# Feeding solar power back into municipal grid: Issues and finger-pointing

I was on your side of this issue before I went back and looked at the circuit model for two power sources driving a load that Jim Wilkins provided. I made a post today in reply to Mho where I went through the math. Sadly, Mho didn't even bother to go through the detailed circuit analsysis I provided. If you look for the post, I encourage you to review the analysis and the math and see what you conclude. The conclusion I came to is HomeGuy and KRW are right. To get power onto the grid, the additional source coming online has to be slightly higher that the voltage on the grid.
Look at it this way. Suppose the utility pole at my house ia at 120.000V. I hook up a power source, be it a generator, PV, whatever on the end of the line inside my house. I place exactly 120.000V on my end of that line. The wire from my power source to the pole has some small resistance, R. With 120.000V on one end of a resistor and 120.000V on the other end, by ohm's law, how much current will flow? Zero.
How do I get current to flow? At least one of three things must happen:
1 - I raise the voltage on my end of the wire slightly.
2 - The load increases on the rest of the system which in turn lowers the voltage at the utility pole outside my house
3 - Whatever else is driving the load reduced it's voltage slightly, which in turn lowers the voltage at the utility pole by house.
Those are the only choices to get a potential difference across the line connecting the source in my house to the grid and only then can current flow and power from my house make it onto the grid. And with option 1, my raising the voltage slightly on the house end, in turn must raise the voltage at the utility pole slightly as well. In other words, I've raised the voltage of the grid at the utility pole at my house. It's a very small amount, but it's real.
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On 4/17/2011 2:20 PM snipped-for-privacy@optonline.net spake thus:
[cut to the chase, i.e., trader4's example:]

All I can say is, good example, but wrong conclusion.
Let's change the example just a little. We'll have a large AC source (call it "the grid"), and a smaller AC source (the solar system's inverter), connected *in parallel*, and then some resistance (the total load of "the grid") after all that, completing the circuit.
In this example, "the grid" and the PV inverter are operating at *exactly* the same voltage.
Let me argue this negatively, and see what you say to it: If you're saying that this will not work (i.e., that the PV inverter cannot contribute any power to the circuit because it isn't at a higher voltage), then *no* circuit where you have two power sources in parallel could ever work under the same circumstances.
No electric vehicle would ever work, because the battery cells in them are (pretty close to) exactly the same voltage, so how would each tiny cell (tiny in comparison to the total power of its siblings) ever be able to "push" electricity into the circuit?
That elementary electronics tutorial (Kirchoff's Law, etc.) explains everything you need to know here. And it doesn't require any higher voltage.
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wrote:

Simple: The motor is at a lower potential.

Elementary, sure, but you're still not getting it.

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But what exactly is wrong with this example, which closely conforms to what we have been discussing? We have one end of a power wire connected to the PV array in my house. The other end is connected 100 feet away to the grid, ie the utility pole by my house. The sun is behind an eclipse, the array isn't generating any power. The voltage at the utility pole is 120.000V. The sun comes out. To put power on the grid, current must flow from my house, through that wire, to the utility pole. The wire has some small resistance, R. According to physics in my world, the only way to get current flowing in that wire is to have the end in my house at a HIGHER voltage than it is at the pole. It doesn't have to be a lot higher and it actually will be only slightly higher. But without that difference, tell me how could current ever flow?
The instant it does flow, I'm now pushing current out onto the grid. Assuming the rest of the grid stays the same, ie the load is fixed, the other power sources don't change, that means that the voltage at the pole now increase slightly as well. Net result is the voltage in my house is now say 120.1V, the voltage at the pole is now 120.05V and I have in fact raised the voltage of the grid.

Only if you assume the connection between the two is zero ohms, ie a perfect conductor. That would be akin to declaring the line connecting my house to the pole to be a perfect conductor and changing the model from what it is in the real world. The model that corresponds to what we have and also to your new proposed example is essentially the circuit example that Wilkins provided. The two resistors R1 and R2 represent the internal impedance of the two power sources. If you want to model the grid connecting them, then you could add two resistors, one after R1, the other after R2 to model the resistance of the grid between each of the power sources and the load. It doesn't change the analysis.

You can model a two cell battery with that circuit diagram as well. Let's leave V2 at 20V, making it a 20V cell with an internal resistance of 20 ohms modeled by R2. It's companion cell is V1 with internal resistance of 10 ohms. Do the math and you'll find that with V1 at 13.2 volts, no current flows through the V1 half of the circuit. All the load current comes from V2. Start increasing V1 and only then does current flow through V1 and through the load. The consequences of that are then that the voltage at the load increases, which in turn decrease the current flowing from V2.
In your real battery with cells connected in parallel, R1 and R2, the internal resistances, would be very close or equal. And V1 would be close in value to V2, both would be supplying about half the power. But it doesn't change the application of Kirchoff's Law or the conclusions.

Did you look at the detailed analysis I provided in an earlier post where I went through the analysis of that circuit example? Here it is repeated. Take a look at that circuit and go through it step by step. It is a model of two power sources driving a load.
http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
Voltage source V1 and R1 represent a simple battery, with R1 being the internal resistance of the battery. Same with V2 and R2. For our purposes a suitable model for a PV array and another power source on the other end of the distribution lines.
Let's leave V2 as is at 20V, supplying all the current to the load, with no current coming from V1. You then have a simple series circuit consisting of resistors R2 and R3 connected to voltage source V2. A current of 20/(40+20) = .33A is flowing, which is I2 in the drawing. The only way for no power to be flowing through the other half of the circuit encompassing V1 is if V1 is at the exact same potential as the load resistor. With .33A flowing through the load, R3, you have R3 at 13.2 V. That means V1 must be 13.2 volts. With V1 at 13.2 volts the voltage across R1 is zero and no current flows. So, everything is in balance. V1=13.2V, I1=0, the voltage on the load is 13.2 volts, and I2= .33A is flowing from V2 through R2, R3.
Now, if we want V1 to start supplying part of the power, what has to happen? V1 has to increase ABOVE 13.2 volts. And when it does, the voltage across the load resistor R3 will also increase. As that happens, current will start to flow now from V1 through the load resistor and at the same time the current from V2 will decrease slightly, as the potential drop across R2 is decreased slightly.
The net result of this is that the voltage across the load has increased. Current I1 is now flowing from V1, I2 from V2 is now slightly less and the combined currents of I1 and I2 which together are I3 has increased slightly.
The other ways to get V1 to supply power would be for either the load resistor R3 to decrease in value or for voltage source V2 to decrease.
If you want to more closely model the situation, we could add two more resistors to model the distribution line resistance. A resistor could be added after R1 and after R2. But if you go through the analysis, it doesn't change the basics of the above analysis. For source V1 to supply power, the voltage on it's portion of the distribution system and across the load must increase.
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wrote:

Again, physics doesn't care what you like and don't like. It is.

Well, I guess you could say that "currents" push against each other, but it requires a difference in voltage to have a current. Think of the intersection of two rivers.

You assume wire has zero resistance. Bad assumption.

That is "said" has little bearing on physics.

The biggest flaw is that resistance is not zero and you take what people "say" too literally. Analogies are always flawed. That's why they're called "analogies". ;-)
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Let me say it again, perhaps you'll catch on. Physics doesn't care what you like. It is what it is.

Because, like electricity, water always flows "down hill" - high to low.

OK.
OK.
Voltage is dropped across a resistance. Not all points in the grid go up because your solar cells output more voltage because there is non-zero resistance between all points. If you're generating electricity, your house will be at a *higher* voltage than the pole. If the resistance of the wire were zero this couldn't happen.
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It makes more sense if you think of the inverter as forcing a constant CURRENT and let the voltages be whatever the source (wire etc) resistance makes them at that current.
The grid may or may not act like an infinite sink. The continual load variations will probably swamp out any voltage measurement you might make, so it's reasonable to consider it an infinite sink unless you have a very large inverter. The GTI wants to dump all the current from the array onto the line and will adapt itself to the line voltage, whatever it may be.
If you connect a PV panel to a 12V battery the panel will source as much current as the sunlight produces, at the voltage of the battery even if the panel's open circuit voltage is above 20V. The battery voltage will rise a little because of the IR drop in its internal resistance.
http://en.wikipedia.org/wiki/Current_source
jsw
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On 4/17/2011 7:19 PM Jim Wilkins spake thus:

It *sounds*--and I'm sure you'll correct me if I'm wrong--as if you're agreeing with me, and with Smitty, and others when we say that it is *not* required that the photovoltaic inverter supply a higher voltage in order to transfer current to the grid. (I take this from the last sentence in the next-to-last paragraph, where you say " ... will adapt itself to the line voltage, whatever it may be".)
The arguments against this, with all the pseudo-science being thrown around (most of it by the ones who are also slinging insults) are getting quite tiresome here.
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wrote:

No, the voltage will STILL be higher if you're supplying current to the grid. Wires have resistance. Current sources don't go against physics and really are the same thing as voltage sources (Norton/Thevenin duality). Physics doesn't lie.

You're throwing around the pseudo-science. If you don't like the treatment, you can easily leave.
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On 17/04/2011 20:40, David Nebenzahl wrote:

If you take the voltage drops into account, the voltage at the inverter has to be higher than the voltage at the grid connection point.
This is a real world scenario.
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Sorry, totally incorrect assumption. I guess the real world isn't for you
Load current is shared depending on impedance of parallel source loops.
--------------------
"g" wrote in message If you take the voltage drops into account, the voltage at the inverter has to be higher than the voltage at the grid connection point.
This is a real world scenario.
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You're wrong, 'g' is correct. You can't even figure out how to use a newsreader.

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The inverter output is higher -internally- by the V=IR drop between it and the grid.
In this case I is the independent variable, the array's output, and V is whatever it takes to make I pass through R to get to the grid voltage.
jsw
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On 4/18/2011 4:34 AM Jim Wilkins spake thus:

OK, now we're getting somewhere.
At the risk of igniting another round of sniping here, how does that work, exactly? I assume you're talking about the voltage drop between the inverter and the point where it's tied to the external power lines (= grid), correct? So since it can only "see" its own internal voltage, how does the inverter even know what that voltage drop is? How does it regulate its voltage so that it's equal to the grid voltage at the point of connection?
Or is this somehow self-regulating, where the inverter simply "aims" at what it calculates is the grid voltage, based on the current delivered by the PV system, and the voltage self-stabilizes?
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Take a look at the simple circuit analysis I provided earlier today that is a few posts down in this thread. I drew a model of the situation we have been discussing which is very similar to the example circuit Wilkins provided earlier. And it shows how not only the internal voltage of the PV array must rise, but so too the voltage on the grid at the point of connection. It all follows directly from Krichoff's Law.

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On 4/18/2011 2:20 PM snipped-for-privacy@optonline.net spake thus:

The problem I have with that post is that the ASCII graphics you used are all jumbled and I can't make sense of them (trouble with line lengths, I think). Any chance you can redraw it in such a way that news clients won't scramble it? (Maybe try drawing shorter lines with "hard returns" at the end?)
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wrote:

Make sure it's in a fixed space font.
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Yeah, I just looked at it and it came out looking all jumbled. Here's another try:
1 1 Vg 1 1 1 1 ohm ------R1-- Rw------Rg-------Rg--------Rw --- R2------ ! ! ! ! ! ! PV V1 Rload 100 V2 240v array ! ! ! ! ! ! ------------------------------------------------------------------
Here's a simple model of what we have. V1 and R1 are a Thevinin model of the PV array. R1 is the internal resistance of the PV array. Rw is the resistance of the wire connecting the array to the grid. Rg is the resistance of the grid wire connecting the load two blocks away. That load is driven by another power source in parallel, modeled by V2 and R2, it;s connecting wire Rw. I hope we can all agree that this is a valid model for the discussion at hand.
Step 1: The sun is not shining, the PV is supplying no current. What is the voltage at the grid connection point Vg? We know the only current flowing is from V2. That current is 240/(100+1+1+1) = 240/103= 2.33 amps. That current flows through the Rload resistor producing a voltage of 100 X 2.33 = 233 volts across the load. Since we know zero current is flowing in the left side of the circuit, the current through R1, R2, Rw must be zero. The only way for that to occur is for Vg to be equal to 233 volts. The grid voltage is now 233 volts. The voltage at V1 must also be 233 volts.
Step 2: The sun comes out and the PV is going to put it's power on the grid. The only way for current to start flowing through V1 is for V1 to INCREASE above 233 volts. As it does, the voltage on the grid will slowly increase too. How do we know this? Let's look at the simplest case, where the PV array can supply half the total current. Writing the Kirchoff equation for the right side of the loop we have:
240 = i2 x (R2 + Rw + Rg) + (i1 + i2) x Rload
240 = i2 x 3 + ( i1 + i2) x 100
240 = 100 x i1 + 103 x i2
Where i2 is the current flowing from source V2 and i1 is the current flowing from the PV array. We also know that each source is supplying half the current, so i1 = i2. Substituting, we have
240 = 100 x i1 + 103 x i1
240 = 203 i1
or i = 1.182 amps.
We know that twice that is flowing through the load, so the voltage across the load is 100 x 1.182 x 2 = 236.4 volts
Before we brought V1 online it was 233 volts. Now it is 3.4 volts HIGHER. Intuitively this makes sense, because the voltage there must RISE enough to reduce the current that was flowing from the other power source V2.
Now that we have the voltage across the load, and know that I1 is 1.182 amps we can calculate the voltages at the grid, Vg and what V1 must be:
Vg = 236.4 + i1 x Rg = 236.4 + 1.182 = 237.6 volts
V1 = 236.4 + i1 x (Rg + Rw+ R1) = 236.4 + 1.182 x 3 = 239.9 volts
All the voltages along the way have increased, the voltage at the load, the voltage at the grid, and the voltage at the PV array.
QED
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wrote:

Another try a straightening out the drawing....

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OK, got the drawing straightened out and it plus the analysis are above. Just to add some clarification, I said the only way for the PV array to start delivering power is for it to raise it's voltage, which in turn raises the grid voltage. That assumes that both the load and the other power source remain constant. As I stated in other posts, the other ways for the PV array to deliver power without it's voltage going up would be for the load to increase, ie Rload gets smaller, or for the other power source V2 to decrease in voltage.
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