Solid State Relay Circuit
The circuit for a solid-state relay
is quite simple. Here is the schematic:
The center of the circuit is the
MOC3042 or MOC3041 optoisolator. The difference between these parts is the current
required to trigger the internal LED. When a current source is introduced between V+ and
V-, the 22uF capacitor is charged. The 1K resistor limits the current from the cap into
the optoisolator and causes the LED to illuminate inside the chip. Line voltage flows
through the 22 ohm resistor and into the zero-crossing circuit in the optoisolator. When
the zero-crossing circuit detects a zero crossing voltage and the LED is illuminated, the
gate on the triac in the optoisolator is triggered with the same phase as the phase of the
line voltage. This line phase current is then used as the gate trigger to the alternistor.
This causes the line current to flow through the alternistor and power the load.
Remember, this circuit only works if
line current is fed into pin 6 of the optoisolator. If you hook the circuit up wrong, the
zero-crossing circuit will not detect the voltage zero crossing and will not trigger the
The Lamp Module and the Wall Switch
circuits are very similar. The X-10 chip has an output pin driving a transistor to switch
-15V. This signal in conjunction with a couple of diodes and a couple of resistors is used
to trigger the triac. If we remove the triac, the two resistors and the two diodes, we can
insert the above solid state relay circuit and the output of that SSR circuit can power
small appliances and fluorescent lights. The SSR circuit V- could be connected to the
transistor, since it has a potential of -15V and V+ could be connected to the logical
ground of the Lamp or Wall Switch circuit.