Charging Lithium - Ion Battery With Solar Cell

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About: I study Teaching physics and geography, I enjoy with laws of nature :D Be happy and fun with science.

This is project about charging Lithium - Ion battery with sollar cell.

* some correction I make to improve charging during winter.

** solar cell should be 6 V and current (or power) can be variable, like 500 mAh or 1Ah.

*** diode to protect TP4056 from reverse current should have low drop voltage ("drop out"). I use bad, which take 0,5-0,6 V, which is a lot. You can use Schottky diode, which take only 0,1 - 0,2 V.

Step 1: Material (links Are Affilliate)

1 x Solar cell 6V

Link:6V 1 W

Link: (more cells with different watts)

Link: (more for choosing)

1 x Li - Ion charger board TP4056 (choose board with 4 output - 2 for battery, 2 for connecting device)

Link: (5 pieces, cca 0.20 $ / piece)

Link: (1 piece, 0.29 $ / piece)

1 x Schottky diode (better, 0,1 - 0,2 voltage drop) or 1N4148 (worse, 0,5 - 0,6 voltage drop)

Link:(set of diodes)

Link:(1N4148)

1 x Lithium - Ion battery (18650), I buy 1 poor, you can choose better with capacity around 2000 mAh - 3000 mAh,

Link:Lithium - ion battery

1 x Lithium - Ion battery holder

Link:battery holder

1 x cables, I use internet cables with 6 wires inside or awg 22 wire kit

Links:

quality: set of AWG 22 cables

ethernet cable: ethernet cable (need to cut off 6 wires)

1 x solder tools (station, tin, rosin etc.)

Step 2: Right Solar Cell

* solar cell should be maximum 6V, because TP4056 has maximum input 6V. It is better then 5V.

* current from solar cell (or power) can be variable, because TP4056 "eat" as much as it need. So you can choose 500 mAh solar cell or 1 Ah solar cell.

For Li - Ion battery I choose solar cell with 5V and 160 mA. For choosing solar cell, you must choose:

1. voltage of solar cell 1.5 x voltage of battery, so 3.7V to 4.2 V of Li-Ion is equivalent of 5.55 V to 6.3 V of solar cell.

2. current of solar cell should have 1/10th of capacity battery diveded by 1 hour (for Ni Mh batteries). I use same rule for Li - Ion battery. It is called C - rate rule. So If I have 500 mAh battery, I should choose 50 mA sollar cell. Good Li- Ion batteries have 2000 mAh, so current should be around 200 mAh or 1.2 W.

I use bad Li - Ion battery with measured around 600 mAh. For that, I should choose solar cell with 60 mA peak, or 0.360 W (POWER = CURRENT X VOLTAGE).

Step 3: Lithium - Ion Batteries 18650

I find good website with tests lithium - ion batteries. Mostly there are 3400 mAh maximum.

Here is: http://lygte-info.dk/review/batteries2012/Common18650Summary%20UK.html

Here is some theory of charging them:

https://www.instructables.com/id/Li-ion-battery-charging/

https://www.instructables.com/id/SOLAR-POWERED-ARDUINO-WEATHER-STATION/

Step 4: Circuit

Circuit is simple, but I describe it here.

Connect positive terminal of solar cell to anode of diode. Connect negative terminal of diode to IN+ (input positive) of TP4056. I use diode because of reverse current.

Also connect negative terminal of solar cell to IN- (input negative) of TP4056. Finally connect battery, positive terminal of battery to BAT + of TP4056, similar negative terminal.

Step 5: LED Diodes on TP Board

On board, there are 2 diodes, which also consupt some power. I remove them with knife. Check picture.

Step 6: Calculation of Efficiency

Test you charging, you can connect your multimeter to solar cell, or battery.

Test:

cloudy, with a little sunny 10 mA (output current from TP4056), 24 mA (from solar cell)

cloudy, not direct to sun 0.87 mA (TP4056), 5.1 mA (solar cell)

sunny, direct sun 26 mA (TP4056), 89 mA (solar cell)

According pveducation.org website, you can calculate direct solar radiation in kW. Just fill your home lattitude and longtitude. And remember time, because radiation during day vary. I got around 1 kW/m2.

So, solar cell give me 89 mA, and 5V, so it gives 445 mW, or 0.445 W. Surface of solar cell is around 70 cm2 (basically only small lines make energy, so around 30 cm2).

Solar cell output = 0.089A x 5 V = 0.445 W

TP4056 output = 0.026 A x 4 V = 0.104 W.

To calculate how much solar radiation fall on 30 cm2 according pv education website, we must convert surface to m2, it is 0. 00 30 m2. Incident radiation is 1000 x 0.003 = 3 W.

Incident radiation = 3W

Efficiency of solar cell = 0.445 W / 3 W = 0.1483 = 14.8 %.

Efficiency of TP4056 = 0.104 W / 0.445 W = 23.37 %

Total efficiency of system = 0.104 W / 3W = 0.034666 = 3.46 %.

So total efficiency is not much, but helps. Do you remember C-rate? For this project, the bigger solar cell is necessary. I test on september, which is average between winter and summer. I use battery for my esp logger, which must survive during winter, summer is good. I will test others solar cells, in future, and show my results.

Step 7: Extra: Thingspeak Graph

I test battery voltage with my esp logger. I got graph on thingspeak. Results are in ADC values, not in voltage. Values 720 is equivalent of battery with 4.07 V. I use bad 600 mA Lithium - Ion battery.

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    7 Discussions

    0
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    samm316

    Question 1 year ago on Step 2

    Hi

    Could you give me some further explanation why you choose solar panel voltage to be 1/5 of battery voltage and solar panel current to be 1/10 of battery current divided by 1 hour.

    Regards

    sam

    2 answers
    0
    None
    Michal Chomasamm316

    Answer 7 months ago

    I hope you understand, and basically choosing solar panel depend on TP4056, not battery.

    0
    None
    Michal Chomasamm316

    Answer 1 year ago

    voltage - input of TP4056 must be higher then 4,2 V so I choose 1.5 times bigger voltage.

    Current - C-rate law. I just check what usually is charging current for lions battery - it is around 500 mA, but smaller currents are good , too. Also, during summer, the battery is charge fully + as temperature increase, the voltage of battery is higher. I dont want bake battery.

    0
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    llaci80

    Question 1 year ago

    hi why not use scotty diode? example 1N5819 5819 1A 40V SCHOTTKY DIODE?
    voltage drop not 0,6-0,8V only 0,1-0,2 V ?

    1 answer
    0
    None
    Michal Chomallaci80

    Answer 7 months ago

    It should be better, and I think if is need to use any diode. I use that diode, and if it not sunny, solar cell e.g. give 5 V, but after diode, votlage is less then 4,5 V, so charging is cut. Not good, I must improve it.

    0
    None
    NicholasP95

    Question 7 months ago

    Hi, can you tell me how to know if the circuit is charging the li-ion battery or not?

    1 answer
    0
    None
    Michal ChomaNicholasP95

    Answer 7 months ago

    It is possible via leds. I remove leds, but also, you can try to measure voltage on that places, where leds were. How, I dont know exactly, but try to use analog pins, or digital pins on arduino/esp8266.