Stuck need help

Discussion in 'The Projects Forum' started by Locruid, Jan 19, 2008.

  1. Locruid

    Thread Starter New Member

    Jan 18, 2008
    I am A school bus tech in Oregon. It was recently brought to our attn that we needed a single Flashing Green Light for when our wheelchair lift doors are open. they currently have a constant red light. So we ordered a few small green LED lights (for automobile) and some Flasher units (typical car style for signal lights). I hooked them up and THEN figured out they dont pull enough juice to cause the flasher unit to initialize the Flash. As a temp backup, I installed a DC motor to cause enough current draw to activate the flash but really dont want to keep the motors in the Buses permanently lol.

    Anyways, it was suggested to get 1/4W Resistors and tie them in. I did that but now the bulb wont even work at all. With resistor no go, without resistor Solid green light.

    Would prefer the CHEAPEST alternative or maybe im not hooking these things up right. I have the ground on the frame and the positive goes to the resistor then to the Flasher unit.

    not very knowledgeable in this area but I can rebuild the 444 or 365 in that bus lol

    ANY help would be appreciated. would like to use the equipment I already have but if one of you professionals can steer me in the right direction I would appreciate it.
  2. eeboy

    Active Member

    Sep 27, 2007
    First off... I don't know much about the flasher so I will assume what you have said about a minimum operating current is correct. If this is indeed the case, a simple resistor should be sufficient.

    However, the resistor needs to be properly sized. I suspect this might be the problem. First, you need to select the resistor value to get the desired minimum current. To do this simply take the maximum voltage present (I assume it's rougly 13.8VDC.. no different than any other auto) and divide it by the minimum current required (R=V/I). The result is the maximum resistor value you would need. Of course you will likely not be able to purchase the exact resistance you have calculated so choose the next closest value that is less than the calculated value.

    Next, you need to make sure the resistor is rated properly for the application. When you choose your resistance you will need to calculate the power dissipation. This is calculated by taking the voltage squared and dividing it by the chosen resistor value (P=V^2/R). This value will be in watts. You will want to choose a resitor with a rating that is greater than this calculated value. It's good to build in a buffer!

    This all assumes that you know that minimum required current value. You could always use rated current for the DC motor in the calculation because you know it works when it's hooked up.
  3. hgmjr

    Retired Moderator

    Jan 28, 2005
    Greetings locruid,

    You mentioned that the resistor wattage was 1/4W, however you failed to mention in your description the value of the resistor.

    I suspect that the motors were placed in parallel (across) your LED to get the flasher unit to work. I think the resistor should also be placed in parallel with the LED. As eeboy has mentioned, some calculation will be needed to determine the optimum value of resistance needed.

    I further assume that the LED is rated for 12V to 18V. Information that came with the flasher unit should have indicated the minimum load the unit needed to operate.

    Can you check and see what that minimum load value is? It may be expressed as ohms or it may be expressed in amps.

    I suspect all that is needed is to determine the proper value of resistance and connect it in parallel and you will be all set.

  4. SgtWookie


    Jul 17, 2007
    If you don't see ratings on the flasher unit, that's OK - you can still figure it out using the motor that's been wired in parallel as a load to make the flashers flash and a multimeter set to read amps.

    Ground one of the motor's leads. Set your meter on the 10 Amps scale (usually you have to move the red lead to a different jack on the meter). Connect the other motor lead to the battery using the meter's leads, and read how many Amperes the motor draws.

    Once you have the ampere reading, you can figure out it's resistance while it's running using Ohm's Law.
    Resistance = Voltage / Current
    R = E/I
    For example, if your motor was using 2.5 Amperes:
    R = 12.7 / 2.5
    R = 5.08 Ohms
    You can round that off (down) and say it's 5 Ohms.
    Now, figure out the power (Watts)
    Power = Voltage x Current
    P = E x I
    P = 12.7 x 2.5
    P = 31.75 Watts
    (that's without the alternator running, let's see what happens when it is running)
    P = 14 x 2.5
    P = 35

    In this case, you will need a 5 Ohm resistor rated AT LEAST 35 Watts.

    Radio Shack carries 50 Ohm 10W wire wound resistors for $1.99 in a 2-pack; you could put 10 of them in parallel to get 5 Ohms, but that's getting expensive.

    Auto Zone carries an ignition ballast resistor, part # AL795 for $4.99. This was for 60s vintage Dodge/Chrysler ignition systems. I don't know what the resistance or wattage rating is offhand, but I DO know they're mighty beefy.
    Here's a pic:
    Notice the convenient mounting tab and 1/4" terminals for wiring. Handy!

    These ballast resistors were for limiting the current through the primary winding of the coil during normal operation of the engine, dropping the 13.8V down to around 8V at the coil. You would have to measure one using an ohmmeter to find out it's resistance. I suspect you'd need a couple of them in series.
  5. hgmjr

    Retired Moderator

    Jan 28, 2005
    Greetings locruid,

    Hopefully you can find out the minimum loading information needed straight off a label on the flasher unit but failing that, there is a good chance that the manufacturer's website will contain the information you need.

    If you are uncertain, you can supply us with a part number and manufacturer for the flasher unit. We can search the Web for the information needed to determine the optimum resistance. It would also be helpful for you to provide the manufacturer and part number for the LED you are using.

    Of course you will what to choose the value of load resistance carefully. There is no need to expend any more current than is needed to get your flasher unit to operate.

    Compliance with ADA regulations is worth the effort. We look forward to assisting you in bringing your project to a successful conclusion.

  6. Locruid

    Thread Starter New Member

    Jan 18, 2008
    WOW, lol wasnt expecting so much help!

    Thanx a ton for the replies.

    The Flasher unit is avail from Napa I'll get that info asap.

    The Light is from our private supplier (Allied Bus Sales) and since it's the weekend and i'm off, Ill have to wait till Monday to get the actual inf from the lights. I do know it's 12V but beyond that, i'll have to wait till Monday.
  7. Gadget

    Distinguished Member

    Jan 10, 2006
    Are you using a mechanical flasher ?
    I would be using something like a 555 or similar.
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    He's using a standard automotive flasher.

    The attached is Siemens Electromechanical Application note for automotive flashers.
  9. hgmjr

    Retired Moderator

    Jan 28, 2005
    I think you are correct Joe.

    I happened across the same document in my search.

    Do you think he has a thermal-actuated flasher? It sure sounds that way since his requires a minimum load to flash. I think the more recent versions are electronic and are less dependent on a minimum load to operate.

    What is your thinking?

  10. Locruid

    Thread Starter New Member

    Jan 18, 2008
    Flasher Unit:
    Tridon 552 DOT. 12V, 32cp, Max 5

    Solico 14v, 1W

    Thats what we are using
  11. rwmoekoe

    Active Member

    Mar 1, 2007
    i agree w/ gadget.

    the world would be better off with conservations such as these.
    build a new flasher using 555 and an addition npn trans (say bd139) at the output.
    and give away the mechanical flasher to someone who needs it more :).
    this conserves the energy (current consumption and all), and the money to buy the power resistor, and the materials used to make the flasher and the power resistor.