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maximize efficiency of a solar panel charging a Li-Ion battery

C

Cosmin3

Hi,

I have a solar camping 36 Led lamp.
The solar panel on top of the lamp is: 0.7W 5V 70x70 mm.
In direct sun light at noon I got ~ 120..130 mA (short circuit) and ~5.2V (open circuit).
But it can't charge the battery...

The battery is a 3.7 V Li-Ion 2000 mAh.
The AC charger "pumps" 400mA ~8 hours to charge it. So the charge efficiency is only ~62%.
I think this can explain why the solar panel can't charge it but I don't fully understand the details...

Can someone show me a schematic of a charger I can place between the solar panel and the battery so I can maximize the power transfer?

I have a 600mA 5V USB converter from 1..5V:

converter.jpg

Would this solve the problem?

Thank you.
 
Last edited:
Chemelec

Chemelec

Your "10PCS" Picture Doesn't Show.
And What is the Open Circuit Voltage of that solar Panel?
Maybe Connect two of them in Series for a Higher Voltage.
 
C

Cosmin3

Just click on name or download the attached file.

I have edited my previous post to show more info.

I tried with 2 in parallel (separated by schottky diodes) >> only a little improvement.
The second one I design it to be connected at the same input the AC charger is connected.

To connect them in series (a long term solution not just testing) I have to modify the case of the lamp and I am not good at this :(
Anyway there has to be a solution with one solar panel - because I already replaced the original solar panel (5 V 60 mA, same problem) with the current version (5V 140 mA).
 
Last edited:
C

Cosmin3

Sorry, problem solved.
It was an incorrect extension (jpg instead of webp).
My advice is to use a better picture viewer instead of the default one...
 
Last edited:
C

Cosmin3

From the seller (I would post link but it may not be allowed):

Description
1.Small PFM control DC / DC boost controller chip
2.Enter any DC voltage of 0.9V ~ 5V, can be stable output 5V DC voltage, output current of 500 ~ 600MA with two AA batteries input to a single AA battery-powered output current 200ma so, for the mobile phone, camera , single-chip, digital products supply. Stable output 5V DC voltage
3.USB female, can direct power to the USB interface device with USB female versatile
4.PCB size: 34 (mm) x16.2 (mm) ultra-small size, installed in a variety of small equipment

HTB14A43KXXXXXbgXFXXq6xXFXXXf.jpg

The solar panel has 5.2 V ONLY in open circuit, it drops below 5 V when charging.
So this would help keeping 5 V no matter the load.

But the downside is this converter has < 100% efficiency...
 
Last edited:
(*steve*)

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I would first connect a diode between the solar panel output and the battery pack and measure the charge current.

any sort of switch-mode converter (that's not MMPT controlled) can get itself into a corner state where it loads the solar panel to a very low voltage and passes only a small amount of charge to the battery.

Also remember that any charging like this will not terminate when the battery is fully charged and can easily damage the battery.
 
(*steve*)

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
If it is just the output of a solar panel, you should get maybe 70-80% of the short circuit current at 80-90% of the open circuit voltage. If the open circuit voltage is 5V and the battery is 4.2V it's a pretty good match already.

If there's already some sort of circuit inside, then all bets are off.

The problem with a simple buck or boost converter is that if more power is demanded than is available, the voltage from the solar panel will fall faster than the current will rise. This reduces the total power available and thus delivered. You can get into a situation exactly the opposite of what you want.
 
C

Cosmin3

Well, because I can't count too much for the sun to shine in exactly the same way for a long time, I
tried to create a "test chamber": an old filament light bolt 100 W + the solar panel inside a dark
color box (10..15 cm distance between them).
The angle of the panel is at 20..30 degrees (to simulate the real conditions).
I replaced the battery with 6 Si diodes in series.

I connected the light bolt to AC power source for a few seconds.
I got these results:

short circuit current (the resistance of the wires and of the multimeter is ~ 1.5 ohm): 91 mA
open circuit voltage: 6.08 V
diodes current: 85.5 mA
diodes voltage: 4.77V


Oh, there are no circuits between the solar panel and the battery on the lamp PCB, only a Shottky diode.

A 80 mA current should fully charge the battery in 8 * 400 / 80 = 40 hours.
But it doesn't.
 
Last edited:
(*steve*)

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
It might, if the sun were a 100W bulb 10 to15 cm from the panel.

The output of a solar panel changes during the day. Depending where you are and how you place the panel, you'll only get the equivalent of 4 to 8 hours of full sunshine every 24 hours
 
C

Cosmin3

I think I haven't explained well: those 80mA, no matter the source (100W bulb or sun on the solar panel, AC charger and so on...) won't charge the battery.

In my first post I said: "The AC charger "pumps" 400 mA ~8 hours to charge it. So the charge efficiency is only ~62%."

It's like, instead of "ditching" 38% from any current, it decides to "ditch" all until it reaches, let's say, 150 mA. After this 100% of the current is used to charge the battery...

My english is not 100% perfect but I hope you understood what I'm trying to say...
 
(*steve*)

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Cosmin3 said:
It's like, instead of "ditching" 38% from any current, it decides to "ditch" all until it reaches, let's say, 150 mA. After this 100% of the current is used to charge the battery...
Click to expand...

That seems unlikely.

Is it just a battery, or is it a battery management system and battery combined?
 
(*steve*)

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
No, a cell like that, even if it is protected, will charge even at low currents.
 
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