Simple low power solar battery charging circuit - Page 1
Forum section: electronics / projects

author: schratterulrich
author_rank: Regular Contributor
timestamp: December 14, 2021, 02:13:06 pm
content: I needed to add a solar panel with 5Wp to a 12 V 2000 mAh battery. I found the following requirements:- Small and simple- Limit the voltage to 13,8 V- When the solar cell does not deliver power, the reverse current from the battery should be < 100 µA- The circuit itself should have low quiescent current- The circuit should be able to handle 500 mA from the solar panelI came up with a solution using a simple LDO linear regulator LT1129 as it states “No Reverse Output Current” in the datasheet. But this is only valid up to 6 V, so I have added an additional diode at the output to limit reverse current.The size of the board is 35 mm x 25 mm. I approximated it to be able to handle 1 W of power dissipation. I did not populate electrolytic capacitors on the board.When the battery is not fully charged and so the output is lower than 13,8 V, the power dissipated by the regulator is, in first order approximation, simply the dropout voltage times the charge current. For 500 mA this is about 170 mW. No problem.But when the voltage of the charging battery has reached 13,8 V the input may go higher, while the output stays constant. Then, with a strong enough panel, the dissipated power could easily exceed 1 W. Here I have added an NTC glued on the SO8 package of the LDO regulator. When the temperature reaches 60°C the diode D2 starts to conduct and the output voltage is lowered by up to 2 V at 100 °C. A lower output voltage will decrease the charge current until a thermal steady state is reached.This is a thermal image when dissipating about 0,8 W. The charge current has regulated to a value that heats up the circuit by a constant value.It is a simple solution that works for my purpose quite nicely. I need it for powering beehive scales.I thought it would be interesting to share because it is smaller than available solutions. Most of available solutions are switcher circuits, which are not needed for such low power panels while others are discrete circuits needing more space and have higher quiescent current.The measured efficiency is about 70 % at 1 mA input current and about 95 % above 10 mA input current.

author: Scottie
author_rank: Contributor
timestamp: December 14, 2021, 04:05:21 pm
content: That's a nice clean and simple solution. You may consider referencing your feedback divider after D1 so you can more precisely define the charging float voltage. Of course this would add 10 uA of always on quiescent current.I love the thermal foldback design.

author: Peabody
author_rank: Super Contributor
timestamp: December 14, 2021, 06:30:32 pm
content: A couple questions.I did not find in the datasheet the 6V limit on reverse voltage. Where is it?When the battery voltage reaches 13.8V, what happens to the charging current? Doesn't it fall off automatically? I have no experience with lead-acid charging, but I just would have thought that if the regulator is producing 13.8V and the battery is at 13.8V, not a lot of current is going to flow, so it shouldn't matter much what the input/output voltage differential is. But of course your thermal image suggests I'm wrong about that.Ok, I lied. A third question. Doesn't this all depend on the panels you select? Couldn't you use panels with a maximum open circuit voltage of around 14.5V or 15V and not bother with the thermal circuit? How do you decide what panels to use for this?

author: schratterulrich
author_rank: Regular Contributor
timestamp: December 14, 2021, 08:56:44 pm
content: I have concluded the 6V limit from Figure 3:The charge current will fall off automatically but not instantly.The simulation, with the battery replaced by a capacitor, shows that below 13,8 V the current is limited by the panel only. It is in CC (constant current mode) for 120 s.Top = Power dissipationMiddle = Charge currentBottom = Voltage of panel / Voltage of batteryIf the output voltage of the regulator is reached, the battery is still charging with the same amount of current but starts to fall off. It is in CV mode (constant voltage mode). The open circuit voltage of the battery is still not 13,8 V at that moment.If the panel can provide a higher voltage than the charging current of the battery at 13,8 V, the input voltage of the regulator will raise.This is the moment when the power dissipation would peak at its maximum.The thermal image was made with a constant voltage load combined with a few ohms in series at the output.I think if the panel has only 15V open circuit voltage, it would already need a very high light intensity to reach 13,8V. Also the maximum power point would be much lower than 13,8 V. But I did not go into depth on this topic.I would choose the panel according to the desired or maximum charge current of the battery as it is not limited by the circuit. The maximum of the LT1129 is 700 mA I think.

author: Peabody
author_rank: Super Contributor
timestamp: December 14, 2021, 10:00:36 pm
content: Quote from: schratterulrich on December 14, 2021, 08:56:44 pmThis is what I'm having trouble with. If we assume that the physical size of the panel is the same in all cases, then the power they can generate should be the same under any given level of illumination. Your panel is effectively 22V OC, 500mA SC. The alternate same size panel would be a 15V OC with much higher current (would it necessarily be 733mA?). I don't know where 13.8V would fall on their respective curves. I haven't found any formal discussion of this issue. Everybody seems to assume that you want the highest voltage panel that works so it will still produce something useful in partial sunlight. But I guess I'd like to see that proved in some way, because it's not obvious to me since it would certainly produce less current in full sunlight.Maybe this is just something people come to with experience, but I would just like to find some kind of formal analysis of the question if one exists.I understand that you are depending on the maximum current of the panel as the limit the regulator will have to deal with. But another regulator might work with higher current. And a 15V panel might make all the thermal limiting unnecessary.And I don't think MPP is relevant here. It would be if you were using a switching regulator, but this is a linear regulator, so it's current that matters, not power. The panel will provide as much current as it can at the lowest voltage that still works (battery voltage plus dropout), and anything beyond that will just be dissipated as heat.

author: schratterulrich
author_rank: Regular Contributor
timestamp: December 15, 2021, 06:47:18 am
content: When I started looking for PV modules, I also wondered why the voltage is so high for 12 V modules. But it seems to be a standard to use 36 cells for charging 12 V batteries.I did a little googling and found that there may be an alternative. Search for "self regulating solar panel". They use 30 cells to get a Voc of 15 V. However, this does not seem to suceed or not work.MPP may not be directly relevant in my application but if it is assumed to be slightly higher than the battery voltage, there wouldn't be to much loss in efficiency as the current stays nearly constant below the MPP point.On the other side if the MPP point would be slightly lower than the battery voltage the loss in efficiency could be quiet high as the current fells of rapidly above that point.