low voltage ac to dc converter answer asap!!!

Discussion in 'General Electronics Chat' started by programmer, Jul 10, 2009.

  1. programmer

    Thread Starter New Member

    Jul 10, 2009
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    hi, im working on a little project that requires around 6 volts dc 100 ma. i found a transformer for 6 volts and 130 ma which will work, but problem is, is 6 volts ac not dc. so does anyone have the schematics for a simple ac to dc converter that might work?
    please answer ASAP!!!

    thanks in advance,
    programmer
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You could rectify the AC to DC, buffer it with a large capacitor and stabalize it with a low drop regulator like the LT1086.
    See the datasheet of the LT for more info.

    Greetings,
    Bertus
     
  3. bertus

    Administrator

    Apr 5, 2008
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    Hello Alberto,

    The voltage at the capacitor may be to low for the LM317 to regulate.
    That is why I recommended the LT1086.

    Greetings,
    Bertus
     
  4. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    Does your project require a regulated voltage?? If not just stop at the big capacitor.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    The attached isn't fancy, but should do the trick.

    The 11DQ03 diodes are Schottky diodes rated for 1.1A, 30v. You could use 1N5817 Schottky diodes instead. Schottky diodes have a much lower forward voltage than standard diodes, which is important in your application. The downside is that they will be destroyed much more quickly than standard silicon diodes if they receive high reverse voltages.

    The Zener provides regulation. R1 supplies current to the Zener. Q1 follows the voltage across the Zener, and it's emitter is about 0.7v lower than the Zener voltage, depending upon the load. At 100mA load, there will be about 50mV ripple with C1 being 2200uF, which is less than 1%. Don't go much smaller, or you'll wind up with lots of ripple.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    You don't even really need the transistor. A 6.1V diode and a 10Ω resistor would work fine.

    [​IMG]
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Bill, I simulated your circuit with a 100mA resistive load.

    Even using Schottky rectifiers, ripple was 926mV p-p. I don't know if our OP could accept over 15.4% ripple.

    Increasing the capacitor back to 2200uF decreased ripple to 329mV, or roughly 5.5%. That's still well over six times the ripple I observed in my circuit simulation.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    I have trouble believing almost 1V ripple, what the cap doesn't get the zener would. I've build these circuits. As long as the zener is conducting the ripple should be miniscule.

    Still, the 1000µF cap was an afterthought. In the world of power supplies larger is better.

    Using conventional diodes I got 7.2 V across the cap, with Schottky's it would be around around 8.0V. Is that what you get?
     
  9. millwood

    Guest

    it wouldn't work for 100ma load current.
     
  10. millwood

    Guest

    one approach is to put some forward biased diodes in serial to lower the voltage down to 6v. each diode drops .7v.

    or if you are brave, you can use an amplified mosfet to drop more (Vgs=3-4v for regular mosfets, and less for logic level mosfets).

    or you can add a npn / n-channel mosfet.

    let me see if I can put something together quick.
     
  11. millwood

    Guest

    here it is. using two forward-biased silicon diode to drop 1-1.5v.

    ripple is about .8v - not that great.
     
  12. millwood

    Guest

    this will get you there.

    it uses a zener to stabilize the base of a bjt and allow the bjt to be the regulator - so you can handle large current.

    the zener's voltage should be about Vout + .7v. I used a 4.7v zener + a 0.7v forward biased diode to get me that value.

    ripple is about 4.5v - 4.6v Vpp. not great, but not bad either.
     
  13. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The TS has requested an 6 Volts AC to 6 Volts DC conversion.

    My sugestion with the LT1086 needs 7.2 volts at the capacitor to work.
    So the use of Schottky diodes and a 2200 μF capacitor should be ok.

    @millwood, your voltage is getting to low.

    Greetings,
    Bertus
     
  14. millwood

    Guest

    i just realized that you need 6vdc.

    here is the closest i can find in my sim.

    obviously, the higher you push towards the supply, the higher the ripple is going to be.

    in this case, it is about 0.5v. you can cure it by increasing C1: C1=2200uf reduces ripple down to 0.2v.

    you can add a choke in the output as well.
     
  15. millwood

    Guest

    something like this will never work, and the reason is quite simple.

    zeners work well if the current going through it doesn't vary a lot. since the load is hanging off the zener, that means the load cannot suck away too much current, usually no more than 1/10th of what the zener can take.

    zeners, typically 1w or 2w, usually take 10 - 30ma. that means you have no more than 1 - 2ma for your load -> while the load needs 100ma.

    a set-up like yours will burn up in no time: the rectified ac is about 8.4v (6*1.3 - 1.4). with a 6.1v zener, the current going through that 10ohm resistor is (8.4-6.1)/10=230ma. since the load takes away 100ma, 130ma of current will go through the zener.

    a typical do41 zener like 4735 drops 6.2v at 41ma. pushing 3x of that through this little guy is going to be bad.
     
    Last edited by a moderator: Jul 12, 2009
  16. SgtWookie

    Expert

    Jul 17, 2007
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    If you'd replace the MUR120s with 1N5817's, you would lose less voltage across the bridge, which would drop ripple significantly. I'd already determined that 2200uF was about optimal. Also, try replacing your 2N5550 with a 2N2222.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    That's the thing; the Zener's current varies wildly.
    Up to a point; then you reach that "diminishing return" thing. ;)

    I get an average of about 7.6v across the 2200uF cap. Low is about 7.42v, high around 7.75v. I didn't save the simulation I did of your circuit.
     
  18. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Well, let's see.

    "at 6.1v, the zener is dissipating 6.1v*130ma=8w."

    should be:

    at 6.1v, the zener is dissipating 6.1v*130ma = .8w.

    If the load drops to zero, then the zener will dissipate 6.1v*230 mA = 1.4w which a 2w zener with proper heatsinking can handle.

    The 10Ω resistor will dissipate (.230 A)^2 * 10Ω = .53 watts; not a problem.
     
  19. millwood

    Guest

    you got that right. my math error. a 2w zener wouldn't be able to do it: it doesn't have enough current rating for it.

    a 5w zener (1n5341, rated 6.2v@200ma) will do.
     
  20. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    A 2 watt zener will be able to do it.

    I know of no limitation to the current a zener can handle other than dissipation.

    What reason do you know that would limit a zener's current carrying ability other than power dissipation?

    See the attachment. A 2 watt, 6.2 volt zener can handle 292 mA.; at least these can.
     
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