LED Project

Make ten groups of three LEDs each; connect in series three LEDs in each group.

Then connect the ten groups in parallel to your battery.

Your question is singularly inappropriate.

First, do not put them all in parrallel. Find out how many you should put in series, then parallel up sets of those. Your schools Ohms law class should supply you with the value for the resistor. But remember to figure out the power? Otherwise someone will get hurt when it burns up. Make sure the power resistor banks rating is at least 4 times the power dissipated. And make sure you have a big enough fan to cool it!

Be careful!!

I will suggest that you need a DC to DC to pump-up the supply.

As you can calculate that the led is 2V, supply of 6V so you can drive only 3 led in serial. Then you need to parallel of 10 rows of led.

I dont think your supply of 6V can supply this, even it can I dont think the battery will last.

Another way they drive the leds is by switching so fast that the power to the led is only half, but you need some controller to do this.

Each parallel string should have a current limiting resistor or LED damage can occur. If your LED's have a 2.0V drop for 20mA forward current, I suggest this as the cheapest and simplest circuit.





Total battery current will be 15 x 20mA = 300mA.
A charge pump circuit with more LED's in series will provide higher efficiency, but circuit cost will also be higher. Good luck!

I do agree with your design.

Don't know if the battery can last, since it want to be cheap.

I think the battery will not last more then a hour.

http://www1.duracell.com/oem/primary/alkaline/mn1500.asp

For the circuit posted in #6, I believe 4x AA batteries in series could provide acceptable LED brightness for "approximately" 4 hours. However, this should be tested specifically in the desired application.

I started with AA batteries as they usually have a low(=good) cost/joule rating. Larger batteries would be a better choice as they have more capacity and a lower internal impedance (less internal loss). Smaller batteries (like coin cells) would be a poor choice since they will waste most of their energy in internal heating due to the 300 mA current required by the example circuit.