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Discussion in 'General Electronics Discussion' started by screwball, Jan 5, 2013.

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  1. BobK

    BobK

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    Jan 5, 2010
    No, you would use a constant current of 900ma if they are in series.

    Whoops, Kris go there first.

    Bob
     
  2. screwball

    screwball

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    Jan 9, 2012
    I'm totally lost on what to do now :(

    I'm just after the most basic circuit i can make it that works that would cost little to build if that makes sense,

    I'm coming towards the decision of using 3 LEDs, light 1 led for lowest brightness, 2 for 2 brightness and 3 for brightest and then of course no LEDs lit for OFF,
    That would be fairly simple? Also i was thinking each LED is 10W of power or so, could i do this without resistors? a 10W rated resistor will be huge and expensive, thats if they exist :rolleyes:
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I don't think it's a good idea to use extra LEDs to get different brightness levels.

    A boost converter with current regulation is not large. The largest part is the inductor. And there are lots of ICs that do all, or most, of the work. You would still be best to use a micro to read the pushbutton and cycle through the brightness levels.

    A boost converter with constant current output will not need a current limiting resistor. This is good because it reduces losses and heat dissipation, and size as well.

    Try Googling boost converter LED driver constant current. I will have a look as well, but not today; I have some important stuff to do.
     
  4. screwball

    screwball

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    Jan 9, 2012
    Could you explain the reason behind the extra LEDs for different brightness levels just to learn,

    I will look it up :)
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The issue with high power LEDs is that using resistors as current limiting devices is generally not done. Depending on the voltage you're running it from, the resistor could be dissipating anything from 5W to 50W (yes, it's not that hard to get resistors up to 50W and beyond).

    The main issue is heat. Unless you also want this to be a space heater, heat is something you want to minimise.

    The easiest way to do what you want is with a constant curent source. You need one that is capable of 900mA and rated for delivering between 10W and 30W (wider is OK).

    You simply connect the three LEDs in series to this power supply. All three lights lit.

    To turn a light off, you can simply short it out (yes, that works as long as you NEVER short out all three). THIS ONLY WORKS WITH A CONSTANT CURRENT SOURCE AS DESCRIBED ABOVE.

    Three of these LEDs being driven from 12V isn't going to work (since each one required about 12V) So, again, start with a higher voltage, or have three constant current devices.

    What is your source of 12V anyway?

    If you're doing this for 3 light levels, it isn't going to work because your eye has a non-linear response to light levels. You may only notice a small change in light levels between 1 and 3 LEDs (obviously not the case if you're illuminating 3 different areas).
     
    Last edited: Jan 11, 2013
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I don't know what you're asking me. You suggested using multiple LEDs, not me. As Steve explained, two LEDs don't look twice as bright as one LED (assuming the currents are the same). This is a big reason why using one, two or three LEDs for different brightness levels is not a good idea. The best way is to use as many LEDs as you need for the maximum brightness you want, connected in series, and use a switching supply with output current regulation to drive a certain current through them. Change the current to vary the brightness.

    Steve emphasised my point about avoiding current limiting resistors. They are just a guaranteed way to waste power and generate heat. A switching converter has some loss, and generates some heat, but with a current limiting resistor, this is intentional! And it will usually be worse.

    Steve also points out that you can use a buck converter if your input supply voltage is significantly more than the maximum forward voltages of all the LEDs summed together. Buck converters are more common than boost converters, but either would be suitable. A buck converter is somewhat more compact, which is significant considering that you want to draw up to 900 mA from it.

    I suggested a boost converter because I wasn't sure what your supply voltage was. With a boost converter, though, the supply voltage MUST be less than the total LED forward voltage, just like it MUST be higher for a buck converter.

    Thanks Steve for pointing out the output power required. It's a significant amount of power, and many buck and boost converters you can find online are not powerful enough.

    You COULD use a linear current source, as long as your power source has a significantly higher voltage than the total LED forward voltage, but these waste power as heat, just like a series resistor does. That's why I suggest a switching converter, either buck or boost depending on the voltages involved.

    You can convert a switching regulator with voltage output into a current regulator. Just monitor the LED current through a small-value resistor at the bottom of the LED string, amplify that voltage, and feed it into the voltage feedback pin. By varying the gain of the amplifier, you will vary the LED current.

    If you can give me a full description of the power source and your LEDs, and decide on the LED current and the number of LEDs you want, I can draw up a design for you using a National Semiconductor Simple Switcher or something similar.
     
    Last edited: Jan 11, 2013
  7. screwball

    screwball

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    Jan 9, 2012
    Hi, Thanks,
    I see where this is going,
    Also supply is 12V 6800mAh Li-ion battery,

    I will look into it anyway as said and design a circuit and see what you think?
     
  8. screwball

    screwball

    89
    1
    Jan 9, 2012
    Came across this

    [​IMG]

    Work or not work?
    Im thinking work but there is a downfall, there is a resistor in series(heat issue, high wattage resistor)?
    But i guess its a starting point?
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That's a linear current regulator, controlled by a PWM signal. I mentioned this option in my earlier posts. It will work, but Q2 will dissipate heat and waste power.

    The power dissipated in any component can be calculated from the Power Law: P = V I, where P is power in watts, V is voltage across the component in volts, and I is current through the component in amps.

    In this circuit, Q2 is used as a "controlled resistor", to take up the "slack" voltage, i.e. the voltage difference between the LED string voltage and the power supply voltage and ensure that the desired amount of current is forced through the LED string. (Actually, R3 will also have about 0.6V across it, and at 900 mA, will dissipate about half a watt. R3 is used as a current sense resistor in this circuit. But most of the power will be lost in Q2.)

    This is the third and final time I will say this. If you want to avoid significant wasted energy (and therefore significant heating), you need a switching converter - buck or boost, as I described in my earlier post. A simple current limiting resistor, or an active current limiting circuit like this one, will dissipate significant heat and waste significant power.
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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  11. screwball

    screwball

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    Jan 9, 2012
    I know you've said and ive read and im still not 100% on what everything is etc so i guess i keep referring back to the things i do know, apologies,

    I looked at the thread, I see it says 3021/3023 driver, im reading on the datasheet that Peak Forward Current:80mA, thats the figure i need for knowing if the driver is going to take the current, dont i need 900mA?

    I tried looking on digikey.com, i searched driver and then it comes up with a list of sub categories, I dont have a clue which one to look in? Plus they're damn expensive :O Looks like i got no option lol
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    The data sheet I have doesn't mention 80 mA anywhere. It says the BuckPuck is available in four maximum current ratings of 350 mA, 500 mA, 700 mA and 1A. The actual output current can be controlled in several ways. It should be easy to set the maximum output current to 900 mA, then control the actual output current through a control pin. You would still need some logic (microcontroller, probably) to cycle through the different brightness levels. The part number you want would be 3021-D-I-1000.

    USD 17 doesn't sound too outrageous considering how much trouble it will save. If I wanted to make lots of these things, whatever they are, then I would have second thoughts, but for a one-off it seems like a pretty good solution. But obviously that's up to you.
     
  13. screwball

    screwball

    89
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    Jan 9, 2012
    Thankyou, I see, the reading i got was not from the actual datasheet so that explains that figure, I can now find the correct part, I will either go for this or play with a different cheaper solution, see which works best,

    I'm not sure if this is just a one off or not, If it goes good I will be making a few for members of a small organisation (mostly hobbiest level sport),

    The heat issue isnt that an important factor now, (i know regardless heat produced will be less efficient etc) but im looking at the best option (usually most costly) and the "will do the job option (cheaper) if you have any ideas?

    I really appreciate your help on this and apologies for any trouble i have caused by misunderstandings ;)
     
  14. TedA

    TedA

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    Sep 26, 2011
    Screwball,

    Perhaps you could tell us a bit more about what you are trying to accomplish. I realize that you initially asked about transistor selection, but that was not the entire question you need to have answered.

    Some things we would like to know include:

    What is your power source and voltage? It matters whether you have a regulated DC supply, a lantern battery, or a cigar lighter plug.

    Does it matter if the design wastes power? ( If the LED is on for one second every second week, and the power comes from the local utility, the answer is likely no! )

    What is the forward voltage of your LED when it is on? Indeed, can you point us to a data sheet for this part?

    How much time and trouble are you ready to expend on this project? Sometimes the quick solution trumps the "good" one.

    You did mention in a later post that you plan construction of one to a few examples. ( Foxcon will not be devoting a production line…) This is important, because it means the design should be biased toward easy to implement and conservative ratings for your parts, rather than absolute lowest parts cost.

    Chances are, a buck converter circuit ( a type of SMPS ) would be what you want to use to power the LED. It might be implemented using a microcontroller alone as the controller, though there are advantages to adding a dedicated SMPS controller IC to the design.

    An important point so far overlooked ( I think ):

    Your schematic, posted back on pg. 2 (01-06-2013, 02:11 PM), shows things connected-up in a bad way. The transistors are shown connected as source followers, driven by what I suppose are fairly low gate voltages. The LED may barely light up at all, since the source voltage will be lower than the gate voltage by at least the threshold voltage of the FET. You may just get eight levels of off.

    I believe you want to use a grounded source configuration. Put the LED and resistor(s) between the positive supply and the drain(s) of the FET(s).

    Ted
     
  15. screwball

    screwball

    89
    1
    Jan 9, 2012
    Thankyou :cool:
     
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