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Convert 120v led to 12v

Discussion in 'General Electronics Chat' started by jjpaul, Jan 8, 2013.

  1. jjpaul

    jjpaul New Member

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    I have a 120v 1.5 watt led light strip (only 2 led's) that I want to run directly from a 12v battery charged by a solar panel. I don't want to power it thru an inverter.

    What do I need to remove and add. How to wire? Resistor size? I have KOA Speer 1.0 ohm Thru-Hole Resistor Carbon Film 1/4W 5% CF14-J resistors and I have 12v switch to replace ac switch. Thanks for any help.
     

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  2. panic mode

    panic mode Well-Known Member

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    what you have is classic non-isolated power supply using capacitor (big brown part on the left side of the PCB) to drop the voltage.
    to run them on DC you will just need a series resistor. to estimate value of resistor it would be good idea to take current reading and forward voltage drop of the LEDs
     
  3. jjpaul

    jjpaul New Member

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    Thank you for your reply. Forgive my ignorance but are you saying to put the resistor in series with the two led's and bypass the PCB? Thanks
     
  4. dave

    Dave New Member

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  5. panic mode

    panic mode Well-Known Member

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    yes, that's all you need but the resistor need to be sized (resistance and power)
     
  6. unclejed613

    unclejed613 Well-Known Member

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    R=(Vs-Vf)/I where R is the required series resistor, Vs is the supply voltage, Vf is the total forward voltage drop, and I is the diode current. so for a pair of LEDs that operate at 3.5V forward drop (each) at 100mA, and a 12V supply... the total forward drop of two diodes in series is 7V, so subtract from 12V and you have 5V 5V/100mA=50 ohms. the power dissipation of the resistor should be 5V*100mA=0.5W. with a safety margin, and using a standard value from the E24 series of resistors gives you a 47 ohm 1 watt resistor. the current will be slightly higher, but not enough to be an issue, although if you want slightly less current use a 51ohm 1 watt resistor. you will find the 47 ohm more commonly available.
     
    Last edited: Jan 10, 2013
  7. dr pepper

    dr pepper Well-Known Member Most Helpful Member

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    Another effcient and scavenger way to do it would be to get a mr16 12v 1 or 2w led lamp, break it open and take out the pcb and connect it up to the leds in your circuit.
     
  8. jjpaul

    jjpaul New Member

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    Thanks for all of the replies, I appreciate it.

    I see the 47 ohm resistors at radio shack, I'll try them. I don't have mr16 leds at that size but I think I'll get a couple and compare light output to my strip and take apart to see how they're made then add to another strip and see how it works out.

    I picked up these strips cheap and are great for what I want them for except that they are 120v.

    One last question.
    I have wire and switches for 12v but would the existing switch and lamp type wiring after removing the plug end be ok to use with the 12v system? The cord is 5' in length and I could get away with as little as 18" of length if needed.
     
  9. dr pepper

    dr pepper Well-Known Member Most Helpful Member

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    Yep I'dve thought so.

    Running the leds at a lower voltage might pull more current, but as long as the cable is 0.5mm2 or more it'll be fine.
     
  10. unclejed613

    unclejed613 Well-Known Member

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    after re-examining the picture you posted, i recommend re-calculating for 25mA at first. those LEDs are smaller than i thought at first glance. so you should go with 200 ohms as a starting point, then see how the brightness compares to when they were running off the line supply.
     
  11. jjpaul

    jjpaul New Member

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    Thank you everyone.

    I started with a 200 1/4 w resistor and compared it to the electric version and it seemed to be 50-60% less bright...I ended up using a 51, 68 seemed slightly less bright 51 may be slightly more bright...very hard to tell. The housing at the resistor end gets warm. The electric one doesn't get warm at all.
     
  12. Diver300

    Diver300 Well-Known Member

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    You seem to have the correct circuit.

    A resistive dropper will get warm. You will get as much power dissipated in the resistor as in the LEDs. That is inevitable with that circuit.

    If you use something like this http://www.farnell.com/datasheets/1636609.pdf you can reduce the input power, but it is much more complicated.
     
  13. unclejed613

    unclejed613 Well-Known Member

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    capacitive droppers don't dissipate (much) heat as it's a reactive rather than a resistive component and tends to act more like an AC current source.
     

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