LED tails...ad nauseum

Discussion in 'General Electronics Chat' started by doug3460, Oct 27, 2008.

  1. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    Question: if I place a diode (i.e., 1N4004) coming off the brake light wire, & another diode (same type) coming off the running light wire then connected to a resistor (i.e., 68Ω), then connect both these to stock 12V LED running lights, will they brighten when the brake is activated?

    Background: I'm working on my motorcycle. I want to convert 12V 'plug & play' LED truck clearance reflectors to run/brake lights. I have added the LEDs to the saddlebags on my motorcycle.

    I'm not positive about the resistor size since I have no data on the lights except they're 12V but I can play w/ the resistance to get what I need.

    I have searched the net extensively, including this site. The 3 threads that come up on this site really only gave one solution: Sgt Wookies, but that is not doable short term because a) I have no shottky diodes; b) I have no pnp transistors. :D
     
  2. SgtWookie

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    Jul 17, 2007
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    That might work.

    Can you make a schematic of what the circuit board looks like? Or take a photograph of both sides of the board and post it?
     
  3. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    Thanks for the reply. I made a schematic in Eagle (which I linked to from this site, tyvm), so it's taken me a bit to learn the tutorial. Hope this helps.
     
    Last edited: Oct 27, 2008
  4. SgtWookie

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    Can't read your Project1.sch. I have an older version of Eagle Layout Editor.

    Export it as an image to a .PNG file. 100 DPI resolution would be good.
     
  5. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    working it as we type. lol. thanks for your patience.
     
    Last edited: Oct 27, 2008
  6. SgtWookie

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    I'll have to look at it tomorrow. Can't keep my eyelids propped open any longer!
     
  7. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    no problem. apologies it took so long. couldn't figure out how to get the file size down to uploadable. I'm a night shifter, so will catch-up after next few nights of work. Appreciate the help. :)
     
  8. SgtWookie

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    Oh, well you've just drew the connections you described earlier. I thought you were going to draw a schematic of the inside of the LED running light? :confused:
     
  9. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    Apologies for delay in getting back. As I stated, I can't get the sealed unit apart & the website offers no help. I can describe what's there: two part unit - the base contains a rubber non-polarized female plug w/ one lead coming out of it. The ground is a simple strap to a mounting screw. I've replaced these for my application w/ a two wire trailer harness. The light unit has the male plugs coming out the bottom & under the lens are 6 LEDs, 4 diodes which I cannot identify & 2 small resistors. This is their arrangement:

    resistor diode diode
    LED LED LED LED LED LED
    diode diode resistor

    So, we have 2 arrays of 3. I don't have the knowledge to understand how the diodes are in the circuit to permit the plugs to be non-polarized, but I'm confident that's their purpose.

    Assuming standard LEDs in the red-orange range, AlGalnP type, the typical values would be 2.2V, 20mA. That would mean the resistors in the unit are something like 330Ω, 1/8 watt, which would be right for their physical size (I can't read the bands thru the colored reflector).

    So, if I want to drop the Amps to, say 15mA for my running light function, Ohm's Law tells me I need a 440Ω resistor & that can handle .081A (1/8 watt). Nearest resistor I have to that is 470Ω, but that will do fine.

    I hope this helps you review my plan. I realize this is way below your level of work, but it's my stepping stone: next project is to attempt a combination flasher & pulser circuit for the brakes without using a PIC. :D I figure it's some sort of combination of monostable & astable gate arrangement. I'm getting there, only been doing this a few weeks.

    Thanks for the help & review. I realize how valuable time is.
     
  10. SgtWookie

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    Jul 17, 2007
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    I have some red-orange, AlGalnP type Luxeon LEDs; they give Vf's of 2.2v (min), 2.6v(typ) 3v(max) @ 70mA.

    I suggest that you measure the current through the lamp assembly as it is now.
    The safe way to do this is to use a 1 Ohm resistor in series with one of the power leads, and measure the voltage drop across it, using a 13.8v (nominal) voltage supply. Since I=E/R, and R=1 Ohm, the voltage across the 1 Ohm resistor is the current flowing through it; eg: 20mV = 20mA. I suspect that the LEDs in the light are the Luxeon type, unless they are extremely cheaply made.

    Let's assume for the moment that the LED's Vf is indeed 2.6v each. The diodes are the tough part; I don't know if they're Schottky or standard rectifiers. Let's assume for the moment that the diodes will drop about 0.7v across themselves.

    In a full-wave bridge rectifier, two diodes will be in series with any current flow. Since each diode has a Vf of 0.8 (ballpark) the diodes in total will have a 1.6v drop.

    If the LEDs are indeed grouped in two strings of three (which is also my best guess, given that there are two resistors) then there will be 3 x 2.6v = 7.8v drop across them.
    So, for the total Vf across all LEDs and diodes, we wind up with 7.8v + 1.6v = 9.4v.

    Let's look at the supply side for a minute. When the vehicle is running, the alternator will be powering the system and charging the battery. If the battery is really depleted, the alternator may be putting out around 14.5v; if battery is charged and no accessories are running, the alternator may be putting out around 13.4v. Let's go with 13.8v as a typical value.

    So, we have a 13.8v supply, and the LEDs & diodes have a total of 9.4v drop.
    13.8 - 9.2 = 4.4v left over to drop across the current limiting resistors.
    Now is where the actual current flow would come in handy. But let's make some guesses on what the values of the internal current limiting resistors might be.
    If the LED current is 70mA, then Rlimit = 4.4v/70mA = 4.4/0.07 = 62.9 Ohms, 308mW
    If the LED current is 25mA, then Rlimit = 4.4v/25mA = 4.4/0.025 = 176 Ohms, 110mW
    etc.
    When selecting a resistor wattage, rule of thumb is twice the actual wattage required, then select a resistor that is rated >= to that result.

    Since you're adding diodes in series, you're going to decrease the voltage available to the circuit from 13.8 down to about 13. This will change the amount of current flowing through the LEDs. Run the numbers the same way I did to see the new results.
     
  11. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    Excellent help! I'll go examine it with your suggestions now. I'm off the next few nights & intend to have this up & running soon. I do believe these are cheaply made because the LEDs are the T1-3/4 size. I'll get back with the results.

    By the way, I re-read my post & realized I forgot to correct for the resistors already in the lamp. What I meant was the resistor I would add would have to total that 440 ohm target if I want to shoot for around 15mA. Anyway, let me explore it & thanks again.
     
  12. SgtWookie

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    OK.

    I forgot to mention that if each string had 25mA running through it, then the total current observed would be 50mA.

    The reason they use separate current limiting resistors is because the Vf of LEDs can vary up towards 10%, even in the same batch. A graph of the distribution of Vf for a given current for a batch of the same type LEDs would look like the traditional bell (Gaussian) curve.

    Try values from roughly 47 Ohms to 120 Ohms. I suggest using 1 Watt resistors, because you'll likely be insulating them using shrink tubing. The shrink tubing will slow the dissipation of heat.
     
  13. doug3460

    Thread Starter Active Member

    Oct 19, 2008
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    I was called-in to work for a couple nites, so couldn't reply. Lights are now installed & functioning fine :cool:. Here's the numbers:

    Measured current was .018. That means the resistors they probably used are 330Ω, 1/4 watt (doubling .101 & rounding up). LEDs in that range average 2.1 Vf. Doing the math meant I should have been able to connect the unit to a 9v battery. I did and it worked.

    I ended up going with 150Ω, 1/2w resistor. That put the run mode around 10mA & brakes around 15mA. Of course, I'd need to go to <9mA run mode to get a really big difference, but this suffices.

    I realize the lights are cheaply made, but it was their dimensions, sealed unit & reflector lens in addition to cost that made them appropriate. Lots of red lensed LED lamps out there - not a lot with reflector lenses in the size I needed.

    I picked-up on some things you wrote & did further study on diodes, rectifiers and semi-conductors in the AAC volumes. I'll be starting my flashing/pulsing circuit project soon & hope to have a schematic together by the weekend. But that's a different thread :D. Thanks again for your help & advice. It was exactly what was needed. :)
     
  14. SgtWookie

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    Jul 17, 2007
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    Glad to hear of your success! :cool:

    Be careful of those cage drivers out there. ;)
     
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