Solar Based Handheld Led Light

[PRIYADHARSHINI]

Hi,

I have 12V ,1.5W hand held solar power panel small cell charger.I would like to make LED emergency light with Battery backup.My doubts are

1.How can i choose the rechargeable battery for my application.?

2.How do i choose the LED combination(series or parallel)?

3.How do i calculate the current limiting resistor values if the series or parallel combinations choosed?
Pls suggest me the circuit ideas.


Thanks in advance

 

[Externet]

All the effort, time, parts, making a proper housing where all fits will cost more than :

-----> https://www.aliexpress.com/wholesale?SearchText=solar+flashlight&opensearch=true

 

[PRIYADHARSHINI]

hI Externet,

I have seen this already.But i want to make my own.Its my home project.

Thanks in advance

 

[Externet]

A solar panel of about 12cm x 10cm could still be called handheld; and nine AA size Ni-metal cells also. There is cell holders for nine AA cells.

The LEDs can be better series-parallel, each series string sum to be fed with 12V, the number of parallel branches you want. The Vf voltage/color of the LEDs tell how many per string. Probably 3 for white.

The series limiting resistor for each series string should be 12 minus the Vf drop of the series stringed LEDs divided by the string current.

Bat(+)----------------|>|----------|>|---------|>|----------/\/\/\/------------(-)
Several equal branches paralleled as this one ^
The solar panel could be directly connected to the battery, with a diode in series. Add a switch.

/\/\/\/ for 3.4V Vf LEDs ---> 12 - 3.4x3 = 10.2 divided by 0.02A = 510Ω

The hardest part in any project is a proper housing. A clear acrylic box may work.

 

[PRIYADHARSHINI]

A solar panel of about 12cm x 10cm could still be called handheld; and nine AA size Ni-metal cells also. There is cell holders for nine AA cells.

The LEDs can be better series-parallel, each series string sum to be fed with 12V, the number of parallel branches you want. The Vf voltage/color of the LEDs tell how many per string. Probably 3 for white.

The series limiting resistor for each series string should be 12 minus the Vf drop of the series stringed LEDs divided by the string current.

Bat(+)----------------|>|----------|>|---------|>|----------/\/\/\/------------(-)
Several equal branches paralleled as this one ^
The solar panel could be directly connected to the battery, with a diode in series. Add a switch.

/\/\/\/ for 3.4V Vf LEDs ---> 12 - 3.4x3 = 10.2 divided by 0.02A = 510Ω

The hardest part in any project is a proper housing. A clear acrylic box may work.

Thank you so much for your reply.I understood what you mentioned.My doubt is how i can choose the battery with respect to solar cell.Regarding its charging time,charging current,Battery voltage etc

 

[PRIYADHARSHINI]

Hi Externet,

I have attached One schematic with 6V /5W solar panel recharging 4.8v/800mA battery.


1.How can we select the battery properties according to selected solar panel
2.How can we make the battery recharging time less?
3.Shall we select the battery voltage equal to the Solar panel voltage ?

Pls guide me to do this
Thanks in advance

 

[Externet]

1.- A solar panel can work recharging any battery chemistry.
2.- Less charge time is done with higher charging current, which means more sunlight or more current yielding panel.
3.- Charging battery voltage + series diode Vf drop should be less that panel voltage output.

 

[Audioguru]

Your circuit has almost nothing to control the battery charging current except R2. Di you calculate minimum and maximum charging current through R2? The solar panel will produce a very high output power when it directly faces the sun at noon in the middle of summer on a sunny day and produce almost no power when these things are the opposite. Then the battery might explode or take forever to charge.

Nobody makes an LED that is exactly 3.6V. Some will be 3.3V and soon burn out and others will be 3.9V and be very dim in your circuit unless you buy thousands of LEDs, test them all and pick one (if there is one) that is exactly 3.6V. Oh, you also must keep its temperature from changing somehow because its voltage changes (and therefore its current changes in your simple circuit) when its temperature changes.

How will you keep the battery voltage from being almost 6V when it is fully charged and being 4V when it is running low? The LED will soon burn out or be very dim.

Two simple calculations should be done:
1) 3.3V LED with a 6V fully charged battery is (6V - 3.3V)/3.9 ohms= 692mA. Will the LED survive?
2) 3.9V LED with a 4V battery is (4V - 3.9V)/3.9 ohms= 26mA if the LED conducts with such a low voltage. Dimm?

An LED is usually fed from a resistor with a value high enough so that the wide change of voltage from an LED part number and from the wide range of battery voltage does not affect the current very much, or a constant current circuit is used.