1 Ah of battery capability looks a little bit abnormal to me, but let’s assume for usefulness.
Even so, you also require to make positive that your charging circuit reaches the minimal voltage.
The voltage fall on a solitary photovoltaic mobile is really small (dependent on the mobile sort), so it is necessary to set numerous in collection.
If not, you not only need to be equipped to decreased the voltage, but also elevate it, so you have a buck-raise converter. They are frequently not really effective.
In any case, let’s assume a correctly matching voltage of the photovoltaic cells.
You even now have to provide at minimum all around 14mA of existing to cost these types of a battery in 100 hours.
The rule of thumb is that you have to set in about 1.4 moments the sum of ability to charge a battery with a certain amount of electricity.
14 mA at 3.6 V (regular lithium mobile voltage) = 50 mWatts.
However, we will calculate how large your photovoltaic panel should really be.
A relatively typical total of gentle is “60 Watt bulb over the eating table”.
That is a 6 to 7 Watt LED lamp on an spot of 1.5 m ^ 2
So to light-weight up 1 cm ^ 2 you want about 50 mWatts of vitality which you put into the LED lamp.
Having said that, these types of an LED lamp is only 30 to 35% economical (if you currently have a superior a person).
So you essentially have about 50mW of mild at 3cm ^ 2 on your desk.
So get reasonably efficient photovoltaic cells that reach 20% performance and therefore 15 cm ^ 2 of photovoltaic cells are demanded to arrive at a charging existing of 14 mA in a moderately very well-lit space.
This with optimum positioning and orientation of the panels.
In brief, with 1 mA of “quiescent current” you you should not have a lot time to recharge.
Nevertheless, several DC / DC circuits simply attract 5 – 10 mA of “leakage latest” if you are not thorough.
Calculation correction. I do not have to do all the things by coronary heart, especially not with a beer
That is a 6 to 7 Watt LED lamp on an space of 1.5 m ^ 2
So to mild up 1 cm ^ 2 you have to have about 50 mWatts of vitality which you put into the LED lamp.
1.5 m ^ 2 is 15000 cm ^ 2
So you have 7Watt / 15000cm ^ 2 = .5mWatt for each cm ^ 2
In brief, the relaxation still desires to be altered down by a component of 100.
So you do not need 15cm ^ 2 of panels, but 1500cm ^ 2 or a lot more than 2 A4 sheets to get 14mA of charging present.
[Reactie gewijzigd door TD-er op 16 augustus 2022 23:32]