A question that is asked all the time is "How do i work out what resistors are needed". Firstly we will assume you are using DC voltages. There is a few things you need to know. What is the overall source voltage to the string? ie 24v How many LED's in a section of series connection? ie 6 What is the Forward Voltage (Vf) for each LED? ie 3.1 What forward current will be used? ie 20mA Things to remember: 1. Forward Voltage (Vf) varies for each colour. Generally Reds are from 1.9 to 2.1, Green/Blue will range from 3.0v to 3.4v. 2. Leds in one colour can have different Vf figures even withing the same batch, especially true of really cheap LED's. 3. Accurate resistor values can only be found by actually measuring current flow. 4. All Vf figures in specifications are usually for a 20mA current. 5. Actually Forward Voltage drop will vary based on the current flowing. Once you have answers to the above questions the simple way is to use the LED Calculator over at http://led.linear1.org/led.wiz to get a starting value for you. The LED series/parallel array wizard will give you nice diagrams and clearly show series parallel LED setups. What about those 24v 100 count red LED strings we can get, just go to the wizard above and enter 24v, 2.1v, 20mA, 100 Leds then look at the 10x10 array that is shown at the bottom of the page. This is the series parallel connections that would be in the string of lights. I call each set of 10 series LED's a "SECTION" and the complete set a "STRING". You will notice that the entire string draws 200mA. One thing that can be highly confusing is when looking at a string of say 24v Big-W Blue leds that are 120 leds and have 10 LEDS per section and NO resistors. The astute reader will have already realised that Blue LED's have a Vf of 3.0v to 3.4v (at 20mA) and that 10 x 3.0v = 30v about 6v higher than the nominal 24 rating. Measurements clearly show that at 24v these strings only draw a few mA and are just glowing. The 20mA / section is not reached until the voltage is raised to 31v-32v which clearly matches expectations based on forward voltage specifications. The manufactuer in this case is clearly using a lower voltage than the total Vf to do the current limiting rather than use resistors. Cheers Phil

thought i might add some more to this topic Warning: Technical discussion under the pretty picture Terms. Vf â€“ Voltage forward drop. The amount of voltage dropped across the LED at a certain current. If â€“ Current (I) forward. The current flowing in the circuit when power is applied DC â€“ Direct Current. Technically the resistor does not have a Vf specification. The voltage drop across the resistor is dependant of the â€œRâ€ the resistor value and â€œIâ€ current in the circuit and is calculated with V=I*R. Different colour LEDâ€™s are manufactured from different semiconductor materials, this leads to different Vf values for different colours. All Vf figures are based on a nominal test If and for 3/5/8/10mm LEDS this is 20mA. For half watt LEDâ€™s this will be 100mA and for 1W LEDâ€™s this could be 350mA to 400mA. Red is usually 1.9v to 2.1v Blue is usually 3.0v to 3.4v White is usually 3.2v to 3.6v Green can be manufactured with two different materials. The Green most of us will use is 3.0v to 3.4v The green often found in indicator leds will be 1.9v to 2.1v The fact that the LED Vf will vary from LED to LED for completely accurate resulted the Vf needs to be measured. Examples: RED LED DC = 12 If = 20mA Vf (1-2-3) = 1.9 R = ?? Voltage drop needed across resistor = 12 â€“ (3 x 1.9) = 6.3v R = v / I R = 6.3v/20mA = 315 Ohms If Vf of Leds equals 2.1v Then Voltage drop needed across resistor = 12 â€“ (3 x 2.1) = 5.7v R = 5.7v/20mA = 285 Ohm BLUE LED DC = 12 If = 20mA Vf (1-2-3) = 3.0 R = ?? Voltage drop needed across resistor = 12 â€“ (3 x 3.0) = 3.0v R = v / I R = 3.0v/20mA = 150 Ohms If Vf of Leds equals 3.4v Then Voltage drop needed across resistor = 12 â€“ (3 x 3.4) = 1.8v R = 5.7v/20mA = 90 Ohm Now lets look at what happens when you select say the 90 ohm resistor above for your blue then get a new batch of LEDâ€™s that have a Vf of only 3v, the resistor will now drop 3v across it at 12v and by Ohms Law I = V / R I = 3 / 90 = .03333 = 33mA (a significantly higher value than the desired 20mA) Cheers Phil[/attach]