Solar Cell with Joule Thief

Thread Starter

ouzcoskun

Joined May 12, 2016
32
I have used a small joule thief with old solar cell. Its current increased from 0.15 mA to more than 20 mA. But I can only add them in parallel, and the voltage and the current increases. Do you have an idea to add them in series ? SOlar Cell&Joule Thief.jpg SOlar Cell&Joule Thief1.jpg SOlar Cell&Joule Thief2.jpg SOlar Cell&Joule Thief3.jpg
 

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BobTPH

Joined Jun 5, 2013
8,813
What are the specs of the solar cell? What are you trying to power (current and voltage?)

But most of all, what are you trying to do?

Bob
 

Thread Starter

ouzcoskun

Joined May 12, 2016
32
Thank you very much for your interest.
I just want to charge 2 x AA or 2 x AAA NiMH batteries.
 
Last edited:

mvas

Joined Jun 19, 2017
539
I have used a small joule thief with old solar cell. Its current increased from 0.15 mA to more than 20 mA. But I can only add them in parallel, and the voltage and the current increases. Do you have an idea to add them in series ?
Research "MPPT" = Max Power Point Tracking
MPPT is a circuit that extracts Maximum Watts from the Solar Cells.
Watts = Volts x Amps
You need to maximize Watts, not just volts individually, and not just amps individually.

The Joule Thieve may, or may not, be operating the Solar Cells at Max Power Point.
There are many combinations of Volts and Amps , where the Watts is below Max Power.
 

Thread Starter

ouzcoskun

Joined May 12, 2016
32
Research "MPPT" = Max Power Point Tracking
MPPT is a circuit that extracts Maximum Watts from the Solar Cells.
Watts = Volts x Amps
You need to maximize Watts, not just volts individually, and not just amps individually.

The Joule Thieve may, or may not, be operating the Solar Cells at Max Power Point.
There are many combinations of Volts and Amps , where the Watts is below Max Power.
Thank you.
 

Thread Starter

ouzcoskun

Joined May 12, 2016
32

BobTPH

Joined Jun 5, 2013
8,813
Because no circuit can put out more power than it takes in.

1.5V at 16.5 mA = 24.75 mW

2.3V at 23.16 mA = 53.3 mW

Bob
 

Thread Starter

ouzcoskun

Joined May 12, 2016
32
Because no circuit can put out more power than it takes in.

1.5V at 16.5 mA = 24.75 mW

2.3V at 23.16 mA = 53.3 mW

Bob
Ok. I see. Maybe it is increasing the efficiency of the solar cell. You know a joule thief is working like a buck booster.
So do you have an idea about adding the cells in series ?
 

BobTPH

Joined Jun 5, 2013
8,813
I am not entirely sure what you are asking. Are you trying to charge multiple cells in series, or do you have multiple solar cells you want to place in series.

Your solar cell will take 133 hours to charge 1 AA 2000 mAh. Charging two would take twice that.

Bob
 

Bernard

Joined Aug 7, 2008
5,784
The pictures are nice but meaningless unless we know how everything is connected. Assuming 3.88 V is from 2 cells in series OC, open circuit, then under load V might be about 2.9 V , ready to charge 2 batteries in series.
Disconnect charging when battery V = 3 V, about 2 weeks later ? Might be best to add a blocking diode between solar & batteries if leaving connections over night.
 

Thread Starter

ouzcoskun

Joined May 12, 2016
32
I am not entirely sure what you are asking. Are you trying to charge multiple cells in series, or do you have multiple solar cells you want to place in series.

Your solar cell will take 133 hours to charge 1 AA 2000 mAh. Charging two would take twice that.

Bob
Yes. Multiple solar cells with a joule thief on each, and I want to add them in series.
 

Thread Starter

ouzcoskun

Joined May 12, 2016
32
The pictures are nice but meaningless unless we know how everything is connected. Assuming 3.88 V is from 2 cells in series OC, open circuit, then under load V might be about 2.9 V , ready to charge 2 batteries in series.
Disconnect charging when battery V = 3 V, about 2 weeks later ? Might be best to add a blocking diode between solar & batteries if leaving connections over night.
Thank you. I have added also the circuit diagram. As below.SOlar Cell&Joule Thief4.png
 

BobTPH

Joined Jun 5, 2013
8,813
Well, that is definitely not the way to do it.

Connect the - of the top one to the junction of the diode and capacitor in the bottom one.

Connect the load from the - of the bottom one to the junction of the diode and capacitor in the top one.

Like this:
upload_2019-6-13_16-40-59.png

Bob
 
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