IC1 = 74C 14
A very charge coupled AC relay circuit can be built using very few components. The main parts of the circuits involves a couple of power misfets, a couple of CMOS gates, a bridge rectifier, that’s it!
The function of the circuits is basically to switch a load up to 400watts operated at mains AC through small DC input triggers.
You will require the following parts to build it:
R1 = 220 K, 1/4 watt
R2 = 100 K, ¼ watt,
Brdge diodes = 1N4007,
D5 = 1N4148,
C1, C2, C3 = 100 pF,
MOSFET Q1 and Q2 = IRF 640,
Begin the assembly by soldering the IC first, this gives you clear picture of the layout and the positioning of the rest of the passive parts around the IC.
After fixing the IC, start linking the resistors and the capacitor to the IC pin outs as per the shown schematic.
Connect the diodes as shown in the figure and finally fix the mosfets preferably at the edge of the PCb and integrate them to the IC circuit.
You mosfet relay circuit is complete. You can now integrate this assembly to any other circuit which requires a solid state load handling application and can deliver the required amount of input triggers to the above circuit.