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How to use transistors to amplify current.

Discussion in 'General Electronics Discussion' started by exception, Jun 17, 2010.

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  1. exception

    exception

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    Jun 17, 2010
    Ok, this is gonna be long so here goes. I design and build glowsticks with high powered leds. Not the little 5mm ones but the big 10mm .5w ones. I wire three in parallel and use an a23 12v battery to power them. They are ridiculously bright. The problem I keep running into, is that the battery gets extremely hot. The specs on each bulb is 2.4- 2.6 volts max at 100ma. Im going for maximum brightness so I dont use a resistor as they barely light at all. Eventually Id like to upgrade to the 1w 300ma leds so each stick is 3w. Is there a way to use a transistor to amplify the current to higher levels so i have more to play with that way I can use a resistor to prevent the battery from getting hot? Keep in mind, I dont want to modify the case itself anymore than I already have, and due to fitment issues I cant wire them in series. Any help would be appreciated.
     
  2. jackorocko

    jackorocko

    1,284
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    Apr 4, 2010
    Current has to come from some source, in your case that source is the battery. So I don't see how using a current amplifying circuit will help. You can't just materialize current out of thin air.
     
  3. Mitchekj

    Mitchekj

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    Jan 24, 2010
    Poor LEDs... I'd worry more about them getting hot than the battery. Though the battery is getting hot because you've got TONS of current flowing, I'd wager.

    Three LEDs in parallel, with 12V and no resistor... so each LED is getting 12V. Wow, there's a reason they're "ridiculously bright." They will also be ridiculously short lived. I'm amazed they've lasted long enough for you to even view more than a few photons.

    I don't see how a "fitment" issue could prevent you from wiring them in series, but that's how you should be doing it. Wire them in series, ane USE a current limiting resistor, as a bare minimum. (By the bye, did you just pick a resistor value at random and declare that it didn't work since they barely light? Sounds like you've gone about this wrong.) I would advise regulating your 12V, or better yet, regulating the current. But it may be a bit too much to ask here.
     
  4. exception

    exception

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    Jun 17, 2010
    I used the resistors that came with the leds. Im sure I could find a way to make series work, but then I'd lose brightness and parallel was much easier so I stuck with it. Plus, I think the resistors are designed for a car batteries 12 volts, not a small a23 remote battery. Any resistor I use is gonna decrease the light right? I want to get maximum brightness with maximum battery life. I was under the assumption that when you wire more than one led in parallel the voltage gets combined and the current stays the same,and series combines the current and the voltage stays the same. I thought that the voltage drop was causing the battery to heat up. So its actually the current that's causing the battery to get hot?


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    Last edited: Jun 17, 2010
  5. jackorocko

    jackorocko

    1,284
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    Apr 4, 2010
    In series, the voltage drop across all components will add up to Vin and the current remains the same across all components.

    In parallel, the voltage across each component is equal to Vin, and the current across each component is a fraction of the total current draw, Iin

    Absolutely
     
  6. exception

    exception

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    Jun 17, 2010
    So, how do i keep the battery from getting hot without losing brightness?
     
  7. florinanghel

    florinanghel

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    Jun 14, 2010
    If things were that easy, we'd all power our home appliances from 12V batteries. :) You want maximum brightness, maximum efficiency and minimum heat. You won't get any of these unless you make a compromise.
     
  8. jackorocko

    jackorocko

    1,284
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    Apr 4, 2010
    I'm with mitch. Your design is flawed and if you change the design to a proper one you will lose some brightness. Maybe the internal resistance of the battery is the only thing that has saved your LED's thus far.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I guess that the current is limited by the internal impedance of the battery.

    Hahahaha -- look here -- the battery is rated for 0.5mA continuous with peaks of 5mA. No wonder it gets hot. And the internal impedance almost certainly limits the current to something that is quite likely safe for the LEDs.

    Note the capacity of 33mAh, of which I'd guess you would get no more than 1/4 of that in this application. My guess is that the battery is totally dead in under 5 minutes -- is that right?
     
  10. exception

    exception

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    Jun 17, 2010
    Actually, no. If I put the battery in and leave it, the sticks will light for over twelve hrs. But maximum brightness for only a few minutes. I usually don't put them in and leave it though as I turn it on and off as I use it.
     
    Last edited: Jun 17, 2010
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    So even less than 5 minutes :)

    place them (the LEDs) in series. They will probably be brighter for longer. The initial brightness may be lower, but you will get more energy out of the battery.

    The best solution would be a switch mode regulator, but I suspect that's way beyond what you're going to want to do. It would give you almost constant brightness until the battery can no longer produce enough voltage. It may be bright for 10 to 15 minutes before pretty suddenly going out.

    Have you measured the current from the battery at several points over the life of the battery. It would be interesting to measure it each minute for the first 10 minutes, then approx every 5 minutes until the end of the first hour, then every 30 min to an hour thereafter.

    I think you might find the graph interesting. I suspect that after 12 hours or so the current would be some fraction of a mA.
     
  12. exception

    exception

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    Jun 17, 2010
    So, then, the bulbs aren't even running at maximum brightness are they?? Since they're getting enough voltage but not enough current??
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Voltage isn't the issue. Current is.

    Measure the current and tell us what it is. That's the only way we'll know.
     
  14. exception

    exception

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    Jun 17, 2010
    And how exactly do i do that? I have a meter, but when I try to use the dca it doesn't read anything.
     
  15. florinanghel

    florinanghel

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    Jun 14, 2010
    Are you sure you have set it up correctly?
     
    Last edited: Jun 18, 2010
  16. exception

    exception

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    Jun 17, 2010
    When the batteries are at 9 volts, the reading when the meter is set to 200ma dca is 42.0and dropping a point every second. Its a cheapo radioshack meter.
     
    Last edited: Jun 18, 2010
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, that means your LED current starts out at around 14mA and drops *rapidly*.

    Incidentally, 14mA is an acceptable current for normal (i.e. not high power) LEDs.

    I would tend to suggest you've never seen a bright LED if you think that a 14mA makes for *bright* :)

    If you measure the voltage across the battery when the LEDs are connected, you will find that it is closer to 3.5V than 9.

    Oh, 9V -- that is with a technically flat battery. Try it with a new battery it should start much higher -- above about 90 mA would almost justify the use of higher power LEDs (but only temporarily). I would imagine the voltage across the battery would fall to maybe 4V when new and under load (it's just a guess).

    I guess the main point is how precipitously the current is falling. This is a *really* good indication that the poor battery is way overloaded -- you've almost put a dead short across it (the difference is negligible) . It gets hot because more power is actually being dissipated in the battery than in the load.
     
  18. exception

    exception

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    Jun 17, 2010
    So since the leds are rated at 2.6v @ 100mah,how do I achieve this power level? I need to raise the current correct?
     
    Last edited: Jun 19, 2010
  19. markus.dnd

    markus.dnd

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    Jun 17, 2010
    characteristics and datasheets

    I suggest "exeption" to get a datasheet of the leds and check out the current rating of the battery.

    Then calculate if and how can the battery supply the leds and after that there are all these other things to do like calculating the needed resistor for led, maybe a little power supply unit etc.
    Usually the shop that sells the leds has good idea on characteristics of the led and batteries have it written on them.

    Maybe you can post them even here after you have tried to calculate all the needed things bu your self. So we could be more help an you could learn more :)

    Of course i just am kind of guy who likes to know as much as possible when i am building so i suggest the same for others.

    Markus
     
  20. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I suggest Markus read the entire thread ;) I looked up the specs for the battery in my first post in this thread. The most important phrase from the datasheet is "the battery is rated for 0.5mA continuous with peaks of 5mA".
     
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