ripple voltage for capacitor filter

Thread Starter

vvkannan

Joined Aug 9, 2008
138
Hello,
My text says the peak to peak ripple voltage of a capacitor input filter is

I/(f*c) (I -> dc load current ,f->ripple frequency ,C->capacitance)

But it doesnt provide the exact derivation.How do we get it?

Thank you
 

mik3

Joined Feb 4, 2008
4,846
Assume that the ripple voltage is a triangular wave and that it takes half the period (T/2) to fall from the upper peak to the lower peak.

Voltage across the capacitor=charge/capacitance

thus

Vc=Q/C (also is the ripple peak-peak voltage Vr)

Q=current*time=I*T/2

thus

Vc=Vr=Q/C=(I*T)/(2C)

T=1/f

thus

Vr=I/(2*f*C) and not I/fC
 

mik3

Joined Feb 4, 2008
4,846
Thank you mik3.
i saw it as I/fc in malvino's 'electronic principles' and thats why the doubt.
I have the same book and it is correct. If you read the following lines it says that for a full wave rectifier f=120 Hz and for a half wave rectifier f=60 Hz. This is the same as the formula I told you because in the formula I told you set f=supply frequency (60 Hz for example) and not twice the supply frequency. Also, the formula I told you is true for full wave rectifier only.
 
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