Geoff Schultz wrote in
:

You can't produce more power
than the panels are generating.


Ok, here's a little electricity for everyone....

This is the schematic of a solar panel:


--[cell]---[cell resistance]--

Now, what this controller with the variable inverter load in it is doing
is simplified into this circuit:

|--[cell]---[cell resistance]---|
| |
| |
|-[controller input resistance]-|

In any series circuit like this where the SOURCE has a series resistance
(impedance in AC and RF circuits), Load POWER is "optimized" to a peak
level whenever load impedance (resistance) EQUALS source impedance
(source resistance). Moving away from this optimum power peak in either
direction of load impedance creates LESS POWER in the load, in this case
the controllable inverter's panel input. If the inverter's loading of
the panel can be made to electronically change its load resistance on the
panel to the same resistance as the panel's own built-in resistance,
which is quite substantial, Yes, the power generated will be greater than
if the panel were simply connected through a blocking diode to the
batteries. Very ingenious.

This is NOT magic. It's math. The source voltage is 19VDC, that's what
the cell puts out due to the number of cells in series, a function of
silicon junction voltage. Just for demonstration of this effect, let's
say the panel cell resistance (which varies, by the way with cell
temperature quite a bit) is 10 ohms. Two simple electric calculations
are needed:
Power = current squared times resistance.
Current = Voltage divided by resistance.

Let's check 3 points...varying the load resistance...to see what power
develops in the load. Let's check a load R of 10 ohms, same as source,
for a reference. Total resistance is 20 ohms. 19V/20 ohms = .95A
circuit current. Now square that (.9025) x the 10 ohm load resistance =
9.025 watts "output" for the inverter to send to the battery, minus its
losses, of course, in conversion.

OK, now let's change the load resistance to 8 ohms (lower). Total
resistance is 18 ohms. 19V/18 ohms = 1.055A circuit current. Now square
1.055A (1.113) times 8 ohms = 8.9 watts...POWER OUT DROPPED OFF from our
peak.

OK, now let's change the load resistance to 12 ohms (higher). Total
resistance is 22 ohms. 19V/22 ohms = .8636A circuit current. Now square
..8636 (.7458) times 12 ohms = 8.94 watts...POWER OUT DROPS OFF THIS WAY
TOO!

So, if the panels resistance right now equals the controller's input
resistance, right now, the maximum power output peak of the panel will be
realized. This can be accomplished with a little electronic trickery in
IC regulators, probably custom made for this purpose, so that at any
panel resistance, the IC can sense and adjust the controller's load
resistance on the panel to maximize panel output. With switching power
supplies now in the 99% efficient range, this is very feasable.

By the way, this is the exact same reason we strive to make a radio with
an output impedance of 52 ohms connected to coaxial cable with an
impedance of 52 ohms connected at the top of the mast to a 52 ohm
antenna....because it creates maximum power transfer from the transmitter
to the antenna. (The radio is actually designed to match certain types
of antenna's natural impedances, not the other way around.) It even
works the other way on receive....so we use a 52 ohm input receiver, too.

This completes today's electrical lecture. Please read pages 324 through
468 in your textbook and complete the workbook section 3-6 to hand in by
tomorrow's class. The workbook answers are 40% of your grade in this
course. (God, they all looked like they could kill me when I used to
tell 'em this just before the bell rang.)....(c;

Larry
--
I just can't stand it when I don't know why something does what it does!

I had to learn how our ship's steam turbine plant worked, to the dismay
of the ratings in the engine room, even though I was an Electronics Tech.
If you need power, give me a little time to fire the boilers and get the
pressure up and we'll go!