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Setting up idiot led's...

Discussion in 'LEDs and Optoelectronics' started by Korishan, Apr 1, 2017.

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  1. Korishan

    Korishan

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    Sep 2, 2015
    Hi,

    I have a work truck that has had an independent ground line ran from wire cluster under the dash to the taillights. The previous owner had done this so they could tell when one of the bulbs blew by having the ground line run through a filament bulb before going to the lights.

    I have upgraded the taillights from filament to leds. The leds lights don't draw enough current to power the filament bulbs on the dash. I was thinking of replacing the filament bulb with a led. However, I know I can't/shouldn't run the ground through the led to power the taillights. But, that I could use a transistor (thinking an NPN) that would turn on the led idiot light when there was current being supplied to the taillights.

    The problem is, I can't quite figure out how to set this up. I am new to the electronics field, and have watched several youtube videos on npn, pnp, mosfets, etc, and I'm still fuzzy about it all.

    I think I need an npn transistor that has a resistor bridging base/collector (?) with the emitter going to the led. When power is supplied, the transistor would turn on providing the negative supply to the led. Am I thinking correctly here? If so, what size resistor would i use?

    Btw, the led idiot lights are 12V automotive grade, so they have a resistor built in the unit.

    I think that's everything.

    Thanks,
    Kori
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    A common circuit for detecting current flow is this:
    upload_2017-4-1_18-6-53.png

    The current though the taillight (taillightcurrent) develops a voltage drop V=R*I across the bottom resistor R1. When the voltage across R1 reaches 0.7 V (approx.), transistor Q1 will turn on and the 'idiot'light (modeled by LED D2 with current limiting resistor R2) will turn on.
    When there is no taillight current, the 'idiot' light will be off.

    The equation for R1 is given as R=1.2*0.7V/taillightcurrent (current in Ampere). 0.7 V is the (approx.) turn-on voltage of an NPN transistor (Q1). The factor 1.2 takes into account a certain variation in taillightcurrent and in sensitivity of Q1.
    R1 needs tp be rated for a power of P=0.7V*taillightcurrent.

    The choice of transistor Q1 depends mainly on the current through the 'idiot' light. You need to chose a transistor with a maximum allowed collector-emitter current higher than the current required by the 'idiot' light.
     
  3. duke37

    duke37

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    Jan 9, 2011
    I would add a resistor (1k?) in series with the transistor base to limit the transistor current otherwise an accidental large load current will go through the transistor and damage it.
     
  4. Korishan

    Korishan

    10
    0
    Sep 2, 2015
    Thanks for the info. That makes sense. I was kind of thinking of something similar, but couldn't quite get my thoughts on paper and know the proper values/equations for the other parts.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The circuit given above requires a voltage drop of at least 0.6V. If this is a problem (and it might be more so for a 12V system than a 24V system), then a circuit like this requires 0.1V (or less) to be dropped across the resistor.

    Current Flow.png
     
  6. Korishan

    Korishan

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    Sep 2, 2015
    steve: How come you're using 2 diodes in series? D1/D2 and D3/D4
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I'm doing that to bias the bases at a constant voltage above the ground. Assume no current is flowing, Vbe of both transistors is about 0.65V and the bases are at 1.3V, so the voltage across R5 is 0.65V (and the current is therefore 6.5mA.

    If everything is nicely balanced, the current through both R2 and R3 is equal and thus the voltage across the LED is 0V.

    When current flows through R6, the voltage at the base of Q2 becomes higher than the voltage at the base of Q1 by the voltage across R6. Because the emitters are the same voltage, this changes Vbe for both transistors.

    Because transistors are VERY sensitive to the voltage across the base/emitter junction, this causes Q2 to turn on more, increasing the voltage at it's base so that it's Vbe is around 0.65V. Because the voltage at the base of Q1 has not changes, the increase of the voltage at its emitter causes it to turn off.

    This action causes the voltage at the anode of the LED to rise above that of the voltage at the cathode. at some point this imbalance is great enough to turn on the LED.

    Practically speaking, this circuit needs some tweaks, increasing R2 and R3 and decreasing R5 is probably a start.

    Current Flow.png
     
    Last edited: Apr 2, 2017
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    :)
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Probably far too complex, and not protected against spikes, but this will detect current flow with more than about 15mV drop across the sense resistor.

    Current Flow 2.png
    With the values as shown, the LED turns on fully with about 150mA being drawn by the bulb. R6 can be reduced for higher current loads.

    If you want to play with it, this might help.
     
    Last edited: Apr 2, 2017
  10. Korishan

    Korishan

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    0
    Sep 2, 2015
    Greatly appreciated! I like the various options and descriptions of each one. Gives me some things to work on and try out. Thanks again! :)
     
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