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-12Vdc Circuit design

Discussion in 'General Electronics Discussion' started by Danno, Feb 19, 2016.

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  1. Danno

    Danno

    65
    2
    Aug 31, 2015
    Hi,
    I have a question on the design of part of a power supply, on the attached is a snipped of the power supply circuitry, there are two design variations.
    What I am interested in is the -12Vdc.

    I have shown caps and resistor for -12Vdc circuit but left all others out for clarity.

    In design A the -12Vdc is routed through a 7912 regulator and D9 & D10 and back onto the respctive 15Vac windings, all makes perfect sense.

    In design B the 7912 has been done away with and D9 & D10 now go back to the 9Vac windings.

    My question is how can design B produce a regulated -12Vdc supply?


    Thanks.
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Short answer: It can't.

    9V (RMS) is ~12.7V (peak) - Vpeak=sqrt(2)*Vrms for a sinusoidal waveform.
    Voltage drop across a diode is ~0.7V, therefore Vout = 12.7V-0.7V = 12 V - completely unregulated. The output voltage will vary with load current and/or mains voltage.
    The reason circuit A uses the 15V windings (not the 9V) is, that the 7912 requires the input voltage to be considerably higher (I think more than 3 V but I'm too lazy too look up the datahseet) than the output voltage. From 15V (RMS) the input to the 7912 is ~20V, more than enough for good regulation. Note, however, that the voltage drop across the 7912 is 21V-12V=9V. Multiply this voltage with the output current on the -12V rail to see the power dissipated by the 7912 (e.g. at 0.5A, P=0.5A*9V=4.5W!). You'll need a big heatsink to protect from burning up.
     
  3. Danno

    Danno

    65
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    Aug 31, 2015
    Hmmm, I wonder why they changed the design, maybe the regulators were failing, they are on a big heat sink.

    I understand the peak voltage but when I measure it with a multimeter it reads -12V, surely the multimeter is reading RMS?
     
  4. Danno

    Danno

    65
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    Aug 31, 2015
    No forget that it's DC, so not RMS, I'm confused again LOL.
     
  5. Danno

    Danno

    65
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    Aug 31, 2015
    I still don't understand how it's getting -12V, surely if you rectify 9V RMS you will get 8.3Vdc?
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    No, surely not. The smoothing capacitor stores the peak voltage of the rectified AC. The peak voltage of a sinusoidal AC is sqrt(2) * V (rms).

    Without the capacitor the voltage is not truly DC, it is DC superimposed with AC, the RMS of which is the same as for the pure AC.
     
  7. Danno

    Danno

    65
    2
    Aug 31, 2015
    Running it through a circuit simulator it shows -8.3Vdc, or I'm I missing something
     
  8. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Most likely your AC source is rated by peak-to-peak voltage and you have entered the RMS voltage. That would be true if you are using LTSPICE. To model a 9V RMS AC source in LTSPICE you have to enter 12.7V as the voltage.

    Bob
     
    Harald Kapp and Arouse1973 like this.
  9. AnalogKid

    AnalogKid

    2,571
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    Jun 10, 2015
    Note that just because a schematic says 12 V, that doesn't mean it is 12.00002836 V all the time. It might be close enough under load, driving a circuit that doesn't care about a little ripple.

    ak
     
  10. dorke

    dorke

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    Jun 20, 2015
     
  11. Danno

    Danno

    65
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    Aug 31, 2015

    Ah yes the penny has dropped, I was working it as RMS but as you say it's peak voltage, thanks
     
  12. Danno

    Danno

    65
    2
    Aug 31, 2015

    You are quite right, simulator requires peak voltage entered and I was using RMS
     
  13. Danno

    Danno

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    Aug 31, 2015
    One other quick question guys please, why would they have used two diodes in parallel on each leg of TX winding, why not just one on each?
     
  14. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    2 diodes in parallel = 1/2 current per diode. Lower currrent diodes are less expensive plus by distributing the power dissipation between 2 diodes, you may not need a heatsink.
     
  15. dorke

    dorke

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    Jun 20, 2015
    In general this is bad practice,very bad practice!
    should be avoided at all cost...
    there is a thermal runaway danger i.e diode destruction.:eek:

    Unless the diodes are matched to perfection and are at the same temperature .
    i.e the diodes are on the same die and same package.
     
  16. Danno

    Danno

    65
    2
    Aug 31, 2015
    Diodes are IN5402, which can handle 3A, maximum current of PSU is 2.5A on the 5Vdc, so it doesn't appear to need two diodes.
     
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