Resistor advice. Simple LED onto switch

Discussion in 'General Electronics Chat' started by jc0r, Oct 15, 2013.

  1. jc0r

    Thread Starter New Member

    Oct 15, 2013
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    0
    Hi all

    Im just getting started in the electronics world and am currently making a 24v battery pack. Everything is complete on the pack however im stuck on the final hurdle, and i would have thought this would have been the easiest aspect of the project!

    All i want to do is add an LED so that when the pack is turned on, the LED illuminates. I understand i must have a resistor in place on the hot side before it meets the LED to drop the voltage, however after frying a few LED's already, im lost. Using an LED that has a voltage of 2.2 and a max current of 25mA, the calculator told me i need a 1W 1200 Ohm resistor. This failed. When i put it on the multimeter, i could only get a reading on the 200 ohm setting, to which it read 10.0. Having checked the colours, it would appear the shop gave me a 10 ohm resistor instead.

    I do have a working example on a battery pack someone else made, the resistor looks to be GREEN BLUE RED BROWN which would make it 5.6k ohms. The picture below is it taken out and tested on the multimeter on the 20k Ohm setting.

    http://s13.postimg.org/twuyicn07/20131015_182429.jpg

    There is another LED which appears to have a diode also connected. I believe this LED turns off when the battery level is low. The resistor on this one appears to be RED BLACK BROWN BROWN. It measured 0.08 on the 20k setting and combined with the diode, they had a resistance of 3.38. Here is a picture

    http://s13.postimg.org/k0tviph87/20131015_182355.jpg

    Could anyone please explain any of this to me please. What do the reading mean that i have found on the multimeter. I am after the correct resistor to power a typical 2.5v 25mA LED. Do these resistors need to be absolutely precise?

    Many thanks for any help
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,380
    494
    Ok, I don't know if this is right, but I saw this formula here on the forums the other day: Resistor in series with LED = (Voltage supply - LED voltage) / LED current

    So if I am reading your info right:
    (24 V-2.2 V)/25 mA=872 Ohm
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,090
    3,027
    The resistor does not need to be precise unless you want to run right up to the specified max current, which is a bad idea anyway. The LED will last much lounger at, say 5-15mA. I find this plenty bright, and you could even consider putting 2 or 3 in series if you want more light at the lower current.

    Anyway, the calculation is ohm's law: voltage drop across resistor ∆V = current I in amps time resistance R in ohms. ∆V is battery voltage minus the Vf of your LED(s), about 22 in this example. Current is, let's say, 0.01A. So the resistor calculates to 22/0.01 = 2,200Ω. That happens to be a standard value. You could also try a 2,700Ω and a 1,800Ω to see how they look. You could go down to 1,000Ω for maximum brightness (and reduced life) at 22mA.
     
  4. jc0r

    Thread Starter New Member

    Oct 15, 2013
    6
    0
    Thank you for your explanation. How do the resistors in the pictures compare. The values the multi throws out, are they not accurate. When looking at resistors, should I only look at the ohms of resistance or is there a difference between 2w 1k and 1w 1k?
     
  5. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    829
    Yes, there's a difference.

    After you calculate the required resistor value, you need to calculate the wattage rating. This is the voltage dropped across the resistor multiplied by the current through the resistor. In the example above, you are dropping 22 volts across the resistor. The current (used to calculate the resistor value) is 0.01A. Hence, the wattage is:
    22V * 0.01A = 0.22W
    Always double the calculated rating for safety, hence a 1/2W resistor should be sufficient.

    In your question, the answer depends on the voltage and current of the circuit the 1kΩ resistor is in.
     
  6. jc0r

    Thread Starter New Member

    Oct 15, 2013
    6
    0
    Ok that's all makes sense to me. What doesn't is the leds multimeter readings taken from the working example in my original post. You state i should be using a roughly a 2.2k but the resistance on those is way higher at 5.49k and 3.38k. Also, I have tested both on a flat battery and the red one illuminates but the green one does not. That's doesn't make sense to me as the red led has more resistance? Surely if anything it should be the other way round?
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,380
    494
    Make sure LED legs attached to right polarities. Maybe green one was attached wrong.
     
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