AC to DC Converter

Discussion in 'General Electronics Chat' started by vindicate, Aug 3, 2009.

  1. vindicate

    Thread Starter Active Member

    Jul 9, 2009
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    So If I were to want to build a AC to DC converter all I would need is a Transformer, Rectifer, Capacitors and a Regulator correct? Or is there more to it than that?

    Also with a Center Tap transformer, I assume you can either choose to use or just leave the center tap unconnected depending on the voltage you want correct?
     
  2. t06afre

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    Perhaps some fuses ;). But yes you are on the correct track
     
  3. vindicate

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    What about this circuit?

    http://www.discovercircuits.com/PDF-FILES/18VPS1.pdf

    Seems like their calculations are wrong. The transformer converts the 120VAC to 20VAC. After the diode drop their should only be 18.6V left. Then with the LM350T you lose 1.25V. So the max it could ouput is 17.35V.

    Is my logic correct?
     
  4. t06afre

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    Not quite you are doing the mistake as here http://forum.allaboutcircuits.com/showthread.php?t=26178 The voltage over the 4700uF cap. will be (20-diode drop)*1.41 volt max. You will also have some ripple depending on how much you are loading the power. Do not forget a large heat sink for the lm350.
     
  5. vindicate

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    So if I have this right, if the Transformer puts out 20VAC then after the Cap is charged it will actually be 26.22 volts output(before the regulator)?
     
  6. t06afre

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    Yes If you have no load, so will that be correct. At least a very god approximation.
     
  7. vindicate

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    What happens when there is a load? How far can it drop?
     
  8. t06afre

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    Last edited: Aug 4, 2009
  9. vindicate

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    But if I put a regulator on it then I don't have to worry...correct?
     
  10. t06afre

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    Then you have a regulator the effects of the ripple will almost be removed. but not 100%. You said that you regulator had a voltage drop equal to 1.25 volt. Now you have the ripple formula. And you can find the lowest voltage (on regulator input) at max load. Then subtract a few volt for safety margin. I remember I often used 5 volt as minimum regulator input/max output difference, as a thumb rule
     
  11. vindicate

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    What about my center tap question :)
     
  12. vindicate

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    Also if all you need is a transformer, rectifer, filter cap and regulator, how come when you open up a Wall-Wart(like from a cell phone) they have so much other stuff in them? Is it because they are switching(I think most cell chargers are these days) supplies?
     
  13. SgtWookie

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    If your transformers' secondary winding is 10v-0v-10v, you leave the center tap disconnected. You just connect the "ends" across the full-wave bridge rectifier.

    If your transformers' secondary winding is 20v-0v-20v, then you can ground the center tap, and use a single diode on either end of the secondary to create a full-wave rectifier.
     
  14. SgtWookie

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    Yes, cell phone/laptop charger accessories are generally switching supplies, which require more components than a simple linearly-regulated transformer supply.

    However, typical "wall-warts" just contain a transformer, a pair of diodes, and a thermal fuse - perhaps a capacitor. They aren't regulated, and ripple at their rated current will be quite high.
     
  15. SgtWookie

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    The 1.25v is the nominal Vref; or the nominal difference in voltage between the OUT and ADJ. It may vary from a low of 1.2v to a high of 1.3v and still be within specifications.

    What you are looking for is the dropout voltage. A typical value for dropout voltage is 1.7 for the LM317 and LM350, however you really need to look at the graph in the datasheet.

    Go to http://www.national.com (National Semiconductor's site)
    and download the datasheet for the LM150 / LM350.
    Look at the plot for "Dropout Voltage", bottom left corner of page 3.
    You'll see that depending upon output current and temperature, the dropout voltage might vary anywhere from 0.9v to 2.6v, but typically be between 1.3v and 2.2v - assuming you're using a suitably-sized heat sink.
     
  16. vindicate

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    So with the second example would that then be 10V ouput?

    Ok I'm looking at the chart:

    [​IMG]
    What I don't get is the Input-ouput Differential sidebar. Why does it only go up to 3V? What if I put in 30V and out 10V? How do I calculate dropout then?
     
    Last edited: Aug 6, 2009
  17. SgtWookie

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    No. That transformer has 20v-0v-20v in the secondary winding, or 40v when measured across the ends of the winding. If you grounded the center tap and used one diode on either end, you would have ripple 20v output (less the diode drop, of course).



    The "dropout voltage" is the minimum voltage drop across the regulator at a given current and temperature. You can always increase the drop across the regulator by decreasing R2's value (ADJ to ground) - but you can't decrease the dropout voltage below the specified minimums.

    Note that as you decrease R2's value to decrease the regulator's output voltage, the power dissipation in the regulator will increase.

    For example, if the regulator's input is 20v, and you have it adjusted to provide a 15v output with a load current of 1A, then the regulator is dissipating (20v-15v) x 1A = 5v x 1a = 5 Watts of power, and the load will be dissipating 15 Watts.

    If you then decreased the regulator's output voltage to 10v with a 1A load, the regulator would then be dissipating 10 Watts of power, and the load would dissipate the other 10W.

    You can see that at higher levels of current output, power dissipation in the regulator will become a real problem; particularly if you try to regulate to a low voltage.

    If you have a load that requires 3A at 5v, and your regulator's input is still 20V, then your regulator's power dissipation is (20v-5v) x 3A = 15 x 3 = 45 Watts! :eek: You would need quite a large heat sink to dissipate that much power.
    If you re-adjusted the
     
  18. vindicate

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    Jul 9, 2009
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    So what you are saying is if you are dropping enough voltage already then you don't have to worry about the min dropout voltage?

    I thought the dropout voltage was in addition to what you wanted to drop. But I think my problem came from mixing in some thoughts on fixed regulators.

    So when a transformers lists ouput as 9VCT or 6VAC what does that mean the output will be after rectification?

    Does 9VCT mean 9VCT center tapped? Meaning if I don't use the center tap and use the 2 end outputs it would be 18V out?

    What bout 6VAC?
     
  19. SgtWookie

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    Something like that.
    Just don't expect to have 20v in and get 19v out with a 3A load. You won't get it.

    The 78xx series have a typical dropout of 2v. That varies with load & temp, of course.

    It depends on how you rectify them, what current you're drawing, and even what kind of rectifiers you use. Schottky rectifiers typically have a lower Vf than silicon rectifiers.

    9VCT could either be wired to put out 4.5v or 9v, less the rectifier drop(s), *1.41. The 6v transformer would need a full-wave bridge rectifier; thus would have two diode drops.

    I take that to mean 9v end-to-end, which would be 4.5v from an end to the center tap.
     
  20. vindicate

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    Jul 9, 2009
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    So does that mean each end would be 4.5V and the center 0v?

    Also, how do current rating work on a transformer. I see alot of them(on jameco anyways) listed with a VA Cap. How does that work into Max Current Output?

    From what I have read the lower the voltage out the higher current output you should have, since transformers don't waste any power.
     
    Last edited: Aug 6, 2009
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