Single Phase Inverters on 208 3 Phase

wwhitney said:
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
Click to expand...
Usually, the way you'd make anything close to an idealized current source in practice, is with a voltage source that dynamically adjusts its output.

I'm not aware of any kind of source that natively acts approximately as an ideal current source, without a control loop in some form or another to help it adapt to changes in the impedance of its output circuit.
 
Carultch said:
Usually, the way you'd make anything close to an idealized current source in practice, is with a voltage source that dynamically adjusts its output.

I'm not aware of any kind of source that natively acts approximately as an ideal current source, without a control loop in some form or another to help it adapt to changes in the impedance of its output circuit.
Click to expand...
Well, a shorted PV module acts a lot like a current source, as ggunn mentioned upthread. But I can't think of any native AC sources.
 
jaggedben said:
Well, a shorted PV module acts a lot like a current source, as ggunn mentioned upthread. But I can't think of any native AC sources.
Click to expand...
A "current source" as I understand the term, means that it produces the same current, that is independent of load impedance. At least up to a maximum allowable output voltage.
 
Carultch said:
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.


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.
Click to expand...
If we go in the opposite direction and have very little load and higher voltage on the grid, do the inverter's rise in DC input voltage demand an even higher voltage rise at the inverter to keep sending current to the battery management?
 
wwhitney said:
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
Click to expand...
I think you have to abandon the idea of an ideal current source because such a thing does not exist. Perhaps it is better to say a GTI "looks like" a current source or "acts like" a current source. Consider that an ideal voltage source with a resistor in series is equivalent to an ideal current source with a resistor in parallel. A real life current source adjusts its output voltage to maintain the specified amount of current flow. Thus for a GTI, pick a moment in time and the inverter is producing whatever voltage it needs to "push" its specified amount of current into the grid. Here is something to think about: An ideal current source would increase the voltage until an arc is created and the specified current flows through the arc.
 
Last edited:
Carultch said:
A "current source" as I understand the term, means that it produces the same current, that is independent of load impedance. At least up to a maximum allowable output voltage.
Click to expand...
Yeah, and I believe that is a tolerably accurate description of a PV module once the load drops the voltage below a certain point on the I-V curve. It's not an ideal current source of course, and not across its entire possible voltage output. But it acts a lot like that below a certain completed circuit resistance.

PS, I agree with electrofelon regarding inverters.
 
bellington said:
If we go in the opposite direction and have very little load and higher voltage on the grid, do the inverter's rise in DC input voltage demand an even higher voltage rise at the inverter to keep sending current to the battery management?
Click to expand...

Ideally, very little load should mean the voltage at the grid is at nominal, not higher.

To my understanding, it is true that the circuit between the solar inverter and the 'grid' needs to handle the voltage drop. Your project should be designed to handle max voltage drop from load or max voltage rise (aka drop in the other direction) from backfeed, whichever is going to be greater in your system. Another reason to hire an engineer.

Utility grids have a lot of equipment on them to regulate voltage at various places along the line. Some of it doesn't work well bidirectionally, or so I've read.

(The DC inverter input voltage is not really part of this. There's a lot of power electronics in between DC and AC in a grid tie inverter that. It can do different stuff in different situations.)
 
Thank you so much for all the discussion. I appreciate your comments. After my attempts to absorb all I can from the discussions, let me describe what my brain thinks needs to happen to feed current back into the 480-volt 3-phase grid.

A single (or multiple) 3-phase inverter(s) (not the single-phase inverters we were discussing) connected to a wye secondary would be the best option for what I need.

With a 3-phase inverter connected to A, B, C, N, and G, I see two options, if it operates by creating 3 separate voltage waveforms:

(1) Treat each transformer coil individually (waveform applied to A-N, B-N, and C-N) and provide a voltage waveform of greater magnitude than that of each grid coil voltage A-N, B-N, C-N, or

(2) Treat the three coils as pairs A-B, B-C, C-A and provide voltage waveforms of greater magnitude than the two pairs combined and switch the current on each cycle such that it moves from A to B, then A to C, then C to A, then B to A, (I may not have correctly stated the timing, or direction) and repeat this process for each two-leg pair. Although this allows my head to visualize what I think should happen, it seems this option would be less efficient than option 1.

The ability to have one waveform transferred from A-B to A-C, reversing, and repeating for each pair of legs solves the problem that my head has been having with mismatched waveforms.
 
bellington said:
The ability to have one waveform transferred from A-B to A-C, reversing, and repeating for each pair of legs solves the problem that my head has been having with mismatched waveforms.
Click to expand...
To be clear, the problem is only in your understanding.

Your option (2) works as well as your option (1).

And 3 separate single phase inverters in a balanced arrangement A-B, B-C, and C-A will work just as well as your option (2).

In fact, consider the case where all inverters involved have plenty of DC input so their AC output is pegged, and you have either (a) 3 such single phase inverters, or (b) a single 3 phase inverter of equivalent AC capacity. Then if you look at the voltage and current waveforms on an AC feeder connected to the inverter(s), you won't be able to tell the difference between (a) and (b). They will behave identically.

Cheers, Wayne
 
electrofelon said:
Here is something to think about: An ideal current source would increase the voltage until an arc is created and the specified current flows through the arc.
Click to expand...
Some CT devices will do that if you try to disconnect them when there is a lot of current flowing through the conductors they are measuring. It can be very dangerous.
 
electrofelon said:
Yeah , it may seem kinda counterintuitive on the surface, but a CT is a step up transformer.
Click to expand...
Sort of. The type that has no internal resistor is a virtual current source that sends current proportional to the current being monitored to a calibrated resistance in the monitoring equipment. The higher the resistance the higher the voltage, and for an open circuit the voltage can get dangerously high. They should be installed with a shorting switch across their terminals so that they can be serviced without having to shut down the monitored circuit.
 
electrofelon said:
Well it is, but the secondary is shorted during normal operation.
Click to expand...
I'm not sure what you mean. During normal operation the "secondary" of that type of CT (if that's what you are calling the output) isn't shorted; it's routed through a calibrated resistance in the monitoring equipment which measures the voltage across the resistance to calculate the current in the conductor being monitored. It should be installed with a shorting switch so that the CT never sees an open circuit when the monitoring equipment is being serviced. V = IR and I is fixed, so when R is "infinite" V gets very high.

This has gotten far afield of the original subject, but an advantage to using this type of CT is that since it is a current source the leads from the CT to the monitoring equipment can be as long as they need to be because the Vd in the leads isn't a factor.
 
wwhitney said:
To be clear, the problem is only in your understanding.

Your option (2) works as well as your option (1).

And 3 separate single phase inverters in a balanced arrangement A-B, B-C, and C-A will work just as well as your option (2).
Click to expand...
This is where my understanding is facing the greatest obstacle. If 3 separate inverters are forming a load-derived resultant waveform to place back across A-B and transfer it efficiently, then it appears to me that the inverter for A-B would need to swap with the A-C inverter at different times in each wave cycle, more closely replicating what a 3-phase generator would do.
wwhitney said:
In fact, consider the case where all inverters involved have plenty of DC input so their AC output is pegged, and you have either (a) 3 such single phase inverters, or (b) a single 3 phase inverter of equivalent AC capacity. Then if you look at the voltage and current waveforms on an AC feeder connected to the inverter(s), you won't be able to tell the difference between (a) and (b). They will behave identically.

Cheers, Wayne
Click to expand...
 
Carultch said:
It does create a voltage, just not an independent voltage.
Click to expand...

Thanks for sticking with me. I realized that my confusion came from the point of view that the second source is in series with the first source (which, if you delete all the loads, they are in a loop so it is sort of series). But the second source is in parallel with the first source (because the loads are connected to both sides of both sources, not just one side of one source and one side of the other source).

Cheers, Wayne
 
bellington said:
This is where my understanding is facing the greatest obstacle. If 3 separate inverters are forming a load-derived resultant waveform to place back across A-B and transfer it efficiently, then it appears to me that the inverter for A-B would need to swap with the A-C inverter at different times in each wave cycle, more closely replicating what a 3-phase generator would do.
Click to expand...
Your continued discussion of "waveform" without referring to current or voltage really makes me question whether you are conflating the two. They are separate, and you really need to consider both. So please label them "current" or "voltage" in your discussion.

The primary source generates the voltage waveform(s). The inverter(s) (single phase or 3 phase) see those voltage waveform(s) and create their own matching internal voltage waveform(s). They use their voltage waveform(s) to push out current waveform(s) into the wiring that are in sync with the voltage waveform(s).

So as to "it appears to me that the inverter for A-B would need to swap with the A-C inverter at different times in each wave cycle, more closely replicating what a 3-phase generator would do," that happens automatically as a consequence of the above. The 3 phase primary source generates 3 voltage waveforms that are 120 degrees apart. So the A-B inverter sees a voltage waveform 120 degrees apart from the voltage waveform the C-A inverter sees. That means the A-B inverter pushes out a current waveform 120 degrees apart from the current waveform that the C-A inverter pushes out.

Cheers, Wayne
 
bellington said:
This is where my understanding is facing the greatest obstacle. If 3 separate inverters are forming a load-derived resultant waveform to place back across A-B and transfer it efficiently, then it appears to me that the inverter for A-B would need to swap with the A-C inverter at different times in each wave cycle, more closely replicating what a 3-phase generator would do.
Click to expand...
What is a load derived resultant waveform? Are you talking about generators or grid tied PV inverters? A three phase PV inverter does not "closely replicate" the behavior of three identical single phase inverters; some (possibly all) three phase inverters are actually three single phase inverters internally.
 
ggunn said:
A three phase PV inverter does not "closely replicate" the behavior of three identical single phase inverters; some (possibly all) three phase inverters are actually three single phase inverters internally.
Click to expand...
But as I understand it, there is still a difference in behavior between a 3 phase inverter and 3 AC-coupled single phase inverters. In that in the 3 phase inverter the single phase subunits would share a common DC bus, so they'd each use 1/3 of the available power to jointly create a balanced output. While the 3 separate single phase inverters would have separate DC inputs and may not be able to jointly create a balanced output.

Cheers, Wayne
 
wwhitney said:
But as I understand it, there is still a difference in behavior between a 3 phase inverter and 3 AC-coupled single phase inverters. In that in the 3 phase inverter the single phase subunits would share a common DC bus, so they'd each use 1/3 of the available power to jointly create a balanced output. While the 3 separate single phase inverters would have separate DC inputs and may not be able to jointly create a balanced output.

Cheers, Wayne
Click to expand...
Of course. When I said "identical single phase inverters", I meant identical in all respects. Also of course, in the design phase I only consider the AC maximum output current of each inverter irrespective of DC, and if the single phase inverters are not identical or if they are not evenly distributed, I calculate the maximum current on each phase and size everything to whichever is the highest.
 
You must log in or register to reply here.
Top