Joule thief the ONLY option?

Discussion in 'The Projects Forum' started by SuperToby, Oct 13, 2013.

  1. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Hey guys I'm a first time poster with little EE experience. I recently assembled a pulsing LED circuit using the schematic below. I added four more LED lights (wired in parallel) and tried to power the circuit using four AA (NiMh) rechargeable batteries (wired in series) but the batteries don't provide enough voltage. Is the joule thief my only option if I MUST use these LED's and batteries:

    Battery specs:
    AA NiMh Rechargeable
    1.2 V - 2500 mAh

    LED specs:
    3.3 - 3.8 FW voltage
    20 mAh forward current
     
  2. Experimentonomen

    Member

    Feb 16, 2011
    331
    46
    The joule thief is the simply way of driving a LED, it relies solely on the high voltage inductive kickback and will burn out the led over time as these spikes can be in the hundreds of volts.

    You might find this kinda driver in the cheap noname chinese garden lights with a solar panel on top.

    In actual applications a little smps using a pwm controller ic is used to provide a steady current of desired value to the led(s).
     
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  3. #12

    Expert

    Nov 30, 2010
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    I disagree. The problem is putting LEDs in parallel. They do not work well like that. They don't share properly.

    Second, you did not label your chip so we can not tell if it can deliver enough current for 4 LEDs, each with its own resistor, and those put in parallel.

    It goes something like this:
     
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  4. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Thanks for our response.. It's a 555 timer: TLC555CP (see attached image)
     
  5. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Thanks! My main problem is trying to power 5 LED's (3.3 - 3.8 FW voltage
    20 mAh FW current) using four AA NiMh Rechargeable batteries (1.2 V - 2500 mAh).
    Is this even possible?
     
  6. #12

    Expert

    Nov 30, 2010
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    First page: Sink up to 100 ma (typical). What is 20 ma times 5 LEDs?
     
  7. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Got it thanks! I can cut back to 4 LED's no prob.
    My main issue is still trying to power 4 LED's (3.3 - 3.8 FW voltage
    20 mAh FW current) using four AA NiMh Rechargeable batteries (1.2 V - 2500 mAh).
    Is this even possible?
     
  8. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    The current of battery is enough, why you don't want to using the batteries in series and connecting the LEDs in parallel and then you don't need the Joule Thief circuit, the LED connection will be like as the #12 attached.

    What's the reason to make you to using the Joule Thief circuit?
     
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  9. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Thanks for your response Scott. When I replace the 9V battery in my LED circuit with four 1.2V AA batteries (wired parallel 4.8V) the LED doesn't light up.. even with only 1 LED.
     
  10. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    excuse my ignorance.. I'm not an electrical engineer.. I meant wired in "series" in my last post. I wired the four 1.2V AA batteries in series to give me 4.8V but for some reason it doesn't light up the LED.
     
  11. #12

    Expert

    Nov 30, 2010
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    I think you broke something.
    There is nothing wrong with the voltage or current available.
    Start looking for some part that went bad while you were experimenting.
     
  12. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,394
    1,606
    There are no spikes on the order of hundreds of volts in an inductive kickback converter. In fact, the output voltage of these things is... exactly the forward voltage of the LED being driven, no more, no less.

    It must be exactly the LED voltage since as connected the LED itself provides voltage regulation.

    And BTW, the cheap no name Chinese garden lights now all contain an application specific integrated circuit (ASIC) as the controller to put a serious lowball on the parts count.

    These things are just flyback boost converters, another type of SMPS.
     
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  13. ronv

    AAC Fanatic!

    Nov 12, 2008
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    The problem you have is that your power supply is almost the same voltage as your LEDs now. Try it with 15 ohms in series with the LED instead of 150.
     
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  14. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    The cap is swinging from about 1.6V to 3.2V, so you are running out of juice. Other than using a darlington NPN to drive a second NPN transistor pith poor gain control, I would go to PWM , many choices here on AAC.
     
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  15. #12

    Expert

    Nov 30, 2010
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    I overlooked the connection to pin 2.
    Thanks for picking out that I was off track.
     
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  16. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Okay, thanks for the feedback!
     
  17. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    Thanks ronv I'll def try switching out the resistor.
     
  18. SuperToby

    Thread Starter New Member

    Oct 13, 2013
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    I'm not familiar with what you're talking about Bernard.. I'm new to EE. Thanks anyway for the advice.
     
  19. WBahn

    Moderator

    Mar 31, 2012
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    First, the LED spec you keep talking about is NOT 20mAh, it is 20mA. A mAh is a measure of charge, namely the total charge that has passed a point if 1mA is allowed to flow continuously for 1 hour or 3.6A of charge is allowed to flow for 1 second, or any other product of current and time that comes to 1mAh. The LED spec is talking about the instananeous current that is flowing, as in 20mA is 20mC of charge flowing past the point each second.

    Second, the circuit you posted has two LEDs in it. But all of your descriptions talk about it as though there is just a single LED in the circuit that you then put four more in parallel with. Well, which LED in the circuit did you put the four other ones in parallel with? What did you do with the other LED in the circuit?

    That circuit was designed (meaning that the component sizes were chosen) assuming a supply voltage of about 9V. When you cut the supply voltage in half, you can expect to have to resize the components at the very least. But beyond that, if you are using LEDs that can have a forward voltage of up to 3.8V and you are powering it from a supply that is just 4.8V, you don't have much overhead to work with, especially when you consider that the base-emitter junction of a BJT transistor is going to be aboug 0.6V to 0.7V. Now you have well under half a volt to work with, and that's assuming that the output at Pin 6 can get all the way to the supply voltage and that the supply voltage really is all the way up at 4.8V.

    Instead of taking a 555 circuit that is designed to do something that is somewhat similar to what you seem to want to do, how about taking a step back and start by telling us what the basic problem is that you are trying to solve and the constraints you have to live within.
     
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  20. ronv

    AAC Fanatic!

    Nov 12, 2008
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    I think this will do what you want. I missed that the transistors were not on the output of the 555. But they are now.;)
     
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