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How to get constant 5.2V from ~1-7 Volts? Answered

I have multiple solar panels (3V, 6V). I want to build a charger for bicycle, to keep my battery powered lamps (DIY) battery alive. Currently, I use my DIY lamp with a USB Power bank (5,2V , approx. 50k mAh), but my lamp contains 39 LEDs, and they are ultra-bright LEDs, so they drains the battery fast. So I thought if I connect some solar panels to it, I can charge power bank during daytime.

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-max-

3 years ago

I kinda doubt your power bank is 50 AH, or 50,000 mAH. (kmAH is not a measurement, the 'k' and 'm' cancel out.) That would imply that it can output 260WH, or about 100 watts for 2.6 hours, that is a lot of energy, even if you do take losses into account. 2.6 hours is a reasonably long time and 100W LEDs are insanely bright. I think the powerbank is more like 5AH, or 5,000mAH.

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Sp1k33-max-

Answer 3 years ago

On the side of the power bank, or Chinese friends wrote the following informations: Capacity: 50.000 mAh, 3,7V/48Wh (MAX). Input: DC 5V=1A(MAX) Output1/2/3 DC 5V=1A(MAX)

BTW I don't think that this power bank could hold up that much current, its not that big (14x6x2 cm) or heavy (around 250 gramms) :D

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-max-Sp1k33

Answer 3 years ago

I made a mistake on the last calculation, often the advertised capacity is just for the 3.7v lithium ion cell. It would be 185WH, still a lot, and about 148WH after 3.7v --->5.2v conversion, which is still way too high a number for me to believe.

48WH ideally will run a 10W LED for 4.1 hours, which is a long time. The CREE XM-L2 LEDs can be up to 170 lumens per watt, but average around 100 lumens per watt at 10W, or about 1000 lumens. I recommend using this LED or some other high efficiency LED (CREE is one of the best IMO)

One thing you could do is hack the charger to get access to the lithium battery inside to connect a proper LED driver directly to the cell(s). Most likely the charger uses a few smaller lithium ion cells in parallel and will have a voltage range of 3.1 to 4.2 volts. There are many LED drivers that are designed for flashlights to operate off of a single 18650 lithium cell, get one designed to power the XM-L LED or equivalent and this should work. You can also pick up the 3400mAH panasonic cells pretty cheaply nowadays, those are (or at least used to be) some of the best 18650s money can buy.

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Sp1k33-max-

Answer 3 years ago

Actually, I use the LEDs with 5.2V. I know they are not designed to hold that much voltage, but they can (even for hours)

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-max-Sp1k33

Answer 3 years ago

5.2V across an LED die will instantly burn it out. Your LEDs must have a resistor in series with them, and that is a lot of lost power. The only way an LED can operate at higher voltage is if it is a multi die LED, with multiple LEDs wired in series inside of the actual LED package. Those are generally special SMD high intensity LEDs designed for operation with AC and a capacitive dropper.

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Sp1k33-max-

Answer 3 years ago

You want me to take a pic of it? :D pure 5.2v, no resistors, no pararell connections. Previously some LEDs died of course but these are not (2 dollars / 100 pcs from ebay)

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-max-Sp1k33

Answer 3 years ago

Sure! If you are using thin wires for power the wires may be dropping the excessive voltage. I can power my 35W 4000 lumen CREE XHP70 LED which is supposed to be driven with 6V off of a 2S LiPo which when fully ccharge can output 8.4V. The difference in voltage is what is lost when 5A of current flows through the LED.

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Sp1k33Sp1k33

Answer 3 years ago

And most of the time only 30 LEDs are working, because 9 LEDs are for visibility, if I don't want too much brightness (like in the city area)

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Sp1k33-max-

Answer 3 years ago

BTW my future plans are: buying CREE LEDs, and build a battery pack for them.

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-max-Sp1k33

Answer 3 years ago

39, 5mm LEDs should take about 3.7W of power if each LED draws 30mA and has a 3.2V drop. Or if you are using a simple resistor to drop the voltage from 5.2v to that 3.2 volts, then the power used in total will be 0.03A*5.2V*39 = 6W. Your 48WH battery, after 20% losses should power that for 6.3 hours. But like I said, I recommend using better components to get better efficiency, and WAY more light output than ~100 lumens w/ 39 crappy 5mm LEDs.

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-max-

3 years ago

buck converter can be used to step down voltage with minimal losses, and boost converters can be used to step up voltage with minimal losses. YOu can also get buck-boost converter that do both. Keep in mind these voltage conversions can at the cost of current, and power loss. By stepping up the voltage, you reduce the current that can be delivered compared to the current feeding the boost converter.

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Sp1k33-max-

Answer 3 years ago

Sadly I can't charge USB power bank with more or less then 5.2V. Thx for your help, I'll look after these converters.