> I have an LED that needs 1.6Volts to light up. and I have a 9 volt battery. > now I see every calulation to go from Volts, watts, resistance, amps, etc.. > What I have not seen is what the end result is. It I have a 9 volt battery > and a 40ohm resistor then I end up with a amp of 0.225. now if I take the > amps and multiply it by 40ohm I end up with 9 volts. Ok I think I got Ohms > Law. Now how do I get the value of the Volts after the resistor is applied > to the line ? An LED is specified at a particular current, not voltage, usually 20mA for common LEDs. So, you have to figure out what resistor to put in series with the 9V battery and the LED so that 20mA will flow. The battery will produce 9V, and you say the LED will have 1.6V accross it. That leaves 9V - 1.6V = 7.4V accross the resistor. Now you know the voltage and the current so you apply Ohm's law: R = 7.4V / 20mA = 370 Ohms. > Ok now is it better to have a PICs chip connected to the ground pin of an > LED or is it best to have the PIC control the voltage to the device? will > either solution result in a voltage leak or drain? All else being equal, use the PIC pin to sink, not source, voltage. This means the current will flow from the power supply, thru the dropping resistor, thru the LED, and thru the low side driver of the PIC output. The low side drivers are a bit beafier than the high side drivers. This also means you write a 0 to the port pin to turn the LED on. I don't like putting 20mA thru a port pin because it is getting close to the spec limit. For simple circuits, I usually just run LEDs at 10mA. There are also ways of making an external transistor do the heavy lifting, but that is a bit advanced for where you are now. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads