RECTIFIERS

Discussion in 'General Electronics Chat' started by helpmeplease, Feb 20, 2006.

  1. helpmeplease

    Thread Starter New Member

    Feb 20, 2006
    2
    0
    hi new to the website and rely need some help .....


    a half wave rectifier operatin from a 50hz 24volt AC supply is connected to a 100 Ω load
    caluclate the value of the smoothing capacitor required to maintain a peak to peak voltage ripple at 2volts

    PLEASE HELP!!!
     
  2. dragan733

    Senior Member

    Dec 12, 2004
    152
    0
    U ≈ Umax- ∆ U/2=Umax- ∆ Q/(2C)=Umax-I/(2Cf)=Umax-U/(2RCf)
    U ≈ Umax-U/(2RCf)
    from here
    2RCf ≈ U/(Umax-U)
    C ≈ U/( ∆ URf), where ∆ U=2(Umax-U)
    C ≈ 24/(2*100*50)=2400 μ F
    C ≈ 2400 μ F

    Regards
     
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