October 3, 2022

Taylor Daily Press

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Adidas headphones have photocells and can be charged in any type of light – picture and sound – news

1Ah of battery capacity seems like a lot to me, but let’s assume that for the sake of convenience.
However, you should also make sure that the charging circuit reaches the minimum voltage.
The voltage drop across a single PV cell is very low (depending on the cell type), so you have to put several in series.
If you don’t, you should not only be able to lower the voltage, but also up, so you have a buck converter. They are generally not very effective.

Anyway, let’s assume exactly identical voltage to the photovoltaic cells.
You still have to provide at least 14 mA of current to charge such a battery in 100 hours.
The rule of thumb is that you have to put in approximately 1.4x the amount of energy to charge a battery with a given amount of energy.
14mA at 3.6V (standard lithium cell voltage) = 50mW.

However, we will calculate what size the PV panel should be.
A fairly standard amount of light is a “60 watt incandescent lamp over the dining table”.
Any 6 to 7 watt LED lamp on a 1.5m^2 . surface
Therefore, to light 1 cm2, you need approximately 50 megawatts of power that you put into an LED.
However, this LED is only 30 to 35% efficient (if you have a really good lamp).
So you actually have about 50mW of light at 3cm2 on your table.
Then take reasonable efficiency PV cells that achieve 20% efficiency then you need 15cm2 of PV cells to achieve 14mA charging current in a well lit room.
This is with optimal placement and orientation of the panels.

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In short, with 1mA of “quiet current” you don’t really get much time to charge.
However, many DC/DC circuits easily draw 5-10mA of “leakage current” if you’re not careful.

release:
Account correction. You don’t have to do it all by heart either, especially not with beer :)

Any 6 to 7 watt LED lamp on a 1.5m^2 . surface
Therefore, to light 1 cm2, you need approximately 50 megawatts of power that you put into an LED.

1.5 m^2 is 15,000 cm^2
So you have 7W/15,000cm^2 = 0.5MW per cm^2
In short, the rest still needs to be adjusted down by 100.
Next, you need not 15 cm2 of plates, but 1500 cm2, or more than two sheets of A4 size to get a charging current of 14 mA.

[Reactie gewijzigd door TD-er op 16 augustus 2022 23:32]