Single Phase Inverters on 208 3 Phase

[ggunn]

I was talking about connecting a delta voltage source to the wye side of a transformer. (Forget about grid tied inverters and current sources.) Do I get a true neutral point?

Grounding has nothing to do with the question.

Yes, a voltage source is not a grid-tied inverter. I mean a generator or an off-grid inverter or something.
I guess if your answer remains the same I'll take your word for it?

This has gone off into the weeds a bit, but if you were to connect three voltage sources phase to phase that are exactly the same voltage and phase locked to be 120 degrees apart to the primary side phase conductors of a wye primary transformer, I think (but I do not know for sure because I have never tried anything like this) that the center of the wye would probably be a neutral for that array of sources.

 

[ggunn]

This may hold a record for the most active post ever, only second to the debate of pre twisting when installing a wire nut.

Not even remotely close. The Music Game thread in the Campfire Chat subforum is on post #17,070 and my first contribution was #4.

 

[BackCountry]

Not even remotely close. The Music Game thread in the Campfire Chat subforum is 854 pages long.

I stand corrected.

I have never ventured into the campfire chat; however, I will briefly stop by shortly — while I pre twist and call it “split phase.”

 

[Bill_F]

And again.... OOOOO MY!
And, I can't believe that several of our intense, knowledgeable, recognized members have contributed SO heavily, of their time, education, & experience, just going at it!
I fully admit - I'm having a tough time keeping up!
But I thank u guys again, as I will be re-reading, so as to try to continue learning!
Thanks!
Bill

 

[bellington]

First, expressing all numbers as decimals hides their meaning. So I'm going to edit my quote of your post by replacing decimals with their relevant symbolic expressions. The sqrt(2) factor comes from the ratio of peak voltage to RMS voltage.

The decimals are time and the voltages listed are those that occur at those times. I don't need RMS; I'm using actual instantaneous voltages as represented by the corresponding sine waves.

Yes, A-N can stay 0,

If current is running through A-B from N-B voltage, and the resultant voltage being fed back into A-B, then at the time A-N should be at zero voltage, there are two other sources pushing electrons through it. Electrons moving in the same direction from two other sources, demands that there is a voltage rise on A-N at that specific time.

and if A-B = N-B as you've correctly stated, it must be zero. When we define voltage between two points, the resulting values have to satisfy certain rules. The most fundamental is that adding up the voltages around a loop is always 0. So for voltages, A-B = N-B + A-N, always. When A-B = N-B, that leaves us with A-N =0.

I'm not sure how you conceive of voltage difference, but it's a difference of electrical potential energy. So an analogy for gravity near a small area of our planet's surface is just height difference. If the height difference from B to A equals the height difference from B to N, then A and N have to be at the same height.

[And just like we can define an absolute height if we pick a zero reference, say that mean sea level is height 0, the same works for voltage. Often we choose N as the zero voltage reference; then we can just talk about the voltage at A and at B, etc. But for the behavior between two other points in the circuit, like A and B, we still have to take the difference of those values to get the relevant voltages.]

Also, the voltage between two points is different from the current between two points. Just because one point has higher voltage than another doesn't mean that electrons are necessarily moving from the higher voltage to the lower voltage. That happens for a resistive load, but this discussion requires going beyond that simple case.

I didn't check the rest of your decimals to see if you have the correct values or not, as I think your first paragraph already provided enough for discussion.


I think you may be misapprehending the difference between voltage and current. With a grid-following inverter, the voltages in the entire system are determine by the grid (or in a blackout, by a grid-forming inverter elsewhere in the system). All of your discussion earlier in your post was just about voltage waveforms, and how they are related A-B, N-B, and A-N.

The grid-following inverter can't change any of those voltages; all it can do is pump out current as required to transfer power. So the system voltages are all fixed, and we just need to think about the current. When the inverter is connected A-B, it pumps out current as the exact negative of what a water heater connected A-B would do. If there happens to also be a water heater connected A-B, the current the inverter pumps out can be thought of as going directly to that water heater, end of story (as long as the water heater's current demand is no smaller than the inverter's current output).

It seems a voltage difference is required to push the current at the correct time. That current must have an entrance into the circuit and a return. Entrance into leg A and return into N provides such a path. Entrance into A and return through B requires A-B to be in the same phase.

But in the case where there are no other loads between the inverter and the transformer, then the current passes back through the transformer. The relationship between the resulting primary side current and the secondary side current the inverter is creating is exactly the same as the relationship between the primary side current and the secondary side current for a water heater connected A-B, except for a minus sign in the case of the inverter.

In other words, the single phase grid-following inverter is just a "negative water heater".

That's the part that doesn't work if you look at each wave at 30 degree intervals of time.

Cheers, Wayne

 

[wwhitney]

The decimals are time and the voltages listed are those that occur at those times. I don't need RMS; I'm using actual instantaneous voltages as represented by the corresponding sine waves.

On the decimals, I'm just saying that 1/360 is more meaningful in general than 0.0027778. On the RMS, I was just explaining why the formula I decoded some of the decimals into included a factor of sqrt(2). In neither case was I saying that your decimals weren't accurate.

If current is running through A-B from N-B voltage, and the resultant voltage being fed back into A-B,

Current and voltage are distinct, there is no resulting voltage.

Electrons moving in the same direction from two other sources, demands that there is a voltage rise on A-N at that specific time.

No, current doesn't have to be in phase with voltage.

It seems a voltage difference is required to push the current at the correct time.

No, I would not say that. If you have a voltage difference, that voltage difference by itself will tend to push current. What the inverter is able to do, using the power available from its DC side, is to push out current in opposition to the voltage difference it sees, thereby adding even more energy to the electrical system.

That current must have an entrance into the circuit and a return. Entrance into leg A and return into N provides such a path. Entrance into A and return through B requires A-B to be in the same phase.

No, current doesn't have to be in phase with voltage.

Cheers, Wayne

 

[wwhitney]

That's the part that doesn't work if you look at each wave at 30 degree intervals of time.

Sure it works out. You seem to like graphs of sine waves, but I can't type those.

Fortunately we have a great short hand for this. I can write X ∠ Y, where X is a magnitude, and Y is a phase angle in degrees; that's called a phasor. In the context of 60 Hz AC, X ∠ Y means the sinewave y(t) = X*sqrt(2)*sin(2*pi*(60*t - Y/360)). So if you want to see some of these values as graphs, you can go to your favorite graphing platform and plug in that equation, putting in the values of X and Y.

With that established, we've been talking about A, B, N as 3 conductors of a 208Y/120V system. For the voltages, we have:

V(A-N) = 120 ∠ 0
V(B-N) = 120 ∠ 120
V(N-B) = -120 ∠ 120 = 120 ∠ 300
V(A-B) = V(A-N) + V(N-B) = 120 ∠ 0 + 120 ∠ 300 = 120 * sqrt(3) ∠ 330

[Now it may not be clear how I did that very last step of adding the two phasors together when their phase angles differ. But if you check with your graphing platform, you will see the result is correct at the level of adding sinewaves point-wise. The upshot is that we can associate X ∠ Y with a point in the plane (X cos Y, X sin Y), and then add the two phasors coordinate-wise, and turn the result back into a phasor. Equivalently for a phasor, we can draw a line on the plane that starts at the origin and proceeds for length X at an angle Y CCW from the x-axis; to add two phasors, we just start our second line at the end of the first line, rather than at the origin; and the sum phasor is the line from the origin to our final destination.]

Again, these voltages are fixed by the primary source (the grid or a grid-forming inverter), and our loads or grid-following inverters won't change them (for this idealization where we ignore conductor impedance).

Then if we put a resistive load across A-B, the current through it will be in phase with V(A-B), this is a special case where that is true. If the resistance is 12 ohms, then the current is 10A, and taking current from B to A like we take voltage, we have I(A-B) = 10 ∠ 330.

With the wye secondary, that same current waveform travels around in a loop through the load, through the coil A to N, and the other coil N to B. So we'll have I(A-B) = I(N-A) = I(B-N) = 10 ∠ 330.

So look at the coil A-N. The current I(A-N) is the negative of I(N-A), or -10 ∠ 330 = 10 ∠ 150. While the voltage is 120 ∠ 0. They are 150 degrees out of phase. More than 90 degrees, which means that the coil isn't consuming energy from the circuit like the load would be, but it is instead putting energy into the circuit. It's one half of the source of energy for this load.

Likewise for the coil B-N. The current I(B-N) is 10 ∠ 330, while the voltage is 120 ∠ 120. Again 150 degrees out of phase, again because the coil is providing energy to the circuit, rather than consuming it.

If we replace our resistive load A-B with an inverter A-B that is pushing out 10A to add energy to the circuit, everything is identical, except we have a negative sign on all the currents. Now I(A-B) = -10 ∠ 330 = 10 ∠ 150, or the current is 180 degrees out of phase with V(A-B) (same thing as negative here). While in the coils, the current will be 30 degrees out of phase with voltage. Because in this example, with no other loads in the circuit, the secondary coils will be "consuming" the energy put out by the inverter and "regurgitating" it back out on the primary side, for loads there.

Cheers, Wayne

 

[jaggedben]

If current is running through A-B from N-B voltage, and the resultant voltage being fed back into A-B, then at the time A-N should be at zero voltage, there are two other sources pushing electrons through it. Electrons moving in the same direction from two other sources, demands that there is a voltage rise on A-N at that specific time.

I can't hardly make any sense of that. However, I'm pretty sure that where you refer to electrons* moving in the 'same direction from two other sources' you're getting one of the phases backwards. Take a look at the generic three-phase graph on Wikipedia. Note that whenever A-N is at zero, B-N and C-N are the same amplitude in opposite directions. So instead of adding together they cancel out. In fact, at all times in this graph, two waves cancel out the remaining one. (*By the way it's electric charge that's moving. Not electrons. Amperes is coulombs per second.)

I'll grant that following the direction conventions when you add a solar inverter is genuinely complicated and confusing, even to professionals.

That current must have an entrance into the circuit and a return.

True, but there's lots of options for that. For a grid-tied solar inverter connected to A and B and putting in current at A (for a half cycle), that current can go through a load or through the grid to the N, B, or C conductors on the system. And the current coming into the inverter on B can come from N, C, or A through a load or through the grid. I think one problem you're having is that you keep trying to reduce the possible current and voltage flows in a three-phase system to only two phases. (For example, where I quoted you above in this post, you don't mention C.) That third phase is always going to be there to balance out the voltage and current from a single-phase grid-tied inverter connected line to line. (We will grant that if you suffered a phase loss, things would stop working correctly. That's related to the reason the code requires that three-phase grid tied systems detect the phase loss and cease exporting power on all three phases.)

 

[winnie]

Ok this thread is going in circles, which means that some basic concept has been missed.

To @bellington : Remember that there are _two_ features of the system to consider, the voltage waveform and the current waveform. To the first order approximation, the grid (or the BESS) _defines_ the voltage waveform measured at all the various nodes. For the simple system we are discussing, the grid defines V(A-N), V(B-N), V(C-N). V(A-B) and the others are derived and in fixed relation to the three defined values.

If the inverter were not present at all, the grid defined voltages would still be present. The inverter does not need to create the various voltages present in the system.

The single phase inverter sees and responds to V(A-B). It generates current I(A-B) which is in phase with V(A-B). The inverter is 'pushing electrons' in phase with V(A-B). The current being supplied by the inverter is out of phase with V(A-N) and V(B-N). The current is only in phase with the portion of the system that the inverter 'sees'.

You keep expressing dismay that the single inverter output V(A-B) cannot create the necessary 'waves' matching V(A-N) and V(B-N). This is simply not a problem. The voltage is created by the grid. The current supplied by the inverter on those legs is out of phase with the grid voltage. The power transfer will depend on the phase angle difference. Because of the phase angle difference, there will be a power factor and more current flow to deliver the same power.

In the case of a single A-B inverter connected to a wye secondary, the phase difference between the inverter output current and the transformer coil voltages is 30 degrees. This gives a power factor of sqrt(3)/2. As a reminder from the start of the thread, power factor doesn't mean energy lost, but does mean capacity loss.

What if you have an inverter connected A-B and loads connected A-N, B-N? The power from the inverter will never be able to supply all of the current consumed by the loads. The inverter is not in phase with the loads, so some current must come from somewhere else, meaning the grid. Take an even more extreme example: inverter connected A-B, load connected C-N. Exactly 0 inverter current supplies the load. Instead inverter current (and power) goes out to the grid (or the BESS), and grid current (and power) supply the load.

What happens when you have 3 single phase inverters connected L-L supplying a Wye transformer? The inverters taken separately would generate current that is out of phase with the transformer coils. But taken together they balance out, I(A-B) and I(A-C) combine at terminal A to make current in phase with V(A-N).

Finally, some of the stuff that is bugging you and we are dismissing? It is probably real, but as a second or third order effect. Don't let perfect be the enemy of the good (or good approximate understanding).

Jon

 

[ggunn]

Ok this thread is going in circles, which means that some basic concept has been missed.

This is why I have encouraged the OP to train himself up on the fundamentals of three phase power before trying to do a deep dive into the mathematical minutiae. He keeps coming up with ideas of why a a three phase inverter through a delta-wye transformer cannot work without a significant loss of power when we all know that it does indeed work and VA in = VA out.

 

[bellington]

Would it be accurate to say that 3 single phase inverters connected A-N, B-N, and C-N to a wye secondary, or A-B, B-C, and C-A on a delta secondary can inject matching in-phase currents provided by in-phase voltage?

 

[wwhitney]

Would it be accurate to say that 3 single phase inverters connected A-N, B-N, and C-N to a wye secondary, or A-B, B-C, and C-A on a delta secondary can inject matching in-phase currents provided by in-phase voltage?

What does "provided by in-phase voltage" mean to you?

A single phase inverter connected X-Y will typically inject current in phase with the voltage X-Y (up to a sign; when the voltage X-Y is at its peak, the current injected will be at its minimum). So if you had left on that phrase "provided by in-phase voltage," the answer would be certainly.

Cheers, Wayne

 

[winnie]

Would it be accurate to say that 3 single phase inverters connected A-N, B-N, and C-N to a wye secondary, or A-B, B-C, and C-A on a delta secondary can inject matching in-phase currents provided by in-phase voltage?

Yes.

More generally, a balanced set of 3 inverters (Wye or Delta) can generate in phase current for any balanced 3 phase source (Wye or Delta)

Jon

 

[wwhitney]

Yes.

Careful of that final phrase "provided by in-phase voltage". Which makes me think the OP wants to think of a grid-following inverter as a voltage source, which isn't really correct.

Although come to think of it, I'm not really clearly on how this grid-following inverter is able to "inject current" in phase (up to sign) with the voltage applied to its terminals. I mean, I usually think of an applied voltage as driving current through a load. But this grid-following inverter is somehow able to drive current without creating voltage?

Cheers, Wayne

 

[bellington]

What does "provided by in-phase voltage" mean to you?

I meant that the current would be the result of the voltage that caused it to flow.

A single phase inverter connected X-Y will typically inject current in phase with the voltage X-Y (up to a sign; when the voltage X-Y is at its peak, the current injected will be at its minimum). So if you had left on that phrase "provided by in-phase voltage," the answer would be certainly.

Cheers, Wayne

 

[Carultch]

Careful of that final phrase "provided by in-phase voltage". Which makes me think the OP wants to think of a grid-following inverter as a voltage source, which isn't really correct.

Although come to think of it, I'm not really clearly on how this grid-following inverter is able to "inject current" in phase (up to sign) with the voltage applied to its terminals. I mean, I usually think of an applied voltage as driving current through a load. But this grid-following inverter is somehow able to drive current without creating voltage?

Cheers, Wayne

It does create a voltage, just not an independent voltage.

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.

 

[ggunn]

I meant that the current would be the result of the voltage that caused it to flow.

Sort of. Look up the differences between a voltage source and a current source. Grid tied PV inverters are current sources.

 

[jaggedben]

Would it be accurate to say that 3 single phase inverters connected A-N, B-N, and C-N to a wye secondary, or A-B, B-C, and C-A on a delta secondary can inject matching in-phase currents provided by in-phase voltage?

Leaving aside the last bit regarding how the inverter behaves as a current source...

I think the answer to your question is essentially yes... from the loads' point of view. Going back to posts 143 and 146, 'in phase' is a matter of direction convention. The inverter connects in parallel to the grid; which means, wherever the inverters output current reaches a node where it flows to a load, the current will flow to the load in-phase with the current from the grid. (Well, for a resistive load. A load with power factor may shift things.) So, if you set up a scope on the load circuit, you will see no difference in current flow and you will not really be able to tell whether the current comes from the grid or the inverter.

However, as a matter of convention, if you set up the scoping on the inverter circuit, expecting to measure power to flow from the grid, the current will be 180 deg out of phase because power is flowing backward. (These mean the same thing in this case.) If you want to measure power flow on the inverter circuit as a positive number, you need to flip your CT around to have current in phase with voltage. But, again, when the power from the inverter meets a node where it flows to a load, it 'switches direction' and is in phase with the grid current.

So it is critical to state in any scenario where you want to ask whether inverter output is 'in phase' where you are measuring and what direction of power flow you consider positive. I think part of the reason we're going in circles is not doing that.

 

[bellington]

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?

 

[ggunn]

It does create a voltage, just not an independent voltage.

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.

Only insomuch as the Vd requires. If R=0 (Vd = 0) it still works fine.