Neutral Current Calculation

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Also, I got to thinking, if the output currents on the two legs are always equal, there is really no need to have a neutral connection.

I did a little research (very little) and discovered that some inverters don't even have a neutral connection. Others have one, but it is only for voltage sensing purposes, not to carry load, as evidenced here:

http://www.xantrex.com/web/id/1746/docserve.aspx

This validates that both legs of the inverter produce the same current, and that the system neutral carries only the unbalanced load current.
 
080714-1135 EST

ELA:

Since the 2400 watts of load equals the available power from the inverter there is no power supplied by the transformer primary. This means the transformer secondary is coupling the excees energy from the left side to the right side in the above example.

If your meter charges you $0.11/KWH and pays you $0.04/KWH, then you lose on your own high cost power. However, it eliminates the need for expensive batteries. Apparently in Michigan standard meters are used and thus the buy and sell rates are the same.

I have heard that in Germany you may get paid more for what you sell than the cost of what you buy. Is that true?

.
 
gar said:
Since the 2400 watts of load equals the available power from the inverter there is no power supplied by the transformer primary. This means the transformer secondary is coupling the excees energy from the left side to the right side in the above example.
Click to expand...

It just dawned on me DOH! In the diagrammed situation of 5 buy and 5 sell, there would be no current flow in the primary winding! The overabundance of current on the left side of the inverter is actually applying its energy to the 15 amp load on the right side of the circuit.

Thanks for that revelation, gar!
 
I understood the concept of being able to run the meter backwards and actually sell power back to the utility (if the inverter output was greater than the load). I just didn't like the terms off sell on one leg and buy on the other leg.
Since in this example the inverter is putting out the same as the load requires we are neither buying nor selling with respect to the utility.
 
crossman said:
Agreed. But the concept of "buy" on one leg and "sell" on the other seems to be a valid means of discussing what is happening in the system.
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Why does reading this remind me of the movie Trading Places? ;)
 
crossman,
I think I can adapt to the talk about "buy and sell on different legs" as long as it is understood that it is with respect to some device doing the monitoring (possibly the inverter?) and not the utility.

I do not know much about these inverters and so I am still struggling with your diagram at the utility side where the two current arrows point towards each other and each have 5 amps. Wouldn't this imply zero volts across the secondary? Yet there is 240V at the other end across the loads?
 
ELA said:
I do not know much about these inverters and so I am still struggling with your diagram at the utility side where the two current arrows point towards each other and each have 5 amps. Wouldn't this imply zero volts across the secondary? Yet there is 240V at the other end across the loads?
Click to expand...

The whole thing is kind of strange!:confused: Still, I am convinced that my thinking is correct.

I think a good way to look at it would be this: Since the primary has no current, there is nothing being induced into the secondary from the primary. Therefore, the secondary winding on the left hand side of the diagram is simply acting as a load in respect to the inverter. There is still 240 volts across the two windings.

Darn... I am still confused:confused:
 
Perhaps the OP should have supplied a diagram and some input on how these work?

Even with no current through the primary if you put 5 amps through the secondary that implies a voltage and with the arrows as shown there would be no potential difference across the outside of the secondary?

I wonder if the inverter deals with dividing the currents somehow?

Perhaps this is more complex than can be explained by a simple diagram?
I am eager to learn.
 
ELA said:
crossman,
I think I can adapt to the talk about "buy and sell on different legs" as long as it is understood that it is with respect to some device doing the monitoring (possibly the inverter?) and not the utility.

I do not know much about these inverters and so I am still struggling with your diagram at the utility side where the two current arrows point towards each other and each have 5 amps. Wouldn't this imply zero volts across the secondary? Yet there is 240V at the other end across the loads?
Click to expand...

There's still voltage across the transformer secondary winding, and typically several volts higher than otherwise when "selling". This spring I'd see the output voltage as high as 260VAC when "selling". Right now I'm lucky to see 240VAC when "buying", and that might rise to 250VAC early in the day before everyone turns on the A/C.
 
080714-1448 EST

ELA:

Consider this:
A 12 V battery. This can sink or source current.
A variable power source (you might call it a power supply) that can sink or source current.
A resistor between the battery and the variable power source. This is the internal resistance or impedance.

When the variable source voltage is greater than 12 V current flows from the source to the battery, when the source voltage is lower than the battery current flows from the battery to the source. But all this while the ideal battery remained at 12v.

.
 
Gar,

Ok, that makes sense to me after some thought.

The 5 amps through some small internal transformer impedance produces a small voltage, that in one leg adds to the 120V source and in the other leg subtracts from the 120 source. Since those two small source impedance drops cancel across the entire secondary you are still left with the net source voltages adding to the 240V.

I was only looking at the canceling effect of the internal impedance drops and had ignored the fact that the source voltages were still present.

Thanks
 
080714-1947 EST

crossman:

The Xantrex manual has the following statement:
In order to operate, the GT inverter must have grid power available and connected. It will not provide backup power if the AC grid fails.
Click to expand...
This corresponds with the no neutral requirement, except voltage reference.

If you are not tied to the grid, then neutral must be connected and you can expect current in the neutral. Otherwise there would be no control of the 120 voltages, just the 240.

.
 
080714-2014 EST

tallgirl:

If your main panel bus voltage goes to 260 when selling it probably means you have a light load at that time and maybe the grid open circuit voltage is also high.

This could present a problem of overvoltage on those devices that are always on.

This is a point I had not thought about. It might mean we need more voltage tolerant devices, or local voltage regulation. There are local phase shift regulators made for refrigerators that may be effective. Incandescent lamps are a real problem without whole house regulation.

.
 
gar said:
080714-2014 EST

tallgirl:

If your main panel bus voltage goes to 260 when selling it probably means you have a light load at that time and maybe the grid open circuit voltage is also high.
Click to expand...

Are you guessing?

No, what it means is that the voltage drop in the service conductors has to be overcome. The greater the "sell" current, the higher the output voltage, up to the 262VAC limit. This is because Vdrop = Isell * Rservice-lateral.

If the grid open circuit voltage is high, the voltage drop in the service lateral would be added to that and push the output voltage over the 131VAC per leg limit and selling would stop.
 
gar said:
080714-1947 EST

crossman:

The Xantrex manual has the following statement:

This corresponds with the no neutral requirement, except voltage reference.

If you are not tied to the grid, then neutral must be connected and you can expect current in the neutral. Otherwise there would be no control of the 120 voltages, just the 240.

.
Click to expand...

No, it means that it is a GRID-TIE inverter and not a GRID-INTERACTIVE inverter. These are two different species of inverters.
 
080715-0631 EST

tallgirl:

From your point of view as a user of this system if your bus voltage at your main panel is at 260 V, then you will have shorter life of many devices in your home than if this bus voltage was the nominal 240 V. The devices that won't show much difference are ones designed with switching regulators that operate from about 95 to 270 V input.

Obviously the source impedance from your main panel bus to somewhere in the system where we can define a constant voltage source (the primary side of your pole transformer is a useful point) and the current fed into the grid will determine your main panel bus voltage.

If the service is 200 A and a 3% impedance, then at 40 A there should be about 3/5% voltage change from backfeeding 40 A.

Maybe more specific infromation on your system is needed.

.
 
gar said:
080715-0631 EST

tallgirl:

From your point of view as a user of this system if your bus voltage at your main panel is at 260 V, then you will have shorter life of many devices in your home than if this bus voltage was the nominal 240 V. The devices that won't show much difference are ones designed with switching regulators that operate from about 95 to 270 V input.
Click to expand...

Put a voltage monitor on an unloaded panel someday. 260VAC is the upper limit for an unloaded 120/240 service. The nominal range is 228 to 252VAC.

The voltage on the output side -- grid interactive inverters "sell" on their input terminals -- is typically a few volts below that.

As I said, it sounds like you're guessing. A voltage drop of 4 or 5 VAC is typical in the service conductors. So, if the transformer secondary is 250VAC, the main panel might be 245VAC. In order to sell, that 5VAC drop has to be overcome twice -- once so that the inverter inputs are at the same voltage as the transformer secondary, then a second time so the current will flow from the inverter to the transformer. This has nothing to do with transformer impedance at this point. It also has nothing to do with the load at the panel since the panel load is negative (by definition) when the inverter starts to sell.
 
crossman said:
With everything set up properly, you can actually sell energy back to the utility.
Click to expand...

Correct. I typically use about 5KWH by the time the inverters turn on. During the day I'll make 10 or 12KWH so that by sundown my consumption is close to or less than 0KWH and the meter has been going backwards most of that time. When the sun goes down the meter starts going forward again.

As far as I know, they produce the maximum current which is provided by the amount of sun at any given time on both legs.
Click to expand...

That's pretty much correct, except that there is no requirement that there be two inverters, or that the input of both be the same. So, if I have a toaster oven on L2 (I think it's on L2 ;) ) and the master inverter, on L1, is outputting 8A, and there is no other load on L1, L1 is -8A, L2 is +12A, and N is 20A. If there were a single inverter, solar noon, 2800WDC array, cool spring day, and I toasted a bagel, it would be L1 = -16A, L2 = 12A and N = 28A.

If it is a bright sunny day, and no one is home, and 95% of the load is turned off, why not go ahead and let the inverter produce maximum amps on both legs and sell the energy back to the utility? In effect, the meter turns backwards.
Click to expand...

And if it's a cool spring or fall day, the meter goes backwards really fast. The rest of the time it just goes backwards.
 
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