LED indicator circuit problems

Discussion in 'The Projects Forum' started by gtitim, Jan 21, 2009.

  1. gtitim

    Thread Starter New Member

    Jan 21, 2009
    3
    0
    Hi, i found this site whilst looking for a solution to the following problem.

    The purpose of the circuit is to indicate when electronic VIS motors on my car are turning on and off, so i know they are working correctly.

    I have built the following circuit as per this diagram:

    http://s268.photobucket.com/albums/jj23/turbotim511/?action=view&current=VIS2.jpg

    I have however changed the resistor (bottom left) for a 2k1 resistor as the designer recommended it.

    The problem i am having is that I get 12vthrough the LED and diode, but 0v after the resistor which means the circuit is not complete.

    How do i test the resistor to check i did not burn it out when i soldered it in, as i am no expert. I have a multimeter but no idea which setting i would need it on, or where to place the probes to check it.

    If i can master this, I hope to be able to design more gadgets for the vehicle, but am rapidly losing faith in my ability!!

    Any help is massively appreciated.

    Tim
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The illustrated circuit is pretty cruddy. Each diode and LED should have a separate resistor. 1k3 would be a good value for 12 volts, but any value from 1k0 to 1k6 will work.

    The earth side of the resistor should be at 0 volts (your chassis ground). Applying +12 to the anode of each LED will enable current through the resistor that will illuminate the LED. You want individual resistors, as the circuit shown will just get dimmer as each LED comes on.
     
  3. gtitim

    Thread Starter New Member

    Jan 21, 2009
    3
    0
    Only two inputs will be on at one time, so i am not worried about how bright the LEDS are.

    The LEDS have inbuilt resistors, the extra resistor is purely to prevent spiking to the ECU as a bit of overkill protection.

    I was just unsure if the resistor had been burnt out or was working as I test wired the circuit across the battery on the car by putting the earth side of the resistor to ground and then connecting each input to 12v in turn, but no LEDS lit up. I then checked the LEDS by grounding the circuit before the resistor and the LEDS lit up, so i couldnt work out if that was correct or if the resistor had failed
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The diagram calls for two CJ86Y LEDs, saying they are MAPLIN part numbers.
    MAPLIN does not carry CJ86Y LEDs, but they do carry CJ68Y, which are 3mm 12v yellow LEDs.
    The CJ67X are 3mm 12V green LEDs
    Since they are rated for 12v, they should not need external current limiting resistors.
    I'm not quite sure why the 1N4007 diodes were included, unless the originator was attempting to compensate for the additional 0.7v that an auto battery would be above 12v. However, when the engine is running, the alternator might put out as much as 15v, so that would put the LEDs at around 25% over their voltage rating.
     
  5. gtitim

    Thread Starter New Member

    Jan 21, 2009
    3
    0
    so in other words, being a complete novice and electronics idiot - i don't need that extra resistor? The LEDs used are the green and yellow 12v 3mm MAPLIN ones.

    If the circuit were to fail, it wont damage the ecu by backfeeding through the inputs, or through the earth?
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    It would short to ground (earth).

    But, using resistors in series with LEDs that have either internal resistors or internal regulation would limit the current through the LEDs to such an extent that they may not light at all.

    Let's just look at a 1k resistor across 12.7v.
    Current = Voltage / Resistance, or I=E/R.
    So, I = 12.7/1000 = 12.7mA (milliamps). That would be enough current to light a standard LED. However, you're using LEDs that already incorporate current limiters of some sort.
    But now, you've increased the resistance outside the LEDs to 2k.
    I = 12.7/2000 = 6.35mA. If the LED has a 1k resistor internally, the current will be closer to 4mA.
     
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