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How to stop LEDs lighting at low current

Discussion in 'LEDs and Optoelectronics' started by VoodooRoller, Apr 25, 2013.

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  1. VoodooRoller

    VoodooRoller

    9
    0
    Apr 24, 2013
    Hi everyone,

    I'm trying to design a circuit for some LED DRLs (daytime running lights) I'll eventually be installing in my car. I'll be tapping in to the grounding wire for the electronic decay/fade-out of the interior lights in order to get them to fade in with the interior lights when I unlock the car - this bit I'm fine with.

    However the electronic fade-out for the interior lights is controlled by a transistor deep in the body control module. The problem is that this transistor was designed to control filament bulbs, so when the transistor is in an "off" state it actually leaks some current through; not a problem for filament bulbs but it very dimly lights up my LEDs which will be on the outside of my car.

    I only know basic electronics so my question is whether there a component--or group thereof--out there that will only let current through if it reaches a certain threshold?
    For example, if there is 1mA flowing the component will break/stop the circuit, but if there is 2mA flowing the component allows it through.

    I made a test circuit with the Falstad java applet with just one LED here:
    http://tinyurl.com/vr-drl-circuitry

    NB: The capacitor is there to keep the DRLs powered between me shutting the door and turning on the ignition, and I've modified the value to exaggerate the fade-in/fade-out effect.

    The thing I'm looking for will need to go (I think) just before the diode on the DRL line. I've set the transistor to show a current that lights the LED in the applet but not the filament bulbs. This is not the actual value in the car. The transistor is controlled by both the door being physically opened and the door being unlocked remotely.

    Thanks for any help, I'm quite stumped with this one. Even though it's just solving an aesthetic problem, it will bug me to no end if it's left how it is!

    -VR


    Edit: I've added a screenshot of the circuit for those who cannot access the applet.
    [​IMG]
     
    Last edited: Apr 25, 2013
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
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    Jan 21, 2010
    Place a resistor in parallel with your LED(s) so that it drops less than the Vf of the LED at some small current.

    For example if you have a LED with a Vf of 3.5V, and you want it to go out totally when the current drops below 2mA, place a resistor in parallel with it of the value (3.5/0.002) = 1750 ohms (use 1k8).

    Be aware that at very low currents the Vf is lower, so a lower value resistor (say 1k in this case) would be better.

    Hopefully this provides enough information for you to work with.
     
  3. VoodooRoller

    VoodooRoller

    9
    0
    Apr 24, 2013
    Hi Steve, thanks for the reply.

    I put a resistor in parallel with my LED but it didn't seem to affect it unless it had a small value, in which case the current just bypassed the LED and made it dim during normal operation.

    I also tidied my diagram up and made it easy to simulate the door opening/closing, and upon playing around with different ideas I think I may have come up with a solution:
    http://tinyurl.com/vr-drl-circuitry-v2

    In practice it looks like I'll need to choose my relay's activation current very carefully, but in theory it seems to work.

    Here is what should happen in the simulation:-
    1. Car is unlocked remotely, transistor switches from .73V to .75V and activates parallel DRL relay
    2. Interior lights and DRLs fade-on
    3. Door is opened, no change
    4. I get in, door is closed, transistor switches from .75V to .73V and deactivates parallel DRL relay which in turn disconnects the DRLs from the battery
    5. Interior lights fade out and DRLs switch to capacitor supply (30-60s of power intended for final build)
    6. Ignition is turned on, DRLs switch back from capacitor supply to battery
    7. Drive etc.
    8. Ignition is turned off, DRLs switch to capacitor supply
    9. Door is opened, transistor switches from .73V to .75V and activates parallel DRL relay
    10. Interior lights fade-on, DRLs now supplied by battery again
    11. I get out, door closes, no change for 30 seconds (this is controlled by the BCM somehow)
    12. Car is locked remotely, transistor switches from .75V to .73V and deactivates parallel DRL relay which in turn disconnects the DRLs from the battery
    13. Interior lights fade out and DRLs switch to capacitor supply
    14. DRLs fade-out completely after 30-60s by draining the capacitor


    Can anyone see any inherent problems that would arise by using a relay in this way?

    Thanks in advance,
    -VR


    Edit: I've added a screenshot of the circuit for those who cannot access the applet.
    [​IMG]
     
    Last edited: Apr 25, 2013
  4. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Is anyone besides me curious why this transistor doesn't completely turn off?

    Chris
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
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    Jan 21, 2010
    Heh.

    I think the circuit needs to be traced a component or so back from the base of that transistor.

    Also VoodooRoller, the circuit as shown does not have a resistor in parallel with the LED. Where did you put it ?
     
  6. galaxy

    galaxy

    28
    2
    Nov 3, 2012
    In the second cct, the transistor is shown with 0.73V locked and 0.75V unlocked.
    So a 0.02V difference ?

    Doesn't sound right to me. Maybe looking at the wrong transistor. As (*steve*) said, maybe needs to be re-traced.
     
  7. VoodooRoller

    VoodooRoller

    9
    0
    Apr 24, 2013
    I cannot fathom as to why the transistor doesn't completely turn off, but it is deep inside the car's Body Control Module so there's no way I'm opening that up (costs several hundred pounds).

    As I mentioned, the transistor value is not true to real life, rather I chose values in the simulator that mimic and exaggerate the real life operation.

    I would just like to add that the 'leaky' current is tiny - you can only see the LEDs illuminate in near pitch darkness.

    I put the resistor in parallel over the LED and current-limiting resistor. It has just occurred to me that you might have meant put it across just the LED, however in practice the DRLs will arrive in a sealed module and so this wouldn't be possible.
     
    Last edited: Apr 27, 2013
  8. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    Uh, something odd in your schematic. I was going to tell you exactly what but you'll get more satisfaction from finding it your self. It has nothing to do with your leakage problem though. At least that I can see.

    Hint: The wiring of the relay you added and DRL.

    If anyone else answers this before VR attempts a go at it you'll be doing him a disservice.

    Chris
     
  9. VoodooRoller

    VoodooRoller

    9
    0
    Apr 24, 2013
    Thanks Chris, I completely agree that it is better to discover things by yourself rather than be given the answer straight away.

    The only thing that sticks out to me is that the relay coil is connected directly to its own switch. Would that be a problem?
    It works in the simulator when the coil is set to an activation current of 10mA over 12V, and there'll be still be current flowing through the coil when the switch is closed.
    With the 'leaky' transistor there will also be a small amount of current flowing through the coil when the transistor is closed (3.2mA in the simulator), but I'm going to check the actual value in the car and choose my relay carefully so that it is open with this small current.
    I'll try to get some readings from my car in the next few days.

    In practice I'm hoping to get the relay's resistor (the 1k in the diagram) to as high a value as possible to make sure the majority of the current is flowing through the DRL LEDs, since effectively the resistor is in parallel with the LED when the relay is closed. Sapping power from the LED is something I want to avoid, as DRLs need to be as bright as possible during operation.

    I'll also probably add a few more diodes after the resistor and down by the ignition relay just to be safe from unwanted current flow.

    If any of this isn't the problem you were thinking of then I'll mull it over for a while longer.
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
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    Jan 21, 2010
    Actually, noting the position of the extra 1k resistor in the 2 diagrams, I note that if it was intended to be what I suggested you add, it doesn't (it's connected to a totally different place, AND it is also likely to raise eyebrows for other reasons.

    YES, the 1k resistor should be placed directly across the LED to cause it t turn off more completely.
     
  11. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Take another look at where your LEDs are getting their power source.

    Chris
     
  12. VoodooRoller

    VoodooRoller

    9
    0
    Apr 24, 2013
    There is only 1 diagram with the extra 1k resistor and you're right, it does not depict your suggestion.

    Here is what I tried first:
    [​IMG]
    That was wrong and didn't work.
    (Note that the LED in my simulation is 2.5V hence the 1.2k resistor)

    Here is what I realised you meant:
    [​IMG]
    Which, while it works to bring the LED below 1mW in "off" mode, it is impractical to accomplish in the real situation since there will be over 100 surface-mount LEDs spread across 4 sealed modules.

    I also noticed that putting a resistor across the LED in this way will sap the power by about 5mW in the "on" state; however upon further inspection this is also the case with the 1k resistor in my new relay circuit, which is annoying.



    Through the relay coil... is that a problem? I thought that as LEDs are low-power it shouldn't.

    Would this be better?
    [​IMG]
    http://tinyurl.com/vr-drl-circuitry-v2-1

    In fact, why didn't I do that in the first place? Haha!
    It also seems to solve (or lessen the effect of) the power being sapped from the LED.

    Thanks Chris
     
  13. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    VD, you did it! You found the errant wiring on your own. A gold star is bestowed upon you.

    I do have a request and some observations though. First the request...

    (1) Please name the components in your schematic, like R1, D2, K1, etc.
    (2) Place an asterisk next to any component that's not part of the OEM design. Example: *K1
    (3) The Base Emitter voltages shown on your print don't make much sense. Did you actually measure this?

    Observations & Suppositions:

    Few power controlling devices are linear these days. When a design has a choice (some don't) of linear vs digital, digital wins hands down because dissipated power from the controlling device is greatly reduced. PWM is a form of digital technology in that active component states are either on or off and that's where I'm going here. I could be very wrong but I'm guessing that the Base of that Transistor is receiving a PWM current not a linear decaying current.

    Chris
     
    Last edited: Apr 28, 2013
  14. VoodooRoller

    VoodooRoller

    9
    0
    Apr 24, 2013
    *takes a bow* thank you, thank you.

    I shall make sure all relevant components are aptly labelled in future diagrams.

    The base voltages were specifically chosen by me to mimic the operation and exaggerate the undesirable effect to a visual level for easy diagnoses during tests.
    0.74V was the threshold in the simulator in which the DRL was visibly on but the interior lights were not. This value is actually irrelevant, so much so that I'm blocking out the base voltage values in subsequent diagrams to avoid any further confusion on the matter.

    Again I don't know why the transistor doesn't turn off fully - all I know is that it doesn't. I can't change this so I need to focus on a solution that doesn't involve the transistor; it probably only lets a few uA through anyway but it's enough to notice if you really concentrate on the LEDs.


    My next step is adding in yet another relay to the sidelight circuit that dims the DRLs when the sidelights are on (conforming to UK regulations on DRLs).
    No problems here but I thought I should post it anyway as it finishes off the thread nicely (although if you can see any obvious problems please let me know!)

    [​IMG]
    http://tinyurl.com/vr-drl-circuitry-v3

    A big thank you to everyone has helped me in my troubleshooting for this project :)

    -VR
     
    Last edited: May 5, 2013
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