Touch delay light switch circuit (5)

Another connection method is the touch delay light switch circuit , which has only two external lead wires, directly
replacing the ordinary switch and connecting to the lighting circuit.
    Normally, the transistors VT1 ~ VT3 are all in the cut-off state, f: reaches about 9V DC voltage, the thyristor
VT-I is turned off, and the lamp E does not
light up. At this time, the luminescent tube I.I_D
lights up and can indicate the position of the switch
for the convenience of the driver.
When the human hand touches the electrode HM
, the leakage current
of the human body flows into the base of VT1 through R ,
and Vl, 1, V], and 2 are quickly turned on.
The DC voltage of about 9v obtained by the joint regulation of I, F, D and vs is passed through rE. vri, VT2, R. Feng connects the base of VT3
to turn on VTS, so transistor VT4 obtains the trigger current and =LF passes, and the lamp F is powered and emits light. Delete the same,
C. The stored electric bamboo is quickly released through VT1 and VT2. After the human hand leaves M, VT1 and VT2 return to
the cut-off state, but the 9V positive power supply passes through R. The emitter junction of VT3 and the door panel of VT-1 charge the capacitor C.
This charging current keeps the gate VT-1 open, so the lamp E can still continue to emit light. When (,,
after the charge is fully charged, VT3 is cut off and VT4 loses the trigger current. When the alternating current crosses zero, it is turned off, the light goes out, and
the circuit returns to its original state.
    The delay time of this circuit is mainly determined by (1., R . is determined by the charging time constant. Changing their values ​​can adjust
the duration of the lamp lighting. The capacitor C and the transistor VT3 require extremely small leakage current. If the leakage current is large
, VT4 will always be turned on and cannot be turned off. .There are no special requirements for other components.