Pete the Elaner Posted August 1, 2020 Share Posted August 1, 2020 I'm quite happy to deal with voltages & resistances, but selecting the correct resistances for LEDs is something I am less familiar with. The supply I would like to use is 12v dc. I have some LEDs with a Vf (typical) 2.1v, Vf (max) 2,5v & Vr (max) 5v. Max forward current is 30mA. I guess vR is irrelevant if I ensure the polarity is correct? I would like to use 4 LEDs & I expect the best way is to connect them in parallel. Is this correct? If so, what resistances should I use? Since diodes do not react like resistors & the voltage is more significant, should 4 in parallel be treated the same as 1 when deciding on a series resistor? That just doesn't seem right to me. Link to post Share on other sites More sharing options...
KalKat Posted August 1, 2020 Share Posted August 1, 2020 I would start with 1K resistance EACH LED , possibly increasing that if they are too bright. Basically you subtract Vf from the supply voltage then divide by the current to give the resistance .....simples! Emma 1 Link to post Share on other sites More sharing options...
smokebox Posted August 1, 2020 Share Posted August 1, 2020 There are LED calculators available online if you search for them. However, for most applications 1k (1000 ohms) is a good starting value. Many, probably the majority, are listed as having 30mA as the maximum value but if you run them at the maximum, they won't last very long and it's better not to exceed about 20mA. Many LEDs will operate at a current as low as 5mA or less, though they will be much dimmer. 1 1 Link to post Share on other sites More sharing options...
meil Posted August 1, 2020 Share Posted August 1, 2020 try here: http://led.linear1.org/led.wiz 1 Link to post Share on other sites More sharing options...
RMweb Premium Ian Morgan Posted August 1, 2020 RMweb Premium Share Posted August 1, 2020 Rather than parallel, you can put the LEDs in series. 4 red LEDs will have a combined forward voltage of about 9V, so you only need to drop 3V across the resistor. Modern hi-intensity LEDs operate on about 5mA, and older LEDs needed 10 or 20mA. For 5mA, the resistor would be 3/0.005 (ohm's law) so 560 or 680ohm are the nearest commonly available values. Three white or blue LEDs in series would also have a combined forward resistance of about 9V, so the same resistor value would do. 1 1 Link to post Share on other sites More sharing options...
smokebox Posted August 1, 2020 Share Posted August 1, 2020 Or this one for multiple LEDs. It can even draw out a wiring diagram or schematic diagram https://ledcalculator.net/ 1 1 Link to post Share on other sites More sharing options...
H2O Posted August 1, 2020 Share Posted August 1, 2020 1 hour ago, Pete the Elaner said: I'm quite happy to deal with voltages & resistances, but selecting the correct resistances for LEDs is something I am less familiar with. The supply I would like to use is 12v dc. I have some LEDs with a Vf (typical) 2.1v, Vf (max) 2,5v & Vr (max) 5v. Max forward current is 30mA. I guess vR is irrelevant if I ensure the polarity is correct? I would like to use 4 LEDs & I expect the best way is to connect them in parallel. Is this correct? If so, what resistances should I use? Since diodes do not react like resistors & the voltage is more significant, should 4 in parallel be treated the same as 1 when deciding on a series resistor? That just doesn't seem right to me. There are a number of ways you could configure 4 LEDs with advantages and disadvantages for each. The most (electrically) efficient way is to have all 4 in series and one resistor. To calculate the resistor in this case think of the 4 LEDs as one LED with 4x the forward voltage but the normal forward current (4 x 2.1V and say 20mA in the above example). For 12V this would be (12 - 8.4) / 0.02 = 3.6 / 0.02 = 180 Ohms (but I'd use 220 or more to be on the safe side). You could have all 4 in parrallel and 1 resistor. For this resistor calculation think of the LEDs as having the normal forward voltage but 4x the forward current (so 2.1V and 80mA). For 12V: R = (12 - 2.1) / 0.08 = 9.9 / 0.08 = 124 Ohms, again probably 150 Ohms or more to be safe. Big disadvantage is if one LEDs fails the others will take more current and also fail. You could have one resistor per LED, this is less efficient and needs more resistors but has the advantage of being able to adjust the brightness of each LED. Here the resistor value would be (12 - 2.1) / 0.02 = 9.9 / 0.02 = 495 Ohms, so use 560 Ohms or higher. For all LED calculations use the typical values rather than maximum to make sure components last much longer. 1 Link to post Share on other sites More sharing options...
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