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Electronics => Electronics => Topic started by: z.s.tar.gz on January 02, 2010, 04:32:33 PM
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Here's my situation: I have 4 solar panels wired in series that output on average ~15V @ 0.25A. I need to use a dc/dc converter to output 7.8V @ 1.1A.
I'd prefer to use this (http://www.sparkfun.com/commerce/product_info.php?products_id=317) from sparkfun with free day coming up and all, but I don't really understand how it works or how to use it.
I have read the datasheet multiple times, but it's just that: data. I'd really appreciate a short explaination, as google doesn't really come up with anything.
Here (http://www.sparkfun.com/datasheets/IC/MC34063A.pdf) is a direct link to the datasheet.
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Hi,
Here's my situation: I have 4 solar panels wired in series that output on average ~15V @ 0.25A. I need to use a dc/dc converter to output 7.8V @ 1.1A.
15V/0.25A (=3.75W) and you want to convert it to 7.8V/1.1A (=8.58W) - might be a bit hard ;)
Why not connect them in 2 by 2 and then make do with the power they're able to give?
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I was thinking about doing that, but I've already epoxy'd them down. I guess I can cut up the wires if I have to.
So if I connect them all in parallel, I should be able to get out ~4V @ 1A, which is close enough to charge one of my batteries at 1C or both somewhere between 0.5C and 1C.
Also, if I only charge at .5C, I'll only need ~2.1W, right? Or is cutting both in half actually a fourth? Regardless, I can get around 4 watts out of my solar panels in direct sunlight, if not a little more.
This means that I can probably charge at 0.5C with only a regulator and some monitors?
http://www.sparkfun.com/commerce/product_info.php?products_id=527 (http://www.sparkfun.com/commerce/product_info.php?products_id=527)
That's an adjustable regulator. The question is: What happens to the current that gets put into a linear regulator? Does it stay the same, or does it drop some?
Also, does the 3% charge current thing still apply if it's charging at say 0.45C?
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The question is: What happens to the current that gets put into a linear regulator? Does it stay the same, or does it drop some?
The data sheet (follow the link on the page below) can answer that:
http://www.electronics-lab.com/articles/LM317/ (http://www.electronics-lab.com/articles/LM317/)
But briefly page 4 states min, typical and max current. Max current it will put out is 3.4 amps. This might ruin your calculations a bit.
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If you wired the panels in 2 sets of 2 ie. A->B series, C->D series, then AB -> CD parallel, you could get ~7.5V @ 0.5A i believe (series doubles voltage, parallel doubles current) someone please correct me if I'm wrong.
-HyperNerd
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If you wired the panels in 2 sets of 2 ie. A->B series, C->D series, then AB -> CD parallel, you could get ~7.5V @ 0.5A i believe (series doubles voltage, parallel doubles current) someone please correct me if I'm wrong.
-HyperNerd
No, you are quite correct.
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That's my plan, as that'll be very close to 0.5C.
As far as the whole 3.4A max output, the batteries I'm going to be charging only have 1.1A at 1C, and I doubt that my solar panels can even reach that if all wired in parallel.
Overall though, that will work? I can use that adjustable regulator to charge my batteries at 0.5C? (with proper monitoring of course)
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Hi,
Also, if I only charge at .5C, I'll only need ~2.1W, right? Or is cutting both in half actually a fourth?
If 1C is 8.58, then C2 (=C/2=0.5*C) is half that, so it takes 4.3W to charge with C2 (and that's forgetting losses).
The question is: What happens to the current that gets put into a linear regulator? Does it stay the same, or does it drop some?
The current stays the same (apart from the very small regulation current << 1mA), but the voltage drop - around 3V on a regular type.
Also, does the 3% charge current thing still apply if it's charging at say 0.45C?
3% of the 1C rate, in your case it's 33mA.
But... With such limited available power, you have to minimize your losses and make it charge as much as possible, so a SEPIC converter/regulator is what would be best .
Something like the LT1513 (http://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1037,C1078,C1089,P1517), which can be had in a through hole mounting, TO220 style housing and which can be coupled as a SEPIC with an input range of 2.7V to 25V, maintaining the voltage to 1% when in CV mode.
That one and your solar panels in 2x2 coupling would give optimum charging in your case.
There are several other switcher ICs that can be used for SEPIC topology converters as well, but most is SMD nowadays.
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Actually, because I'm using LiFePO4 batteries, so 1C = 2*3.9 = 7.8
Sorry if I didn't specify that ahead of time. I've always found linear regulators to drop about 1.5V, but I think I'll check out what you're saying as the less drop the better.
And it seems like everything is smd nowadays.
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Yes, you can charge your batteries at 0.5C, but since the voltage is only 7.5V or up to 8.5V, you can't use the LM317 because it need the differential between the input voltage and the output voltage is approximately 3V. you need a Low Dropout Adjustable Regulator like LM2941.
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Yes, you can charge your batteries at 0.5C, but since the voltage is only 7.5V or up to 8.5V, you can't use the LM317 because it need the differential between the input voltage and the output voltage is approximately 3V. you need a Low Dropout Adjustable Regulator like LM2941.
Thanks! That seems right up my alley. http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=LM2941CT-ND (http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=LM2941CT-ND) is the link (so I can find it later)
And at only a dollar eighty too!
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Hi,
And at only a dollar eighty too!
The LT1513 that I linked to is a much more efficient solution and it can be had for free.
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The con is the LT1513 need a two windings on a common core Inductors: COILTRONICS CTX10-4
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=513-1200-1-ND (http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=513-1200-1-ND)
which cost USD$3.20
But pro is LT1513 can Charge with Input Voltage Higher, Equal to or Lower Than Battery Voltage is a plus points.
and you can ask for samples ;D which free!!!
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So I've run into a major roadblock: the solar cells don't put out 250mA, they put out 29mA. This is an error in reading it the first time.
As you can tell, even if I wired all in parallel and only charged one battery at a time, it would still take 28 hours. Looks like it's time to find some new solar cells!
http://www.solarbotics.com/products/scc3766/ (http://www.solarbotics.com/products/scc3766/)
Double the voltage, +10mA, and half the size. I wish I had gotten these first.
With 8 of these, I can get 352mA if they're all in parallel. This means only 9 hours for both batteries as opposed to 28 for one.
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Hmmm,
That is an improvement, but I'm sure you can find a way to charge your robot's batteries in less time.
If the robot moves on wheels, then how about dynamos on the wheels to generate power, or if it is a flying bot then mini turbines?
These would help increase battery life at the least so you wouldn't have to recharge so often.
-HyperNerd
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But wouldn't adding dynamos to the wheels create more resistance and therefor require more power from the motors?
How could you attach them to a servo anyways?
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Through a gear or belt system?
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Oh yeah brainfart there...
Even so, it's break even at best (and even that's impossible)
I guess it could increase efficiency, but it would be awefully sporadic.
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Hi,
If the robot moves on wheels, then how about dynamos on the wheels to generate power, or if it is a flying bot then mini turbines?
The laws of thermodynamics still apply and over-unity machines are still "illegal" under those laws! ;)
billhowl <- You don't need to use that specific core, just find one with reasonably similar specs. in your trashcan err, project bin.
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OK then,
Thermodynamics 1 - 0 Me ;D
But you could still use turbines or something else due to wind :D
-HyperNerd
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Not going to go that fast, thanks anyways.
Now I've got a new problem: I've found some really good cheap solar panels, but they're high amp ones.
They're rated ~0.5V@4A. What can I use to actually increase the voltage in exchange for amps? A regulator won't work, as I need to step it up, not down. I've calculated that it can be converted to ~8V@250mA, which is more than 4 time what I'm capable of getting now.
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Have you seen the mightyboost or whatever they are called?
They are used to boost 1 cell up to 5v, they consist of a boost/buck converter along with some caps and stuff, i think sparkfun have them and there are a few instructables on them.