Quote:
Originally Posted by Not Sure
So in other words, what you are claiming is that no matter how many volts that were hooked up to a 12v battery, as the multimeter reads 12v at the battery terminals, then only 12v is (ever) coming into the battery?

NotSure,
I realize your post crossed mine. The problem is that you don't master the basics of electricity. A voltage is never "coming into the battery". It is the current that flows
through the battery and the power that
goes into the battery.
Compare it with this: you boil a pot of water. No matter how high you turn the flame up, the water temperature will never exceed the boiling point of the water (100 degrees Celsius at 1 bar atmospheric pressure).
Quote:
I suppose you're next going to tell me that if one were to rig a 12v battery to attract lightning, when the lightning bolt hits the battery and in the nanosecond that the multimeter reads '12v' before the entire works explodes in a ball of flames..... that only 12v was going into the battery?

At that moment in time the voltage was indeed 12V. Your meter doesn't lie. But I understand the way you think. The difference with the lightning bolt is that it is much more powerful and it is able to transfer an enormous amount of power and thus take the voltage up into the thousands of volts. The difference with your panel is that it is
low power and that 0.7A output current at 12V is nothing more to the battery than a tickle.
You approach was to go up in voltage to "kickcharge" the battery. That is okay and the way those "fast" chargers work. But they also have the power rating to deliver the current that is needed.
I realize that I didn't explain the battery internal resistance and Ohms first law enough. So here it is, hold on:
V = I x R. R=resistance, V=voltage and I=current. The maximum output of your panel is 43.2W at 0.6A and 72V. Now, we can rewrite Ohms law to isolate R (high school math again, divide each side by I):
R = V / I. So, we fill in the panel values and get R = 72 / 0.6 = 120 Ohms. This means that if we connect a 120 Ohm resistor to your panel, you will measure 72V over the resistor and a 0.6A current flowing through it, which calculates to a power transfer of 43.2W.
The resistor has become a 43.2W heating element.
Now... what do you think happens when we connect a resistor with a lower value, say 100 Ohm??
With less resistance, more current will flow. So, we rewrite Ohm's law to isolate I:
I = V / R and that becomes 72 / 100 = 0.72A. So, what is the power transferred:
P = V x I = 72 x 0.72 = 51.84W. OOOOPPPPPPPPPSSSSSSSSSS!!!!!!!!!!!!!! the panel can't deliver that much power...
Houston, we have a problem! But the panel can deliver that 0.72A current, so what happens if we connect it anyway? Now, hold on to your seat, we just graduated to the next class.... We have to determine the boundaries within which we can work and those are the specs of the panel. We know this for sure: R=100 and Pmax = 43.2. We will have to come up with yet another formula that has P, I and R. Here we go:
P = V x I and
V = I x R
So, we can replace V in the first formula with
I x R. When we do that, we get P =
I x R x I and that leads to
P = I^2 x R. (I^2 is my notation for "I squared"). Now we have to isolate I: I^2 = P / R so
I = sqrt(P / R) (sqrt = square root). Here we go: I = sqrt(43.2 / 100) = sqrt(0.432) = 0.657A
What does this mean? It means that
a 43.2W power source can deliver an absolute maximum current of 0.657A through a 100 Ohm resistor. That is where it's power can take it and no further. Ha! now we know a maximum current so we can put that into the V = I x R formula:
V = 0.657 x 100 = 65.7V !!!!!!!!!!!!!!!! So, the voltage has dropped down to 65.7V. And now that we know that, we can calculate the power transfer: P = V x I = 65.7 x 0.657 = 43.165W and that is less than 43.2W. We have
lost efficiency and the reason is that we didn't connect the optimal
impedance value. Lesson:
for maximum power transfer, the inputimpedance of the load must match the output impedance of the power source. If there is an impedance mismatch, less power is transferred.
Now, before we get to the grand finale, read this:
Internal resistance  Wikipedia, the free encyclopedia
Okay, let's take the battery for a 1 Ohm resistor.
I = sqrt(P / R) = 43.2 / 1 = 43.2A. Now, we hit a boundary big time because the panel can only supply an absolute maximum of 0.8A (into a 0 Ohm (zero Ohm) resistor). It ain't fair but it's the best the stupid panel can do... we have to live with that. So, instead of working with that promising 43.2A value, we have to replace it with the disappointing 0.8A value. V = I x R = 0.8 x 1 = 0.8V. So, if you connect a 1 Ohm resistor to the panel, you will see 0.8A and 0.8V. When you disconnect it the panel will jump back to high 72V+ vopltage and zero amps. When the resistor is connected, a power transfer of P = V x I = 0.8 x 0.8 = 0.64 W takes place. That's almost nothing!!!! This is the reason that you can short a solar panel without things blowing up.
I previously wrote that it ain't fair that the panel is limited to 0.8A but there is a good reason for that: this is caused by the
internal resistance of the panel itself! This resistance
limits the current just like when you connect a resistor to it's output terminals. This is all too complex for engineers to understand so they came up with another trick for us: we replace the panel with two components: a perfect power source connected in series with a resistor. This is called the
replacement diagram. The resistor is the internal resistance of the panel... we can't eliminate it but we can think of them as separate components!! AHA shout the engineers!!! Now we can put Ohm's law to it again. So, when we short the panel, we don't short the power source... it's just like when we connect a resistor to it! The current flows through that resistor and the output power dissipates in it as heat. In real life, the panel heats up because this resistance is inside the panel. That resistance must also be significant because we would get very high currents otherwise, like when shorting a battery. And now we start realizing that all we calculated before, like with that 120 Ohm resistor is wrong.... because there was another resistor!! we're back to scratch... such is the life of the engineer ;)
But we don't give up. We start to understand that the Open Circuit Voltage of the panel is the voltage of the actual "perfect power source" in there because when no current flows, no voltage is
lost over that internal resistor. And when we short the panel, we actually connect that internal resistor to that perfect source. Hey, wait, we have enough values to do some math again:
R = V / I = 98 / 0.8 =
122.5 Ohm. YES! we know the internal resistance!!! And ain't that awfully close to that 120 Ohm we worked with before??!! And I really hope this opens your eyes now: when we connect a 120 Ohm resistor to this panel, we have a good
impedance match and that is why we get maximum power transfer. So now, you can also know what that MPPT controller does: it offers a 120 Ohm load to the panel to "make it" output maximum power, while on it's own output, it creates a 1 Ohm impedance to get maximum transfer to the battery. Yes, it really is that simple.
The battery, like the panel, is a power source with an internal resistance. They tell us it is about 1 Ohm. So, we have
12.5V and 1 Ohm. Now the panel:
98V and 122.5 Ohms.
Before we connect all that I need to explain about the
collision that is about to take place. I am going to use a rather weird example: we have two pipes that are aimed at each other. we pump water through both so that the two streams collide in mid air. If we increase the pressure on one of the two pipes, the
collision will move closer to the other pipe, right? If we keep increasing pressure, we can even prevent water coming out of the other pipe or, taking it further, we make the water in the other pipe reverse flow and push it back into that pipe.
When you connect to power supplies positivetopositive and negativetonegative you get the same with pressure translated to voltage. So, the higher voltage supply will discharge into the other one.
But here we are with that nasty internal resistance again. Think of it as a restriction in the pipeline... it will reduce the flow (reduce the current).
The higher voltage supply must overcome the voltage of the other supply first. When the voltages are equal, we have a standoff. This is what happens when you connect batteries in parallel. If there is a difference in voltage, it is the value of that difference that determines the resulting current... together with any resistance in the circuit.
So, the panel is 98V and the battery is 12.5V. The difference is 85.5V We can draw a replacement circuit of one power source of 85.5V in series with two resistors of 1 and 122.5 ohms or just one resistor of 123.5 Ohms. Now, it is easy to calculate the current:
I = V / R = 85.5 / 123.5 =
0.69A and guess what, that is right on target and matches the output curve in the panel's IV graph.
Next class:
So with that 0.69A current, what happens in the battery? The current flows through it's internal resistance and a voltage will be present over that resistance. It is V = I x R = 0.69 x 1 = 0.69V. That voltage gets added to the battery voltage (the "internal resistor" is in series) so on it's terminals we will see 12.5 + 0.69 = 13.19V. Now the panel... V = I x R = 0.69 x 122.5 = 84.525V This time, the voltage gets substracted from the panel voltage (the current flows the other way) so 98  84.525 = 13.475V. Holy Moly, that is just above the voltage on the battery and that is why the panel charges the battery and the current flows from the panel to the battery.
And only now I can answer NotSure's question: what happened with those 72V of the panel when it gets connected to the 12V battery and we only measure 12V at the terminals????!!!!!
The voltage difference can be found over the internal resistance inside the panel. Just like you can take two equal 12V light bulbs and connect them in series to a 24V battery:
each bulb will have 12V over it.
If you understand all above, you can have a job tomorrow anywhere in the world ;)
And really, I am just a retired engineer and nowhere near as smart as the professors who tried to program me with all this info. And I really believe that at least 80% of the members of this forum can comprehend this when they take the time. If you do, you will never need to call for help anymore while working on the
electrical system on your boat!
cheers,
Nick.