Re: Schematic Help!
Here are the calculations for the LEDs. They can be run in series. Remember unlike a regular light bulb, Diodes are devices that use current, not voltage. According to the datasheets, the diode maximum current is 30mA which is .03 amps.<br />You have to use a resistor to limit the current even if you use a voltage regulator. The regulator would keep the voltage constant so you would not have to worry about whether your supply voltage is 12, 14 or 16 volts, but you have to worry about power dissipation in the regulator wich is the voltage dropped by the regulator times the current of the circuit. The regulator would be better for a normal bulb, but is not needed with an LED. Each LED should drop .7V in the circuit.<br /><br />16V - .7V = 15.3V<br /><br />15.3V/.030A = 510 ohm resistor I would go higher than this. The resistors are +/- 10% of their value for regular resistors. .<br />15.3V x .030A = .46W of power dissipated by the resistor<br /><br />This is almost 1/2 a watt of power. You would need a 1/2 watt or 1 watt resistor<br /><br />(Note if your maximum voltage delivered is only 14 V then<br />14V - .7V = 13.3V<br />13.3V / .030A = 443 Ohms<br />13.3V x .030A = .40W<br /><br />Now the Datasheet does not confirm the .7V drop but this a standard and cannot change unless the diode is made from non-standard LED materials.<br /><br />If you use Diodes in series, then just add in a .7V drop for each diode.<br />16V - .7V -.7v -.7V = 13.90V<br /><br />Remember, you cannot change the intensity of the LEDs, they are either off or on. The only way to change the intensity is to use a circuit to turn them on and off extremely quickly. By varying this, your eyes interpret this as a less light because the LED is turned on only for a fraction of a each second.<br /><br />Schematic check my calculations to make sure I did not do anything wrong. It has been 12 years since I got my electronic degree and I have used it only rarely.
