How To Make Dc Power Supply ?

Discussion in 'General Electronics Chat' started by abhimanyu143, Aug 29, 2014.

  1. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    After some reading I have made this calculation
    component
    transformer
    diode rectifier
    filter capacitor
    R load

    1. transformer
    primary voltage Vp= 230 Vrms ac
    secondary voltage Vs= ?
    primary turns Ns = 5
    secondary turns Np= 1
    primary current Ip =?
    secondary current Is ?
    R load = 50 ohms
    formula for transformer
    Vp/Vs=Ip/Is=Np/Ns

    secondary voltage

    Vp/Vs= Np/NS

    230/Vs=5/1

    Vs = 46V rms Ac

    secondary current


    Is =Vs/Rload

    Is= 46/50= 0.92 amp rms Ac

    primary current


    Vp/Vs=Is/Ip
    230/46=0.92/Ip

    Ip= 0.184 amp rms AC

    transformer with load

    primary voltage Vp= 230 RMS ac
    secondary voltage Vs= 46 volt Rms Ac
    primary turns Ns = 5
    secondary turns Np= 1
    primary current Ip = 0.184 A rms Ac
    secondary current Is = 0.92 A rms Ac

    2.rectifier
    If we connect single rectifier diode then

    We know following for rectifier

    Input voltage = 46 Vrms Ac

    Input current =0.92 Arms Ac

    I have to find out output voltage, current for rectifier


    Output current for rectifier


    Idc= Vs/RL= (46-0.7)/50=0.906 dc


    Output voltage

    Vdc = Idc* Rload=0.906*50= 45.3 vdc

    3.filter capacitor
    now we know input voltage and current for capacitor
    Idc= 0.906 dc
    Vdc =45.3 vdc

    I want to do some calculation for capacitance
    how to choose capacitor value ?





     
    Last edited: Aug 29, 2014
  2. #12

    Expert

    Nov 30, 2010
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    Sqrt2 C Er F = I where Er is the allowable, peak to peak ripple voltage, F is the frequency after whichever rectifier design you used, C = capacitance, and I = current to the load.
     
  3. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,570
    2,381
    Voltage out 46vac x 1.414 = 65vdc.
    Max.
     
  4. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    I know only frequency 60 Hz and load current Idc= 0.906 dc
    how to determine peak to peak ripple voltage ?
     
  5. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    I don't know what are you telling ?
     
  6. #12

    Expert

    Nov 30, 2010
    16,348
    6,836
    Use the math I gave you. sqrt2 = 1.414 60 Hz = F if you didn't use a full wave rectifier. It's 120 Hz if you did. Either you decide the ripple voltage you want to allow or you decide the capacitor size. The answer comes out of using the formula.
     
  7. #12

    Expert

    Nov 30, 2010
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    He is telling you that the RMS voltage of a sine wave will reach a DC peak of 1.414 times the RMS value after it is rectified (minus the voltage lost in the diodes).
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    The first figure might help you understand the peak voltage that the filter capacitor charges to, and how the value of the filter capacitor effects the ripple. The second figure shows why it is better to use a full-wave rectifier...
     
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    Last edited: Aug 29, 2014
  9. MaxHeadRoom

    Expert

    Jul 18, 2013
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    ...And if you want 46vdc as a final result, that would be 46vdc X 0.707 = 35vac required.
    Max.
     
  10. cl10Greg

    Active Member

    Jan 28, 2010
    49
    0
    I am not sure how set in your design you are, but have you looked at any of the various power supply designer web tools? TI has one, Fairchild, Power integrations, etc...
     
  11. ISB123

    Well-Known Member

    May 21, 2014
    1,239
    527
    Use 4700uF cap at 80V.
     
  12. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    primary voltage Vp = 230 v ac
    peak voltage of secondary=65.004 V Ac
    Rms voltage of secondary = 46 V ac
    output of rectifier = 65.044-0.7=64.344 V ac
    main frequency F= 60 Hz
    load current Idc= 0.906 dc
    peak to peak ripple voltage Vr= 2 v

    using formula for capacitance

    C=I load /2*F*Vr
    C=0.906/2*60*2
    C=0.906/240

    C=0.003775 F
     
  13. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    If my capacitance value is correct then I want to add regulator Ic
    how to do some calculation for regulator Ic ?
     
    Last edited: Aug 30, 2014
  14. #12

    Expert

    Nov 30, 2010
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    C = Iload /1.414 * F *2
    C = .906/1.414 * F * 2
    C = 53388 uf

    Read the datasheet for the regulator you want. It tells all about which capacitors to use and how to adjust for different output voltages. They are often a little different from brand to brand, so pick one and read about it. Small capacitors are used to keep the regulators from oscillating. Your big capacitor is finished.
     
    abhimanyu143 likes this.
  15. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    what is 1.414 ? C=I load /2*F*Vr Is it wrong ?
    If I connect regulator I think I need 2 small capacitor
     
  16. ISB123

    Well-Known Member

    May 21, 2014
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    Last edited: Aug 30, 2014
  17. bertus

    Administrator

    Apr 5, 2008
    15,649
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    Hello,

    Have a look at the attached PDF.

    Bertus
     
  18. #12

    Expert

    Nov 30, 2010
    16,348
    6,836
    On page 8, that PDF goes right into talking about Vpeak, but I can't find where it explains that Vpeak = the square root of 2 times the RMS value of the voltage. Transformers are usually labeled in RMS voltages. ps, the square root of 2 is 1.4142136 according to my calculator.
     
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