1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Unregulated power supply