1-phase motion detector, how does it work?

Discussion in 'General Electronics Chat' started by somlioy, Mar 6, 2012.

  1. somlioy

    Thread Starter Member

    Jul 28, 2010
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    Hi

    Me and some of my colleagues had a discussion about how 1-phase motion detectors work. They're used for controlling lights etc. Connected like this:

    [​IMG]

    How do they power the electronics inside the detector? :confused:

    Normally we would use two phase in and get both out to the lights.

    Talking about 230v 50hz mains for the records.
     
  2. Brownout

    Well-Known Member

    Jan 10, 2012
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    With an internal power supply.

    Only one phase needs to be switched. Instead of connecting to "N" it would connect to "L2"
     
  3. somlioy

    Thread Starter Member

    Jul 28, 2010
    22
    3
    Last edited: Mar 6, 2012
  4. Brownout

    Well-Known Member

    Jan 10, 2012
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    Good question. Maybe it uses the load to drop the line voltage at a ver small operating current.
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
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    The bulbs provide enough current for the internal supply to work.
     
  6. somlioy

    Thread Starter Member

    Jul 28, 2010
    22
    3
    Yeah, but when the switch/triac is closed you have the same mains potential on both side of the switch. Tho I could see how that would work if the lights had to go off for a while in order to power the caps before it could turn on the lights again. But I think the lights can be constant on if someone is in the detection area. Will check out that, but if someone got some ideas. Please tell.
     
    Last edited: Mar 6, 2012
  7. Brownout

    Well-Known Member

    Jan 10, 2012
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    It would take only a few volts to operate the electronics. Thus, a small value resistor inline with the load would provide enough voltage without affecting the power delivered too much.
     
  8. somlioy

    Thread Starter Member

    Jul 28, 2010
    22
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    Wouldnt that be a seriously big resistor consider the detector can take a resistive load up to 350W?
     
  9. Brownout

    Well-Known Member

    Jan 10, 2012
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    Depends on what you mean by seriously big. If, for example, the electronics required only 1V ( a reasonable assumption) then the resistor would be 350W/230=1.5W.

    Or, you could just use a couple diodes to get ~1.5V, no matter what the current is.
     
  10. somlioy

    Thread Starter Member

    Jul 28, 2010
    22
    3
    I might have missunderstood you or? Do you mean a resistor in series with the load? Why do you divide 350W by 230V and get 1.5W as a result? Wouldnt that be current as a result, 1.5A?
     
    Last edited: Mar 6, 2012
  11. Brownout

    Well-Known Member

    Jan 10, 2012
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    I assumed that the electronic would work on 1V.

    I=350/230 Load current ( same as resistor current in series )

    P(resistor)=V*I = 1*350/230 = 350/230. to a good approximation.

    But, a couple series diodes would provide a near constant voltage for the electronics. This would not work when the load is switched off. So, the diodes would need to be switched for a bleeder resistor divider when the load is off.

    Not saying I know how it works, but this is one possibility.
     
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