Joule thief

Discussion in 'General Electronics Chat' started by boatsman, Dec 8, 2014.

  1. boatsman

    Thread Starter Senior Member

    Jan 17, 2008
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    I have an emergency lamp consisting of 120 white leds that is connected to the mains 220vac. The battery inside is a sealed lead acid battery 6v 2.8ah. The lamp no longer works and the battery voltage now reads 4.45v. There is a switch by means of which either 60 leds or 120 leds can be lit. As the cost of a replacement battery is almost that of a new emergency lamp I would like to build a Joule thief circuit to run off the remaining voltage and current in the sealed battery. I would appreciate any practical advice concerning winding the ferrite toroid and current limiting resistors.
     
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
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  3. takao21203

    Distinguished Member

    Apr 28, 2012
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    its difficult to make a joule thief which has output of 340 VDC@ 20mA.

    You need input current of at least 1 Amps.
     
  4. boatsman

    Thread Starter Senior Member

    Jan 17, 2008
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    I don't think I can recharge the battery so when the battery is completely dead I'll use the leds somewhere else.
     
  5. boatsman

    Thread Starter Senior Member

    Jan 17, 2008
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    How do you get the figure of 340vdc @20mA? The input voltage from the secondary of the transformer is 9v at 300 mA .
     
  6. wayneh

    Expert

    Sep 9, 2010
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    I wouldn't invest any effort in that battery. Once they start to fail, there is very little life left.

    I would invest a little time to confirm that the charger is performing as it is supposed to, so it is not responsible for the poor battery status. Maybe try charging the battery outside of the device.

    And I'd shop around for more battery options. It shouldn't cost that much.
     
  7. takao21203

    Distinguished Member

    Apr 28, 2012
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    change it to lithium, its 8 volts instead of 6, the inverter probably wont bother about that.
     
  8. ian field

    Distinguished Member

    Oct 27, 2012
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    Emergency lights of this type often overcharge the battery and gas away the distilled water, then they get neglected because they don't work anymore and become sulphated.

    You can usually pry off the cover plate to get at the rubber vent caps and fill with more distilled water - not so easy, is getting any charge current to flow.

    Not sure what the moderators will have to say about my remedy, as the circuit is live and must be treated with respect! - a wattles dropper consisting of a series capacitor, which must be a filter type that is specified for connection directly across the mains, fed into a bridge rectifier. The capacitor must be in the live lead and one AC terminal of the bridge to the neutral lead.

    The wattless dropper (capacitor) and the bridge rectifier give a reasonable approximation to a constant current - if the battery doesn't draw any, the voltage is limited to the mains peak voltage.

    Sometimes it works, sometimes it doesn't - YMMV.
     
  9. takao21203

    Distinguished Member

    Apr 28, 2012
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    A little bit on the weird side- these lead batteries are gel type and arent meant to be for adding distilled water. I am not sure if this works.
     
  10. ian field

    Distinguished Member

    Oct 27, 2012
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    Sometimes it doesn't - but it can be cheaper to try than just giving up and buying a new battery.
     
  11. takao21203

    Distinguished Member

    Apr 28, 2012
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    they are not meant to be for refilling i looked on wikipedia. There is no facility to open/reclose them.
    And these batteries dont cost THAT much. As i suggested- use two lithium batteries.
     
  12. boatsman

    Thread Starter Senior Member

    Jan 17, 2008
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    Thanks everyone for replying. The transformer is rated at 9v 300mA 2.7w. I am not attempting to change or recharge the sealed lead battery. I just want to be able to light up the 60 or 120 white leds. Each led is rated at 0.1W and I want to build a Joule thief to run them off 4.5 v.
     
  13. ian field

    Distinguished Member

    Oct 27, 2012
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    Switching to lithium isn't trivial - with 2 cells and no charge balancing chips it can be dangerous!
     
  14. wayneh

    Expert

    Sep 9, 2010
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    It's possible that its existing circuit shuts it off at 4.5V to prevent excessive battery discharge. Perhaps you could bypass or alter that. Can you get at the PCB?
     
  15. Dr.killjoy

    Well-Known Member

    Apr 28, 2013
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    Please just buy the right battery and be done with it... The reason is that battery might be worst off than you think..So you spend all this time and money building this circuit for only the battery to die or lose another cell and at witch point you are back at the begging again with a bad battery when you could have just bought a new battery and be done with it..
     
    #12 likes this.
  16. boatsman

    Thread Starter Senior Member

    Jan 17, 2008
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    Sorry, I'm not making myself clear. I need a Joule thief circuit so I can utilize the leds. If I can't run it off a rundown battery then I want to run it off a transformer/power supply at 4.5 volts dc or anything up to 12 volts dc.
     
  17. Dr.killjoy

    Well-Known Member

    Apr 28, 2013
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    You have to remember when you use a Joule Thief that the extra voltage has to come from some where ... So it means you have to pull extra current to make up for the extra voltage and I am not sure but I think that transformer will not be up to the task ... Also it will drain the half dead battery alot faster..
    Please Correct me if I am wrong guys ...




    Thanks
    Jay Sr
     
  18. wayneh

    Expert

    Sep 9, 2010
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    I'm not sure that's true. But anyway, look on e-bay for a DC-DC converter. I believe you can find one that you can set to output 6V when fed any input voltage within some wide range. As long as you get one that can handle the current draw of your LEDs - which you must determine - this will be an efficient and cheap solution.
     
  19. boatsman

    Thread Starter Senior Member

    Jan 17, 2008
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    If I use a12v 2a dc power supply would that be sufficient to power the circuit of a Joule thief? According to my calculations the 120 leds are drawing a total of 12w, so a 12v 2a output should easily cover it. I just thought that the Joule thief would be more economical.
     
  20. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    A half dead battery is unlikely to be able to provide enough current to drive a Joule thief circuit to do what you want.
    If you want to drive the LEDs from another battery, then sourcing a suitable DC-DC converter would probably be cheaper than trying to build your own.
     
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