LED resistor connected to 12V Battery:
current is typically around 10mA. When current flows throw the
LED, electric voltage of about 1.6V will develop between its pins
(between the LED's Anode and Cathode).
connected in example (A) below, in order to calculate
the resistor value, you need to subtract 1.6V from 12V and
divide the result by 10mA (using Ohm's law) resulting in a value
of 1040 Ohms. In this example, a resistor
of 1K Ohms will suffice.
resistor that suits your application, please visit EID's resistors
switch can be added between
the resistor and the battery so as to control the LED (see above drawing
B). When the switch is closed, current flows and the LED light
will be illuminated. In the open position, there is no current
flow through the LED the LED
will not be illuminated.
you change the battery to 6V battery the resistor value will
change as follows:
the formula R= (VBatt-VLED)/ILED,
the new Resistor value is R=(6V-1.6V)/10mA=440 Ohms. In this
instance, a resistor of 560 or 680 Ohms will suffice.
the LED's Cathode pin, usually the short pin, to the negative
battery terminal. Connect the appropriate serial resistor between the
LED's Anode, usually the long pin, to the positive terminal on