Motorcycle LED Tail/Brake Light - HELP HELP

Discussion in 'The Projects Forum' started by foeslts, Nov 22, 2009.

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  1. foeslts

    Thread Starter New Member

    Nov 20, 2009
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    Hello,
    Me and my dad are working on restoring his old Honda motorcycle. We are trying to make the turn signals and tail/brake light with leds. We both have a basic understanding of electronics and are willing to try any suggestion that you guys might have.

    We have the basic design for both the turn signal and tail/brake light. We can't figure out how to have the taillight drive our 40 leds at at a reduced brightness and then when the brake is applied have the same 40 leds get brighter. Do we use a relay, diodes, etc.........?????:confused: ANY help will be much appreciated.

    These are the specs for our designs:

    Turn Signal
    - (4) 180 Ohm, 1/8 watts resistors
    - (16) - 5mm ORANGE LEDs
    Forward Voltage (V) : 1.8 ~ 2.2
    Max Continuous Forward Current : 30mA
    Max Peak Forward Current : 75mA



    [​IMG]



    Tail/Brake Light
    - (8) 56 Ohm, 1/4W watts resistors
    - (40) 5mm RED LED
    Forward Voltage (V) : 1.8 ~ 2.2
    Max Continuous Forward Current : 30mA
    Max Peak Forward Current : 75mA


    [​IMG]
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    You are going to fry your LEDs if the electrical system puts out anything over 13.8v.

    What is the system voltage when the engine is off?

    What is the system voltage right after the engine starts, and is running at a fast idle?

    What is the system voltage after the battery is fully recharged with the engine running?
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The best way to have two different brightness levels is to PWM for dim, and run DC for bright. A 555 multi and a FET will do the PWM; OR a signal to the FET gate for full bright.
     
  4. k7elp60

    Senior Member

    Nov 4, 2008
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    I agree with SgtWookie. I would replace the 180 ohm resistors with 270 ohm resistors. Redue the brake light to only have 4 LED's in series and also us 270 ohm resistors.
    Use the existing circuit that is posted to cause the LEDs to get brighter on the braking.
     
  5. ftsolutions

    Active Member

    Nov 21, 2009
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    Does your bike only have (2) wires running to the rear tail light? On my BMW, I've got (3) wires - for running, ground, and brake. If you've got (3) wires like this, you could simply have every other LED (or row of LEDs( wired through to the running light line, and the other rows wired to the brake wire for 1/2 the LEDs on during running operations, and all the LEDs on for braking.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    I said above that you'll fry your LEDs if you exceed 13.8v.

    Let me explain this assertion.

    Here's your schematic for the red LEDs:
    [​IMG]

    You specified:
    In this situation, you should not exceed the 30mA continuous current, ever.
    The Max Peak forward current is for when LEDs are being operated on a low duty cycle, with a very brief ON time.

    Let's assume that the average Vf (forward voltage) with 30mA current will be 2v.
    So, 5 x 2v = 10v dropped across the LEDs.
    If the engine is not running and the battery has a good charge, the electrical system might measure 12.7v.
    So, 12.7-10v = 2.7v.

    You're using 56 Ohm resistors in each string.
    2.7v / 56 Ohms = 48.2mA - you've already exceeded the maximum continuous current rating by a large margin.

    As a matter of fact, the battery would have to be discharged down to 11.68V before the LEDs would receive the 30mA they were designed for.

    11.68v-10v=1.68v; 1.68v/56 Ohms = 30mA.
     
  7. foeslts

    Thread Starter New Member

    Nov 20, 2009
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    Thanks for the quick replies.
    I just bought a book on basic circuits so hopefully soon I won't be making or asking dumb questions/mistakes.

    k7elp60
    thanks for the diagram, how could I modify it to incorporate the 3 wires the bike has going to the taillight (ground, running light, brake light).

    ftsolutions
    The bike has 3 wires going to the taillight (ground, running light, brake light). I thought about wiring every other led, but I really want the usefulness of all 40 leds.

    SgtWookie
    I had just assumed that a 12 volt battery put out a constant 12 volts - sorry newbie mistake - i read up on batteries last night so I understand the concept a little better.

    When designing the light, I used one of those led/resistor calculators. I had just assumed that it would come up with an optimal configuration - the easy way is usually the incorrect way to go.:D

    I have read through the basics of leds, but your last post actually makes things really clear.

    [​IMG]

    How does this one look?

    Thanks again for everyone's help with this. Electronic circuits are really interesting. I look at all the stuff people are doing on this site and it makes me excited. I am ready to dig in.

    Thanks,
    Paul
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Trust me, they aren't "dumb questions". It's something very new to you, and you're bound to make mistakes. We will try to help prevent those little mistakes on paper/pictures from becoming circuits that experience a melt-down. :eek:

    OK, the first thing you need to understand is the Vf range for the specified current. This a very typical error that new people make, and it can quickly lead to burned-out LEDs.

    It's been specified as Vf = 1.8v to 2.2v @ 30mA.
    You used the highest voltage specified. This is a mistake. You would be better off to use the average of the two numbers, eg:
    (1.8v+2.2v)/2 = 2v

    If you measured the Vf of 100 LEDs, you would find that about 70% of them would measure very close to 2v; your average or typical Vf. The remainder would be spread anywhere from 1.8v to 2.2v.

    Sheer random distribution gives you a pretty good chance of getting strings of 4 series LEDs that will have about 8v total with a 30mA current.

    The next thing is the electrical system voltage. A fully charged lead-acid battery will generally measure somewhere between 12.6v and 12.9v when it's internal temperature is around 25°C; depending on some internal chemistry. As the battery discharges, it may fall as low as 11.5v before the battery is considered completely discharged.

    However, when the engine is running, it's a different story. If you start it up and the battery is heavily discharged, the generator/alternator might put out up to 14.5v to get that battery charged. This is where your calculation will run into real trouble; with 4 in series and 180 Ohm resistors, you'll have:
    (14.5v - (4x2v))/180 = 6.5/180 = 36.1mA

    You are now operating the LEDs at more than their rated current, and their life will be short.

    Let's increase the resistors to 220 Ohms.
    (14.5 - (4x2v))/220 = 6.5/220 = 29.5mA - just under the limit.

    Let's see what happens to the current when just running off the battery; partially discharged at 12.4v:
    (12.4 - (4x2))/220 = 4.4/220 = 20mA - they will be about 2/3 normal brightness.

    Let's try again with just 3 LEDs in series per string.
    Rlimit >= (14.5 - (3x2v))/30mA = 8.5/0.03 = 283.333 Ohms.

    Here is a table of standard resistance values: http://www.logwell.com/tech/components/resistor_values.html
    E24 values (the green columns) are typically available locally to hobbyists; you'll generally have to order E48/E96/E192 values.

    You could order a 287 Ohm resistor, or just go with a 300 Ohm resistor. Let's see what you'll get with 300 Ohms:
    8.5v/300 Ohms = 28.3mA
    You probably wouldn't be able to tell the difference in brightness between 28.3mA and 30mA.

    Now let's see what happens with a battery run down to 12.4v:
    (12.4v - (3x2v))/300 = 6.4/300 = 21.3mA - still a significant drop in brightness, but not nearly as bad as before; we're at 75% brightness instead of 66%.

    Also, you have better protection in case the input voltage exceeds 14.5; maximum LED current of 30mA wouldn't be reached until system voltage hit 15.
     
  9. foeslts

    Thread Starter New Member

    Nov 20, 2009
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    wow - that makes alot of sense. i had no idea that the voltage of a battery would vary so much - but it makes sense.

    I know this is down the road a little in the learning curve, but could a capacitor be used to even out the voltage?
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Not really. A lead acid battery is sort of like a really, really huge capacitor - but not really.

    Capacitors have a very different discharge curve than a battery does. With a resistive load, a capacitor discharges very rapidly at first, but takes a long time to become fully discharged.

    A battery will maintain within about 10% of it's charged voltage until it is very nearly completely discharged; then it's voltage falls rapidly.

    In order to charge a 12v lead-acid battery, you need to charge it at a somewhat higher voltage; about 13.8 to 14.5v. Ideally, the charging current will be limited, but most automotive alternators don't limit the charging current.

    That's why you really don't want to jump-start a car from another; because very heavy currents will flow. You risk damaging the batteries and the alternator.
     
  11. foeslts

    Thread Starter New Member

    Nov 20, 2009
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    hmm good to know.

    So how to I go about dimming the leds for the running light and making the leds brighter when I brake. I have a couple of designs and thoughts, but I am still unclear on how to go about it.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    To go any further, we need to confirm a bit more about your wiring.

    Was the original brake light a 1157 type bulb? Two filaments, one for park, one for brake?

    Most older auto-type systems used one ground (to the frame) and one 12v wire for parking/running lights, another 12v wire for (a) brake light(s).

    Is yours like that?
     
  13. legac

    Well-Known Member

    May 4, 2005
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    Hi
    I attach the schematic of tail/brake light. There are two LM317 which serve as constant current sourse. The tail light is using AC source from generator via 1 ampere diode rectifier and 4700UF filter. The brake light is from 12vdc.Two other diodes serve as isolators. Resistor is calculated for about 100 mA/tail and 260mA brake.
    LEDs are of 10 lines, each line comprises 4 of 3mm HB LED ( 2,2v/30ma)
    If both tail/brake light are from 12 vdc, then diode and C input are not needed.
    The circuit works well.
     
    Last edited: Nov 24, 2009
  14. SgtWookie

    Expert

    Jul 17, 2007
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    Here's a simplified version of what you can do:

    [Schematic removed; we no longer provide assistance for such projects.]

    Ignore the 14v symbol, S1 and it's connecting wiring; that was just for testing.

    I only put in three strings of LEDs. R5 and the 6v battery represent the load another 10 strings of three LEDs would have.

    D99 is a rectifier diode rated for 3A. It allows current to flow from the parking light switch to the taillights, but prevents the brake light switch from powering the parking/running lights. It will drop about 0.7v across itself when current is flowing through it.

    Rpark is a 30 Ohm resistor rated for 2 Watts. The value doesn't have to be exact, but it should be somewhere between 20 and 30 Ohms.

    When the brake switch closes, 14v is applied to the LED array current limiting resistors; each string of LEDs gets about 28mA.

    When the brake switch is open and the parking switch is closed, roughly 12mA flows through each string.

    You may need to adjust the value of Rpark somewhat, but somewhere between 20 Ohms and 30 Ohms should be about right.

    [Schematic removed; we no longer provide assistance for such projects.]
     
    Last edited: Aug 14, 2011
  15. foeslts

    Thread Starter New Member

    Nov 20, 2009
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    SgtWookie
    Thx so much for the diagram. I just bought a breadboard, a bunch of resistors and some leds. It's time to experiment - woohoo.

    I have a question about resistors.
    Are (3) 1/4 Watt, 100 ohm resistors in series the same as (1) 1/4 watt 300 ohm resistor.
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    Wheee!

    Three 1/4 Watt, 100 Ohm resistors in series will have the same resistance as one 300 Ohm resistor.

    However, the power will be split between the three 100 Ohm resistors; so they will be able to dissipate 3/4 Watts instead of just 1/4 Watts.

    Also, you will have to make many more connections. The more connections you have to make, the more complicated, large, expensive and less reliable the circuit will be.

    I would not suggest building the entire array on a breadboard; perhaps three strings would be enough.

    You will have to adjust the value of "Rpark" upwards to scale it to the size of your string array.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    By the way, Legac's circuit is not without merit.

    However, it's somewhat dangerous. If strings start to fail, the regulator will attempt to keep the same current flowing through all of the other strings. This will result in a "domino effect" or "China syndrome", where suddenly all of the strings "pop" due to excessive current.
     
  18. foeslts

    Thread Starter New Member

    Nov 20, 2009
    7
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    here it is, please let me know if you guys have any suggestions, or ways to make this thing any better.

    [​IMG]

    Thx,
    Paul
     
  19. SgtWookie

    Expert

    Jul 17, 2007
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    Other than D1 being backwards :eek: it looks fine to me.

    You have to get D1 fixed, or your running lights won't work unless your brake is on.

    [eta]
    Not sure about R7, R21, R27 yet.

    I'll have to re-read the thread.

    [eta]
    R7, R21, R27 should be 430 Ohms, not 470 Ohms. Otherwise, the LEDs may have a noticeably different brightness.
    The resistors for these will be dissipating a good bit of power.
    (14v - 2v) x 30mA = 12 x 30mA = 12 x 0.03 = 0.36W, double for reliability, that's 0.72W.
    If they will only be on for a short period of time (like turn signals) you could get away with 1/2 Watt, but not for the brake light. Use 1W for reliability.

    Note that your flasher may not work properly after wiring up the LEDs. This is because the LEDs will not draw enough current to heat up the element in the flasher. You will need to wire a power resistor in parallel with the LED array to draw enough current to cause the flasher to flash, or replace the flasher.
     
    Last edited: Nov 28, 2009
  20. foeslts

    Thread Starter New Member

    Nov 20, 2009
    7
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    SgtWookie,

    Thank you so much for all of your help. I have to order most of the resistors (most places in town carry very few options), but when they come in I will start this project ASAP. I think I will order from Newark.com.

    http://www.newark.com/vishay-bc-components/pr02000203009jr500/metal-film-resistor/dp/53M5241

    http://www.newark.com/vishay-bc-components/pr01000104300jr500/metal-film-resistor/dp/53M5172

    http://www.newark.com/multicomp/mcrc1-4g331jt-rh/carbon-composition-resistor/dp/72K6194


    Who do you use when ordering resistors, etc... ?

    Thx again for all your help.
     
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