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Feeding solar power back into municipal grid: Issues and finger-pointing - Page 11

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Posted by g on April 12, 2011, 4:47 pm
On 11/04/2011 23:17, harry wrote:

So, you don't increase current by raising the voltage, but you increase
current by having a higher potential.

Now, difference in potential is voltage?

Posted by David Nebenzahl on April 12, 2011, 6:44 pm
On 4/12/2011 9:47 AM g spake thus:

No, no, no: increasing the current doesn't increase the potential
(that's voltage). It increases the *flow* of electricity (= current), at
least the maximum possible current. But that's not the same thing as
potential difference.

Example: Let's say you run your house off 12 volt batteries (just for
illustration). The *potential* of your power circuit is 12 volts
(assuming the batteries are fully charged, and they'll actually be
closer to 13.something, but let's call it 12).

Now let's say you add some more stuff to your house and find that your
lights are going dim because the battery can't provide enough *current*
(= amps) to the load. So what you do is add another battery in parallel
with the first one. This doubles the available current (= amps), but it
does *nothing* to change the voltage; it remains at 12 volts (nominal,
as explained above). This is true no matter how many batteries you add
*in parallel* with each other. But each battery increases the
*available* current (= amps) you can draw from your power source.

Notice that adding more batteries does not "push" more current through
the system; it increases the amount of current that can be "pulled"
(drawn from) the batteries.

Which is exactly the situation when you connect your photovoltaic system
to "the grid". It increases the *available current* to the grid. It does
not change the voltage of the grid; there's no need for it to be at a
higher voltage than (but it needs to be at about the *same* voltage as)
the grid.

Yes. Please refer to any good basic guide to electricity for more details.

The current state of literacy in our advanced civilization:

   wan2 hang

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

Posted by Bruce Richmond on April 13, 2011, 1:50 am
If you do a little research I think you will change your mind.


At the bottom of that page you will find this link


More specificly I wrote, "forcing power back into the grid".  Power is
watts or KW.  That's volts times amps.  The current will only flow if
there is a difference in voltage.

It takes very little voltage difference to flow a lot of current when
the "load" is the grid.

Yes, I do.  Now let's see if you can understand this.

Take two 12 volt car batteries with one discharged to 11 volts.
Connect them in parallel and check the voltage.  It will be some value
between what they measured seperately.  While they are connected like
this current is flowing from the charged battery to the discharged
battery.  Power is being forced into it raising its state of charge.
If the two batteries had been at exactly the same voltage there would
have been no current flow.  Now take 8 AAA batteries connected in
series to give 12 volts.  Disconnect the charged battery and connect
the AAAs to the 11 volt battery.  Measure the voltage.  It will be for
all practical purposes unchanged from 11 volts.  The AAA cells are
charging the bigger battery but they are so small compared to it that
they seem insignificant.  That is how your PV system looks to the

The voltage on the grid can vary by up to + or - 10%.  It is usually
kept withing + or - 5%.  Take a volt meter and check the voltage at
your wall outlet.  Check it several times during the day and you will
find that it varies.  Don't try to claim that it doesn't, check it and
you will find that it does.  As loads are put on the grid it drags the
voltage down.  The power company responds by generating more power to
bring the voltage back up.  As loads are taken off the voltage will
climb, and it is brought back down by producing less power.

The point here is that there is no such thing as the grid not being
able to accept the power you have produced.  As long as you are
connected you can always force your KW in.

Posted by Home Guy on April 13, 2011, 2:27 am
 Bruce Richmond wrote:

I don't think the issue is whether or not you can force current into the
grid via your 120/208 VAC service connection.

The question is:

a) does your power source need to overcome the instantaneous line
voltage in order to achieve a flow of current (answer:  yes, and to the
extent that your power source has the capacity to do so, you raise the
output voltage as high as you can, because if you don't - then you have
excess capacity that is not going to make it out to the grid and hence
you won't gain revenue for the entire potential of your generating

b) by raising the voltage on your local 120/208 grid, can your local
stepdown transformer adjust it's own operation by sensing that higher
voltage and reduce it's own output voltage in an attempt to regulate the
system back down to the desired setpoint? (answer:  I don't know -
probably not.  The neighborhood stepdown transformers probably weren't
designed to compete with sources of current being connected to their
distribution outputs).

c) So if the voltage on your local 120/208 grid is being raised slightly
because of your PV system and it's desire to push as much current back
into the grid as it can generate, then will this actually reduce the
amount of current that the regional sub-station is sending to your local
step-down transformer?  (answer:  the substation probably doesn't have a
direct line to your local stepdown transformer, and any alterations it
can make to it's output voltage is probably seen by many step-down
transformers including yours that are all wired to the same circuit.  So
in reality it's doubtful that the regional substation would even sense
that your PV system has raised the local grid voltage).

d) So your PV system is raising the local grid voltage, and you're
probably pushing out 40 amps at 120 VAC or 20 amps at 240 VAC on a sunny
summer day.  So what is that extra juice doing?  Well, it's flowing
through the compressor motors of 10 to 20 of your neighbor's AC units -
whether they need it or not.  Because you've raised the local grid
voltage slightly, that translates into a few extra watts (maybe 250
watts for each house that's fed from the same stepdown transformer).  So
all the fridge compressors and AC compressor motors, lights - all linear
loads are going to blow away that extra line voltage as heat - instead
of useful work.

Nuf said?

Posted by Bruce Richmond on April 14, 2011, 12:58 am
Then you should look back and see that is precisely the issue being
discussed here.  You wrote, "But still - you can't push more
electricity onto a network than the load is asking for (given that
your invertors are functioning correctly I guess)."

David wrote, "That second statement is correct: you can't "push"
electrons into the grid. But it doesn't matter *how* your inverters
are working; it's a basic law of physics."

I have explained in simple terms (without getting into power factors,
phase shifting, pulse width modulation, etc) the physics behind how
you can push your power onto the grid whether it is asking for it or

You do not raise it up as high as you can, you raise it just enough to
do the job.  The ignition coil in your car raises the 12 volts of its
battery to many thousands of volts to force a spark across the gap of
its sparkplugs.  You don't need thousands of volts to feed power into
the grid.

Go back to the example I provided using two batteries.  As I said,
when they are connected in parallel the 12 volt battery charges the 11
volt battery and the voltage across them will measure somewhere
between 11 and 12 volts.  Connect a light to the batteries.  You will
measure a slight drop in the voltage but it will still be over 11
volts.  That means the 11 volt battery is still being charged and all
the power to light the light and charge the 11 volt battery is coming
from the 12 volt battery.  Connect more lights (load) to the batteries
and you can drag the voltage down so that it is just over 11 volts.
So long as the voltage across the two batteries is higher than the
stand alone voltage of the 11 volt battery all the current going
through the lights will be coming from the 12 volt battery.  And it
doesn't matter that the 12 volt battery has been dragged down to
within a small fraction of a volt over the 11 volt battery, the lights
see 11+ volts.  Can you see now how the inverter can pump its power
into the system?  By having its voltage just a bit higher than the
transformer, but well within the normal range of the line voltage, it
can take over feeding the local water heaters, cooking stoves, air
conditioners, lights, etc.  No additional controlls are needed to
reduce the current coming from the transformer.  The voltage
difference takes care of it.

As you saw above the current flow through the transformer will be
reduced.  The substation sees that as a reduced load and will behave
the same way it would any other time the load goes away.  No
additional controls are needed.

Yes, you have said enough to make it clear how little you know.

When an electric motor is running it produces a back EMF that counters
the flow of the current.


That AC compressor requires the same power it did before.  Keeping it
simple that power is volts times amps equals watts.  Divide the power
required by the higher volts now provided and you will find fewer amps
are required.  The wasted heat is proportional to the square of the
current.  So raising the voltage means there will be less waste heat
from the motor.

With resistance heaters the higher voltage flows more current, so you
get more heat, which is what you wanted anyway.

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