Ripple voltage in Rectifier circuit

Discussion in 'Homework Help' started by cicada207, Nov 1, 2015.

  1. cicada207

    Thread Starter New Member

    Nov 27, 2014
    2
    0
    Hi....
    I have a problem when calculate Ripple voltage in half/full wave rectifier
    [​IMG]
    Following formula of Ripple voltage
    Vr\doteq\frac{Vom}{fRC}
    (HWR)
    e.g [tex]V_s= 9V [/tex]
    R= 1k (ohm)
    C = 22uF
    => [tex]V_om=V_sm-V_\gamma =12.7 -0.7 = 12V
    =>V_r = 10.9V
    =>V_omin = V_om-V_r=12-10.9=1.1(V)[/tex]
    BUT when I simulate circuit in Proteus
    [​IMG]

    Result [tex]Vomin= 5.8(V)[/tex][plain]

    please help me explains this thing!
    Thanks.
     
    Last edited: Nov 1, 2015
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Well, the answer is quite simple. Your equation gives you only an approximate value for Vr (worst case ever). Because your equation simply assumes that capacitor is discharging via constant current and do not take into a count the diode conduction angle. Also there is no single equation that will give as the exact ripple value.
     
    Last edited: Nov 1, 2015
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