Single Phase Inverters on 208 3 Phase

[jaggedben]

Thank you. Maybe that helped a little. If I had an inverter firing at 180 degrees out of phase with the resultant wave that a 208 load would see, I get no current, but also no power lost. With emphasis from ggunn of VA in = VA out, even though the 208 resultant wave has those period of time where they cancel each, the electrons aren't really in a battle, they just don't move.

To add to what Wayne said...

When you have voltage and current waveforms that add to a positive number, and you flip the current waveform upside down as the inverter does, the result is a negative number which represents power flowing in the opposite direction.

(I prefer to think of the inverter as inverting the current instead of shifting it 180deg. The end result of those mathematical formulas is the same, but the inversion is a little better description of what the inverter actually does.)

That's not correct. Your 208V inverter does push out current 180 degrees out of phase with what a 208 load would draw. So that if you connect a 208V resistive load in parallel with your inverter, and the load wants to draw the same current that your inverter is pushing out, those two currents add up to zero from the point of view of the main voltage source.

To clarify, the current that the inverter pushes out is 180 deg out of phase with the current that the inverter would draw if it were a load. But it is in phase with the current drawn by a load connected in parallel to both inverter and grid.

 

[jaggedben]

The exception to this statement is that your purple resultant waveform, even reversing the polarity of measurement of B-N to N-B on your graph, the resultant wave still has 60 degrees of time that the resultant wave being fed back into the secondary is pushing electrons in the opposite direction of A-N and another 60 degrees of time the resultant wave is pushing electrons in the opposite direction of N-B. The primary side of the transformer doesn't replicate this battle, only the results. Those results feeding back through the primary will be less than what we put into the secondary.

Going back to this one more time at the risk of talking too much and making a possibly oversimplified statement....

That 60deg of time in which the grid is pushing current on A-N in the opposite direction from the solar inverter is precisely the reason that a load on A-N sees 120V instead of 104V and draws current from the solar and grid combined in phase with the A-N voltage (if it's a resistive load). So the load doesn't think that anything is out of the ordinary.

 

[wwhitney]

To clarify, the current that the inverter pushes out is 180 deg out of phase with the current that the inverter would draw if it were a load. But it is in phase with the current drawn by a load connected in parallel to both inverter and grid.

That's a question of direction conventions.

If the load and the inverter are both connected from A to B and we use the A-B convention for both, then the current through the load and the current through the inverter are negatives of each other (180 degrees out of phase if you like).

But if we want to think of the load and the inverter as a little loop, we might think of the load as going from A to B and the inverter going from B to A. Then the current through the load equals the the current through the inverter.

I find these direction conventions the most difficult part of thinking about 3 phase, keeping them all consistent.

Cheers, Wayne

 

[ggunn]

IMO and with no offense intended to the OP, I believe you guys are talking a bit over his head. I spent 20+ years as an electrical engineer in the semiconductor business where everything is low voltage DC and signal; when I made the transition to solar I found myself a virtual babe in the woods when it came to AC power and especially three phase. In the beginning I harbored some of the same misconceptions as the OP seems to have.

I humbly suggest that the OP work from the bottom up rather than the top down. Do a deep dive into the nuts and bolts of three phase power with the understanding that you will likely more than once get to the point where you think you totally understand it only to find out that you really don't. It still happens to me.

 

[wwhitney]

IMO and with no offense intended to the OP, I believe you guys are talking a bit over his head.

Maybe, but in that case, the short answer to the OP is "power is still transferred efficiently even when the current and voltage are 30 degrees out of phase. It's not a source of significant power loss."

If the OP is happy with that, great. If the OP wants more, we can try to address their concerns. But this is a sufficiently mathematical topic, it's hard to address misconceptions without invoking the underlying mathematics.

Cheers, Wayne

 

[jaggedben]

That's a question of direction conventions.
...

Or just semantics. Or, where we choose to look at the current.
If our direction convention is based on the normal direction of current when power is flowing from the grid, then the same current that flows out of phase with the grid from the solar inverter switches direction and becomes in phase with the grid if/when it reaches a node where it flows to a load. If we look at on the inverter circuit it's out of phase, if we look at it on the load circuit it's in phase. Same current. That is what I was trying to get across.

 

[jaggedben]

Also it's really sort of different from a phase shift as we mean when we talk about three phase. It's more simply an inversion that represents the change of direction. In that sense it is always 'in phase'. But that's a semantic statement not a mathematical one. (And one of the record breaking threads on this forum for number of replies was about how to describe that with respect to split-phase.)

 

[jim dungar]

This is where I don't agree,

Then there is no point in continuing this discussion.

.... at least for one wave passing through two phase angle legs of a transformer's secondary coils.

You are using incorrect terminology.

If we are using our resultant wave and sending it back through A-B, then we are pushing electrons backwards through B-N most of the time.

There is no need to even think about electrons and how they flow.

 

[ggunn]

Also it's really sort of different from a phase shift as we mean when we talk about three phase. It's more simply an inversion that represents the change of direction. In that sense it is always 'in phase'. But that's a semantic statement not a mathematical one. (And one of the record breaking threads on this forum for number of replies was about how to describe that with respect to split-phase.)

Uh, oh...

 

[ggunn]

Maybe, but in that case, the short answer to the OP is "power is still transferred efficiently even when the current and voltage are 30 degrees out of phase. It's not a source of significant power loss."

If the OP is happy with that, great. If the OP wants more, we can try to address their concerns. But this is a sufficiently mathematical topic, it's hard to address misconceptions without invoking the underlying mathematics.

Oh, yes, I agree totally; that is why I encourage the OP to train himself up on the fundamentals of three phase power, especially if he plans to work with PV on three phase services.

 

[electrofelon]

A closely related topic that I got thinking about while skimming over this thread: I got thinking about the KW of inverters when run on 208 vs 240. My recollection was that inverters have a "hard" current limit, thus the KW will be reduced at 208 operation. This is indeed the case for SMA sunny boy inverters. Interesting though Solis and fronius inverters provide for more current and the same KW at 208V operation. So it depends.

 

[Carultch]

A closely related topic that I got thinking about while skimming over this thread: I got thinking about the KW of inverters when run on 208 vs 240. My recollection was that inverters have a "hard" current limit, thus the KW will be reduced at 208 operation. This is indeed the case for SMA sunny boy inverters. Interesting though Solis and fronius inverters provide for more current and the same KW at 208V operation. So it depends.

On transformerless inverter designs, as is the norm most recently, you are correct. If an inverter has a flexible voltage setting by firmware, it's generally a fixed current limit, and a power rating that derates proportionally to grid voltage. This means a 6 kW inverter at 240V will effectively become at most, a 5.2 kW inverter at 208V. What is ultimately happening, is the internal relay settings are adjusted to the voltages that define the operating voltage window, and the inverter will produce whatever voltage is needed to follow the grid.

Historically, when string inverters had integrated transformers about 10+ years ago, it used to be that there was a tap you change on the transformer, so that power rating would be fixed instead, and the current would scale inversely proportional to voltage.

In concept, an inverter manufacturer could have transformerless inverters with firmware current limitations that are a function of the voltage setting, but this is less common.

 

[electrofelon]

On transformerless inverter designs, as is the norm most recently, you are correct. If an inverter has a flexible voltage setting by firmware, it's generally a fixed current limit, and a power rating that derates proportionally to grid voltage. This means a 6 kW inverter at 240V will effectively become at most, a 5.2 kW inverter at 208V. What is ultimately happening, is the internal relay settings are adjusted to the voltages that define the operating voltage window, and the inverter will produce whatever voltage is needed to follow the grid.

Historically, when string inverters had integrated transformers about 10+ years ago, it used to be that there was a tap you change on the transformer, so that power rating would be fixed instead, and the current would scale inversely proportional to voltage.

In concept, an inverter manufacturer could have transformerless inverters with firmware current limitations that are a function of the voltage setting, but this is less common.

But 2 out of 3 inverters that I mentioned were constant power at both voltages.

 

[Carultch]

Thank you. Maybe that helped a little. If I had an inverter firing at 180 degrees out of phase with the resultant wave that a 208 load would see, I get no current, but also no power lost. With emphasis from ggunn of VA in = VA out, even though the 208 resultant wave has those period of time where they cancel each, the electrons aren't really in a battle, they just don't move.

I have a Geogebra resource that shows visuals for the math behind this concept.

Interphase Load

Interphase Load

www.geogebra.org

You can see it in either vector view or waveform view. I find it is easier to understand in vector view, after being accustomed to phase vector representation. The waveform view shows what really is happening, with the voltages and currents as functions of time.

 

[bellington]

I have a Geogebra resource that shows visuals for the math behind this concept.

Interphase Load

Interphase Load

www.geogebra.org

You can see it in either vector view or waveform view. I find it is easier to understand in vector view, after being accustomed to phase vector representation. The waveform view shows what really is happening, with the voltages and currents as functions of time.

Thank you. I appreciate the waveform view better. The current aspect was my greatest concern. Still working that through my head. Those 60 degrees of time on each leg represent reduced current flow as compared to an in phase situation. So, for 1/3 of each cycle I am not moving as many electrons as I would be if using the same inverter on a single phase leg of 240. Am I losing energy? No, I'm just not moving it as effectively as I would be on a single leg of 240. I would move more electrons and transfer more power per second using a single leg as opposed to using two legs. VA in still equals VA out, single phase 240 is 7.6 kW and two legs 208 is 6.6 kW. I'm not technically losing that 1 kW; I'm just not using it.

 

[bellington]

But 2 out of 3 inverters that I mentioned were constant power at both voltages.

The ones in question are current limited to 32 amps. So at 240 7.6 kW and 208 6.6 kW.

 

[Carultch]

Thank you. I appreciate the waveform view better. The current aspect was my greatest concern. Still working that through my head. Those 60 degrees of time on each leg represent reduced current flow as compared to an in phase situation. So, for 1/3 of each cycle I am not moving as many electrons as I would be if using the same inverter on a single phase leg of 240. Am I losing energy? No, I'm just not moving it as effectively as I would be on a single leg of 240. I would move more electrons and transfer more power per second using a single leg as opposed to using two legs. VA in still equals VA out, single phase 240 is 7.6 kW and two legs 208 is 6.6 kW. I'm not technically losing that 1 kW; I'm just not using it.

Glad my visuals helped. I figured that would be much easier to read than if I filled this page with equations.

From the point-of-view of the inverter, the interphase voltage it receives, is in-phase with the current it produces. It follows the 208V interphase voltage and produces current is in-phase with it, assuming it doesn't intentionally support reactive power. Given a 7.6 kW inverter at 240V, that derates to 6.6 kW at 208V, it never produces the "missing" kW or kVA on the AC-side in the first place. It's output is current-limited to 32A, and 32A * 208V is about 6.6 kW at unity power factor.

It's when you switch perspectives, and examine the line current and phase-to-neutral voltage, that you get this 30 degree phase shift between voltage and current, and a difference between kW and kVA. Once you balance it with other inverters, the reactive power of the shared phases will add up to zero.

 

[bellington]

This is more or less correct. If we think of the inverter connected A-B as pushing current back through A-N and N-B, then yes, for 1/6 of each cycle, the current it pushes though A-N will not be carrying energy back upstream but the energy flow will instead be like a load. Same for N-B, although a different 1/6 of the cycle.

But all that means is this setup uses a bit more current to transfer power than a setup with individual A-N and B-N inverters. The net power transfer through the A-N coil will be 86.6% as much as it would be if the current through the A-N coil were in phase with the voltage (with a negative sign for power backfeed).

This 100% vs 86.6% does not represent a loss of power generated.


The primary side coils will always see current and voltage in exactly the same phase relationship to each other as in the secondary coils. That's how transformers work.

The primary coils only see the results of actual current flow. When the voltage waveforms opposed each other, the resultant current flow would be reduced and induce a different resultant voltage waveform into the primary. So the primary would not see those smaller opposing voltages, it would only see something less in the greater amplitude wave as it increased toward each peak.

What you see outside the transformer depends on the transformer configuration. If it is wye-wye, then you'll see a current between primary lines A' and B' exactly comparable to the current the inverter is pushing out between B and A. The "battle" will be fully replicated. (Not a battle, just a power factor less than 1).

But if the transformer is delta-wye, the story is more complicated, and I think (not 100% sure) there would be current on A', B' and C'.

Some quantities will be less or will be more, but power transferred will still be the same.

Agreed. When those electrons are confused by opposing voltages, they don't fight and create heat, or fall out of the conductor, they simply do not accomplish as much as they could have accomplished with waves that were in phase.

Cheers, Wayne

 

[wwhitney]

Those 60 degrees of time on each leg represent reduced current flow as compared to an in phase situation. So, for 1/3 of each cycle I am not moving as many electrons as I would be if using the same inverter on a single phase leg of 240.

60 degrees is 1/6 of a cycle, not 1/3.

But the relevant figure is cos(30) = 0.866, or 86.6%. So you're transferring only 86.6% as much energy as you could using the inverter on 240V. Which we already knew and is reflected in the ratio 208V / 240V = 86.6%.

Cheers, Wayne

 

[bellington]

60 degrees is 1/6 of a cycle, not 1/3.

Makes little difference really, but I was considering 60 degrees on each leg, for a total of 120.

But the relevant figure is cos(30) = 0.866, or 86.6%. So you're transferring only 86.6% as much energy as you could using the inverter on 240V. Which we already knew and is reflected in the ratio 208V / 240V = 86.6%.

Cheers, Wayne

Thank you again for you patience!