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goboatingnow · 22-10-2009, 05:22

Quote:

By the way, I've also taken completely dead used automotive batteries that have been sitting for several years dead, hooked them up to the 72v solar charger, and was able to bring them back to life in about 16 hours charging time.....again, approaching 12.5v as their new 'full' charge (measured after resting following charging, of course). Put that one in your pipe and smoke it. (It kinda goes against the grain of everything you've ever been led to believe about batteries, huh.)

Stuff and nonense, any old battery will accept a recharge and display a terminal voltage, what matters is the amount of energy stored and available. You need ot do discharge tests and I expect the capacity of that battery is a fraction of what it should be. Terminal voltage is completely misleading

**s/v Jedi** · 22-10-2009, 06:10

Not Sure: you are obviously clueless about the whole issue and walk over laws of nature like if they would make way for you and your starter batteries. Some facts: you do not charge your 12V battery at 72V, period. You might measure 72V at the solar panel when it is disconnected but not when you connect it to the battery. Your battery pulls it down to the 12V level, which puts the panel at a very low performance point of it's output curve and I estimate the current at 0.5A (20% under panel spec, even at short circuit (0V) you only get 0.8A) at 12V service. So, when your magical system has been charging for an 8-hour day, you have put the fabulous amount of 8 x 0.5 = 4 Ah into the battery. If you call that running it dead down before charging you are right but that is because you never fully charge it or it doesn't take a charge anymore (it is empty all the time). To charge an 80Ah battery from empty you need well over 200 hours of full sun without any power-draw. That battery must be powering a very low-power system. Also, your view of acceptance-rate is seriously flawed. You sound like those people who don't believe man from Earth actually walked on the Moon. You will dismiss any evidence and scientific information while dreaming up your own laws of nature and never give in. ciao! Nick.

Not Sure · 22-10-2009, 13:08

Quote:

Originally Posted by s/v Jedi

Not Sure: you are obviously clueless about the whole issue and walk over laws of nature like if they would make way for you and your starter batteries. Some facts: you do not charge your 12V battery at 72V, period. You might measure 72V at the solar panel when it is disconnected but not when you connect it to the battery. Your battery pulls it down to the 12V level, which puts the panel at a very low performance point of it's output curve and I estimate the current at 0.5A (20% under panel spec, even at short circuit (0V) you only get 0.8A) at 12V service. So, when your magical system has been charging for an 8-hour day, you have put the fabulous amount of 8 x 0.5 = 4 Ah into the battery. If you call that running it dead down before charging you are right but that is because you never fully charge it or it doesn't take a charge anymore (it is empty all the time). To charge an 80Ah battery from empty you need well over 200 hours of full sun without any power-draw. That battery must be powering a very low-power system. Also, your view of acceptance-rate is seriously flawed. You sound like those people who don't believe man from Earth actually walked on the Moon. You will dismiss any evidence and scientific information while dreaming up your own laws of nature and never give in. ciao! Nick.

Actually, the panel measures 90v+ when open circuit. Output is 72v, and of course when connected to the battery what you will be measuring is the battery voltage ...which you can literally watch increase every few seconds on a multimeter when the panel is connected. Naturally the old formerly dead batteries aren't taking a full charge, and I already stated that their 'full charge' is around 12.5v or so.

Frankly, I don't give a rat's azz what you choose to believe or not believe. I'm telling you what happens. And for the record, this panel is putting out close to 100% of its rated capacity, which is excellent (and surprising) for a 20 year old amorphous (thin film) solar panel. They're not supposed to do that either, but there it is nevertheless.

Actually, it's quite funny to see the comments range from "You'll blow yourself up!" to "It's not putting out hardly any charge!". Clearly I'm not the one who 'doesn't know what they're talking about', and just as clearly the truth lies somewhere in the middle of the two extremes.

**s/v Jedi** · 22-10-2009, 13:41

NotSure,

You agree with what I wrote: the voltage is the battery voltage so 12-14V. The panel can deliver an absolute maximum of 0.8A at short-circuit so the output for charging the battery is at best 14 x 0.8 = 11.2 W. (power equals voltage times current). If you want more power out of the panel you have to connect it to a 48V battery bank, or use a MPPT controller that can take 100V input and create a 12V charger-output from that.

Also, after 8 hours in the sun (one day), you get 6.4Ah into the battery which is nothing really.

You approach of the truth must be between two extremes is a polical one, not a scientific view.

cheers,
Nick.

hellosailor · 22-10-2009, 15:40

You know, Nick, I've been playing around with batteries and circuits for maybe 45 years. Some formal education, some less formal. And the one thing I've found? You can often beat or exceed a spec, in some way, for some time, at some point. But when you're trying to do the fuel load for a 72-hour race and including electrical charging and a budget...funny thing but you'll either go black or silent or both when you try to make those "blue moon" numbers work instead of the ones supplied by the battery makers, charger makers, Sandia Labs, all those nice folks who have acid burns in their white lab coats and can repeat their experiments and conclusions all day every day, all year long.

Somebody wants to put a dying battery into some kind of use and keep it out of the recycling bin for an extra couple of years, great. More power to him. I know when I spec a circuit, I can relay on the performance all the way from one end to the other, and squeeze a bit more out when I really need it.

Listen, there are still some 2000 portable strontium piles missing inthe former USSR. If we go looking now....surely there are a few still up for sale?

Not Sure · 23-10-2009, 16:58

Quote:

Originally Posted by s/v Jedi

NotSure,

You agree with what I wrote: the voltage is the battery voltage so 12-14V. The panel can deliver an absolute maximum of 0.8A at short-circuit so the output for charging the battery is at best 14 x 0.8 = 11.2 W. (power equals voltage times current). If you want more power out of the panel you have to connect it to a 48V battery bank, or use a MPPT controller that can take 100V input and create a 12V charger-output from that.

Also, after 8 hours in the sun (one day), you get 6.4Ah into the battery which is nothing really.

You approach of the truth must be between two extremes is a polical one, not a scientific view.

cheers,
Nick.

So, is your claim that a 72v 45 watt solar panel directly hooked up to charge a 12v battery puts a similar amount of juice into the 12v battery as a 12v 5 watt solar panel directly hooked up to a 12v battery? Because that is essentially what you are claiming here. Is that your position?

**s/v Jedi** · 23-10-2009, 20:40

My position is one that Mr Ohm defined for you and me to accept, sorry if you don't like it.

cheers,
Nick.

bobfnbw · 24-10-2009, 07:40

Quote:

Originally Posted by s/v Jedi

My position is one that Mr Ohm defined for you and me to accept, sorry if you don't like it.

cheers,
Nick.

OUCHy

Actually the whole thread is getting kind of pointless.
Its like the difference between democrats and republicans.
But y'all go right ahead, cause its a source of amusement for some.

Not Sure · 24-10-2009, 08:55

Quote:

Originally Posted by s/v Jedi

My position is one that Mr Ohm defined for you and me to accept, sorry if you don't like it.

cheers,
Nick.

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?

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?

**s/v Jedi** · 24-10-2009, 09:02

Sorry for getting a little excited there. I'll make up by explaining it in detail in this post

About beating the specs: this is normal when the specs are from a decent manufacturer. When you produce devices, incl. semi-conductors, you work within a tolerance range. The specs for the final product are a standard that each unit must meet. This means that most units beat the spec and the average is well above the spec. Units that just not meet the spec often come on the market under a different brand name or under a "seconds" label.

On to the solar panel. Every panel (or array of panels) have an I-V and P-V graph like you find in the attached image.

The upper one is the I-V graph. The vertical axis shows I, the output current, and the horizontal axis shows V, the output voltage. Mr. Ohm defined that P (power) = I x V. So, when the voltage is 0 (a shorted panel), the output current is at it maximum, but the power generated is zero because there is no voltage. At the panel's maximum voltage, the output current is 0 (a disconnected panel) so even though we have a high voltage, the power is still 0 because there is no current. Somewhere in the graph we find a point where the output is maximum: this point is called Pmax. The current at that point is Imp where "mp" stands for Maximum Power.

The lower graph shows power output set against voltage. Here we see Vmp which stand for the voltage at Maximum Power output. You also see Voc as the maximum voltage and "oc" stands for Open Connection, a disconnected panel. Isc is the maximum current and "sc" stands for Short Circuit.

The interesting thing is that you can see that the output current is pretty stable until we reach 70% of Voc. Because the voltage increases along the X-axis, the output power shows a pretty linear increasing graph-line until that 70% of Voc is reached, after which it tops out at about 75% of Voc and quickly falls down.

The way an engineer looks at the I-V graph is by surface area. You can see the dashed lines that meet at Pmax.... the rectangle they form with the axis is the retangle that has the biggest surface area that you can draw within the area. So, the power output has a linear relationship with that surface area in the graph. Pretty easy way of finding that Pmax point. (engineers always have tricks ;-)

Now we look at the panel of NotSure. It's Vmp is 72V and Imp is 0.6A. This means that the panel can generate Pmax = 72 x 0.6 = 43.2W. The panel Voc is 98V and Isc is 0.8A.
You can see that the graph fits this panel (all panels have the same graph). when we take 75% of 98V we get 73.5V which is close to the spec'ed Vmp of 72V. Also, Imp at 0.6A and Isc at 0.8A fit the graph.

When we connect this panel to a partly empty 12V battery, the battery is a very low resistance and will allow a lot of current to start flowing (it is a very high "load"). The panel can't provide that much power resulting in it's output voltage dropping like a brick until it stabilizes at a value just above the battery voltage. It will provide a current between 0.8A and 0.6A (let's take 0.7A) as limited by it's curve in the graph at a voltage of, let's say 12.5V. That voltage in the graph is way left on the X-axis, at about 1/6th of Vmp. When you imagine the dashed lines at that point in the graph you will immediately realize that power output is minimal. We can calculate is too: 12.5 x 0.7 = 8.75W.

Everyone reading this will also understand the new MPPT controllers. They load the panel lightly while measuring it's output voltage and current and calculate the power. Next they load it a bit more and check if the power goes up or down. If up, they loadit a little more again... but if down, they reduce the load. After some measurements they find that magical Pmax spot in the curve and transfer maximum power from the panel to the controller. This power is converted into a battery-charging source which can be at a much lower voltage. All the while they keep changing their load to the panel just a bit to check if they still are at Pmax... and correct if not. That is why they are Maximum Power Point Tracking.
When you connect a MPPT controller between NotSure's panel and battery, it's input voltage would measure 72V, it input current 0.6A (a power transfer of 43.2W) and it's output voltage 12.5V and output current 3A (a power transfer of 37.5W). The difference is the consumption of the controller itself and conversion losses (which appear as heat). So, instead of 0.7A into the battery, he would get 3A into the battery, a huge improvement, with the same panel.

However, if he would have connected a 21V panel (as rated for nominal 12V battery) without the controller, his point of voltage operation would have been much closer to Vmp and it would work almost as good (within 30%) as the 72V panel with controller.

So, back to the 72V panel connected to the 12V battery: the question many readers will have is "will the battery explode?!" The answer is disappointing: nope. The maximum power that the panel can generate isn't enough for that. As the 0.7A charge current keeps flowing, the battery slowly charges, which increases it's internal resistance which increases the voltage. When the voltage reaches 15-16V, the electrolyte will be boiling and the battery heating up. But the couple of Watts of power going into the battery will probably not even be enough to get it to boil (try to boil a pot of water with a 10W heating element... won't work). The battery will easily radiate it's heat into it's surroundings with a slow bubble of the electrolyte as the only result. The voltage will probably not even reach 14V.

The important thing to remember is that law that P = I x V and these are the measured values of I and V. You can't just take Vmp and multiply that with Imp (maximum power output) and state that you put that into your battery. You have to measure the voltage and look that up in the I-V curve to find the corrosponding I (and you can measure I to confirm it) and calculate P from those two values.

This is the way it is. It's quiet easy to understand, high school math and accepting Ohms law and not being afraid to read a graph. And it's just silly to state it's all a lie and a choice to believe it or not.

ciao!
Nick.

Not Sure · 24-10-2009, 09:30

Quote:

Originally Posted by s/v Jedi

Sorry for getting a little excited there. I'll make up by explaining it in detail in this post

About beating the specs: this is normal when the specs are from a decent manufacturer. When you produce devices, incl. semi-conductors, you work within a tolerance range. The specs for the final product are a standard that each unit must meet. This means that most units beat the spec and the average is well above the spec. Units that just not meet the spec often come on the market under a different brand name or under a "seconds" label.

On to the solar panel. Every panel (or array of panels) have an I-V and P-V graph like you find in the attached image.

The upper one is the I-V graph. The vertical axis shows I, the output current, and the horizontal axis shows V, the output voltage. Mr. Ohm defined that P (power) = I x V. So, when the voltage is 0 (a shorted panel), the output current is at it maximum, but the power generated is zero because there is no voltage. At the panel's maximum voltage, the output current is 0 (a disconnected panel) so even though we have a high voltage, the power is still 0 because there is no current. Somewhere in the graph we find a point where the output is maximum: this point is called Pmax. The current at that point is Imp where "mp" stands for Maximum Power.

The lower graph shows power output set against voltage. Here we see Vmp which stand for the voltage at Maximum Power output. You also see Voc as the maximum voltage and "oc" stands for Open Connection, a disconnected panel. Isc is the maximum current and "sc" stands for Short Circuit.

The interesting thing is that you can see that the output current is pretty stable until we reach 70% of Voc. Because the voltage increases along the X-axis, the output power shows a pretty linear increasing graph-line until that 70% of Voc is reached, after which it tops out at about 75% of Voc and quickly falls down.

The way an engineer looks at the I-V graph is by surface area. You can see the dashed lines that meet at Pmax.... the rectangle they form with the axis is the retangle that has the biggest surface area that you can draw within the area. So, the power output has a linear relationship with that surface area in the graph. Pretty easy way of finding that Pmax point. (engineers always have tricks ;-)

Now we look at the panel of NotSure. It's Vmp is 72V and Imp is 0.6A. This means that the panel can generate Pmax = 72 x 0.6 = 43.2W. The panel Voc is 98V and Isc is 0.8A.
You can see that the graph fits this panel (all panels have the same graph). when we take 75% of 98V we get 73.5V which is close to the spec'ed Vmp of 72V. Also, Imp at 0.6A and Isc at 0.8A fit the graph.

When we connect this panel to a partly empty 12V battery, the battery is a very low resistance and will allow a lot of current to start flowing (it is a very high "load&quot

. The panel can't provide that much power resulting in it's output voltage dropping like a brick until it stabilizes at a value just above the battery voltage. It will provide a current between 0.8A and 0.6A (let's take 0.7A) as limited by it's curve in the graph at a voltage of, let's say 12.5V. That voltage in the graph is way left on the X-axis, at about 1/6th of Vmp. When you imagine the dashed lines at that point in the graph you will immediately realize that power output is minimal. We can calculate is too: 12.5 x 0.7 = 8.75W.

Everyone reading this will also understand the new MPPT controllers. They load the panel lightly while measuring it's output voltage and current and calculate the power. Next they load it a bit more and check if the power goes up or down. If up, they loadit a little more again... but if down, they reduce the load. After some measurements they find that magical Pmax spot in the curve and transfer maximum power from the panel to the controller. This power is converted into a battery-charging source which can be at a much lower voltage. All the while they keep changing their load to the panel just a bit to check if they still are at Pmax... and correct if not. That is why they are Maximum Power Point Tracking.
When you connect a MPPT controller between NotSure's panel and battery, it's input voltage would measure 72V, it input current 0.6A (a power transfer of 43.2W) and it's output voltage 12.5V and output current 3A (a power transfer of 37.5W). The difference is the consumption of the controller itself and conversion losses (which appear as heat). So, instead of 0.7A into the battery, he would get 3A into the battery, a huge improvement, with the same panel.

However, if he would have connected a 21V panel (as rated for nominal 12V battery) without the controller, his point of voltage operation would have been much closer to Vmp and it would work almost as good (within 30%) as the 72V panel with controller.

So, back to the 72V panel connected to the 12V battery: the question many readers will have is "will the battery explode?!" The answer is disappointing: nope. The maximum power that the panel can generate isn't enough for that. As the 0.7A charge current keeps flowing, the battery slowly charges, which increases it's internal resistance which increases the voltage. When the voltage reaches 15-16V, the electrolyte will be boiling and the battery heating up. But the couple of Watts of power going into the battery will probably not even be enough to get it to boil (try to boil a pot of water with a 10W heating element... won't work). The battery will easily radiate it's heat into it's surroundings with a slow bubble of the electrolyte as the only result. The voltage will probably not even reach 14V.

The important thing to remember is that law that P = I x V and these are the measured values of I and V. You can't just take Vmp and multiply that with Imp (maximum power output) and state that you put that into your battery. You have to measure the voltage and look that up in the I-V curve to find the corrosponding I (and you can measure I to confirm it) and calculate P from those two values.

This is the way it is. It's quiet easy to understand, high school math and accepting Ohms law and not being afraid to read a graph. And it's just silly to state it's all a lie and a choice to believe it or not.

ciao!
Nick.

So your claim is that if a 12v battery were rigged to attract a bolt of lightning, the battery wouldn't explode, the charge would be reduced to 12v, and the electrolyte would simply bubble slightly.

Nobody is stating that 'it's all a lie'. It just isn't the constant that you make it out to be.

**GordMay** · 24-10-2009, 10:15

Quote:

Originally Posted by Not Sure

So your claim is that if a 12v battery were rigged to attract a bolt of lightning, the battery wouldn't explode ...

A lightning bolt will have millions to BILLIONS of watts of energy, not 45W.

**s/v Jedi** · 24-10-2009, 11:33

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 "kick-charge" 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 re-write 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 input-impedance of the load must match the output impedance of the power source. If there is an impedance mis-match, 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 positive-to-positive and negative-to-negative 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 stand-off. 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 I-V 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.