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Like so many else trying to charge a power bank from a solar panel:

I've got a solar panel claiming 18V and 280 mA, so 5W and a 2800 mAh power bank expecting USB in so 5V 1A. The simple "circuit" I've made is simply a 5V, 2.5A step-down voltage regulator connected to the power bank and the panel.

My problem is that the voltage drops to 0.8-1V when in direct sunlight and connected to the power bank. I don't know what charging circuit the bank has (a cheap super market bought one) has but it seems like it's trying to take as much as it can up to 1A and that brings the panel on its knees. The 1V I get is too low to make it charge.

Is there something I can put in between the panel and the bank to either limit the maximum current the bank gets or something charging up and letting out when it's enough to power the bank?

I know the regulator is completely over dimensioned but I had one at hand I assume it won't be the problem. I'm also hoping to get my hands on a more suitable charging IC and battery but before that I'd like to get this prototype working.

*edit: What I know about the power bank is that it's got USB in and USB out. It says 3.7V and 2800 mAh so I guess it's a single cell li-ion in there. I don't know the tolerance of the bank, I've only got access to 5V and the panel. So I know it charges at 5V and doesn't at 1V. I'm getting a lab power supply this weekend so then I can figure that out.

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    \$\begingroup\$ "Maximum Power Point Tracking" \$\endgroup\$ – Ignacio Vazquez-Abrams May 30 '16 at 18:41
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    \$\begingroup\$ What you are trying to do is not so easy and depends on the nature of the power bank. When the maximum power available is enough, you want to provide 5V to the power bank by way of DC-DC converter. When 5V is not available, you want to provide as much current as possible while maintaining the panel at Vmpp. What is unknown is, what does the power bank do when Vin starts to fall below 5V? Many chargers automatically back off when this happens (instead of crashing Vin down to zero). \$\endgroup\$ – mkeith May 30 '16 at 18:56
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    \$\begingroup\$ \$18\,\text V \cdot 0.28\,\text A= 5.04\,\text W\$ power from the panel. \$5\,\text V \cdot 1\,\text A=5\,\text W\$ power consumption. This means that your power converter would really have to have 99.2% efficiency if your power bank would really "need" the 1A current. Such power converters do not exist. However, luckily, it's not likely that your power bank will actually need that much current – it will probably work with less, but slower (hopefully, as @mkeith says, no-one knows). \$\endgroup\$ – Marcus Müller May 30 '16 at 19:04
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    \$\begingroup\$ The real problem is that if your panel is rated 5W, then that's what it gives at full brightness – generally, you'll get a lot less. I wouldn't expect to get more than 50% of that during any time of the day, unless you have perfect summer at noon. \$\endgroup\$ – Marcus Müller May 30 '16 at 19:05
  • \$\begingroup\$ I'm fine with slow charging, I guess what I'd like is something that keeps the voltage at a steady 4-5V (or the limit I'm hoping to find out in a couple of days) and keeps the current at bay if the bank gets too greedy. Regarding the MPPT @IgnacioVazquez-Abrams, could LT3652 be an alternative then? \$\endgroup\$ – Jon May 30 '16 at 19:11
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It seems at a fundamental level, the panel puts out less power than the thing you are trying to power wants. That means you can't "convert" your way out of this. You can convert some combination of volts and amps to a different one, but the volts x amps product of the output can't ever be higher than the input volts x amps product.

The first thing you need is proper specs for the thing you are trying to power. Apparently it expects around 5 V. So give it 5 V and measure the current. You can't design a circuit to meet a current requirement if you don't know what that current requirement is. That really should have been obvious.

It sounds like your switcher can deliver sufficient output current when given sufficient input voltage. The problem is that your panel can't supply the necessary power, so the switcher keeps trying to draw more current from the panel. That causes the panel voltage to drop, so the switcher draws even more current, which causes the panel voltage to drop even more, etc. Rather quickly, the panel voltage collapses. It then produces even less power than it could if managed properly.

So what to do? One possibility is to add another panel in parallel. With enough input current capability, the system will at least work in steady state in full sun. However, when the insolation goes away, the voltage will collapse again, and may not be able to recover when the insolation returns.

What you need is a circuit that disables the switcher altogether when the panel voltage is too low. If the switcher doesn't have a shutdown input, get one that does. That could be done with a external transistor, but at your apparent level it is better to drive a shutdown input that is meant for the purpose. Many buck switchers have shutdown (or enable) inputs, so this is not a onerous requirement.

Derive the shutdown signal from a comparator with hysteresis. Find what a reasonable maximum power voltage for the panel is under a bit less than optimum illumination, then set the comparator off threshold a bit below that. Set the compator on threshold a bit below the open-circuit output voltage at medium illumination.

Now connect a big capacitor across the panel. This should be 10s of mF at least, rated for 25 V or more.

What will happen now is that the panel will charge up the capacitor to the comparator on threshold. The buck switcher then makes 5 V and your device charges. The current drawn by the buck switcher will exceed what the panel is producing. That capacitor supplies the remaining current, but discharges in the process. After some time, the capacitor voltage drops to the comparator off threshold. The buck switcher turns off, stops charging the device, and stops drawing input current. The solar panel current now charges the capacitor, and the cycle repeats.

The larger the capacitor, the longer the device will be charged at a time. Depending on how the charger in the device works, it may need some minimum on time to do any useful charging. A larger capacitor will lengthen that on time. It will also lengthen the off time, but that shouldn't bother the device.

Overall, you still aren't getting more power out than in. However, the output power is now in burst of high power with gaps in between. The device charges during those bursts of high power in the way it is intended to work. Obviously the overall charging will take longer, but that's again due to basic physics limited by the available input power.

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First, you need to know more about your solar panel characteristics. 18V and 280 mA doesn't tell you much, is it open circuit voltage and short circuit current or is it values at maximum operating power (which is likely to be) ?

Usually the datasheet will indicate Voc, Isc, Mpp, V@Mpp and I@Mpp, you can calculate MPP (maximum power point) by multiplying V@Mpp and I@Mpp but not Isc and Voc.

Furthermore, you need to regulate the current you're drawing from the current as a current source, if you draw too much current then your voltage would drop to Zero.

So if your battery bank is charging at 1A and your panel Isc value is 280mA, this means under 1kW/m2 of sunlight you will get 0V, so this isn't going to work.

To charge a Li-Ion battery with a solar panel you either need a battery pack with an oversized panel and some sort of MPPT regulator, or a charger in which you can directly set the drawn current from the source.

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What you want is some kind of way to limit the current. There are buck converters which have a potentiometer to do this, which may or may no be sufficient for your purpose, as this is manual labor. If you limit the current, the solar panel voltage will not drop to a unusable level.

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If your 18V and 280mA are respectively Voc (open circuit voltage the solar cell can produce) and Isc (short circuit current), then that means the cell can produce 280mA of current at an output voltage of 0V and when no current is being drawn from the cell there would be 18V at the output. In between these points, the Current-Voltage characteristic (general example of a solar cell) looks like this: enter image description here
You cannot use the solar cell on its own in an operating point 'outside' of the curve.

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The power bank is usually a bank of individual LiIon or LiPo cells in series (or parallel and series depending on the current they output). They have a specific charging profile with specific current and voltage values needed for charging. Without this specific profile in your charge circuitry you could damage the battery. There are specific chips available for this such as this one from Adafruit.

What you need is something like this between the power bank and the solar panel. Even though the solar panel outputs 5W, that maybe a best case estimate and you should do your own tests in varying amounts of sunlight. While the panel may give 18V, you should see if the charger circuitry can accept 18V. If you are looking for sophisticated technology to track and extract the maximum power the panel can output, TI sells specialized chips for charging from solar panels such as this one.

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    \$\begingroup\$ Pretty sure the OP already has a power bank that expects to receive power from USB. He is not trying to design an new power bank. He is trying to interface an existing solar panel to an existing power bank. \$\endgroup\$ – mkeith May 30 '16 at 18:59
  • \$\begingroup\$ The adafruit product mentioned here is a ready-made one which accepts a solar panel on end and a battery on the other. I thought to list a chip from TI just in case he wanted to design one on his own. I know I would. \$\endgroup\$ – electrophile May 30 '16 at 19:02
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    \$\begingroup\$ So all the OP needs to do is buy a new solar panel, strip the battery out of his power bank, buy the adafruit IC... No offense, but it doesn't seem as though you have helped the OP very much. \$\endgroup\$ – mkeith May 30 '16 at 19:32

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