Electronic – How Does Line Frequency Affect Coils of a Contactor

If we assume that N (neutral) is grounded at the incoming electrical connection point somewhere (where exactly varies from country to country) then N is at the same potential as G and we can ignore R2 and N for now.

• L will alternate positive and negative with respect to ground on each mains cycle. On the positive half-cycle the voltage at SEAL will rise until it reaches a little over 5 V. At that point D1 will be forward biased to the +5 V rail and prevent the SEAL voltage rising any further.

• On negative half-cycles D2 will be forward biased almost immediately L drops below zero. This, because it's a Schottky diode (see below), will limit the SEAL voltage to -0.3 V on negative half cycles.

• R2 and N will work in exactly the same way in the event that the L and N connections are reversed as would easily happen with reversible mains connectors used in North America and most of Europe.

Be careful with this type of circuit.

• The G connection would have to be grounded for it to work properly.
• Most small resistors are not rated for mains voltage. The 3 MΩ resistor should be made of two or three resistors in series.
• You'll have mains on the circuit board. Good layout with adequate isolation is required.

I think the purpose of this circuit is to generate a square wave with line frequency (50 or 60 Hz), 5V and 50% duty cycle ...

Correct.

What should be the ratings on the zener diodes?

^{simulate this circuit – Schematic created using CircuitLab}

I think they should be Schottky diodes rather than Zeners although Zeners would work, provided they had a breakover voltage > 5 V.

There are two reasons you may have trouble.

50Hz into 60Hz-designed inductors can always be a problem, depending on how conservatively the original 60Hz design was with respect to saturation. If saturation is more than 20% away at 60Hz, then you'll get away with it at 50Hz. If closer, then the core could saturate, drawing many times more current. You can avoid this by transforming the voltage down to 83% of the rated voltage, so 100v, but of course this will reduce the power available.

Vibratory motors will often use a mechanical resonance between the mass of the armature and the compliance of the spring used to support it, to make up for the poor efficiency of that type of motor. If that's the case with this one, then the 50/60 difference could render it completely ineffective. Try it and see.