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Help needed with a simple LED circuit

Discussion in 'LEDs and Optoelectronics' started by Alan Jones, Jun 26, 2014.

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  1. Alan Jones

    Alan Jones

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    Jun 26, 2014
    Hi all, I hope someone can help me. Please forgive my ignorance, I have no background in electronics, but I can use a soldering iron and have a basic grasp of circuits. I have a motorcycle with a small LED in the instrument cluster which is a shift light. This lights up at a preset point to indicate when to shift to the next gear, the trouble is it is very small and easily missed when going at speed. The shift point is set within the instrument cluster as is the method of illumination, it can either be steady LED or flash. For anyone interested in seeing one in action, here is a short youtube video of someone who has already done this modification.



    I believe the voltage at the LED is about 2.3v, but I will confirm this if/when I pull the cluster apart. My idea is solder a wire to the +ve side and run it outside the cluster and connect this to the base pole of an NPN transistor, something like this one which is rated at 1.5amp
    http://www.ebay.co.uk/itm/321421930...eName=STRK:MEWAX:IT&_trksid=p3984.m1423.l2649

    Question 1. Do I need a resistor between the base source and the base connector. Most of the diagrams I have seen show one, if so how do I work out the size of the resistor required?

    To the collector connect a 12v +ve feed from suitable source (just tap a switched source, there will be plenty around the cluster).

    Connect the emitter to the positive side of a 12 LED car bulb (one similar to this one , rated at 0.3w)
    http://www.ebay.co.uk/itm/Hot-2-x-T...arts_Vehicles_CarParts_SM&hash=item5d4c822d52

    Question 2. Can I just connect the bulb direct or does it need any kind of resistor between it and the emitter. As it is a 12v bulb, I assume it can be connected direct.

    Connect -ve side of bulb to a -ve supply.

    Question 3. Can I use any -ve supply or is their any benefit in using the -ve side of the LED in the cluster, I can't see that it would make any difference.

    The circuit diagram would look something like this
    [​IMG]

    Many thanks in advance for any help you can give me with this
     
  2. Alan Jones

    Alan Jones

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    Jun 26, 2014
    Or thinking about it some more, as it is an NPN, should it actually be connected like this with the load before the transistor and the transistor just providing the switch on the negative side?
    [​IMG]
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Hi Alan, welcome to our forum.

    Your first circuit won't work, the second one has a good chance, provided that:
    • you add a resistor into the base connection of the transistor.
    • -ve = gnd

    At 12V/3W the current through the lamp is 2.5mA. DC gain for a typical NPN small signal transistor is ~100, this means a base current of 25µA is required. Base-Emitter voltage of the transistor is 0.6V, that leaves (2.3-1.6)V=0.7V across the base resistor. 0.7V/25µA=10kΩ. To ensure the transistor is safely on make this 10kΩ.
    Also another resistor from base to gnd to ensure the transistor is off when the 2.3V are not present.
    Plus a zener diode across Collector-Emiter of the transistor to protect it from voltage spikes.
    The full circuit could look something like this:
    [​IMG]

    However,
    if -ve in your diagma is not identical to GND, then this method will not work because it wil create a short circuit between -ve and gnd. In this case you could replace the LED in the instrument cluster by an optocoupler, e.g. an ubiquituous CNY17. You connect the LED of teh optocoupler instead of teh LED in the instrument cluster, you connect the transistor in the same way as Q1 in my diagram. Note that you may or may not need a current limiting resistor in series with the CNY17s LED, this depends on the driving circuit in the instrument cluster. You'd have to measure the current through the optocoupler's LED and ensure it does not exceed e.g. 20mA (this value is somewhat arbitray, I admit).
     

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  4. Alan Jones

    Alan Jones

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    Jun 26, 2014
    Many thanks for the quick reply. So are these the resisters I need?
    http://www.ebay.co.uk/itm/Resistors...mponents_Supplies_ET&var=&hash=item4613de8ee0

    Are these the diodes I need?
    http://www.ebay.co.uk/itm/BZX55-500...mponents_Supplies_ET&var=&hash=item3a91356736

    What wattage should I use, they do 500mw and 1.3w?
    The broad stripe I assume is the cathode and so attached to the collector
    Why 18v? They do lower voltages such as 15v

    Other than that it makes sense and seems pretty straight forward. Will these components generate heat and do I need to consider cooling? How should I mount them? Can I simply solder them all together using bits of wire as long as they are insulated or do I need to PCB mount them?

    Sorry for all the questions but this is my first time using electronics, I have always used relays for this kind of thing in the past, but this seems a much smarter way of doing things. Interestingly a German company markets a product to do this very thing, but the cost is 58 euros with postage, which for something where the component costs add up to a few pounds seems excessive.
    http://www.ebay.de/itm/Schaltblitz-...pt=Motorrad_Kraftradteile&hash=item58b1855252
     
  5. Harald Kapp

    Harald Kapp Moderator Moderator

    11,444
    2,628
    Nov 17, 2011
    These resistors look o.k.

    Use the BCX55C18 at 500mW. They are for overvoltage protection and ideally are never active. A 12V lead battery can easily reach >14V when fully charged. This would pose a risk of the zener diode becoming conductive (the rating is 15Vnom, min=13.8V, see datasheet page 2). A BZX55C18 has a min. zener voltage of 16.8V and is therefore on the save side. A 24V zener could be used, too, the absolute value is not critical as long as it is safely above 14V.

    The power dissipation should be negligible. No cooling required. However, I'd recommend using a piece of veroboard or similar to solder everything together. Without a supporting board, you'd have to insulate all wires and the mechanical stress from the bike's rattling will sooner or later break a wire or a solder joint (you should somehow secure the board, too, e.g. by placing it between some non-conductive foam).

    If you have an electronics shop nearby, get your components there. The *bay's products are sometimes of questionable providence/quality. You don't want to waste time looking for faults in your setup when the fault is with a component.

    Your do-it-yourself project is going to cost you a lot less - provided you take care not to damage any other components in the course of building and connecting it to your bike. Take special care concerning my note on -ve and gnd!

    Good luck :cool:
     
  6. Alan Jones

    Alan Jones

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    Jun 26, 2014
    Thanks, that all makes sense, I will post up to let you know how I get on. If all goes well, I may make a few and offer them up to the ZX10R forum as this is a common problem on this model, not being able to spot the shift light
     
  7. BobK

    BobK

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    1,688
    Jan 5, 2010
    Harald, Wie viele Biere, haben Sie getrunken?

    12V 3W is 250mA, not 2.5mA. The rest of the calculations are therefore wrong.

    But that is not all. The original LED is certainly driven through a resistor, so you cannot just assume 2.3V at that LED and calculate the base resistor from there. It will work only if the current needed is << the current through the LED.

    The base current calculation should be (using beta of 10 for saturation):

    250mA / 10 = 25mA.

    This is approximately the current through the LED, so, I would just disconnect the LED and run the positive wire directly to the base of the transistor instead, checking the current with a multimeter.

    Bob
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Bob, you're almost right. No beer, but a certain tiredness, I admit. 250mA is right.
    Right again, but even at a base current of 2.5mA (assuming a gain of 100 to drive 250mA to the lamp) this should be small enough to warrant the calculation. You may have noted that I made a reference to the unknown driver circuit within the instrument cluster in my post #3. It is definitely reasonable to assume a simple resistor for current limiting.

    And yes, in view of my miscalculation your suggestion of running the current straight through the transistor is reasonable. Depending on the still unknown circuit within the instrument cluster it may even work with the original LED in series to the base.

    Alan: In view of these corrections (sorry) the ximple optocoupler will not work as explained. The collector current is only 100mA max and therefore insufficient for the lamp. You can, however, use the optocoupler to drive an external transistor:
    [​IMG]
    If the current through the lamp is not high enough (250mA), then the gain of Q1 may be very small, reduce R1 from 1k to e.g. 470 Ohm to increase the base current
     

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  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Bob, methinks we both were wrong.

    Alan stated the Lamp is 12V 0.3W which makes 25mA...

    @alan: Bob's advice is sound nevertheless.
    The optocoupler circuit, though, will not need the external transistor as the CNY17 can easily handle 25mA.
     
  10. Alan Jones

    Alan Jones

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    Jun 26, 2014
    Yes, the lamp is stated as 0.3w, here is the specification. I already have the bulbs, so if there is a way to test/prove this I can do that. So if we only need 25ma, can we go back to the simpler design without the optocoupler?

    LED light Color: Red
    LED Color: White
    LED Quantity: 12
    Size: Approx. 4.5 x 2.5cm
    For DC 12V only
    Bulbs type: BA15S

    Power: 0.3W
     
  11. Alan Jones

    Alan Jones

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    Jun 26, 2014
    Hmmm, not sure I believe that bulb is 0.3w, other similar bulbs are 3w so is that 250ma. I have been reading up on Optocouplers and think I understand what they are. So does the circuit with the optocoupler and the NPB transistor provide the gain we need. I assume what we are doing is achieving the gain twice, once through the optocoupler and then once through the NPN transistor. What if I build the circuit, apply 12v from a motorcycle battery, but to test it use a 1.5v AA battery to supply the power to the base on the optocoupler, would that be a valid test? In your circuit diagram, I understand it, but the Q1 NPN transistor is getting a much higher current than previously, do I need to select a different NPN, if so, any suggestions? Also, does it need a resistor into the anode of the optocoupler?
     
    Last edited: Jun 26, 2014
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Using the optocoupler is probably the fool-proof way to go.

    You can test teh circuit with batteries, provided that you add a current limiting resistor to the base of the optocoupler (this resistor is laready in the instrument assembly so you won't need it in the bike).
    .
    No, the relevant curent is emitter-collector and this is the same in both cases, namely 25mA (or 250mA, depending on what your lamp really requires).

    I sugest you build the circuit and test it. If you're not satisfied with the results, we can discuss them and look for optimizations.
     
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