# Solar powered LED light circuit Solar power is one of the extensive renewable energy available in our earth, using solar power efficiently could help us to meet 30% of our energy demands. And that’s the reason we are seeing many solar based products in the market. And today we are about to see the design of a simple solar powered LED light using high power LED which can be used for household purpose instead of primitive lights.

## LM317: LM317 is  an adjustable voltage regulator that can provide output voltage ranging from 1.2 V to 37 V. It is very similar to a normal fixed voltage regulator but provided with ADJ pin to adjust the output voltage obtained from it. A pot or voltage divider must be connected to the ADJ pin which in turn varies the output voltage obtained from the this IC. The formula for calculating the Vout voltage is given by

Vout = 1.25 * ( 1 + R6/R5) as per the above given design.

## WORKING OF CIRCUIT:

The working of the above circuit starts with the 9V solar panel which converts the incident solar power into the electrical energy. A diode 1N4001 was connected to eliminate the risk of reverse flow of current during night time. The input from the solar power is fed into the LM317 regulator IC. A resistor of R5 and R6 was connected to it to obtain the required output from the IC. Let’s do some math

Applying the R5 and R6 values in the LM317 Vout formula

Vout = 1.25 * ( 1 + 1200 / 240 )

= 7.5 Volts.

We will get 7.5 V in the output pin of the IC LM317. The reason we are using 7.5 V to charge a 6 V battery because the property of lead acid batteries. A 6v Lead acid battery will only charge to 100% when 7.3 V is applied to it. So considering the losses we have chosen R5 and R6 to give 7.5 V as output in the Vout pin.

R4 resistor is to limit the input charging current to the battery because a battery must be given current of 1/10 of its Ah rating. Since we are using a 5Ah rated battery the input charging current should be of .5 A or 500 mA. 15 ohm R4 resistor was used to limit the current to 500mA in the above circuit.

Now comes the lighting part, here we have chosen three high power white LED’s rated at 3.5 V at 350mA. Resistors R1, R2 and R3 are used to limit the current provided to the LED’s. So the total current consumption of lighting part will be

350mA x 3 = 1050mA (Since the LED are in parallel arrangement).

The current rating of the battery  is 5Ah which implies that the battery can power the lights for about

Duration = 5000mA / 1050mA

=  4.45 hours.

Thus our battery can power can power up the lights for 4.45 hours before recharging, but it is recommended to leave 25% charge in the battery intact. So the suggested duration of this light usage will be around 3.30 hours.

## NOTE:

• You can also use Red , Yellow and other high power LED’s in the above circuit.
• Do check the spec sheet of your high power LED as wrong usage may drain out the battery.

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laxman

why 6v battery is used to supply 3.5v LED? instead can use lower powered battery? Guest
Burak

What if solar power is not working %100 efficiently. I mean 7V instead of 9V what will be the Vout? wpDiscuz