Posted by *trader4@optonline.net* on April 18, 2011, 9:20 pm

*> On 4/18/2011 4:34 AM Jim Wilkins spake thus:*

*> >> 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 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.*

*> 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?*

*> Gory details, please.*

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.

*> --*

*> The current state of literacy in our advanced civilization:*

*> yo*

*> wassup*

*> nuttin*

*> wan2 hang*

*> k*

*> where*

*> here*

*> k*

*> l8tr*

*> by*

*> - from Usenet (what's *that*?)- Hide quoted text -*

*> - Show quoted text -*

Posted by *David Nebenzahl* on April 18, 2011, 11:01 pm

On 4/18/2011 2:20 PM trader4@optonline.net spake thus:

*> *

*>> 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?*

*>> *

*>> Gory details, please.*

*> *

*> 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.*

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?)

--

The current state of literacy in our advanced civilization:

yo

wassup

nuttin

wan2 hang

k

where

here

k

l8tr

by

- from Usenet (what's *that*?)

Posted by *krw@att.bizzzzzzzzzzzz* on April 19, 2011, 12:50 am

wrote:

*>On 4/18/2011 2:20 PM trader4@optonline.net spake thus:*

*>> *

*>>> 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?*

*>>> *

*>>> Gory details, please.*

*>> *

*>> 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.*

*>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?)*

Make sure it's in a fixed space font.

Posted by *trader4@optonline.net* on April 19, 2011, 1:45 am

*> 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?)*

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

Posted by *trader4@optonline.net* on April 19, 2011, 1:50 am

wrote:

Another try a straightening out the drawing....

*> 1 1 Vg 1 1 1 1 ohm*

*> ------R1-- Rw------Rg-------Rg--------Rw --- R2------*

*> ! ! =*

*
!*
*> ! ! =*

*
!*
*> PV V1 Rload 100 240v V2*

*> 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*

> On 4/18/2011 4:34 AM Jim Wilkins spake thus:> >> 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 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.> 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?> Gory details, please.