It's all sorted by physics - more precisely Kirchhoff's current law.

Consider the way my inverter was installed.  There's a Henley block.  It has 3 wires going into it - the grid mains, the feed to the main house CU, and the new feed to the inverter's mini-CU.

Kirchhoff's current law says that the total current going into a point (in this case a Henley block) must equal the total current going out.  It rather makes sense when you think about it.

Imagine a case when it's moderately sunny, and the inverter's producing 10A.  Suppose I'm cooking, and using 15A in the kitchen.  It follows that the current being drawn from the grid must be 5A, to make the currents in and out of the Henley block add up.

If I finish cooking, and the load in the house drops to 2A, then the inverter must export 8A out to the grid as that's the only place it can go.

Nobody is deciding to do anything.  It just works.

It's all sorted by physics - more precisely Kirchhoff's current law.

Consider the way my inverter was installed.  There's a Henley block.  It has 3 wires going into it - the grid mains, the feed to the main house CU, and the new feed to the inverter's mini-CU.

Kirchhoff's current law says that the total current going into a point (in this case a Henley block) must equal the total current going out.  It rather makes sense when you think about it.

Imagine a case when it's moderately sunny, and the inverter's producing 10A.  Suppose I'm cooking, and using 15A in the kitchen.  It follows that the current being drawn from the grid must be 5A, to make the currents in and out of the Henley block add up.

If I finish cooking, and the load in the house drops to 2A, then the inverter must export 8A out to the grid as that's the only place it can go.

Nobody is deciding to do anything.  It just works.

Okay, thanks for your reply. So your own generation (PV/wind etc) gets preference over the the grid. As other's have said by raising the voltage slightly higher than grid ?