Delayed lights-out pull switch circuit (2)

It is an improved delayed lights-
off cable switch. Its performance is the same as the delayed lights-off cable switch introduced in the above example
, but it has higher working
stability and good reliability.
    When S is closed, the light is on and the electronic circuit on the right is inactive. When the lamp is on, S is turned on, and the 220V AC
electricity is half-wave rectified by the bulb E and the diode VD1, and the voltage is reduced and limited to flow to C. Charging, due to the small dry charging time constant
, G is fully charged quickly. c, 25V voltage regulator tube is connected in parallel at both ends vs. Therefore, C, the commercial voltage at both ends is
clamped at about 25V. This 25V DC voltage crosses the gate cathode of R and VT to the capacitor [1. Allowing electricity, this
allowing electricity current is the trigger iU current of thyristor VT, so VT is triggered to turn on. Half of the alternating current passes
through the bulb, so bulb E emits light for weak lighting. After about tens of seconds, c" is fully charged and VT loses the
trigger current. When the alternating current crosses zero, it turns off and the light goes out. When the light is turned on again, s closes and the light turns on. At this time, the voltage at both ends of s
drops to zero. C. The stored charge is discharged through R. and vr2. Since the discharge time constant is very small, the charge
is quickly discharged to prepare for the delay of the next lights out. The length of the circuit's delayed lights out time mainly depends on (j
charging call For the time constant of the circuit, the data delay time shown in the figure is about 40s. If you want to extend or reduce the time constant J, you can increase
or decrease the capacity of the capacitor C.