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Offline rngd

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Simple solar questions, pls help
« on: 09/12/2007 18:31:42 »
Hi,

I'm not sure if this is the right place to ask this, my name is Melvin, and I'm planning to build a 2 axis solar tracker using a small solar panel (<5W), a lead acid battery, a microcontroller and 2 motors (its not for everyday usage, more for educational purposes). I have read through a lot of the basic principles of solar panels, but I still have a couple of questions for you guys,

1. What effect does a panel's nominal voltage have on the system ? Is a higher nominal voltage selected just so that it can power higher voltage devices ? For example, a 6V panel is chosen over 3V panel because I have a 5V device ?

2. What will happen if a panel's nominal voltage differs from the lead acid battery voltage that it is charging ? For example, a 6V panel charging a 12V battery, or vice versa.

3. For battery ratings, I understand that 4.5 AH means that if 4.5A is drawn from the battery, the battery's capacity will be enough to last for one hour. Does this mean that at the end of the hour the battery will have 0 charge left (100% DOD) ? Or am I misunderstanding something ?

Thanks for your time, I really appreciate it.

edit : typo, 5V not 5W
« Last Edit: 10/12/2007 16:35:28 by rngd »


 

another_someone

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« Reply #1 on: 09/12/2007 18:53:59 »
1)

The voltage is often a matter of matching voltage to requirements, but other factors also have a part to play.  A higher voltage means you have a lower current for the same power, which means that either you have fewer resistive losses for the same wires, or you can use thinner wires (and hence reduce weight) for the same resistive loss.

2)

You need to have a higher voltage coming from the charger than the batteries nominal voltage, otherwise the current will flow in the reverse direction (i.e. the battery voltage will diver current into the solar panel, possibly damaging it, rather than the solar panel driving current into the battery).  You should ideally ensure there is a rectifier between the solar panel and the battery to ensure this does not happen by accident (the rectifier will not guarantee the current going in the direction you want, but it will mean that if it tries to flow in the opposite direction, there will simply be no current flow at all).

3)

Yes (well, almost) - but taking a lead acid battery down to 0 charge is likely to cause permanent damage to the battery, so it is something to be avoided.

The 'almost' is that as a battery gets close to zero charge, so its voltage will drop rapidly, and you will not be able to get enough voltage out of the battery to draw the last bit of charge out of the battery - but that will not be enough to protect the lead acid battery from damage.

Some lead acid batteries are slightly more resilient to being taken down to zero charge than others, but if that is something you expect to do a lot, maybe you want to think if lead acid is the best battery technology to use (although it is by far the cheapest).
« Last Edit: 09/12/2007 18:56:04 by another_someone »
 

Offline techmind

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« Reply #2 on: 09/12/2007 20:56:02 »
1. What effect does a panel's nominal voltage have on the system ? Is a higher nominal voltage selected just so that it can power higher voltage devices ? For example, a 6V panel is chosen over 3V panel because I have a 5W device ?

2. What will happen if a panel's nominal voltage differs from the lead acid battery voltage that it is charging ? For example, a 6V panel charging a 12V battery, or vice versa.

The actual voltage you get out of a photovoltaic panel (as opposed to the nominal voltage) depends on how much sunlight is shining on it, and how much current you're drawing from it. You will be able to extract maximum power (i.e. (current x volts) max) when you're loading it enough to bring the voltage down quite substantially from the open-circuit voltage. You'd need to see the PV manufacturers' data sheet for more detail, but I suspect the optimum load will probably bring the cell voltage down to something like 2/3rds of the open-circuit voltage. In principle with a fancy circuit featuring boost/buck regulators you could extract near-maximum efficiency for all lighting conditions. In practice for your demonstration, you probably won't bother...
The nominal voltage is probably the voltage corresponding to maximum power-transfer with "one (full) sun" shining on the panel. See mnfrs' data.

What you need to consider is the recommended charge and float voltages for the Pb-acid battery you're planning to charge. Ideally you should probably choose your PV cells to have an open-circuit voltage close to, or just above, the recommended float voltage. This will be the voltage the system settles to in bright sunlight when the battery is fully charged. For a 12V Pb-acid, this may be somewhere around 17V - but see the mnfrs' data sheet. You may choose to use a panel with a slightly higher voltage so as to improve charging rate/efficiency in less than full sun - in that case you should probably consider putting some kind of voltage-limiting device (eg a heavy-duty zener diode) across the panel.
In reality I'm guessing you'd probably want at least a 15V nominal output to charge a 12V Pb-acid battery. You probably should use a voltage-limiting device.
A "6V" panel will not push much energy into a 12V battery.


3. For battery ratings, I understand that 4.5 AH means that if 4.5A is drawn from the battery, the battery's capacity will be enough to last for one hour. Does this mean that at the end of the hour the battery will have 0 charge left (100% DOD) ? Or am I misunderstanding something ?
That's the basic concept. As previously said, it's not usually healthy to take any battery down to 0V, but after the hour (in your example) the voltage would fall rapidly and you would have extracted all the "practical" energy from the battery.

Note that owing to charging inefficiencies you typically have to charge a battery with 125% to 140% of the energy you can take out. Depending on the battery chemistry (e.g. for NiCd/NiMH), charging efficiency may improve at low temperatures - mnfrs' data sheets should have a wealth of information.
« Last Edit: 09/12/2007 21:00:44 by techmind »
 

Offline Pumblechook

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« Reply #3 on: 09/12/2007 21:49:23 »
 Presumably there are high efficiency switched mode chargers which can in take a wide range of input voltages and deliver the correct output for a given type of battery. 

I think maybe the output voltage of PVs doesn't change that much with varying sunlight but the internal resitance will drop so that they can deliver more current. 

 

lyner

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« Reply #4 on: 09/12/2007 22:03:00 »
4.5Ah may not mean that the battery will actually give you 4.5A for 1 hr. It will probably give you 0.45`a for 10 hours, quite happily, but when giving high currents, the  efficiency goes down. Modern battery specs give you that sort of information.
 

Offline Pumblechook

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« Reply #5 on: 09/12/2007 22:06:50 »
I am never sure what the Ah spec of a battery means when they say you shouldn't discharge more than 50%..  Is the 'safe' capacity Ah/2 then?

Leisure batteries are better than car batteries for this sort of use.
 

Offline techmind

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« Reply #6 on: 09/12/2007 22:48:08 »
Presumably there are high efficiency switched mode chargers which can in take a wide range of input voltages and deliver the correct output for a given type of battery. 
That's what I meant by "buck/boost" circuits...

I think maybe the output voltage of PVs doesn't change that much with varying sunlight but the internal resistance will drop so that they can deliver more current. 
Indeed, the open-circuit (no-load) voltage won't change that much with light-level, but owing to the internal resistance change, under load the external voltage will change appreciably with light level.

I believe the open-circuit voltage is proportional to the logarithm of the light level, while the short-circuit current is directly proportional to the light level. (Light-level in Lux, or Cd/m2, or sheer photon-count etc)
 

Offline rngd

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« Reply #7 on: 10/12/2007 17:26:06 »
Thanks for the replies, guys.

So basically, if I choose use a 6Vnominal voltage to charge a 6V battery or a 12Vnom to charge a 12V battery, there will be no problems right ? Also, if I use a 12Vnom panel to charge a 6V battery (with regulation) compared to charging a 12V battery, what would be the difference ?

For the third question, I understand now that it will not really be 4.5A for one hr due to inefficiencies when the current is high. So how do I actually determine how much current my solar panel and battery setup can supply to my system (motors) ? Taking for example maybe a 6V panel and 6V 4.5AH battery. The motors will only require short bursts of current after a certain interval (probably 60 mins). I am really having a hard time with this part...


Melvin
 

lyner

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« Reply #8 on: 11/12/2007 18:34:12 »
Assuming you are charging with the appropriate panel voltage, all you need to know is the motor current requirement and how long it will be running per day. (Twice the panel voltage will mean that you will get half the current out for the same area  and you would need a regulator to dissipate the extra power- so match the  panel to the battery)
Your calculation will tell you out how many Ah you will need over a day. (Motor current times time) Assume 8hours for charging each day, say. So your average charging rate needs to the  the estimated total Ah divided by 8. By the end of the day you should be back to square one.
 Give yourself something in hand and buy a battery at least twice what you calculate and get a panel with a bit higher rating than you think you need. Rough and ready calculations but those simple sums will stop you spending much more than you need on equipment. I am sure that someone would be only too pleased to sell you ten times what you really need.
If you find you have got things a bit wrong, you can always bolt on some more cells / batteries.
 

Offline techmind

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« Reply #9 on: 19/12/2007 11:03:25 »
I've just seen an article in Electronics Weekly (12 Dec 2007) "Power path management for solar-charged batteries" which explains quite a bit about solar cells and getting the most energy out of them. You don't seem to be able to get the article from electronicsweekly.com without registering, but Googling the author (Nigel Smith, Texas Instruments) yeilds this article:

http://www.powermanagementdesignline.com/197008555;jsessionid=GQSP5H2BIBERGQSNDLQSKHSCJUNN2JVN?printableArticle=true

which I think is also very helpful. It includes a simplified electrical model of a solar cell, togther with load curves for differing levels of illumination.
 

Offline rngd

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« Reply #10 on: 19/12/2007 11:35:45 »

Ok, thanks guys for all the replies. I'm gonna start with the project's simpler parts and see how it goes.

Melvin
 

Offline Dick1038

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« Reply #11 on: 21/12/2007 19:16:44 »
Tracking the sun is complicated.  Not only does the sun traverse an arc across the sky, but there are clouds and night time to consider.  You will probably need a microprocessor to control the tracking.
 

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« Reply #11 on: 21/12/2007 19:16:44 »

 

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