this circuit is floating and totally isolated from anything else, the mains can be connected to "ground" of the circuit, and all is well.
    But... if you are trying to design and debug a circuit like this, you must connect your test equipment to the "ground"- this causes a direct mains-to-ground short- BANG! Most scopes, and some power supplies connect the chassis ground AND signal ground to the third wire of the power plug!

    It's easy to kill yourself working on non-isolated mains powered circuits- always use an isolation transformer and use extreme caution.
     
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  13. benvolz

    Thread Starter New Member

    Nov 23, 2017
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    Ok i think im getting there :)

    there is no current path on the drawing above and now i think i have found my mistake on my PCB what i have do instead of get my 5VDC from the 230VAC line with some caps and voltage regulator is to have a separate 24VAC supply (as i need it anyway for external control) and then rectify and buck convert it down to 5V and use if for the mico and modem chip. so i have a 0VDC from the rectifier/buck that needed to be connected to 230VAC to get the chip to word correctly. (i have been using a isolated 24vac supply to do my test with) this is were i get the ider of potential difference from. 230vac to 0vdc. causing current to flow and burn traces !

    the only part im still struggling with is if the ground is say at 240VAC (or +120V) then how does the 5V circuit work as i would of thought that current would move from 240VAC to ground to try and bring ground to a equal potential. so does that mean the 5V circuit would only work when the 230v line is in the lower side of the sine wave(-120V) ?

    could you explain more about how the 5V floats on the 230V please ?

    thanks for the head up about using a isolating transformer. im a electrician by trade not to say i wouldn't kill my self ! i do some silly things ! but like to think i should know better.

    thanks you both for your time and help it much appreciated.
     
  14. DickCappels

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    Aug 21, 2008
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    There is no earth ground in the circuit you posted. That is a chassis ground and is merely a common connection point.

    Here is an example of a DC circuit on the AC line. Imagine you have a battery and a flashlight bulb connect so that the bulb is on. Take one lead from the battery and connect it to the power line (don't really do this!). That has no measurable effect on the battery-bulb circuit. The battery-bulb circuit floats on the AC voltage.
     
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