LEDs (that's "ell-ee-dees") are a particular type of diode that convert electrical energy into light. In fact, LED stands for "Light Emitting Diode." (It does what it says on the tin!) And this is reflected in the similarity between the diode and LED schematic symbols: In short, LEDs are like tiny lightbulbs. At 2 volts, the LED is taking 20 mA. If the LED was manufactured slightly differently it might require 2.1 volts or maybe 1.9 volts to push 20 mA thru it. Imagine what happens when two LEDs are in parallel - if they "suffer" from normal manufacturing variations, an LED that only needs 1.9 volts across it would hog all the current. Every few connections, or once a minute or so, clean the tip on the sponge and re-tin it. Otherwise the flux all boils away, the solder gets gummy and doesn't flow well. Repeat the soldering for the second LED. After soldering and cooling off, trim the other leg of the resistors and the LEDs to about 1/2 inch. Let's assume you have a supply voltage of 12v dc. The chosen LED is rated at a Max forward current (Shown as IF) of 20Ma and its working avearage voltage is 2.2v. The sum is then supply V - LED v / forward current. Or 12v - 2.2 divided by 0.20. (12-2.2=9.8/0.02 = 490) Therefore a resistor of at least 490 ohms is required. Value of Resistor = (VSupply - VF) ÷ IF. Where: VSupply = Supply voltage. VF = Forward Voltage. IF = Forward Current. The following formula can be used to calculate the value of power rating of the resistor. Power Rating of Resistor = IF2 × Value of Resistor. Finding the Value of Resistor to Connect with an LED. Solder one end of a second copper wire to the long lead of the red LED. The long lead is the cathode (positive) lead of the LED. Hold the negative side LED/copper wire to the negative terminal of a 1.5 to 3.0 volt battery. Hold the positive side LED/copper wire to the positive terminal of the battery. The red LED will light and will not burn out. The LED voltage drop depends on their colour (1.8 - 4.0V), to make them 12V compatible they need a series resistor which is built into '12V LEDS'. There's not really such thing as a "12V LED". Anything labeled and/or sold as such is really a 'normal' LED with a series resistor 'built-in'. OK, let's do the calculation. A simplified model for a LED is a fixed voltage source in series with a small resistor. Let's pick this LED from Kingbright. The slope is 20mA/100mV, so the internal resistance is 5\$\Omega\$. The the intrinsic LED voltage is 1.9V. Let's assume that the LEDs need 20mA and that our power supply is 5V. The answer is to place a resistor in series with the LED, and allow the resistor to "drop down" the voltage to the LED by 0.3 V. How do we calculate the resistor value? We use Ohm's Law, which states that V=IR, and substitute 0.3V (the voltage drop) for V, and 0.02A (desired forward current) for I. Solving for R gets us 15 Ohms. If you have a 12V source and a 3V LED, why not put three or four of them in series, instead of all of them in parallel with huge power-wasting resistors for the rest of the voltage. If your max current is 1 amp but you're only driving it to 1 milliamp, why not use a smaller, cheaper LED? 8hUmp.