Automatic lighting controlled by telephone signal

Diodes VD8 ~ VDI1 and thyristor VS form the main circuit of the control switch. Isolation diode VD7, stabilizer tube VD6 and
capacitor C4 form a simple resistor step-down voltage stabilizing circuit, which outputs a DC voltage of about 24V for the control circuit. The photoresistor
RG1 and the light-emitting diode LED1 form a photoelectric coupler, which together with the photoresistor RG2 and the iode VT form
a telephone and light control circuit.
    The Ll and L2 terminals are connected in series to the telephone line. When there is no call, no current flows between L1 and L2, and the light-emitting diode (LED)
does not emit light, so the photoresistor RGI shows high resistance, and the transistor VT is cut off and controllable. Silicon vs is also cut off, so the lamp
H does not light up. At this time, the entire circuit is in a static state, and the consumption is very small. It is measured to be less than ImA, so it can be considered that no power is consumed.
When a call comes in or the phone is off-hook and dialed, current flows in between Ll and L2. The phone signal current causes the light-emitting diode LED1
to emit light, and its light shines on the photoresistor RGI to make it show low resistance. If it is during the day, because RG2 has low resistance when exposed to natural light in the room
, the voltage generated by it and (RGl+RI) divided by it is low, which is not enough to turn on VT. Therefore,
even if you make a phone call during the day, the light will not turn on. Will not light up. Only at night, RG2 has a high resistance, and the voltage distributed on it is higher, so
the transistor VT is turned on, and the charge stored in C4 is quickly charged to C2 through VT and VD5, so that C2 is charged to a
voltage of about 20V. At the same time, C4 also supplies electricity between the gate and cathode of the thyristor V through R3, causing the V to open and the lamp
H to light up. After the phone call is hung up, no current flows between L1 and L2, and the transistor vT returns to the cut-off state. However, since
the charge stored in C2 can continue to provide gate current to vs through R3, vs still remains on, and the L乜 light is still on.
Shine brightly. When C2 is discharged, vs loses the trigger current . When the AC crosses zero, it turns off and the light goes out.