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Invert polarity of output DC voltage

  1. Apr 18, 2008 #1
    I'm building a power supply which is shown below :

    [​IMG]


    The tricky part is that i want a dual polarity PSU so i duplicate the circuit but this time i want to invert the output DC voltage as to be a negative one...what is the simplest way to do this? i thought of passing that output to a negative voltage regulator but i haven't tried it though.
     

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  2. jcsd
  3. Apr 18, 2008 #2

    berkeman

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    You have two main choices -- Use a different transformer that has two output windings (like a long winding with a center tap), and do a similar - rail linear regulator branch, or use an inverting DC-DC converter off of the + rail that you have now.

    Can you easily go to a different transformer with two output windings? If not, look at Inverting DC-DC converter circuits, like this:

    http://focus.ti.com/docs/prod/folders/print/tps63700.html

    The problem with doing an inverting DC-DC for the - rail is that it will have switching noise on it, as opposed to your nice clean + rail from the linear regulator.
     
  4. Apr 18, 2008 #3
    couldn't you use a mechanical switch on the output leads?
     
  5. Apr 18, 2008 #4

    berkeman

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    I got the impression from the phrase "dual polarity" that they want both + and - rails available at the same time.
     
  6. Apr 18, 2008 #5
    exactly i want a +ve and -ve output at the same time....by the transformer, you mean i get a center-tapped transformer? note that it is a step-down transformer.

    anyway, i will look at the dc-dc converter you mentioned but i hope anyone can give me more options cuz if one doesn't work i switch to the other immediately...thanks alot though.
     
  7. Apr 19, 2008 #6

    Redbelly98

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    Yes, you want a 24V CT (center-tapped) transformer. You'll get separate 12Vac outputs from the center tap to each of the other output leads.

    For what it's worth, are you aware that 12Vac will produce about 16Vdc after the rectification?

    Good luck & have fun!
     
  8. Apr 19, 2008 #7
    yes my friend but the circuit has been designed to limit the voltage between 5 - 15 volts...hmm so let me get this clear, i duplicate the circuit as it is with no polarity of the anything -the capacitor- reversed and simply connect it with a center tapped transformer, and i will have a negative voltage?

    if its not too hard for you can u tell me how it works?
     
  9. Apr 19, 2008 #8

    Redbelly98

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    Okay.
    Sure. A center-tapped transformer will have 3 output leads: a center lead and 2 other leads, we'll call them A and B.

    Connect the center lead and lead A to your existing circuit.

    Then, connect the center lead and lead B to another circuit that you must build. This other circuit differs from your existing one as follows:
    1. The + output from the bridge will get connected to the "0V" point in your first circuit, and also to the "common" point of the regulator in the second circuit.
    2. You'll need a different regulator that is designed for negative polarity. One option is the LT1033:
    http://www.linear.com/pc/downloadDocument.do?navId=H0,C3,P1239,D3905
     
  10. Apr 27, 2008 #9
    I saw this circuit on the "designer's guide to the l200"...is there anyway to verify that its working before actually building it? and can someone explain how this circuit works if he/she understands it...please please please help.:cry:
     

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  11. Apr 27, 2008 #10

    Redbelly98

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    It's slightly more complex than the original circuit you posted. I'll assume you understand the L200, as it is used in both. There are 2 main differences in this circuit:

    1. The rectification is done differently. However, the input to the L200 is still a DC-with-ripple signal.
    2. They have built 2 identical DC circuits. Connect the low voltage output of one supply to the high voltage side of the other, and call this ground.

    To verify that it works, build it in stages. Start from the left side of the circuit diagram (the transformers). Verify that each stage does what it should before adding on to it.
     
  12. Apr 28, 2008 #11
    but is there anything that indicates that the output voltage of the second one is negative since as you said they are identical...and can you please tell me what does tracking mean?


    Thanks alot man, i really appreciate your help.
     
  13. Apr 28, 2008 #12

    Redbelly98

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    Yes. The fact that they are calling the positive output "ground" means that the negative output can be used as a negative voltage source.

    They seem to have two adjustment potentiometers that track each other, i.e., they both have the same setting at all times. Don't worry about that, just use normal potentiometers and adjust them separately.
     
  14. May 10, 2008 #13
    With regards to the first circuit posted in the first post, can anyone tell me what are the smaller capacitors -100nF- after the 4700mF and the 10mF capacitors for? what do they do?
    i know the 4700 is the smoothing cap but what does the one after it do?? please this is urgent.
     
  15. May 10, 2008 #14

    f95toli

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    The problem with large electrolytic capacitors is that they are "slow" (high ESR values=equivalent series resistance). This means that they are good for smoothing out low frequency rippel but not very good at filtering out higher frequencies.
    Smaller capacitors (polypropylen etc) which have low ESR values and are good for higher frequencies.
    Hence, in real circuits you need mix different types of capacitors in order to get the performance you want (it is not uncommon to have 3-4 types in a single circuit).

    Learning how to choose the right capacitor for the job takes a while and is one of those things that are often not covered in courses.

    Edit: In regulator circuits the main "job" for the small capacitor after the regulator is usually to filter out the high frequency noise comming from the regulator itself (which is in the kHz range).
     
    Last edited: May 10, 2008
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