Resistor question - Page 1
Forum section: electronics / projects

author: thisisliam
author_rank: Contributor
timestamp: May 24, 2019, 06:47:05 pm
content: Hi everyoneI have a 12v DC, 650mA power supply that I'm trying to power a small LED with. A little research tells me I need a 680r resistor... Would appreciate some input from someone a little more versed with this. I have attached the spec chart for the blue LED I'm needing to use.Thanks for the help!

author: ataradov
author_rank: Super Contributor
timestamp: May 24, 2019, 06:52:29 pm
content: Yes, 680 Ohm resistor will work, and will result in ~13 mA of forward current. It is fine for the LED, but if it is just an indication LED, then it might be too bright.

author: thisisliam
author_rank: Contributor
timestamp: May 24, 2019, 06:57:20 pm
content: Quote from: ataradov on May 24, 2019, 06:52:29 pm@ataradov - thanks for the quick reply! It is just an indicator LED - would you suggest a different resistor so it isn't so bright?

author: Zero999
author_rank: Super Contributor
timestamp: May 24, 2019, 06:59:28 pm
content: Yes, a 680R resistor seems sensible and will give 12mA to 15mA, depending on the forward voltage of the LED.As mentioned above, it might be too bright, unless you have a yellow LED, which tend to have a poorer efficiency, compared to the other colours.

author: mariush
author_rank: Super Contributor
timestamp: May 24, 2019, 07:06:53 pm
content: Voltage power supply - Forward Voltage of LED (or sum of forward voltages) = Current x Resistance.So if you have a single RED led with a forward voltage of approximately 2v and you want 10mA (0.01A) of current with a 12v power supply, then :12v - 2v = 0.01 x R ====> R = (12-2) / 0.01 = 10 x 100 = 1000 ohmAnother example, if you have 3 blue leds in series, which have a forward voltage of 3v each, and you want 20mA to go through them :12v (power supply voltage) - 3 leds x 3v (forward voltage of each led) = 0.02 A (20mA current) x R ====> R = (12 - 9) / 0.02 = 150 ohmAlso, power dissipated in a resistor can be calculated with formula P = IxIxR where I is current and R is resistance.So for example, with the 150 ohm resistor above, you have P = 0.02A x 0.02A x 150 = 0.06 watts, which means you can use a resistor rated for 0.125 watts.You typically want to stay below 75% of the resistor's power rating, for safety and reliability reasons and keep the resistor within acceptable temperatures. If the power dissipated was close to 0.1w, it's better to use a resistor rated for 0.25w

author: ataradov
author_rank: Super Contributor
timestamp: May 24, 2019, 07:10:59 pm
content: Quote from: thisisliam on May 24, 2019, 06:57:20 pmModern LEDs are crazy efficient. If it is indoor indicator, then do your calculations for 1-5 mA. At 13 mA that blue led will be impossible to look at directly.So 2 kOhm to 8 kOhm range would work much better. It really depends on your application. Just try different values and see what works best.

author: thisisliam
author_rank: Contributor
timestamp: May 24, 2019, 07:13:16 pm
content: Thanks, guys.I will be powering two blue LEDs with a forward voltage of 3.1 - 3.6 volts. Based on mariush's example a 150 ohm resister would suffice?

author: ataradov
author_rank: Super Contributor
timestamp: May 24, 2019, 07:15:37 pm
content: 150 Ohm was calculated for 3 LEDs. You need to redo that math for two LEDs. If you use 150 Ohm resistor you will get 47 mA, which would not be goo for the LEDs.Also, LEDs are cheap, just try and see what happens.

author: Zero999
author_rank: Super Contributor
timestamp: May 24, 2019, 07:45:26 pm
content: Just to clarify one thing: in mariush's example, the LEDs are connected in series, not parallel, which generally isn't best practise with LEDs.

author: thisisliam
author_rank: Contributor
timestamp: May 24, 2019, 08:11:00 pm
content: Hi guysMy maths skills have never been on point. The LED is built into a push button switch, so changing out the LED if I blow it will be a task and a half. The LED will only be illuminated for a few seconds at a time when pushed. I basically just want to power both switches off of the power supply, preferably with one resistor if I can.This is the math I have:12 volts - 6 (two blue LEDs at 3v) / 20mA (.020) = 300 ohms

author: ataradov
author_rank: Super Contributor
timestamp: May 24, 2019, 08:14:20 pm
content: Yes, except you don't really need 20 mA. Use something like 1 kOhm and you will be fine.And again, make sure that your LEDs are indeed in series.