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Diodes To Drop Voltage

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by ParkingLotLust, Dec 15, 2008.

  1. ParkingLotLust

    ParkingLotLust Member

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    Awhile ago I was talking about a way to drop 12v @ 1.5A to ~10v using the fewest possible components. I decided on diodes, specifically 1N4002's (I had a pack of 100 kicking around). I hooked up 3 in parallel, then those 3 with another 3 in series, than those 6 in series with another 3 to form a 3x3 array kinda deal. Here's a better visual:
    [​IMG]
    Basically, my logic was that three diodes in series would drop (.6v * 3) = ~1.8v and since each one can carry a maximum of 1A, in parallel that would give me 3A of current-handling. However, when I hook up my LED board, which draws a little under 1.5A, the diodes get hot. Im not sure how hot they should get, but I didnt expect them to get hot at all. Am I missing something?
     
  2. Leftyretro

    Leftyretro New Member

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    There is no way to insure the even sharing of the current between the three parallel diode strings, as there will be slight variation in junction voltage drops.
    This is not a good design.

    Lefty
     
  3. ParkingLotLust

    ParkingLotLust Member

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    What would you suggest? Connecting strings of 3 diodes [in series] in parallel (instead of each one)? Or finding higher-current diodes? Im just trying to work with what I have available.
     
  4. dave

    Dave New Member

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  5. Leftyretro

    Leftyretro New Member

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    Well I would use 3, 3amp diodes in series, but maybe someone else has a better idea for you.

    Lefty
     
  6. ParkingLotLust

    ParkingLotLust Member

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    Can you recommend any from Fairchild, preferably with free samples? I found some like this:
    Product Folder - FFP04S60S - 4A, 600V STEALTH II Rectifier
    But Im not sure of the voltage drop across each one. The only other ones I found were Schottky, and if I remember correctly, they have a lower voltage drop than a normal diode, so I would need more of them.
     
  7. BravoKilo

    BravoKilo New Member

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    Dropping Voltage

    Hi ParkingLotLust. May I join this interesting thread and share some insights in what you are trying to do. Silicon diodes drop .6 to .7 volts. That is true. the other fact is that when you parallel them, they may not share the current flow equally. When we parallel transistors in power supplies, we put a low resistance resistor in series which acts as the bottleneck and more or less allows the same current to each transistor. The same principle can be used on the diodes. I would have made 3 sets of 3S(3 in series) and parallel them but I would not have put the jumper across the anodes I saw in your circuit. I am not saying that it will be the correction as the uneven current distribution is still there. Of course I would series a resistor of the same value on each 3S. But we still have a problem when we parallel diodes. The reason why there is a drop is due to the diode acting as a resistor and when we parallel resistors, the sum value drops which is the opposit of what you were trying to do, i.e. when you series, you add up the internal resistance. Making a single 3S with higher capacity diodes will vanish the curernt distribution problem of course. You might want to use voltage regulators. Get a 7810 voltage regulator which can handle 1 amp easily. If you can find one that can handle higher current, better. The LM317 has a 3 amp version in a TO-3 package similar to the 2n3055 or the 2955. You need to google how to set it up. Shout if you need me on that. The 7810 is a three legged VR much like a regular transistor. Let pin is in, middle is ground, right pin is out. If it gets hot,series an appropriate resistor before the input and heatsink the VR too. We can put a bypass transistor to increase the amp capacity but it will be a little complex. The diode way is simpler


    Hope this helps a bit.

    BravoKilo.
     
  8. Mike - K8LH

    Mike - K8LH Well-Known Member

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    Instead of trying to "burn off" that extra energy, have you considered using something as simple as a 3 amp SMPS 'buck' regulator (like a Micrel MIC4576 'adjustable' version)?

    Happy Holidays. Mike

    [​IMG]
     
  9. Willbe

    Willbe New Member

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    Θja = 50°C/w for this diode, so at 0.5A and 1v drop you have 500 mW which gives a 25°C temp rise above ambient. Too hot to touch.

    You may need some current sharing resistance. From the diode V-I forward curve, a ΔI of 1/2 amp for a Δv of 0.1v gives 0.2Ω incremental diode impedance, so use 3 ea. 1Ω resistors.
     
    Last edited: Dec 16, 2008
  10. ParkingLotLust

    ParkingLotLust Member

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    Thanks for the big explanation! I looked up the 3A version of the 317 - the 350, and found that we have some in stock at our school (that they are willing to give away for free) so that will do exactly what I need.

    Id love to try out SMPS IC's sometime, however Im trying to do this for as cheap as possible (ie. free samples), and I dont have any inductors or Schottky's handy.
     
    Last edited: Dec 16, 2008
  11. Mike - K8LH

    Mike - K8LH Well-Known Member

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    I understand you're concern. I consider myself at a financial/economic disadvantage compared to most hobbyists on the forum. Fortunately Micrel provided several samples and one of the inductor manufacturers provided me with several surface mount inductor samples for my next prototype.

    BTW, that's a Schottky diode, not a zener. Here's the parts list for the 1 amp MIC4575 prototype I built;

    Code (text):
    Mouser

    D1 [URL="http://www.mouser.com/Search/ProductDetail.aspx?R=1N5819virtualkey53310000virtualkey821-1N5819"]821-1N5819[/URL]          Schottky diode, 1a               0.09 ea
    [FONT=Courier New]L1 [URL="https://www.mouser.com/Search/ProductDetail.aspx?R=18R683Cvirtualkey58010000virtualkey580-18R683C"]580-18R683C[/URL]         68 uh, 1.35a, radial    Murata   1.42 ea
    C1 [URL="https://www.mouser.com/Search/ProductDetail.aspx?R=140-ESRL25V330-RCvirtualkey21980000virtualkey140-ESRL25V330-RC"]140-ESRL25V330-RC[/URL]   330 uf, 25wvdc, radial  Xicon    0.62 ea
    C2 [URL="https://www.mouser.com/Search/ProductDetail.aspx?R=140-ESRL16V470-RCvirtualkey21980000virtualkey140-ESRL16V470-RC"]140-ESRL16V470-RC[/URL]   470 uf, 16wvdc, radial  Xicon    0.61 ea
    ---------------------------------------------------------------
                                                            2.74
    [/FONT]


    Good luck on your project. Happy Holidays.

    Mike
     
    Last edited: Dec 16, 2008
  12. Hero999

    Hero999 Banned

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    Why not just use the LM317?

    You should use a series resistor anyway with LEDs as they need constant current not constant voltage.
     
  13. ablemicky

    ablemicky New Member

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    drop 12 volts to 10v at 1.5A

    since diodes are too hot to the touch for u why don't you try a resistor instead to drop 2v from 12v the value of resistor that will do that is 2/1.5 = 1.33ohms(3 watts)u can use 1.5 ohms(3 watts) if 1.33ohms is hard to find.believe me your cct is going to work just fine.trust me its ablemicky.
     
  14. ParkingLotLust

    ParkingLotLust Member

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    They already have individual resistors, but I made a mistake in calculating power and the resistors get hot to the touch (1/4w metal film resistors), so I figured instead of pulling all 80+ LEDs I would just drop the voltage.

    And the LM317 provides a max of 1.5A and my LED board pulls JUST under that
     
  15. mneary

    mneary New Member

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    Since it's dissipating 3 watts, beware it will be getting pretty hot. A small to medium heat sink won't hurt.
     
  16. mhdbsm@hotmail.com

    mhdbsm@hotmail.com New Member

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    very nice i don't have enouogh thanks
     
  17. causalitist

    causalitist New Member

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    LM7810 !!!

    one component. fixed 10v regulator. 3 leads: ground, in,out.
    handles 1amp, has thermal shutdown too.

    you could put a small capacitor vin to ground and vout to ground.. but you dont have to.

    simple as it gets!
     
  18. Hero999

    Hero999 Banned

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    With a dropout voltage of 2V that's cutting it fine, if the voltage even slightly drops below 12V the output voltage will fall if a significant current is being drawn.
     

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