Full and half wave rectifier with Capacitor filter problem

Discussion in 'Homework Help' started by matthew798, Feb 23, 2013.

  1. matthew798

    Thread Starter Member

    Jan 16, 2013
    38
    2
    Hey guys.

    I think my answer sheet is mistaken...

    I have the following problem:

    Half wave rectifier, with 47μF filter and 10k load. RMS on primary xformer winding is 120v at 60Hz. Transformer ratio is 8/1. Find Vdc and Vr. (Avergae voltage and ripple voltage)

    120*√2 = 169.705v on the primary winding peak

    169.705 / 8 = 21.21v on the secondary winding peak

    So since it's a half wave rectifier,

    21.21 / ∏ = 6.75v is the average voltage (Vdc)

    And finally,

    Vr = 6.75 / (60Hz * 10KΩ * 47μF) = 239.36mv

    Now my answer sheet says that it should be 752.13mv

    It would seem that they used Vp / (∫*Ω*f), whereas I used Vdc / (∫*Ω*f). I strongly believe that to be incorrect, because using Vp instead of Vdc does NOT take into account that it is running on a HALF wave rectifier... Which effectively doubles the time it takes for the capacitor to start charging again.

    Thanks guys!
     
  2. #12

    Expert

    Nov 30, 2010
    16,257
    6,763
    I believe your mistake starts at Vp/Pi
     
  3. matthew798

    Thread Starter Member

    Jan 16, 2013
    38
    2
    Can you elaborate? Would it be Vp/2 instead? Is that not for calculating Vrms with a half wave rectifier?
     
  4. matthew798

    Thread Starter Member

    Jan 16, 2013
    38
    2
    Never mind I understand it now!
     
  5. #12

    Expert

    Nov 30, 2010
    16,257
    6,763
    That's the rules on the Homework forum. I can't tell you the answer, but I can point you at it.
     
  6. padoh

    New Member

    Aug 29, 2013
    1
    0
    dear matthew798

    i agree with what u have written. the formula doesnot account for the difference in the ripple half wave and full wave rectifier. can u plz help?
     
  7. donpetru

    Active Member

    Nov 14, 2008
    186
    25
    In the case of half-wave rectifier with capacitive filter, the ripple voltage (Vr) is:

    Vr = I_avg / f * C

    where I_avg - is the average current through the diode, which is equal to current drawn by the load; f = 60Hz, C = 47uF.
    Please note, in this case, the RMS current in the secondary winding of the transformer is much higher as I_avg (usually I_rms = 2.5 * I_avg).

    You say the transformer ratio is 8/1, so this mean 120/8 = 15Vrms in the secondary winding. We know consumer value as 10k. Then the rectified voltage is U = 0.95 * Us where Us is 15V. Now, let's calculate:

    I_avg = I_rms / 2.5 = U / 2.5 * 10k = 0.95 * Us / 2.5 * 10k = 0.57mA.

    Vr = I_avg / f * C = 0.57 * 10-3 / 60 * 47 * 10-6 = 0.202V = 202mV.
     
Loading...