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Can EMI or similar affect the flickering of a neon indicator?

AJJewsbury

OK, bit of an odd question for the new year...

I was in a relative's attached garage when the LED batten light started flickering/flashing/going off for several seconds - I've got to go back and work out why and fix it (for the moment I'm presuming either a faulty electronic ballast/PSU or a local loose connection - it's on the same lighting circuit as half the house and no problems have been reported elsewhere). What I noticed that seemed very odd though was the neon indicator on the boiler FCU (also in the garage) seemed to flicker and turn off/on exactly in time with the light - boiler is on a quite separate circuit all the way back to the CU and the boiler itself didn't seem to be at all upset. No reported flickering anywhere else in the house either. I'm reasonably convinced that the neon itself was illuminating/staying dark, and I wasn't just seeing light reflected from the main light.

I'm pretty sure it's a traditional neon, not some LED substitute - in normal conditions it does that kind of slight wandering rolling flicker I associate with old neon lamps.

So can anyone think of any mechanism by which the switching on/off of the LED batten might encourage or discourage the neon on a separate circuit to strike? (or should I start worrying about a common fault and so some very odd interconnections between the two circuits?)

   - Andy.

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    the stripes may be a visible indicator of   the quantisation of atomic energy levels.  The electrons emitted in the very low pressure mercury vapour accelerate along the tube, picking up speed almost ballistically,  increasing their kinetic energy such that  1/2mv2 =e.V, where the 'V' is the voltage drop moved along the tube from the initial start, and e is the electron charge.

     The suggestion is that once it has enough energy to ionize an atom of mercury, then it does so, but loses its kinetic energy in the process, and has to start picking up speed again more or less from rest.

    This leads to any one electron having a velocity over distance (and time for that matter) that is a sort of sawtooth function of position along the tube, and as the ionization is at maximum where the velocities are highest, bright  ring spacing that are the tube length divided by the no of times the ionization voltage divides into the total tube voltage . (it takes about 10eV to get an outer electron fully off a mercury atom, and the tube voltage is about 10 to 15 times that so a dozen stripes perhaps )

    In a normal 50Hz sine wave situation, the stripes all smear out, and the rings are only then visible on fast camera or with DC discharges, and clear rings are only visible in situations where there is not a high current density so not too much collision broadening, and the supply is either DC, or a square wave of the same peak voltage in both directions, or pulsed at a frequency that matches a sub multiple of the transit time to accelerate between the bunches.

    Mike.

    edit don't confuse v for velocity and V for voltage

    PS

    The electron starts from rest (near enough) so the kinetic energy gained is given by ½mv 2 where m is its mass and v is its speed.

    So we can say that: ½mv 2 = eV

    The mass of the electron is m = 9 × 10-31 kg

    The electronic charge is e = 1.6 × 10-19 C

    For an electron gun with a voltage between its cathode and anode of V = 100V without collisions, the electron will have a speed of about v = 6 × 106 m/s and 10V is about v = 2 × 106 m/s 
    Its all pretty quick.

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  • 0

    the stripes may be a visible indicator of   the quantisation of atomic energy levels.  The electrons emitted in the very low pressure mercury vapour accelerate along the tube, picking up speed almost ballistically,  increasing their kinetic energy such that  1/2mv2 =e.V, where the 'V' is the voltage drop moved along the tube from the initial start, and e is the electron charge.

     The suggestion is that once it has enough energy to ionize an atom of mercury, then it does so, but loses its kinetic energy in the process, and has to start picking up speed again more or less from rest.

    This leads to any one electron having a velocity over distance (and time for that matter) that is a sort of sawtooth function of position along the tube, and as the ionization is at maximum where the velocities are highest, bright  ring spacing that are the tube length divided by the no of times the ionization voltage divides into the total tube voltage . (it takes about 10eV to get an outer electron fully off a mercury atom, and the tube voltage is about 10 to 15 times that so a dozen stripes perhaps )

    In a normal 50Hz sine wave situation, the stripes all smear out, and the rings are only then visible on fast camera or with DC discharges, and clear rings are only visible in situations where there is not a high current density so not too much collision broadening, and the supply is either DC, or a square wave of the same peak voltage in both directions, or pulsed at a frequency that matches a sub multiple of the transit time to accelerate between the bunches.

    Mike.

    edit don't confuse v for velocity and V for voltage

    PS

    The electron starts from rest (near enough) so the kinetic energy gained is given by ½mv 2 where m is its mass and v is its speed.

    So we can say that: ½mv 2 = eV

    The mass of the electron is m = 9 × 10-31 kg

    The electronic charge is e = 1.6 × 10-19 C

    For an electron gun with a voltage between its cathode and anode of V = 100V without collisions, the electron will have a speed of about v = 6 × 106 m/s and 10V is about v = 2 × 106 m/s 
    Its all pretty quick.

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