Solar Panel "Battery Maintainer" Project Ideas - Page 1
Forum section: electronics / general

author: edy
author_rank: Super Contributor
timestamp: August 02, 2012, 04:05:45 am
content: Hello folks,I picked up a couple of what I thought to be cheap solar panels and wondering about what I can do with them. They are the "Blue Planet 1.8W Solar Battery Maintainer" featured here, for about $11.99 (they were on sale from normally $29.99 and according to the website link $19.99):http://www.canadiantire.ca/AST/browse/Green/EnergyConservation/RenewableEnergy/PRDOVR~0112001P/Blue+Planet+1.8W+Solar+Battery+Maintainer.jsp?locale=enI tracked down the "Blue Planet" name and it seems to be the same as a company called "SunForce Products Inc", and the item#: 50012, UPC#: 834319000631 which is located here:http://www.sunforceproducts.com/product_details.php?PRODUCT_ID=65The manual is here:http://www.sunforceproducts.com/prodinfo/manuals/50012_1.8WSolarManual.pdfThe specs written as follows:1.8W/125mAPower Rating up to 1.8WCurrent 125mA at 15VI thought it was a good deal at $11.99 each. I see "two-packs" selling at Costco and Home Depot for about $49.99, so I basically paid $24 like half price for two. The thing is, I didn't buy it to maintain my car battery. I actually want to use it to drive various projects and I'm not sure where to start... For one, is the voltage on these things regulated to come out at a constant amount. What about the current?I'd like it to power a USB device, for example, that means I need to convert the output to 5V. How does it work then with Solar Panels? I'm never used them before. Does the intensity of the light adjust the current, the voltage, or both? How do I make it push out 5V? If the 125mA is high enough, I guess it could power a device as long as it draws less mA.If I plan to connect both solar panels together, in series do I get 24 V? In parallel do I keep 12V but increase my current capability? Please help me understand this stuff. I know it's basic.Ultimately, I'd like to wire it up to power some Arduino or Raspberry Pi type boards and make something run by solar power. The panels aren't strong enough to charge a battery, so if I have a 12V battery being used for my project and it is draining, the solar panel wouldn't be enough to keep it topped up. Unless perhaps I used 2 together, and I wasn't using much current.Well all and any input would be most appreciated! Thank you!

author: edy
author_rank: Super Contributor
timestamp: August 02, 2012, 04:14:17 am
content: So... have I "been had" or are these things any use for some do-it-yourself projects? Is that $11.99 worth it, or is it just a waste of money?

author: HardBoot
author_rank: Regular Contributor
timestamp: August 02, 2012, 04:20:04 am
content: Solar panels generally like running in series to boost voltage, however they have restrictions... can never be sure without the actual panel spec sheet.I'd just run them in parallel to be safe.There's cheap and simple controller chips for solar battery charging, from there a simple low dropout regulator could be used, or a simple switching regulator, to power your project.They're pretty bad panels, no clue what they're worth, $10 I think.

author: edy
author_rank: Super Contributor
timestamp: August 02, 2012, 04:31:48 am
content: Thanks for the reply. Ok, I looked around and someone mentioned the "7805" regulators... Here's a link to the specsheets:http://www.electrokits.com/downloads/pdf/7805-datasheet-fairchild.pdfI will assume if I connect up my 2 panels in parallel, I will maintain 12 V but have 250mA. Actually, the panel itself outputs about 15 V according to their manual (125mA @ 15 V which is why wattage is 1.8W). That is fairly pathetic as far as current goes, but I have voltage going for me. However, in lowering the voltage from say 15V to 5V, is there any way to improve my current?The 7805-series regulators, they seem to be small ICs that will take a range of input voltages and output at 5V. They are pretty tolerant and I think for the amps I'm using, which are quite little, it shouldn't cause a problem. However, as I lower the voltage from 15V to 5V, am I still retaining the same 250mA current (using 2 panels of 125mA each in parallel)? Or can I "trade" the drop in voltage for a rise in current?If that was the case, then if I am reducing my 15V to 5V it is a 3x decrease, then I could increase amperage from 250mA to 750mA... now that would be a 3x increase, and 750mA is good enough to charge up my Blackberry phone.Am I thinking about this all wrong?

author: digsys
author_rank: Unknown
timestamp: August 02, 2012, 05:37:33 am
content: Wire them in parallel, and add an ORring Shottky diode (lower loss) to each +ve, then use thatoutput to charge a 12V SLA battery. That way you are collecting 100% of light energy.You can buy a DC-DC converter module quite cheap, and that will give you 88%+ efficiency.If you need more details, I'll put circuits and part #s together tonight.

author: Balaur
author_rank: Unknown
timestamp: August 02, 2012, 07:07:26 am
content: Quote from: digsys on August 02, 2012, 05:37:33 amYep, that's definitely the way to go ahead.For the DC-DC converter use a standard car charger for your BB. While surprises exist, a reasonably-designed charger should be aswitched-mode PSUand thus ensuring a quite good (maybe >75%) power transformation efficiency.There are switching mode alternatives for the 7805 regulator, such as theV7805series, but frankly, buying a commercial car adapter makes more sense for many reasons.Cheers,Dan

author: G7PSK
author_rank: Super Contributor
timestamp: August 02, 2012, 07:55:12 am
content: I have got one of those solar cells, they contain their own diode so an external one is not required, I use mine to maintain Li Po cells, no regulators or any thing, It just sits in a north facing window withe the cell hooked up to it, it keeps the cell's topped up until I want them.

author: edy
author_rank: Super Contributor
timestamp: August 03, 2012, 01:46:39 am
content: Thanks for the advice. Please, if you have wiring diagrams regarding the parallel-connection and Shotky diode please post!Yes the solar panels have a blue LED in them which shows when it is getting light on it. Not sure if those are the one and only diode in the panel, or just an indicator diode. Either way, it says in the specs that there is a "blocking diode to prevent reverse discharge".I assume that means if you are plugging it into a battery to charge it, obviously you are shorting the battery through the panel itself. If there was no diode, the battery would just drain through the panel in reverse when the light was turned off (or even when the light was turned on). The solar panel with diode is like a one-way door and the sun just helps the electrons move through that door (the way I see it). But then there is a crowd of electrons all piled up on one side of the door. If the door was not one-way valve like a diode, the crown of electrons would push back through the door and even things out again.So far so good.... from what I understand....Connect the panels in parallel, improves the current but keeps voltage at 15V. However, a car adapter is designed to accept 12 V from the cigarette lighter plug. I am assuming that even so, the circuitry on the car-adapter voltage regulator is designed to handle a range of input voltages so even 15 V will still be ok. Yes I do have several adapters (and they are dollar-store items, very cheap and readily available).Ok so solar panels in parallel, wired though the car adapter. Then I have my 5V USB connection and should be able to plug into my devices. What kind of amperage am I going to have with this setup? I will have to try it out! Thanks again for the help.(This may be a good way to power items when going camping?)

author: edy
author_rank: Super Contributor
timestamp: August 07, 2012, 02:27:41 pm
content: Hi folks,I finally had a chance to pull out the multimeter this morning and here's what I got on my panel:voltage 19.5 DC (multimeter was on 20 V setting)amps 9.5-10 average (multimeter was on 20mA setting)For both tests, I had the terminals of the panel connected straight to the COM and the V-ohm-MA terminals of my meter (my V-ohm-MA terminal says 200mA on it, I assume that is the rating to use this one before switching over to the A terminal instead).Certainly the mA is really a bummer here. I would have guessed by the specs that we have 150mA, not 1/10th of that! Something must be wrong. Am I multiplying wrong here? My panel is behind the glass of my sunroof in the car. So perhaps the tint in the sunroof is to blame? ILots of high voltage, but I care more about current than voltage if I want to power any USB devices. I know we talked about using a car-charger as a voltage regulator to pull down the 19.5 V to a more manageable 5V... but what about my current? Any way to boost it?

author: G7PSK
author_rank: Super Contributor
timestamp: August 07, 2012, 03:15:57 pm
content: I think that they have their own boost converter which is why the blue LED flashes at increasing rate until it becomes steady according to the light level, I have not taken mine apart to look inside but I might get around to that, if so I will post a few photographs.

author: digsys
author_rank: Unknown
timestamp: August 07, 2012, 03:26:43 pm
content: For the S/Mode module, you can use something like this -au.element14.com/recom-power/r-78e5-0-0-5/switching-regulator-5v-0-5a/dp/2078564Around 90% efficiency. The low output is definitely due to the sunroof. Any glass will refract light (except direct),and the tinting may be filtering the frequency you're looking for !!! How did you load the cells?

author: G7PSK
author_rank: Super Contributor
timestamp: August 07, 2012, 04:25:06 pm
content: Just opened mine, there is no booster inside despite the information with it when I got it implying there was and the voltage and current pulsing on the output, it is just a self flashing led that takes so much power from the cell that it drops the output in time with the flashing, the output diode is just a in 4001 so lots of loss there, I will most likely remove that to feed a booster.

author: edy
author_rank: Super Contributor
timestamp: August 07, 2012, 06:05:21 pm
content: Thanks, I checked out the RECOM POWER - R-78E5.0-0.5 - SWITCHING REGULATOR, 5V, 0.5A datasheet here:http://www.farnell.com/datasheets/1518729.pdfSomeone mentioned about the LED diode drawing power, but it seems to be for preventing charge reversal because remember this Solar Panel is designed to be plugged into your 12V car battery cigarette-lighter for trickle charging it while you are going away on vacation or something. You don't want the battery to discharge in reverse through your panel when the sun is down.However, if I am to use it for supplying direct current to a device, I can remove the diode if needed. I haven't opened up my panel yet so I am not sure what is in there. Then I could potentially use 2 of the panels joined in parallel into the RECOM POWER regulator, which I assume is still going to be <500mA.I still don't get how the current changes here. The regulator takes an input of 7-28 V and outputs 5 V. However, it doesn't mention anything about current input and what kind of "boost" occurs. The datasheet simply states that it outputs safely between 5-500 mA. So if I give it an input current of 20 V at 100 mA, what is this going to output.... 5 V at also 100 mA?I am also still trying to figure out why my multimeter is showing a current of 9-10 mA (I am assuming, since the display is showing 9.5 on average with the multimeter set to 20 mA setting on the dial) when these panels are supposed to output 1.8 W which on the spec sheet says 15 V x 120 mA = 1.8 W?

author: edy
author_rank: Super Contributor
timestamp: August 07, 2012, 07:25:06 pm
content: I found the following PDF file online, which may be helpful:http://www.national.com/assets/en/appnotes/f5.pdfI think this document helps shed some light on things. Really I am looking here for a transformer since V1 x I1 = V2 x I2 which would mean although my voltage goes down, the current goes up. Yes so a 20V @ 100mA = 5V @ 400mA. If I connect 2 solar panels in parallel then it would mean 20V @ 200mA = 5V @ 800mA which is enough to power quite a number of USB devices.The problem is that I do not have a constant 20V supply from the panel, due to sun/clouds and so on. Without having to resort to a battery (which I use to power my device, and charge up the battery) or using a giant capacitor to act like a battery, how do I take a variable voltage/current from my panel and make it output at 5V... with the variability in the output passing to the current itself. I don't care if my current capacity fluctuates, as long as it is above some threshold needed by the device to run.For example, if the input varies between 10-20V due to sun/cloud and my current also varies 50-100mA (or whatever) based on sun conditions... I want my output to be FIXED at 5V, although the current rating may change... for example, it could support 400mA when full sun but maybe in shaded it would provide 100mA. The point is, the voltage is still always 5V. So if I had a small device needing only 100mA of current to run, it would be able to continue functioning even if the current capacity varies between 100-400mA.However if I am trying to power a device that needs 5V @300mA then I would be "dipping" below 300mA quite often with my solar panel source if it varied between 100-400mA due to sun conditions, then it would shut off below 300mA and only work when there is full sun giving me the 300-400mA current capacity the device needs to draw to function.My understanding is that with a voltage regulator you are controlling the voltage output even though you are getting a wide input voltage range. But does it act like a transformer? For example, if my input was 15-20V and I decide to use a voltage regulator to lower it down to 10V.... so I can have a guaranteed 10V source at varying current. Now, I use a transformer that takes 10V down to 5V, but doubles the current. Or does the voltage regulator automatically boost the current dependent on the "drop" it is performing....For example, if it is 15V @100mA going down to 5V, does the current at that instant become 5V @ 300mA? Then a few seconds later the sun is brighter so 20 V @ 100mA (assuming current stays same?) therefore it transforms down to 5V but that is 4x drop so the current goes 4x higher so 400mA? Or am I thinking this all wrong? (most likely)

author: edy
author_rank: Super Contributor
timestamp: August 07, 2012, 07:40:11 pm
content: Here is another PDF... also quite informative!!! I understand now the pulse-width modulation aspect of how it works!http://www.national.com/assets/en/appnotes/f4.pdfSo to go from say 15 V input down to 5 V output, we are "pushing" our circuit at 1/3rd of the amount of time (duty cycle) using a PWM wave that has 15 V for 1/3rd of the time (switch on) and 0 V for 2/3rds of the time (switch off) so the average over the entire cycle is 15 V x 1/3rd = 5V.The inductor/capacitor combination is there to "smoothen out" the pushes because it slows the change in the circuit voltage and also continues to feed the circuit with current while the PWM is occuring. The output voltage looks like a zig-zag about the 5 V line. The variation in the zig-zag, or "ripple" is 20-30% but usually not enough to cause too many problems. Often the devices used also have their own inductance/capacitance which smoothens things out as well.

author: gxti
author_rank: Frequent Contributor
timestamp: August 08, 2012, 12:52:32 am
content: Quote from: edy on August 07, 2012, 02:27:41 pmDon't connect the leads of your multimeter across a voltage source while in current mode!The meter is acting like a low-value resistor and will draw as much current from the source as it can, potentially blowing the fuse in your meter (hopefully ithasa fuse) or destroying the source. Even if nothing is damaged you're not getting a meaningful reading of how much current the device can put out because the source voltage will plummet and the output power will fall with it.The current mode on your meter is for connecting in series with a load, to measure how much current the load is drawing, not to determine (not directly at least) how much current the source can safely put out. Except in cases of constant current sources like a bench power supply, the load determines how much current flows, not the source. When you connect your meter across the supply, you're measuring how much current the sense resistor in your multimeter draws, which is meaningless and likely to be high enough to blow the fuse.Quote from: edy on August 07, 2012, 07:25:06 pmFirst off, I'm going to assume you're talking about switching voltage regulators, not linear, since you linked to two documents that discuss the switching type. Switching regulators are ideal because they usually get 80% or better efficiency across a wide input voltage range. It's roughly the same concept as a transformer in that the power out is the same as the power in (minus losses as heat). This gives the "fixed current at higher voltage -> higher current at fixed voltage" characteristic that you are talking about in transformers, and would indeed mean that you could draw more current in full sun compared to the shade. But keep in mind that switching converters, solar panels, and USB are all DC power, while transformers are AC power -- so you can't use a transformer to step down power from a solar panel. That's why the DC-DC converters exist.There are actually switching regulators based around transformers but they are more complex, use a special type of transformer at a much higher frequency than a wall wart (around 100khz and up), and only needed when you want to isolate the output from the input. These are the kind you would find in a cell phone charger or computer power supply.If you want something easy, start with a linear regulator (like the 7805 mentioned earlier). It will be inefficient and won't yield more power with higher voltage but it's cheap and really hard to screw up. A simple way to think about how much current you'll get is that since they work by burning off voltage, the current in and current out are almost exactly the same. So if you have 200mA available from the panel, then you'll have 200mA available at the 5V output terminal, regardless of the input voltage. The problem is that all that voltage burned off gives off a proportionate amount of heat so it might shut down if you draw enough to make it overheat. They're all thermally protected though so it's not going to explode. It's easy to calculate how much power it's burning, just subtract volts out from volts in and multiply by the current. The datasheet for the regulator will likely list a nominal figure for maximum power dissipation. Also, what I mean by "available current" in this case is how much current you can draw from the panel before the voltage falls low enough to become unacceptable. The load determines the current, but obviously you can't draw 1000 amps from a $10 solar panel, and with this kind of source that means the voltage drops when you draw too much. The voltage regulator will keep a rock solid 5V output until the input falls to 6-7V (the difference is called "dropout voltage" -- check the datasheet).

author: NiHaoMike
author_rank: Super Contributor
timestamp: August 08, 2012, 01:56:56 am
content: I have a similar panel from Harbor Freight, which I bought about 10 years ago. I'm actually surprised just how effective it is, as in it will generate a surprisingly useful amount of power on a cloudy day. I once used it to keep a 14.4V cordless drill charged, so it would be ready when I needed to use it. Now, the battery has increased in internal resistance so that the drill doesn't work very well, but it works quite well for charging an iPod. Even after the losses from being behind a window and not being at an optimum angle, it still generates enough power to keep an iPod (used about 3-4 hours per day) charged. (The panel is always connected to that old drill battery, but I only connect the iPod when I'm using it.)

author: edy
author_rank: Super Contributor
timestamp: August 08, 2012, 02:05:29 am
content: gxti... Thank you so much for the past reply. I was beginning to worry when I checked again in full sunlight and found my multimeter when attached to the panel was again only showing 10mA current when the voltage reading was around 20V. I knew something was wrong here with the way I was measuring and understanding the current here.I believe the problem is in how I am conceptualizing voltage and current. I think of a waterfall, where voltage is the height of the waterfall and current is the volume of water going over. Obviously the more height and more water falling, the higher the "pressure" of all that water would be to spin a wheel at the bottom (the power).Using the multimeter to measure the voltage gives me the potential separation between the poles of the solar panel. I get that. But a solar panel cannot push out that many electrons. So even though the voltage can be 20 V, the current is very small. Just like static charge can be thousands of volts, when you touch a door-knob and ground yourself the current is tiny because your body just cannot store that many electrons. So the current is short-lived.So when I was measuring the "current" on the solar panels, I was imagining somehow that I was measuring indirectly the volume of electrons or current passing through the meter. But that contradicts the idea that a device only "draws" the amount of current it needs. So it is this misunderstanding that is confusing me.For example, a short circuit draws a huge amount of current because there is no resistance. Since V=IR, then V/R=I and as R approaches 0, no matter what V is, then I approaches infinity. I guess the multimeter has resistance too... so when it is measuring current, it is showing how much current is passing through it. If the voltage I read before was about 20 V, and now my current is displaying as 10 mA, does that mean the resistance in the multimeter is V/I=20/0.010 = 2000 ohm? Or am I just abusing the multimeter here and getting useless results?(By the way I am using a cheap-o electronic one... hope I didn't burn anything out, but it works fine.... hopefully I can't do too much damage at these relatively low voltages/currents)When a device pulls power, it requires a certain amount of current to function. I understand some things require much higher amperage than other devices... the load on a voltage/current source would be higher, so it would draw more power. I guess in the most simplest case, a device with very little resistance would be able to pull more current. A device with extremely high resistance would draw little to no current? So if you have a 5 V source at 100mA, let's say, and you have 2 devices... one with a very high resistance and one with a very low resistance, would the high resistance device function while the very low resistance device end up pulling too much current and risk overheating the source/wires?

author: G7PSK
author_rank: Super Contributor
timestamp: August 08, 2012, 08:01:08 am
content: My photo cell is identical to the one shown by the OP.The led on these panels is for show only it draws power the blocking diode is separate it does not show very well in the photo I put up but it is to the right of the resistor. If you are going to use a DC to DC converter it might be an idea to remove both as the converter will block return current no point in having extra losses on the output of the photo cell.

author: edy
author_rank: Super Contributor
timestamp: August 09, 2012, 03:52:29 am
content: Thanks for the suggestion, G7PSK. I will certainly look at removing the LEDs and replacing them with perhaps just one to be more efficient.I ended buying more of these panels as they were each $11.99, so I have a total of 4. They actually went on sale even more and now $9.99, which is cheaper than these places:http://www.northerntool.com/shop/tools/product_200422045_200422045?cm_mmc=Aggregates-_-Nextag-_-Alternative%20+%20Renewable%20Energy&gt;Solar%20Panels%20+%20Accessories-_-121184http://www.amazon.com/gp/product/B0006JO0KG/The cheapest is Canadian Tire here:http://www.canadiantire.ca/AST/browse/Green/EnergyConservation/RenewableEnergy/PRDOVR~0112001P/Blue+Planet+1.8W+Solar+Battery+Maintainer.jsp?locale=enI figured if I put 4 of these together in a 2x2 arrangement they EXACTLY fit the size of my sunroof of my car! So I can stick them with the supplied suction cups onto the glass, and slide the sunshade over (like in the attached photo).If the specs are correct, then I should be able to get a good amount of current from 4 of these together. If they are pushing out 15 V at 120 mA (1.8W) then 4 in parallel should provide 480 mA and if I step-down the voltage to 1/3rd to 5V (for USB) then I could theoretically increase my mA to 1440 mA. This would be enough to charge up and run my phone, tablet and other small USB devices.I am just trying to resolve how to hook them all up and do the conversion using an appropriate DC-DC regulator.

author: edy
author_rank: Super Contributor
timestamp: August 09, 2012, 02:08:48 pm
content: Thanks for the advice.I plugged 2 of the solar panels together in parallel.... and measured the voltage across. Got about the expected 15-20V in sun. However, after hooking it up to the car-adapter-USB plug (see attached image) I had practically NO output voltage at all. What the heck?When I plug it into my 12V car plug, the LED on the adapter turns on brightly. But when plugged to the 2 panels in parallel, the LED on the 12V-USB adapter doesn't turn on AT ALL.I suspect it has something to do with the LED's on the actual panels... Any thoughts?

author: G7PSK
author_rank: Super Contributor
timestamp: August 09, 2012, 05:38:05 pm
content: They are on sale for a reason, they are total crap for charging car batteries they supply less than the self discharge rate of a car battery. The one I have just keeps a LiPo cell topped up. I put mine on a buck boost converter today and it would not even turn on the led.

author: edy
author_rank: Super Contributor
timestamp: August 09, 2012, 10:39:13 pm
content: Ok so if I disconnect the blinking LED which is supposed to protect reverse charge flow from the panels, will I get a better output? I mean the thing is capable of creating a large enough voltage difference (upwards of 19-20V) across the leads and claims 120 mA. I do not plan on connecting the panel to my battery, so there is no need for the diodes right? I will connect it straight to a load.

author: edy
author_rank: Super Contributor
timestamp: August 11, 2012, 04:22:49 am
content: Ok so I opened up the darn thing and found the solar panel connected to this device (see attached photos). From what I can tell we have our blinking LED and associated resistor. There is also a black component which I believe says N4007 KED... Couldn't find the exact code on Google but seems to be a rectifier diode.There is an S+ and S- written on the board... That is where the solar panel leads connect. They basically bridge across the LED and resistor, hence the blinking LED when the panel is in sunlight. However, current should pass also the other way opposite to the LED/resistor... Through to the B- and B+ which lead to the car's battery. The N4007 KED must be the diode used to protect battery shorting through the panel then, and NOT the blinking LED.Anyone have any thoughts regarding this circuit?So even without this, panel had a 20V reading but the current was not enough to even light up the 12V to 5V USB cigarette plug converter. What is going on here?

author: SeanB
author_rank: Super Contributor
timestamp: August 11, 2012, 05:29:32 am
content: Try adding a capacitor across the output, around 2200uF 25V . This will provide at least enough charge storage so the car charger can turn on. The panel probably will not be able to supply the rated current until it is in direct sun, without any glass between it and the sun, car windows are pretty good at absorbing light. If you have it outside connect the meter across the panel, set to the 10A range, and you will probably get 0.05 showing, and in the car it will be 0.01 only.

author: G7PSK
author_rank: Super Contributor
timestamp: August 11, 2012, 08:10:17 am
content: That black device is the reverse protection diode the led is totally parasitic and is just there to look pretty, it is across the output and just uses up half the power the unit produces in most conditions, to produce full rated power I would think that the cell needs to be under the noon sun on the equator.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: August 11, 2012, 11:18:19 am
content: These crappy amorphous? panels are on sale for a very good reason .They are totallyinadequate for keeping a car battery topped up .I bought exactly the same one about 5 years ago .They only provide about 30 mA in full sun at my latitude (I am in Uk , are you in Canada. ? You will get prolly get about 25 mA full summer sun there ).The self discharge rate of a 50 Ah lead acid car battery is prolly about 25 mAthen on top of that theres the cars standby current consumption of things like clock and car alarm which for my car was about 30 ma .So you can see just to keep a car battery the same will need 50 mA constant charge .I bought one of these 5W monocrystalline oneshttp://www.ebay.co.uk/itm/5-W-Monocrystalline-Solar-Panel-5-Watts-12V-PV-/200753595685?pt=UK_Gadgets&hash=item2ebdd8c525They are about the same area and produce about 3-4 times the output .I get 90 -120 mA in full sun, and about 70 mA on cloudy day.Treat manufacturer's power output rating's on solar panel only as a very very rough guidance ,as they are quoted at very high sun incidence (midday at the equator with cooled panels) .You never get anywhere near these outputs in real life .

author: edy
author_rank: Super Contributor
timestamp: August 13, 2012, 04:33:03 pm
content: Hey folks... An update!I disconnected the LED/diode and connected 1 panel directly to my 12 V car-USB-charger (pictured in a previous post on this thread) in FULL SUNLIGHT, and this time the GREEN LED on the 12V car-USB-charger actually LIT UP! Progress! So I plugged in an iPod Nano. The display on the iPod Nano showed the little "battery icon lightning" symbol showing it was charging!As soon as a I covered up some of the panel in shadow, the iPod battery indicator "lightning bolt" disappeared. When I took my hand away and full sunlight on it again, the lightning symbol came back!Now I don't know if the panel is putting out enough juice, but the iPod seems to think so and it is not complaining (by saying "source insufficient to charge this device") or whatever. I have to try it with my Blackberry phone next to see, as it requires about 700mA charger. The iPod nano (version 1) 2gb is really old, maybe it requires very little mA charging. Not sure. Either that or it doesn't detect a minimal mA and doesn't display any warning.I plan on hooking up my other panel in parallel to hopefully increase the current. I returned the other 2 panels (I bought 4 originally) because I thought it was a lost cause... But maybe there is still some hope!? Still, at $11.99 a pop I don't think I want to invest any more money in this fun little project.... Rather put it into a real panel that does the job.Looking forward on reporting my results once I solder everything up and test the charging capability.

author: edy
author_rank: Super Contributor
timestamp: August 23, 2012, 06:46:46 am
content: I finally tore apart the panels and got rid of the blinking LED circuit. I hooked up one of my panels and read about 10V coming off. The other panel, in the exact same position from the light (an overhead incadescent 60 watt) was reading 2 Volts ! Why such a huge difference?One panel was showing 10 V reading while the other 2 V, same distance from the light. When I brought the 10V panel closer to the light the voltage went up to close to 20V. When I brought the 2V panel closer to the light the voltage also went up, maybe to about 9 or 10 V. I am sure in direct sunlight they will both be up close to 20V, however I am surprised by the difference at the lower light levels.Also one of the wires attached to the back of the panel fell off, I noticed it was soldered with a "blob" to the back of the panel. Didn't look like it was connected to anything, just splatted against the back of the glass on the panel. Perhaps I am not bridging to the proper layer or the solar panel it is shorting itself somewhere.Anyone have any ideas?The back of the panel has a black film which looks to be glued down. When I peel it off, and then rub away the glue layer, I see a silvery finish. If I rub this with sandpaper or scratch it nicely it will go right through to the glass and I can see right through. So I assume the silver finish is the amorphous silicon layer on the glass which is what I need to solder to.I don't think it is my solder job, since I tried the voltmeter probe directly without solder to many other places on the back of the panel and got the same readings. If it was my connection or the layer I was connecting to, I assume it would have changed the voltage reading as I moved my probe around on the back. It was always the same.So could there be a "short" in my panel causing this?One more question, what are the lines visible when looking at the solar panel? There are no obvious wires there, it looks like it is just a gap in the silvery part... because when I peeled back the black backing layer from my panel and looked through, the lines correspond to the areas I could see through the glass clearly, whereas the other part of the panels still looked dark (I assume due to the silvery film and silicon layer stuck on the back of the glass).