can anyone say me how to make 100 IR LED array in PCB?

Question

any one teach me how to design 100 IR LED array, what compontents required. i am using a 12v,car battery for this.where can i design a PCB for this?

Answer 1

To do this design, you need to know the following:

1. Ohms Law: V = IR
2. Power dissipated in a resistor: P = I*I*R
3. The voltage drop of the IR LEDs you will use (typically 1.5 to 1.8V for IR LEDs.
4. The current that you will operate the LEDs at.
5. Diode operation: LEDs are diodes and diodes will have a near constant voltage drop across the anode to cathode that varies very little with the current through them (ie - they do not follow Ohm's law).
6. Tools to layout a PCB. You will need a CAD tool to draw out the PCB and submit it for fabrication, or you can obtain a kit at RadioShack that lets you hand layout your own PCB using an etch resist pen and acid.

I have put two sources on this answer. One is a company that carries a lot of components (Digi-Key) which has fast and reliable delivery. The other is a CAD tool that you can download and use for free to layout the PCB. The CAD tool will generate a quote and is made to be submitted the PCB company that provided it for your PCB. You can also skip a PCB and use perfboard to build the circuit as suggested by another answer...the choice is yours.

As for the design, without knowing the specific LED and power level you are after, I can only provide an example of how I would approach the design problem. One IR LED available at Digi-Key
produces a forward voltage of 1.6V when the forward current is 50 mA. It can handle a maximum of 100 mA, but it is better design to not run it at the full rating. Five of these LEDs in series would drop a total voltage of 5 * 1.6V = 8.0V. A resistor is needed in series with the group of five LEDs to set the current to 50 mA (otherwise, they will draw too much and likely be damaged). The resistor value is found from R = V/I = (12V - 8V)/50mA = 80 Ohms. The resistor will dissipate power P = 0.05A * 0.05A * 80Ohms = 0.2 Watts so a 0.25 Watt resistor or larger should be used for this. You would then need to arrange 20 of these circuits in parallel connected to your 12V battery.

I hope this helps.

Answer 2

You can use a Radio Shack perfboard, 100 IR LED's, 100 330 ohm resistors,. Connect 1 LED (Anode Lead) to 1 330 ohm resistor in series and solder together. Repeat for all 100 LED's and resistors. Connect all of the Cathode Leads together of the LED's. Connect this point to the negative battery voltage. Connect all of the open resistor leads together and connect through a 2 amp fuse and a switch to the positive 12 volts.

Answer 3

http://www.4led-pcb.com/