pranam77 said:Hello every one. Wish you all a happy and prosperous New year 2009.
While experimenting with white LED's i tried to run 60 LED's with 6 volts 4.5 ah battery, and connected as given in the schematic. I have assumed each LED has a forward drop of 3.3 volts as i didint get any specifications from the manufacturer. The led's are of China origin. Too the current was guessed to 20 Ma and a resistor of 150 ohm was connected in sieres of all the 60 LED's. Now when connected to the
battery, total consumption came to exactly 1.015 amphere. Mean it will glow for approximately 2.5 to 3 hours on a 4.5 ah battery. My question is, there any way to reduce the total current without reducing the LED brightness so that i can get more running hours with same battery. Can i impliment some driver circuits so that the voltage is raised and the LED's can be connected in seires? May i expect sugesstions, links, schematics or Ideas in this regard?.
MCMC said:use a serial parallel circuit
you need a 20 pairs of 3 serial led's (60) total, connect 3 led in serial whit one 15 or 20 ohm resistor (test whit one circuit 3 serial led) and then connect this 20 serial led's circuits in parallel, you have a .02 ma x 20 = 0.4 Amp of current is reduced to half, look the attach.
if you add another battery connect it in serial to make 12 volt, and connect 10 pair of 6 leds in serial whit 30 ohm resistor circuit, the total current drop to .2 amp
CMOS said:MCMC said:use a serial parallel circuit
you need a 20 pairs of 3 serial led's (60) total, connect 3 led in serial whit one 15 or 20 ohm resistor (test whit one circuit 3 serial led) and then connect this 20 serial led's circuits in parallel, you have a .02 ma x 20 = 0.4 Amp of current is reduced to half, look the attach.
if you add another battery connect it in serial to make 12 volt, and connect 10 pair of 6 leds in serial whit 30 ohm resistor circuit, the total current drop to .2 amp
There is still a wastage of 0.02 * 0.02 * 150 * 20 = 1.2 Watts. This is not the most efficient way of driving LEDs for lighting. And as pranam77 mentioned, it is not possible to drive 3 LEDs in series each having forward voltage of 3.3V with a 6V battery!
BTW pranam77, did you take a look at switching LED driver's that I mentioned in my previous post ?
Low current red LEDs are 1.8V to 2.0V. Old green ones were 2.2V.MCMC said:I never hear of a led of 3.3 voltage or is a special or dual led.
Audioguru said:Low current red LEDs are 1.8V to 2.0V. Old green ones were 2.2V.MCMC said:I never hear of a led of 3.3 voltage or is a special or dual led.
New blue, new bright green and white low current LEDs are from 3.2V to 3.6V.
The calculation I showed is based on Ohms law itself!!! which accounts to power lost in the resistors in the form of heat and not the total power consumed by complete circuit! Those are called I²R losses.MCMC said:ooohh you forget the ohm law the current in the serial is the same, 3 led in serial like I suggest = 20ma * 6 volt = .12 watts total power, .04 watts per led and 2.4 watts total, compared to 60 led in parallel that is = to .02 * 60 * 6volt = 7.2watts of power in the circuit used by pranam77.
I never ear of a led of 3.3 voltage or is a special or dual led.
I use this led, look the spec of led in the attach the forward voltage is 1.90 to 2.02 volt and 2.4 V max
AlBo said:Hi pranam77 !
Interesting article about LED driving you will find on
**broken link removed**.
See also application notes on national semi web page.
Hai Pranam77, first decide the forward drop of the LEDs you have on hand. then use the formula [Vcc-(n*Vled)]/.02=the resistance that would be needed.pranam77 said:Thank you every one for your valuable sugessions and ideas. This is the LED what i use. I am intending to use it for constucting an emergeny light with 6 volts. Sorry for the blur image. It was taken with my cell phone.
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