Ripple voltage in Rectifier circuit

Hi....
I have a problem when calculate Ripple voltage in half/full wave rectifier

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


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

please help me explains this thing!
Thanks.

 

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.