Single Phase Inverters on 208 3 Phase

bellington said:
Would it also be accurate to say that a balanced set of three single phase inverters connected to a wye secondary to A-B, B-C, and C-A cannot inject both positive and negative current into phases A-N and N-B at the same time?
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What does "inject negative current" mean? Obviously, current cannot flow both directions in a conductor simultaneously.
 
ggunn said:
What does "inject negative current" mean?
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Positive current would be current flowing during the up swing of a sine wave (negative peak to positive peak). Negative current would be current flowing during a down swing of a sine wave (positive peak to negative peak).
ggunn said:
Obviously, current cannot flow both directions in a conductor simultaneously.
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That is my main point.
 
Carultch said:
It does create a voltage, just not an independent voltage.
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OK, but this is just a voltage internal to the inverter, it's not detectable outside the inverter, yes?
Carultch said:
It uses a feedback loop to solve Ohm's law across the conductors leading to it, to provide a slightly higher voltage than the grid provides, following the frequency and phase of the grid voltage. When we speak of "voltage drop" in this context, it's really a voltage rise.
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When you say "slighter higher voltage" than the grid provides, are you referring just to the effect of voltage rise due to the grid impedance external to the inverter? And once you account for that, the voltage across the inverter terminals is otherwise unchanged by the inverter's operation?

Cheers, Wayne
 
bellington said:
Positive current would be current flowing during the up swing of a sine wave (negative peak to positive peak).
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Which sine wave?

If you're talking about the current sinewave, then positive current would mean when the sinewave is above zero, and negative current would mean when the sinewave is below zero. And the notions you are describing would be called "increasing" and "decreasing" current.

Cheers, Wayne
 
wwhitney said:
Which sine wave?

If you're talking about the current sinewave, then positive current would mean when the sinewave is above zero, and negative current would mean when the sinewave is below zero. And the notions you are describing would be called "increasing" and "decreasing" current.

Cheers, Wayne
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It was the best method I have to define the direction of flow of electrons.
 
bellington said:
It was the best method I have to define the direction of flow of electrons.
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The current I records both the magnitude and direction of the flow of electrons. I positive means they flow one way (according to your convention, i.e. if you are talking about I(A-B) and use that to mean from B to A, and are using negative charge carriers, then I(A-B) is positive when they flow from B to A); I negative means they flow the other way.

Current I is different from Voltage V. There are two different sinewaves at each point of the circuit (once you fix a convention for direction of I everywhere, and fix a zero voltage reference point). These two different sinewaves may not be in phase (generally they aren't; a resistive load is a special case where they are).

Cheers, Wayne
 
bellington said:
It was the best method I have to define the direction of flow of electrons.
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Actually the convention for the direction of electrical current flowing from positive to negative was established by Benjamin Franklin way before electrons were discovered. It turns out that electrons instead flow from negative to postive (for example, from the hot cathode to the anode or "plate" of a vacuum tube) because like charges repel and opposite charges attract. And so you'd need to refer to the flow of electrons as an "electron current" to distinguish it from the conventional polarity for current flow. If you are dealing with electrical power as is done on this forum then there is rarely a need to consider the actual direction of electron flow, and so you would use the convention of positive to negative current flow.

https://pwg.gsfc.nasa.gov/Education/woppos.html
 
bellington said:
Would it also be accurate to say that a balanced set of three single phase inverters connected to a wye secondary to A-B, B-C, and C-A cannot inject both positive and negative current into phases A-N and N-B at the same time?
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No that would be inaccurate. As we have already discussed above, there's a portion of the cycle where the A-N and B-N voltages are opposed to each other which is why they add up to 208V and not 120V.

Similarly, each pair of inverters connected to a bus - say A-B and C-A both connected to A - will be out of phase with each other so that the current they put on the A bus is at times an addition of the two and at other times a subtraction. This is why if the inverters output the same current, their RMS current output to the A bus is vector sum of 1.73 times their individual outputs and not the simple sum of both. E.g. if both are outputting 10A the line current on A is 17.3A not 20A.

The same is true for the inverters connected to the B-bus, and the C bus phase shifted 120 and 240 degrees as expected.

Someone may correct me if I'm wrong but I believe the vector sum of the two inverter current outputs on any line bus (e.g. A) will be in phase (from a loads point of view, see above) with the line to neutral voltage on that bus (A-N correspondingly). So if you have a load connected to A-N, the current from the inverters will be flowing in the same direction as grid current at the same time. But also, if it's not in phase, I'm not worried; the grid will supply or absorb the difference. Because the grid maintains the A-N voltage, which is what causes current to flow on A-N.

All of this will also come out with the corresponding phase shifted timing for N-B.

(I'm accepting that you're using 'positive' and 'negative' to indicate opposite directions of current flow, in this case in the series circuit A-N-B.)
 
jaggedben said:
Someone may correct me if I'm wrong but I believe the vector sum of the two inverter current outputs on any line bus (e.g. A) will be in phase (from a loads point of view, see above) with the line to neutral voltage on that bus (A-N correspondingly).
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If the currents are of equal magnitude, that is true; in general it is not.

Cheers, Wayne
 
If 240 single phase is required for some equipment in the building with 134 kW PV, would a 480/277 Y to 240 D 120 4 wire, split leg transformer also work well with balanced groupings of the single phase inverters on each leg of 240 D?
 
bellington said:
If 240 single phase is required for some equipment in the building with 134 kW PV, would a 480/277 Y to 240 D 120 4 wire, split leg transformer also work well with balanced groupings of the single phase inverters on each leg of 240 D?
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You mean a high leg delta? Probably that is worse. First, it will only really work at all if your inverters can operate on 240 without a neutral. Otherwise you'd have to connect the entire system to the split leg, meaning it's not balanced, meaning everything needs to be rated, like, 3 times higher to accommodate the backfeed. That would really be a deal killer. But even if you could balance the output, I have some reasons to question whether that should be done on a high-leg system. I think it had better at least be a closed delta.
 
bellington said:
Why is 240 without a neutral a problem, but not 208?
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It's only a problem for a 240V inverter if the inverter specs say it is; some 240V inverters can run without a neutral and some can't. 208V with a neutral is different because the phase to neutral voltage is 120V but the phase to phase voltage is 208V.
 
Most (but not all) single phase inverters I've worked with would operate at either 240 or 208 with a neutral, because the phase angle would tell them which nominal voltage is the correct one. Some, but not all, could be reprogrammed to work without a neutral.

If you have an inverter that requires a neutral, then in a high leg delta you have two line-line phases that don't have the correct relationship to the neutral. So you can only backfeed the phase that does, and have an unbalanced output. In a 208/120 wye you don't have that problem, because all L-L phases have a consistent phase relationship to neutral.

None of this speaks to the more theoretical questions upthread of whether backfeeding without a neutral could work in theory. It's practicalities of equipment engineering and meeting product standards and I guess manufacturers choices of how they want to support their equipment.
 
jaggedben said:
Most (but not all) single phase inverters I've worked with would operate at either 240 or 208 with a neutral, because the phase angle would tell them which nominal voltage is the correct one. Some, but not all, could be reprogrammed to work without a neutral.

If you have an inverter that requires a neutral, then in a high leg delta you have two line-line phases that don't have the correct relationship to the neutral. So you can only backfeed the phase that does, and have an unbalanced output. In a 208/120 wye you don't have that problem, because all L-L phases have a consistent phase relationship to neutral.

None of this speaks to the more theoretical questions upthread of whether backfeeding without a neutral could work in theory. It's practicalities of equipment engineering and meeting product standards and I guess manufacturers choices of how they want to support their equipment.
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I have built a couple of systems on high leg services with 240V single phase inverters where the A-C inverter(s) used a neutral but the A-B and B-C inverters did not.
 
wwhitney said:
OK, but this is just a voltage internal to the inverter, it's not detectable outside the inverter, yes?
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There will be a very little amount of voltage rise at the service panel, due to the inverter providing power. This reduces the demand and the voltage drop on the conductors from the service transformer, and the utility's distribution network as a whole.

Prior to operation, suppose you start with 230V on a nominal 240V service. Assuming the loads stay exactly the same, it is completely realistic to expect the service panel voltage to move to 232V. It may be 235V at the inverter's terminals to accomplish this, if 3V is the ohmic voltage drop.

wwhitney said:
When you say "slighter higher voltage" than the grid provides, are you referring just to the effect of voltage rise due to the grid impedance external to the inverter? And once you account for that, the voltage across the inverter terminals is otherwise unchanged by the inverter's operation?
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The inverter terminals have to produce the voltage necessary to drive power to flow away from the inverter, instead of toward the inverter. This means, if it is 240V at the service panel, and 5 Volts corresponds to the ohmic voltage drop of the output current and possible output power, then it will have to produce 245V at its terminals, to send the power to the point of interconnection.
 
Carultch said:
The inverter terminals have to produce the voltage necessary to drive power to flow away from the inverter, instead of toward the inverter. This means, if it is 240V at the service panel, and 5 Volts corresponds to the ohmic voltage drop of the output current and possible output power, then it will have to produce 245V at its terminals, to send the power to the point of interconnection.
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I guess I wasn't clear in my question. I understand how voltage rise works in the external circuit if I think of the inverter as an idealized current source.

What I don't understand is how the inverter works as a current source, how it pushes out current. And whether the way it works somehow involves internally creating an even higher voltage that might be detectable at the inverter terminals as some epsilon of additional voltage, beyond what is attributable to voltage rise in the external circuit.

Thanks,
Wayne
 
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