+Vcc and -Vcc Dual Power

Discussion in 'The Projects Forum' started by peter_morley, May 20, 2011.

  1. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I wanted to make a dual power supply so I could use op amps. One having a Vcc and the other -Vcc. And the third terminal would be ground. I'm kinda confused about what the difference between ground and -. I know that with if a voltage is negative electrons flow from ground to balance out everything because the source has a deficiency of electrons. If the voltage is positive electrons flow from the source to ground because there is a surplus of electrons in the source. I sometimes get these backwards but I think i'm right. Well i attached a simple schematic. My design seems weird to me because it only works correctly if I dont have the capacitors hooked to the transformer ground. I also used LEDs as my diodes because I didn't have any diodes hanging around. I don't have much experience with diodes but I came up with the diode circuitry anyways myself so it may be incorrect. Let me know how it looks. i'm sorry if my schematic is really blurry
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Well, you really shouldn't use LEDs as rectifier diodes, as they are usually rated for low current (typical is in the 20mA range) and have a low reverse breakdown voltage (around 5v) and if either specification is exceeded they will become damaged. You really should get some proper rectifier diodes; they're relatively inexpensive. 1N4004 1A 400V diodes are around $0.10 USD each if bought in small quantities.

    The capacitors will cause a high initial current flow that will likely damage the LEDs.

    Keep in mind that the polarity of electrolytic capacitors is important; if you connect them up backwards you will destroy them. You have drawn the cap connected to the negative supply with it's positive terminal to that negative supply; if you grounded the junction of the caps, it would be destroyed.

    If electrolytic caps are used in series back-to-back (negative to negative or positive to positive), they can be used as a single non-polarized cap, but the capacitance is cut in half (if they are equal in uF rating). That's kind of an "emergency use only" thing.
     
  3. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    Are you saying it looks ok if I was using rectifier diodes? Do you see anything wrong with the circuit besides that?
     
  4. simo_x

    Member

    Dec 23, 2010
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    If you want to have a dual power supply you have to use a transformer with center tap like this
    [​IMG]

    The middle line will be the reference for GND, when the sinewave will be at positive voltage above it, at the same time (aproximately) will be negative.

    The most to take care is the transformer to calculate the coil, the turns etc..
    Thake this schematic as reference to have an idea..

    [​IMG]

    As you can see, 1N4007 rectifier diodes are placed after the transformer to obtain a full wave rectified voltage positive above and negative below. Instead of using diodes, you can use also a diode bridge.
    7812 is the fixed positive voltage regulator to obtain in output +12V, same as the 7912 for the negative voltage. Take a look at the 78XX & 79XX serie, or if you prefear a variable regulation, take a look at LM317 & LM337. C1 & C2 are needed to atenuate the ripple voltage.
     
    Last edited: May 20, 2011
  5. CDRIVE

    Senior Member

    Jul 1, 2008
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    Perhaps this will be easier to follow.
     
  6. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    CDRIVE: So the way I have it set up is with the capacitors not hooked to ground but actually hooked to each other. Is this the same as what you have done by flipping the electrolytic capacitors or am I potentially causing a future problem? Nice schematic by the way
     
  7. simo_x

    Member

    Dec 23, 2010
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    No, the way you put the capacitors is different respect to CDRIVE schematics..
     
  8. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    Ok so why does mine work correctly it doesn't make any sense to me that is why I want to know.
     
  9. simo_x

    Member

    Dec 23, 2010
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    Did you put the capacitors as in your first post?
    Did you read SgtWookie's reply?

    Mount the circuit as me and CDRIVE have suggested.
     
  10. CDRIVE

    Senior Member

    Jul 1, 2008
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    Both of your outputs (Pos & Neg) should be referenced to the center tap of your transformer, which is common (GND). The way you had the caps wired would not filter the pulsating DC, as they must be connected to GND too.
     
  11. CDRIVE

    Senior Member

    Jul 1, 2008
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    How are you determining that your circuit is working correctly?
     
  12. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    ok, do you think 470 uF is enough to smooth out the 60 hz signal?
     
  13. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I realized it wasn't working correctly because i hooked it up to a 4520 and still saw pulses so it wasnt wrong.
     
  14. simo_x

    Member

    Dec 23, 2010
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    The capacitor value depends on the ripple voltage you want to atenuate.

    Ripple Voltage = I / [ 2 · frequency · CapValue ]

    2 · frequency because as the rectified output is fullwave, instead of 60Hz you will have 120Hz frequency at output.
     
  15. BillB3857

    Senior Member

    Feb 28, 2009
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    Also, referring to your original drawing, you have one of the caps connected backwards. Make sure you observe the polarity markings on the cap when connecting. Negative to ground for the positive supply and Positive to ground for the negative supply.
     
  16. CDRIVE

    Senior Member

    Jul 1, 2008
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    Cap values are chosen by the following factors:
    (1) Whether it's a half wave or full wave supply. A half wave supply that has to supply the same current as a full wave supply will need twice the capacitance to maintain the same minimum ripple.
    (2) Load current. If the current demands are low you can get away with smaller caps well below 1000uF.

    On the other hand, since electrolytics have become so compact these days I generally use 1000uF caps for currents <= 1 Amp. It could also be that I have a few hundred of them though. :D
     
  17. CDRIVE

    Senior Member

    Jul 1, 2008
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  18. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I set up my circuit the way CDRIVE's schematic is. I don't have an oscilloscope so what is a good way to test if it is direct current. If i put an inductor and a resistor as the load from +Vcc to GROUND I should get no voltage drop across the inductor because an inductor acts as a short when there is DC. Is there any other ways you could think of?
     
  19. CDRIVE

    Senior Member

    Jul 1, 2008
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    You should be able to read your ripple by putting your DMM in the AC mode. With no load on your output you should read about zero. As you increase the load on the supply you will eventually start to measure some ripple.
     
  20. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I just wanted to add one more thing.
    When I have no load at the terminals i get about 9.875V for +Vcc and
    -10.125V for -Vcc. When I place resistors of about the same value across each supply terminal to ground I get the same voltage drop across each but one is pos the other neg. I notice when I place a smaller resistor as the load I get a smaller voltage drop across the resistor. This is probably due to the fact that the LED has some resistance and so there is voltage division involved. I could figure out that LED resistance by finding the right resistor that has an equal voltage drop across the two LEDs collectively and the load resistor. Hmmm do regular diodes have less resistance than an LED? I know I could look this stuff up but I like to get clear answers from you guys.:D
     
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