| Line AC voltage is connected to series circuit of 12uF capacitor, ballast, lamp
Lets assume that if we succeed to start the lamp, it can then keep going in this circuit with leading power factor
An initial pulse for starting the lamp can be made by short circuiting parallel to the lamp, you can try it with an ordinary on/off switch manually. Switch it on for <1 sec and immediately off
The rest of the circuit is supplying 141V DC to a relay (probably with 110V DC or 220V DC coil), which will switch it on and keep it switched on permanently as long as the circuit is powered
This means that in the beginning the relay contact is closed, and it disconnects when the relay is turned on. The connected relay contacts are C and NC (11 and 12 for typical DIN relay)
The 2nd capacitor is there to try to limit sparking in the relay contacts
The problems with this circuit :
Lamp starting by high voltage pulse can sometimes not work. The starter must check if the lamp works, if no then try again. All starting circuits do it : Fluorescent starter (if the fluorescent lamp is not working, the glow lamp will light again), HPS ignitor (it checks the voltage), and so on
This circuit does not. If the lamp did not start in the 1st time, or if it was interrupted by a power blink, it will remain off, and you will have to switch it off and on again
There is no delay in the operation of the relay. If your switch has bad connection or contact jitter, the voltage at turn on will rise slowly. At some moment the voltage may be sufficient to switch the relay, but not sufficient to keep the arc in the lamp, the lamp won't start
The voltage supplied to the relay in this circuit (with 100V AC in) is 141V DC. The closest relay coil ratings i know are 110V DC and 220V DC. For 110V DC, this may be a little high (the relay may overheat over extended use). For 220V DC, 141V may or may not suffice to switch the relay at all
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