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Unregulated power supply

  1. Aug 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Specify ratings and values for the transformer, bridge rectifier and smoothing capacitor for a 230 V unregulated power supply with an output voltage of 12 V, output current of 2 A and a ripple factor of <0.024

    2. Relevant equations
    V * I
    V*sqrt 2
    C= I/2f*Vr
    Id=(1+sqrt2)*IL


    3. The attempt at a solution
    The transformer VA rating should exceed 12*2 so 24VA
    The peak secondary voltage will be 12 * sqrt 2 = 17V
    The value of the capacitor will be: 2/2*50*0.815 = 24,540 micro-farads
    Bridge rectifier will need to be able to handle the peak forward current of: (1+sqrt2)*2=4.83amps
    Power dissipation is 2*2=4watts (assuming 2 volt drop across the bridge)
    Have I approached this is the correct way?
     
  2. jcsd
  3. Aug 18, 2016 #2
    It looks as if your PSU will give 15 volts DC out. 17 volts peak less 2 volts for the rectifier.
    Also could you clarify your capacitor calculation - not sure about it.
     
  4. Aug 18, 2016 #3
    OK thank you I thought that would be the output for the DC.
    This is how I calculated the capacitor value:
    Using C= I/2f*Vr:
    I= 2 amps
    f= 2 x 50
    Vr = ripple voltage which I calculated by: 0.024 * 12 =0.288 V (rms) converted to peak to peak by 0.288*sqrt2*2 = 0.815 V
     
  5. Aug 19, 2016 #4
    Is that the correct methodology?
     
  6. Aug 19, 2016 #5
    Looks correct to me.
     
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