amplifing a.c. or d.c.

Discussion in 'The Projects Forum' started by yash02, Nov 19, 2014.

  1. yash02

    Thread Starter New Member

    Nov 19, 2014
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    Hey all;.
    I am working on my project which is just like faraday's flashlight.
    I m getting a.c. voltage 0.4v only so anybody could help me to amplify it to charge a capacitor to light an led of ratings.
    http://www.ebay.in/itm/50-x-White-L...9?pt=LH_DefaultDomain_203&hash=item3a9bd20d3f
    I also converted a.c. into d.c. using rectifier bridge .
    My problem is that capacitor wont charge beyond 4v and the led uses 4v in less than 1second.
    my capacitor is 450V and 47uF .
    Please someone help.
     
  2. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
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    your going to need to use an op amp, which is going to need a power supply for itself, so its defeating the object.
     
  3. alfacliff

    Well-Known Member

    Dec 13, 2013
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    I dont think an op amp is needed. it is a "shake " flashllight he's talking about, where you have a magnet inside a tube and a coil around the tube, shake the tube and the magnet moves back and forth=, inducing current in the coil which is rectified and used to charge the capacitor.
    have you tried increasing the number of turns? that should increase the voltage out of the coil.
     
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  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Welcome to the forum, but hang on a minute.
    You have a 0.4V AC source feeding a bridge rectifier whose output has a 47uF smoothing cap attached and you're measuring 4V DC on the cap? The numbers don't add up. A rectifier bridge, even if built from Schottky diodes, will drop ~0.6V DC. I suspect you're measuring a 'ghost' voltage on the cap, arising from thermal or dielectric stress effects.
     
  5. wayneh

    Expert

    Sep 9, 2010
    12,155
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    How did you determine that, a meter? Don't rely on that reading - it's probably an artifact.
    "Amplify" is the wrong word - that implies a separate power source and I don't think that is what you mean. "Step up" is perhaps more precise.
    You need a much higher capacitance rating, and you could use a much lower voltage rating. For instance 1,000µF and 10V would give you a longer time.

    You might benefit from a joule-thief circuit - like the ones in every solar LED you see - to keep the LED lit longer while the capacitor discharges.
     
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  6. yash02

    Thread Starter New Member

    Nov 19, 2014
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    thats a probably a good idea of using joule's thief in my project
     
  7. yash02

    Thread Starter New Member

    Nov 19, 2014
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    I also tried and used the wounded coil of a transformer but i got not good result using it too.
    I just opened a transformer(a small one) and removed its iron core the left thing was the coil with 2 input wires and 2 output wires and 1 earth wire on output side
     
  8. wayneh

    Expert

    Sep 9, 2010
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    Well maybe. It will lose some power in the components, but you gain the voltage boost. There is a trade-off and it m ay not always give a net gain.
     
  9. yash02

    Thread Starter New Member

    Nov 19, 2014
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    I did not get the last line you said
     
  10. wayneh

    Expert

    Sep 9, 2010
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    You may or may not see an improvement from adding the boost circuit. Yes, it will allow light to be produced as the capacitor goes to a lower voltage. But it also introduces some power loss that COULD exceed that benefit and actually shorten the total on time. Just try it.
     
  11. yash02

    Thread Starter New Member

    Nov 19, 2014
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    OK THANKS
     
  12. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    The strong your magnet is, the more current you will produce (good 'Neodymium' magnet).

    Also, the geometry of your device makes a big difference. Make sure that the magnetic flux lines are cutting through the coil at the right direction to maximize current flow.

    Post a photo or drawing to indicate the coil winding direction and magnet movement direction if you need confirmation / support.
     
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