Novel delayed lights out lighting switch

When the switch S is closed, the positive half cycle of the alternating current passes through VDI and R
to turn on the thyristor vs. At the same time, both ends of the capacitor C are also filled with
charges. At this time, the positive half cycle of the alternating current supplies power to the bulb H through vs , and the negative half cycle supplies power to the bulb H through the diode VD2, so the bulb H is powered at full voltage at this time. When the light is turned off, the switch S is turned on , and the capacitor C is discharged through the resistor R. This discharge current keeps the thyristor VS on, so there is still a positive half-cycle current flowing through the bulb H, but because S cuts off the VD2 power supply in the negative half-cycle circuit, so the bulb H is in a half-voltage power supply state, and the bulb emits dim light. After about 1 minute , the capacitor C is completely discharged. Due to the loss of gate trigger current, the thyristor VS turns off when the alternating current crosses zero, and the bulb goes out.