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yesterday when i ask about driving circuit, someone told me that it is a little use in optocoupler when i use the same ground everywhere. because so far (iam a green hand in electronic) i just connect the ground to the netral phase, i don't know how to create or connect the ground properly, so how to create a grounding, or how to connect it wisely? from this picture also it has some grounding but not connected one anotherenter image description here

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  • \$\begingroup\$ your schematic has no optocoupler, so there is no "separate ground" possible \$\endgroup\$ – jsotola Feb 28 at 1:37
  • \$\begingroup\$ For DC circuits, ground is 0V. But for AC circuits, neutral is 0V and ground is a separate connection entirely used for safety purposes that is not supposed to carry current unless something goes wrong. What are you talking about exactly? \$\endgroup\$ – Toor Feb 28 at 1:38
  • \$\begingroup\$ one optocoupler that you are familiar with, and that you probably use a lot, is the remote control for your TV ...... i am pretty sure that the grounds of the remote control and of the TV are not connected together \$\endgroup\$ – jsotola Feb 28 at 1:40
  • \$\begingroup\$ An optocoupler has two purpose: (1) it is an isolation barrier between it's input and output side. If you connect a pin on both sides of the optocoupler to ground it is like you are cutting a hole in that barrier. (2) The voltages on the output do not have to be relative/referenced to the voltages on the input at all. Connecting ground on both sides means you are defeating this purpose too by sidestepping it. You might as well not have it. It is like getting an airplane so you can ignore what is on the ground when travelling, but then driving it along the ground anyways instead of flying. \$\endgroup\$ – Toor Feb 28 at 1:45
  • \$\begingroup\$ The circuit you posted does not have an optocoupler. It is using something fancy, but cheaper than an optocoupler. It is called a bootstrap capacitor (C1. C2) and bootstrap diode (D1). Q2's gate can be driven directly with the +12V supply since it's source pin sits at ground and the +12V supply is referenced to ground (0V). \$\endgroup\$ – Toor Feb 28 at 1:55
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Here is one of the simplest opto-couplers that I can think of.

It would be simpler if the flashlight was replaced by the sun.

The receiver indicates the relative brightness of light hitting the sensor.

As light brightness changes, the amount of current flowing in the circuit changes.

Do you see any reason why the two circuits require a common ground to operate correctly?

schematic

simulate this circuit – Schematic created using CircuitLab

Now add some dangerous voltage to the circuit, either by accident or by design.

schematic

simulate this circuit

Now add a common ground.

schematic

simulate this circuit

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