Load sharing between grid and inverter

[lamagra]

There's a question to which I've been looking for an answer but I haven't found a clear one yet so I hope this forum will help.
How is the load shared between a grid-tie inverter and the grid? I understand that the inverter feeds the loads with all the available solar power and imports the remaining balance from the grid if needed or exports the excess if any. I would like to understand the technical details behind this process (how does the inverter have feeding priority over the grid?), because so far I haven't found a clear explanation online. It's been a while since I got into electrical circuits so could it be because of the fact that the grid tie inverter acts as a current source?

Thanks.

[Mike90250]

I don't know, but here's my guess.

The smarter inverters, have all sorts of sensors in them, I know mine measures both Watts and VoltAmps, on each leg. And I'm sure on the board with the transfer relays, it has a amps sensor, and if you are in a Load Shave mode, it only produces enough power to supply the local loads, and not backfeed the grid. It's likely to wobble around a dozen watts - + at any moment, but it averages out in the long run.

[lamagra]

That's how it's pretty much summed up in my mind:

Grid-tie inverter raises its output voltage slightly higher than the grid’s, leading to the fact that the inverter will see both the local loads and the grid as loads and it will output the entire PV power independently, since it acts as a current source. However, the current cannot traverse in 2 directions simultaneously, meaning that the grid will either provide the deficit, if needed, or receive surplus power, if any.

I just need someone to confirm if this is the right way to think about it.

[JFinch57]

There's a question to which I've been looking for an answer but I haven't found a clear one yet so I hope this forum will help.
How is the load shared between a grid-tie inverter and the grid? I understand that the inverter feeds the loads with all the available solar power and imports the remaining balance from the grid if needed or exports the excess if any. I would like to understand the technical details behind this process (how does the inverter have feeding priority over the grid?), because so far I haven't found a clear explanation online. It's been a while since I got into electrical circuits so could it be because of the fact that the grid tie inverter acts as a current source?

Thanks.

Electricity flow and water flow are very similar principles. Water flow equates to electrical current and water head pressure equates to voltage. Think of it as 3 bodies of water, one very large like an ocean (that's the grid) with a level river (pipe) that Y's to two other smaller ponds. One of the small ponds has a variable speed pump that is constantly pushing water up the river with the pump speed proportionally equal to the level of sunlight, that's your solar source. The other pond is your house, and the level goes up and down depending on how much water is getting pumped out of it by your individual circuit loads, and how much water flows into it from the ocean and the other pond (pumping with solar). When there is no sunlight the individual house loads pump water out of the house pond and the water (current) flows from the ocean to your house pond. When there is no load from the house pond (pond is full) and there is sunlight the solar pond pumps water back into the ocean (grid). In-between some goes each way in the Y and it's self regulating. It's that simple!

[Sunking]

Very simple Grid Tied Inverters are CURRENT SOURCES. You got yourself stuck inside a Voltage Source box. Like water current flows downhill to the lower energy state. Either your home load demand will be at a lower impedance than the grid, or the grid will be lower. Current does not care or know the difference.

[lamagra]

Electricity flow and water flow are very similar principles. Water flow equates to electrical current and water head pressure equates to voltage. Think of it as 3 bodies of water, one very large like an ocean (that's the grid) with a level river (pipe) that Y's to two other smaller ponds. One of the small ponds has a variable speed pump that is constantly pushing water up the river with the pump speed proportionally equal to the level of sunlight, that's your solar source. The other pond is your house, and the level goes up and down depending on how much water is getting pumped out of it by your individual circuit loads, and how much water flows into it from the ocean and the other pond (pumping with solar). When there is no sunlight the individual house loads pump water out of the house pond and the water (current) flows from the ocean to your house pond. When there is no load from the house pond (pond is full) and there is sunlight the solar pond pumps water back into the ocean (grid). In-between some goes each way in the Y and it's self regulating. It's that simple!

Thank you for your answer. I've used the water analogy many times to understand electrical concepts but in this case I'm curious about the case when the pv output is less than the demand load. How is it guaranteed that the house pond is being pumped with all the available water from the pv pond and only the difference is fed from the ocean?

[lamagra]

Very simple Grid Tied Inverters are CURRENT SOURCES. You got yourself stuck inside a Voltage Source box. Like water current flows downhill to the lower energy state. Either your home load demand will be at a lower impedance than the grid, or the grid will be lower. Current does not care or know the difference.

Thank you for your answer. From the point of view of loads what you are saying is true. But what i am asking about is: how is it guaranteed that all of the pv production is used by the loads and there is not a portion, no matter how small it is, flowing to the grid since both the grid and the loads are seen as LOAD by the PV system?

[sensij]

Thank you for your answer. From the point of view of loads what you are saying is true. But what i am asking about is: how is it guaranteed that all of the pv production is used by the loads and there is not a portion, no matter how small it is, flowing to the grid since both the grid and the loads are seen as LOAD by the PV system?

To do what you want, the inverter would be installed between the grid and the loads. If you do not want power to flow to the grid, the terminals that the grid connects to could be opened internally, for example. If the inverter is connected to a load center that is also fed by the grid, I think you are correct, all loads would look the same.

[sensij]

Thank you for your answer. I've used the water analogy many times to understand electrical concepts but in this case I'm curious about the case when the pv output is less than the demand load. How is it guaranteed that the house pond is being pumped with all the available water from the pv pond and only the difference is fed from the ocean?

Again, the inverter is a current source. It will at all times be trying to push its maximum available power. If there is not enough load to consume it, and the grid is not available to take the excess, then the voltage would rise and the inverter would adjust its current output down to match the load.

The adjustment mechanism to reduce the current output is to actually reduce the power input from the panels, by shifting the operating point of the panels away from their maximum efficiency.

[foo1bar]

Thank you for your answer. From the point of view of loads what you are saying is true. But what i am asking about is: how is it guaranteed that all of the pv production is used by the loads and there is not a portion, no matter how small it is, flowing to the grid since both the grid and the loads are seen as LOAD by the PV system?

Essentially it's guaranteed by physics.
Familiarize yourself with Kirchhoff's current law:
http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

So - you have 2 scenarios with PV production.

1> PV is producing more than used locally.
2> PV is producing less than used locally.

Scenario 1:
PV produces 10A which flows toward the node we'll call "main panel"
9A flows out toward "local load"
Kirchoff's current law says 1A must flow out of "main panel". So where is that current going to flow? Answer: toward "the grid"

Scenario 2:
PV produces 10A which flows toward the node we'll call "main panel"
12A flows out toward "local load"
Kirchoff's current law says 2A must flow in to "main panel". So where is that current going to come from? Answer: from "the grid"


Current is in 1 direction - it can't flow both in and out on the same wire, so it's not possible to have current from PV flowing out to the grid, while at the same time having current from the grid flowing in on that same conductor.

Now I have simplified things a bit in the above scenarios with an implicit assumption that the local loads are balanced - but really it doesn't change anything, you just would need to look at each leg of the service separately (2 legs for standard US residential service)

[JFinch57]

Thank you for your answer. From the point of view of loads what you are saying is true. But what i am asking about is: how is it guaranteed that all of the pv production is used by the loads and there is not a portion, no matter how small it is, flowing to the grid since both the grid and the loads are seen as LOAD by the PV system?

Voltage seeks its own level just as water seeks its own level. As soon as there is current flow there is voltage drop due to the wire resistance, and as soon as there is water flow there is pressure drop due to the friction in the pipe. The solar pond is constantly pumping water and the volume (current) is based on the available light. The water will flow to wherever there is less pressure just like the current will flow to wherever there is less voltage. If there is no load from the house pond the slight pressure increase has nowhere to go except to the ocean (grid). The ocean, in essence, has a lesser resistance due to the larger pipe. Picture a Y with one leg as a 10" pipe (ocean/grid), one leg as a 4" pipe (house) and one as a 1" pipe (solar). The house starts a load, sucking through the 4" pipe and the 1" pipe is pumping all that it can. 100% of the solar pond production will go to the house pond!

OK, let's theoretically take the whole contraption electrically down to absolute zero where there is no resistance, WTF will happen??? (Hypothetically since the electronic controls won't function at that temperature, but what if they did)

[inetdog]

Voltage seeks its own level just as water seeks its own level. As soon as there is current flow there is voltage drop due to the wire resistance, and as soon as there is water flow there is pressure drop due to the friction in the pipe. The solar pond is constantly pumping water and the volume (current) is based on the available light. The water will flow to wherever there is less pressure just like the current will flow to wherever there is less voltage. If there is no load from the house pond the slight pressure increase has nowhere to go except to the ocean (grid). The ocean, in essence, has a lesser resistance due to the larger pipe. Picture a Y with one leg as a 10" pipe (ocean/grid), one leg as a 4" pipe (house) and one as a 1" pipe (solar). The house starts a load, sucking through the 4" pipe and the 1" pipe is pumping all that it can. 100% of the solar pond production will go to the house pond!

OK, let's theoretically take the whole contraption electrically down to absolute zero where there is no resistance, WTF will happen??? (Hypothetically since the electronic controls won't function at that temperature, but what if they did)

With no resistance, the situation will still be exactly the same. The GTI will produce an amount of current which is determined by the amount of panel power available and the voltage that the grid supplies to the GTI terminals. Current will flow to the load and to the grid or to the load and from the grid in the amount required by the impedance of the load and the voltage of the grid. Which individual electrons end up following one path or the other is irrelevant, since the equations describe only the net current flow.

I think that your original question may have gotten a bit sidetracked by the issue of how some GTIs can be set so as to reduce their output instead of exporting power to the grid. That function requires that the GTI be able to sense (locally or via a remote current transformer or shunt) the current between the grid and your PV plus load system.
As you noted, if the GTI does not have this information, everything appears to it as a single load/source and it cannot limit export.

[Sunking]

Again, the inverter is a current source. It will at all times be trying to push its maximum available power. If there is not enough load to consume it, and the grid is not available to take the excess, then the voltage would rise and the inverter would adjust its current output down to match the load.

Partially true, but not the way it works. A GTI by design generates as much current the panels can produce. If there is not enough load the voltage will spike and shut itself down by design. It is possible to do what you say, but that cost mo money and complicates the design. Example SMA design, and power is extremely limited when the grid is down.

Your every day GTI is a pure current source and that current must have someplace to go. If not it will shut down by design. To do this the GTI voltage must be ever so slightly greater than the utility voltage at the connection point. To do this the GTI is looking for a specific range of impedance to be operating into. If the grid disconnects, the impedance will be way too high.

Current Sources by design have to operate into a narrow rand of impedance. If the impedance is too high, then the voltage goes to the source voltage of the supply. It has been a while since I looked at UL-1741 but the voltage range is fairly narrow of +/- 10%, and frequency is 1%.

[sensij]

Ah, yes, when I said "the grid is not available to take the excess" I didn't actually mean grid down, I just meant that the user desired not to backfeed into the grid. Thanks for catching that. An inverter operating in a zero backfeed mode when the grid is up would have the ability to regulate the output current as I described in the 2nd part of my post, but yes, that capability adds expense.

[Sunking]

Ah, yes, when I said "the grid is not available to take the excess" I didn't actually mean grid down, I just meant that the user desired not to backfeed into the grid. Thanks for catching that. An inverter operating in a zero backfeed mode when the grid is up would have the ability to regulate the output current as I described in the 2nd part of my post, but yes, that capability adds expense.

That would be a hybrid inverter, they cost big buck$, and lower efficiency. It can all be done, how much money do you have?

Sensi FWIW I am experimenting with SMPS right now tinkering with BMS and motor controllers for a start up. Exact same principles.