Zerocross test issue

Discussion in 'The Projects Forum' started by Benengineer, May 6, 2016.

  1. Benengineer

    Thread Starter Member

    Feb 6, 2016
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    Here is the capture I need. Please see the below:
    upload_2016-5-9_15-38-4.png
    This is the zoom out graph when input line voltage (sine wave,yellow) crosses zero, the relay( blue ) crosses 1 voltage.
    upload_2016-5-9_15-40-24.png
    This is the zoom in graph, you can see delay time

    Is this circuit to realize this function, isn't it?

    Thanks
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Don't understand your question. :confused:
     
  3. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
    2
    your simulation result is little different from my capture. Is the circuit output same as requested?
     
  4. crutschow

    Expert

    Mar 14, 2008
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    Is your circuit identical to mine?
    Your output signal looks inverted.
     
  5. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
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    I don't change it. But I found that your simulation output result is little different from me. My output is when line voltage crosses the zero, the relay voltage is falling down crossing 1 volts.
     
  6. crutschow

    Expert

    Mar 14, 2008
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    Since your signal is inverted from mine, then your circuit is not the same.
    My circuit has the output going high when the input goes high.
    If you want similar results to mine then you need to use the same circuit.
    Post your circuit.
     
  7. Benengineer

    Thread Starter Member

    Feb 6, 2016
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    My circuit is at the beginning in this thread. Please read the first post.
    Thanks
     
  8. crutschow

    Expert

    Mar 14, 2008
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    I did read your post.
    My design was to improve the response time of the circuit, which I understood was not satisfactory.
    It switches within 30μs of 0V input.
    Isn't that what you want?
     
  9. Benengineer

    Thread Starter Member

    Feb 6, 2016
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    when my input is high, my output should be low in certain time.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    I didn't know that was a requirement.
    Here's the circuit with the output inverted.
    Notice that the output goes low within 25.7μs of the input zero (465pV) crossing.

    upload_2016-5-10_0-23-53.png
     
  11. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
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    Thank you. I think that is what I want
     
  12. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
    2
     
  13. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
    2
    Crutschow,

    Are you sure C1 and C2 are voltage divider? If they are a divider, C1 and C2 should series, not parallel. Please correct me if I am wrong.
    Thanks,
     
  14. be80be

    AAC Fanatic!

    Jul 5, 2008
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    If you look really careful you'll see that C1 and C2 are taped in the middle
     
  15. crutschow

    Expert

    Mar 14, 2008
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    Yes, I'm sure C1 and C2 are in series, grasshopper. :)
    You are confusing being drawn next to each other in a parallel fashion with being electrically in parallel.
    There is no relation between the two.
    They just aren't drawn in an axial straight line as you are typically used to seeing for a voltage divider.
    You have to look at the connections.
    To be electrically in parallel the capacitors would need to be tied together at both ends and they aren't.
    One goes to the voltage input and one goes to ground with the other ends tied together, as for any proper series voltage divider.
    Okay?
     
  16. Lestraveled

    Well-Known Member

    May 19, 2014
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  17. crutschow

    Expert

    Mar 14, 2008
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    That would insulate the tapped connection. ;)
     
  18. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
    2
    Ok, thank your guys. Your guys are maters.

    Here is what I got:

    Xc2 = 1/2πfc2 = 1/2π*60*0.33x10^-6 = 8038.128Ω

    Xc1 = /2πfc1 = 1/2π*60*10x10^-6 = 265.258Ω

    Input voltage between 120VAC to 277VAC

    Voltage divider V = 265.258/(8038.128+265.258)x120x √2 = 5.43V

    Or V = 265.258/(8038.128+265.258)x277x√2 = 12.54V

    For U2 output voltage is (-3k/100k) * 120x √2 = -5.091V or (-3k/100k) * 277x √2 = -11.752V

    Am I right?

    Since 4N25 forward current is 60mA and forward voltage is 1.3V (in the datasheet), we choose one input as our example; we can figure out 5.43V-1.2V(for 1N4148)=4.23V; then 4.23V – (1.3V-5.091V)= 8.021V;

    Current flow through diode is 8.021/3k = 0.003A, which much less than 60mA. Is this Ok? Please let me know if I am right. Since I am a beginner, please forgive my bothering your guys. Basically, I want to know more design idea from your example.

    Thanks
     
  19. crutschow

    Expert

    Mar 14, 2008
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    Your calculations are okay as far as they go.
    You didn't include the effect of the clamp diode D2, which clamps the negative voltage at the divider to about -0.7V.
    This means the unloaded peak output of the divider is nearly equally to the peak-to-peak voltage of the AC or double what you calculated.
    The output load current reduces that voltage due to the impedance of the capacitive divider so the V+ voltage is about 6.5V for 120Vac input as shown in my first simulation.

    The 60mA forward current for the 4N25 is the maximum current, not the operating current.
    It will operate at any current down to zero.
    So a few mA is just fine as an operating current.

    Your calculation for the LED input current also didn't include the 1.3V forward drop of the input diode which subtracts from the voltage drop across the resistor.
     
    Last edited: May 11, 2016
  20. Benengineer

    Thread Starter Member

    Feb 6, 2016
    133
    2
    If I count D2, the current is about 50nA. It is very small. How do I apply for it?
    What do you mean I didn't include the 1.3V forward drop of the input diode which subtracts from the voltage drop across the resistor 3k?
    Is it 5.23V - 1.3V = 3.93V? I already know the current is about 60mA. Please let me know.
    The other issue is that 0.33uF is so big size and not cost efficiency. Can we use high value resistors to made voltage divider?
    Thanks,
    Ben
     
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