How do I read a solar panel power chart? - Page 1
Forum section: electronics / beginners

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 03:01:25 am
content: There's something I've never understood about the solar panel power chart below. Suppose I connect a6V 10W panel(which has a Vmp of 6.16V, a Voc of 7.39V, and an Isc of 1.71A) to anadjustable linear voltage regulatorwith a Vdrop of 0.5V at a load of 1A, which is 59% of the Isc current.I know that under load, panels settle at a lower voltage than the Voc. And I know that MPPT chargers seek the optimal Pmp at the Vmp. But what voltage should I expect a linear regulator to settle at?I suppose that I look at the current on the Y axis for where it intersects with the red line. So in the chart, since my load is 59% of Isc, I'd expect to find the dot slightly below and to the right of the Vmp, but still not as far right on the X axis as Voc.Did I get this right?

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 01:31:51 pm
content: And if the load's resistance were lowered and the load current demand becomes 2A, would the voltage go to zero?

author: fourtytwo42
author_rank: Super Contributor
timestamp: January 20, 2025, 08:58:42 pm
content: You are missing something important and that is your chart is drawn at just one illumination level, probably 1000W/M2 that is the nominal test value.As the illumination falls so does the current/voltage curver, something like the enclosed pdfThis means your regulator can cause voltage collapse if it tries to draw to much current, so the panel/regulator latch up into a no output state.test_260Wpnl.pdf(104.17 kB - downloaded 27 times.)

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 09:06:14 pm
content: Uh oh. What's the solution--don't use a regulator? If illumination continues to fall, won't the regulator reset?If I shouldn't use a regulator, what do people use for small solar projects? It charges a LiFePo4 cell at 3.5V and that runs an ESP32.

author: fourfathom
author_rank: Super Contributor
timestamp: January 20, 2025, 09:22:45 pm
content: To think about this, just imagine the regulator as a simple variable resistor, connected to ground. Adjust it so the current is 1A and see where that point is on the I/V curve.Forget about regulator latchup or reset -- that depends on the regulator design. And remember that you can't draw more current than Isc.There are charge-controller ICs designed for Li batteries, and they can be fed by an appropriate solar panel.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 09:28:42 pm
content: Quote from: fourfathom on January 20, 2025, 09:22:45 pmI would think the sun going down would reset the regulator, yes? And could you explain what you mean that it depends on the regulator design?

author: eTobey
author_rank: Super Contributor
timestamp: January 20, 2025, 09:58:09 pm
content: Buy a panel, and try it out. Your thinking is right. But as the other said, these are optimal values of the panel. Now take into consideration the following:- Daytime is varying- Height of the sun depends on the location.- High temperatures lower the output of the solarcell- Clouds and trees are bad for solar energy- Solarcells might be worse than advertised- The cells have also to be aligned to the sun for good energy production.- There are two types (among other exotics) of solarcells Mono and PolycrystalinWhat do you want to power? For how long? How often?

author: Peabody
author_rank: Super Contributor
timestamp: January 20, 2025, 10:06:54 pm
content: The battery and its charger are what make this work. With a linear charger, the solar panel in bright sun will charge the battery, and if there's a load sharing circuit it will also power the load directly. With less sun, the panel will contribute as much current as it can at slightly above the battery voltage per its I/V curve - as long as that voltage is below the OC voltage at the present illumination. Panel voltage will not collapse because there's no current draw below the battery voltage. Of course if the panel can't produce enough current at the battery voltage to power the load, then the battery will have to supply what's needed.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 10:07:30 pm
content: Quote from: fourfathom on January 20, 2025, 09:22:45 pmYes, BUT. The battery charger IC I want to use (TI BQ25176JDSGR) is designed for LiFePo4 and specifically calls out solar in the datasheet. BUT. It cannot handle both the input voltage and a high charge current (700mA) without going over the operating temperature in my worst-case scenario (55C ambient, including enclosure solar gain). If a panel goes above 5.7V it lets out the magic smoke.So I had planned to regulate the voltage down to 4.4V before feeding the charger IC, which outputs 3.5V. The worst possible voltage drop of the charger IC is 900mV + 3.5V it comes to 4.4V.The device can also be optionally powered by USB, so I can't have a large voltage drop in the regulator, either; And I can't expect a perfect USB supply or cable, so I want a little bit of forgiveness if less than 5V comes in.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 10:10:29 pm
content: Quote from: eTobey on January 20, 2025, 09:58:09 pmWorsening conditions would actually work in my favor. Trying not to go over operating temperature for the battery charger IC, which is aTI BQ25176JDSGRconnected to a LiFePo4 cell at 3.5V max (and yes there is a load), hence I thought to regulate the input to 4.4V. Connecting a 6.5V panel and trying to charge the battery at 700mA would let out magic smoke.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 20, 2025, 10:43:55 pm
content: I think I'm understanding the chart correctly now; The only thing that's still not clear is what happens to the regulator when the panel's voltage drops from nightfall; Will it lock up and never restart without disconnecting the panel, or will it reset because the panel voltage is too low? I'm thinking the latter, since it is essentially the same as disconnecting the panel.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 21, 2025, 12:49:36 am
content: I think I'll be okay. Theregulatoris aconstant-current typethat has no latching conditions for undervoltage, that I could see on the datasheet. Thebattery chargerhas an undervoltage condition, but it recovers when the voltage rises again.So if there is a cloud, but the charger is still demanding 700mA, the regulator output voltage will drop and the charger IC will go into sleep mode when it drops below Vin_lowv (3.05-3.15V) and stay there until the input voltage rises to Vout (4.4V) + Vsleepz (30-80mV).

author: MrAl
author_rank: Super Contributor
timestamp: January 21, 2025, 11:57:50 am
content: Quote from: cdevidal on January 20, 2025, 03:01:25 amHi,See attachment.This is an approximation to what happens when the insolation level drops.As shown there are two operating currents, i1 and i2.If you are operating with current i1, when the insolation drops to 1/2 of the full value (diagram on the right) you can see you still get power output.If you are operating with current i2, when the insolation drops to 1/2 of the full value (diagram on the right) you can see you do not get any output power.The diagram on the right has a short circuit current about 1/2 of that with full sunlight.The voltage will change a bit too (not shown) and if the voltage happens to drop to a level less than the regulator output, the regulator will start to put out less voltage and less current. Most regulators require at least some overhead voltage too, so that means that if the voltage drops to a level that is equal to the output voltage plus the overhead voltage, the output voltage will drop. For example, if the regulator puts out 10v and requires 2v overhead, the voltage on the output will start to drop when the solar panel voltage drops to 12v (10v+2v).The Max Power Point does not make much difference if you do not intend to operate at the max power point, or you do not have a load that wants to operate at the max power point power level.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 21, 2025, 12:34:50 pm
content: Thank you!!!

author: Peabody
author_rank: Super Contributor
timestamp: January 21, 2025, 03:22:12 pm
content: @cdevidal, would you consider posting your schematic?Also, Adafruit has several solar-powered chargers:https://www.adafruit.com/category/577including one using the BQ25185, which has flexibilty as to the type of lithium battery. Their documentation is pretty good, including schematics. So you might find something useful there. In particular, the Adafruit charger chip choices appear to include output voltage regulation, so whatever the power source is, the output voltage is always under 5V. That would allow for a higher voltage panel as input.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 21, 2025, 05:06:50 pm
content: Quote from: Peabody on January 21, 2025, 03:22:12 pmAttached. Current flows left-to-right, top-to-bottom. Hope it's not too embarrassing, I'm still new at thisQuote from: Peabody on January 21, 2025, 03:22:12 pmThat's interesting, and I didn't know it existed. However, it appears they're not designing for greater than 25°C ambient temps so I don't think their design would help me. It's very similar to the reference design in the datasheet. If I usedthe charger IC I choseinstead of the one they're using, with a panel that is outputting 6.16V and the battery is charging at 3.5V@700mA, the charger IC would be dissipating 1.862W. My charger IC has a junction-to-ambient thermal resistance of 75.2°C/W so that's 140.02°Cbeforeadding the worst-case ambient temperature of 55°C. Operating temperature max is 125°C so it'd be exceeding that on even acoldday.Thecharger IC they usecan adjust the current down when Vindpm drops so that charging is not interrupted, which is such a good feature that I might switch to it; But that doesn't solve the thermal problem trying to take in max current on a bright day. And I'd still need a regulator on the output because it outputs up to 4.5V.I believe I do need an input regulator no matter what.

author: Peabody
author_rank: Super Contributor
timestamp: January 21, 2025, 07:00:05 pm
content: I don't know what the solution would be for the temperature issues. But most of these chargers have internal thermal limiting that reduce charging current when they get too hot. And I guess there's the possibility of a PWM charger chip for LiFePo4, if such exists, although that would be a more complicated circuit. Aside from that, you could reduce the charging current, or use a 5V panel.But I have two possible issues with your circuit. If I understand correctly, you have the ESP32's load being powered in parallel with the battery. If that load current is always greater than 10% of the maximum CC charging current, then the charger will never terminate charging so long as the sun is bright, and that may not be good for the battery. The second problem is that even if charging terminates correctly, when that happens the charger will shut down. And then the battery will have to take over supplying the load even though the sun is still shining brightly.The Adafruit chip has a built-in "power path" circuit which always powers the load from the panel directly if possible, so the termination algorithm is not confused, and the battery never has to take over when the sun is bright.I think I agree that you need a 3.3V regulator ahead of the ESP32. I think a lithium-ion battery would be more suited for that since it would provide more headroom, but I assume you have a reason for using the LiFePO4. I don't know about using a regulator ahead of the charger. Perhaps a carefully selected high-wattage resistor would work as well - just something to dissipate most of the heat.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 21, 2025, 10:39:49 pm
content: Quote from: Peabody on January 21, 2025, 07:00:05 pmOh you are correct. I can certainly charge at a lower rate, or choose a 5V panel--which are not as common. (I fear the ones that exist are merely mislabelled 6V or they have a regulator.) But with an input regulator, I shouldn't need to.Quote from: Peabody on January 21, 2025, 07:00:05 pmI do. That was recommended by the datasheet.Quote from: Peabody on January 21, 2025, 07:00:05 pmIt's well under the maximum CC charge current. The microcontroller runs at about 350mA for one second every five seconds. Average rate would be about 70mA of the 700mA charge current. From the datasheet: "The system load can be placed in parallel with the battery, as long as the average system load does not prevent the battery from charging fully within the safety timer duration. When the system load is placed in parallel with the battery, the charge current is shared between the system and the battery."The output current is up to 700mA when bright, so even the 350mA spike should not prevent it from charging. Pre-charge safety timer is 30 mins, during which very little current goes to the battery.Edit:Oh. And low current goes to the load, as well, so the load would depend upon a low battery. I do check the battery voltage on boot and increase the sleep time when the voltage is too low, so it's probably not a concern. The Adafruit choice is a better chip, and I might switch. But not for this.End editThe fast-charge safety timer is 10 hours, and the places that get that much sun, I don't yet have plans to sell intoQuote from: Peabody on January 21, 2025, 07:00:05 pmCould you explain why you think it's a problem for the battery to take the load? The chip I chose, yes as you said, if it's done charging it shuts down and the battery supplies the load. I don't understand why that might be a problem, sounds like it goes as I would expect.Quote from: Peabody on January 21, 2025, 07:00:05 pmIronically, at first I was happy when I learned about LiFePo4 when I saw I could rid of a regulator ahead of the ESP32--and in most situations, it should work fine. But when I saw the longevity and increased safety, for about the same price as Li-Ion, I decided I liked that chemistry better.I definitely would need an output regulator with Adafruit's choice of chip; With the one I chose, I need an input regulator. Cannot avoid using one with either chip, and both would be about the same size and price.Quote from: Peabody on January 21, 2025, 07:00:05 pmIt's a good idea, but remember too that another goal is to also be able to use USB power, and a resistor chosen for solar voltages would drop USB-supplied voltage too far.Could you explain why you think it's a problem for the battery to take the load when it's done charging?

author: Psi
author_rank: Super Contributor
timestamp: January 21, 2025, 10:54:50 pm
content: Quote from: Peabody on January 21, 2025, 07:00:05 pmDepending on IC you can sometimes put a NTC resistor in circuit somewhere to reduce the load when a temp is reached. Sometimes by simply reducing the output voltage.Or you can put a resistor in series with the output and a PFET across it to detect when resistor Vdrop (current) reaches a limit and start ramping down the output voltage. Basically current regulation.Usually the chips on board thermal shutdown is a last resort at like 105deg C and is not good for the chip's long term survival.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 21, 2025, 10:56:26 pm
content: Quote from: Psi on January 21, 2025, 10:54:50 pmWith an input regulator, I shouldn't need to do any of that--or am I missing something?

author: Peabody
author_rank: Super Contributor
timestamp: January 22, 2025, 03:47:04 am
content: The problem with the battery taking over when it shouldn't need to is that the battery will repeatedly cycle between fully charged and about 0.1V below that, thereby using up the number of charge cycles the battery can provide. So in theory the battery won't last as long as it would if it never had to cycle that way. I don't know if that's a relevant consideration in your situation, but generally, putting the battery through needless charge/discharge cycles reduces its life.My issue about the charger providing load current isn't a problem with charging. It's a problem with termination. The charger terminates after being in CV when charging current falls below 10% of the full CC current. So in your case, that's 70mA. If load current is always 70mA or greater, termination may never take place, which could be very bad for the battery (i.e. - overcharging).In any case, both of these issues disappear with the Adafruit chip. But I'm not sure it could handle your thermal issues. You might have to have both regulators.The resistor is just nice because it drops more voltage, and dissipates more heat, as the current goes up. But it's a bit crude as a solution. But you could have separate resistors for the two sources.

author: MrAl
author_rank: Super Contributor
timestamp: January 22, 2025, 10:38:42 am
content: Quote from: Peabody on January 22, 2025, 03:47:04 amHi there,Sorry to say, but charge cycles do not work like that. Just because you can get say 500 charge cycles with a certain battery that does not mean that if you discharge it by only 0.1v each time that you can only charge it back up to max 500 times. The charge cycles would be for the entire discharge of the cell.So for example if you have a cell charged up to 4.20 volts and you discharge it to 4.10 volts, you should get a lot more than 500 charge cycles. If on the other hand you discharge it down to 2.5 volts, then you are most likely only going to see around 500 cycles before the capacity has dropped to a level that starts to becomes problematic.For some cells like Li-ion it is not wise to charge all the way up though if you want to maximize the life of the cell. The typical levels are 80 percent and 85 percent. You lose some capacity but the battery lasts a lot longer. This functionality is now built into some cell phones but it's provided as an option.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 22, 2025, 02:20:26 pm
content: Quote from: Peabody on January 22, 2025, 03:47:04 amI think I'll be okay with that concern. I'm charging a 6,000mAh cell at 3.5V with protection kicking in at 2.5V, soone percent of SoCis 11.7mV. On the worst-case ofthe longest desert summer day of 7.47 hours of full charging, when discharging at 70mAh average, for a total of 523mA lost during the 7.47 hours. With a 6,000mAh cell and 1000mV to play with, every 6mA consumed drops 1mV, so at the end of the charge cycle it'd lose 87mV. The recharge cycle begins at 100mV, so that cycle wouldn't happen even once.My greater concern with cycle life is the overnight discharge, but even in the worst case it's enough for 46 years' worth of cycles--and by that time, it's somebody else's problemQuote from: Peabody on January 22, 2025, 03:47:04 amYou have a point there. It's not a smooth 70mA but 350mA for one second, deep sleep at 250uA for five seconds. It should terminate given enough sleep time, but I don't know how much time is needed, and I can't find anything in the datasheet, so this is an unknown concern.I asked TI about itand they're very responsive on their user forums so I should get an answer soon.Quote from: Peabody on January 22, 2025, 03:47:04 amTwo different source paths adds another connector, as well. This is for a production device, so cost and user-friendliness are constraints. Cost difference is lower for a regulator, and also reduces accidents so that the user doesn't plug into the wrong socket. I could use different connector types if I still wanted that, but panels with USB-C cables attached (for security cameras) are off-the-shelf, abundant, and cheap.

author: Peabody
author_rank: Super Contributor
timestamp: January 22, 2025, 04:16:23 pm
content: I understand that 0.1V is not a "full" charge cycle, but I think the fact remains that a battery that is repeatedly discharged and recharged by any amount is not likely to last as long as a battery that doesn't experience that. And I would also suggest that most production designs you see these days will incorporate a power path (aka load sharing) circuit so no load current ever passes through the charger. With solar, such a circuit made up of discretes can be complicated, so if the charger IC includes it, that can be a significant benefit.Can we back up and look again at the choice of the LiFePo4? You're going to be powering an ESP32 at 3.3V with a battery that tops out at 3.5V. Even with a direct feed, with no regulator, that seems to not provide very much allowance for battery discharge. Do you have a feel for the percentage discharge state when the ESP32 stops working? If you used a lithium ion or polymer cell, pretty much the entire charge would be used up before the ESP32 craps out. Of course you would need a regulator, but I just wonder if you could get away with using a smaller battery and still have the same or better backup performance. Also, charging at 4.2V would reduce the thermal problems because the drop would be smaller. Well, just wondering about that.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 22, 2025, 04:54:05 pm
content: By the way, I appreciate what you're trying to do, and I'm not trying to be obstinate on this. It's just that switching chips means a not-inconsequential change requiring several hours of rework. If I have to, I have to. But I'd rather not if I don't have to, and I'm not yet convinced I have to.Quote from: Peabody on January 22, 2025, 04:16:23 pmYou would of course be correct if it re-entered a recharge cycle. But again, I don't believe it'll ever re-enter a recharge cycle. See my last comment above, where I calculated from the worst case.Quote from: Peabody on January 22, 2025, 04:16:23 pmI'm sure many production designs do that, yet I noticed that both chips were made by TI and released at exactly the same time. One with a power path, one without. Both specifically list solar in their datasheets or sales documentation. This implies to me that TI thinks there is more than one way to skin this cat.Quote from: Peabody on January 22, 2025, 04:16:23 pmAccording to the datasheet,they handle down to 3.0V. There have been reports of them being stable at lower voltages, but let's suppose it really is 3.0V. That's10% SoC for a LiFePo4, so I am able to utilize most of the capacity. It's not a good idea to go down to 0% SoC, anyway.Quote from: Peabody on January 22, 2025, 04:16:23 pmWould be about the same, because I am regulating to 4.4V for the other chip due to the voltage drop within the IC.

author: Peabody
author_rank: Super Contributor
timestamp: January 22, 2025, 05:06:56 pm
content: As long as you've thought it through, I have no problem. In any case, I'm sure you will be testing it thoroughly at some point to make sure it all works as expected.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 22, 2025, 05:20:25 pm
content: I really appreciate you pushing back, because I've never done this before. Making a product for sale has been the hardest thing I've ever done--harder even than having childrenAnd our conversation can be something other newbies can learn from.

author: cdevidal
author_rank: Regular Contributor
timestamp: January 22, 2025, 05:55:42 pm
content: TIresponded: "The deglitch timer for termination is 30ms. When the charge current decreases to 70mA for 30ms it will set charge complete. This should work with your 1 second wake and 5 second sleep."

author: Peabody
author_rank: Super Contributor
timestamp: January 22, 2025, 10:26:19 pm
content: That sounds very good. Well let us know how it all turns out.