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Silly LED question

Discussion in 'LEDs and Optoelectronics' started by mentholboy, Feb 7, 2011.

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  1. mentholboy

    mentholboy

    3
    0
    Feb 7, 2011
    I've built a perspex box with 5 red leds for the footwells in my car. I've wired them in parallel, the forward voltage is 1.9v and the forward current is 20ma and there running off of 2AA batteries. I put the numbers into a resistance calculator which told me to get a 12ohm resister. All wired up and turning on fine, but the batteries are getting hot and i can't work out why. Also occasionally when i flip the switch, the leds are veryyyy dim and gradually get brighter. What am i doing wrong?
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,397
    2,777
    Jan 21, 2010
    first, putting LEDs in parallel is Baaaad. You should give each LED it's own resistor.

    Secondly, the initial voltage was probably significantly higher than 3V (possibly up to 3.4V), and the forward voltage drop may be lower than 1.9V leading to a higher than expected current.

    But those factors would not cause the battery to heat up. It is more likely that you have a 1.2 ohm resistor instead of 12 ohms.

    The huge current would make the LEDs very bright, but also damage them. One of the symptoms of a damaged LED is that it no longer turns on at low currents or until it gets hot. This may explain why the LEDs start dim and get brighter.

    I would...

    1) measure the value of the resistor
    2) measure the current from the batteries
    3) measure the voltage drop across the LEDs
    4) use an individual resistor per LED.

    Actually, what I would do is...

    1) Place the LEDs in series, choose an appropriate resistor, and run them from 12V
     
  3. mentholboy

    mentholboy

    3
    0
    Feb 7, 2011
    ok, done it in series now and its all good, the batteries are still getting abit warm when its on, but as soon as i turn the switch off they heat up like mad! Im so confused
     
  4. security_man

    security_man

    10
    0
    Feb 10, 2011
    You are saying that when you turn off the switch the batteries get much warmer than when the switch is on. Well that doesn’t makes much sense to me, if the switch is off, then the power on all the circuit components should be zero, Well the only thing I can remember to advise you is, check if there is any short circuit in the circuit. Check if there is any low resistance path that you possibly aren't aware of. Use a multimeter to measure the resistance of your circuit, also measure the internal resistance of your battery. Check the voltage drop in the battery terminals when you observe the battery heating. If you don't have a multimeter, buy one, you can buy a good multimeter for about 20€, multimeters are great, you can measure a lot of things with it, they usually have a special mode for checking connectivity/short circuits, that emits a noisy beep when a low resistance is measured.
     
  5. davenn

    davenn Moderator

    13,597
    1,875
    Sep 5, 2009
    obviously the switch isnt switchin g the batter out of cct. sounds like its possibly putting a short cct across the battery

    time to see a cct for what you are doing/have done :)

    Dave
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,397
    2,777
    Jan 21, 2010
    Yeah, the circuit should go...

    one end of battery (say +ve) to switch
    other side of switch to resistor
    other end of resistor to anode of LED
    cathode of LED to other side of battery (-ve)

    This is a simple 1 LED circuit. Is that like what you have?

    Or do you have both ends of the battery going to the switch?
     
  7. mentholboy

    mentholboy

    3
    0
    Feb 7, 2011
    all sorted now, i was being a tool and put both ends of the battery into the switch :D thanks for the help guys :D
     
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