This is a solid state relay circuit that uses an optocoupler and a MOSFET to control the output connected load. The input side of the circuit is accepting a small control voltage. this control voltage ranges from 5V to 12V. This voltage is powers the LED inside the optocoupler IC of PC817 through a 470Ω resistor.
When this LED is lights up, it activates the internal phototransistor of the optocoupler. This step is important because it provides electrical isolation between the low voltage control signal and the higher voltage that powers the output load.
On the output side of the optocoupler, the phototransistor is connects to the gate Pin of an N channel MOSFET of BUZ71A. A 10kΩ resistor is connected between the gate and ground and this will ensuring the MOSFET is stays off when no signal is present.
When the optocoupler turns on, it drives the gate of the MOSFET to high and allows the MOSFET to conduct. This setup effectively uses the MOSFET as a switch that can handle larger currents and voltages than the optocoupler itself could manage.
The load, such as a motor, lamp or LED strip is connected between the positive supply voltage (+6V to +24V) and the MOSFET’s drain terminal. The source of the MOSFET is connected to the ground. When the MOSFET is switched on by the gate signal, The current will flows through the load and powers it. When the input control signal is removed, the optocoupler turns off and the MOSFET gate is pulled low by the 10k resistor, and the load switches off.
Lets we see the In summary, this circuit works like a relay, but instead of moving parts, it relies on semiconductor devices. Because of the optocoupler, the control side and the load side are electrically isolated, which improves safety.
Compared to a mechanical relay, this solid state relay is faster, quieter, and longer lasting since there are no contacts to wear out. It is useful in many applications where a low power control signal needs to safely control a higher power load.
