Smoothing capacitor

Thread Starter

cdennis414

Joined Dec 19, 2012
15
Hi everyone
I have to design a power supply for my assignment but i struggle to find a right equation for smoothing capacitor... There are a few different equation i have found but they all giving me different result.
The most common equation i came across is C=I/f*V and f is equal 100 Hz.
But the one i have used is from a book. I have attached a pictures from the book. What i can't understand is why in they calculations the frequency timed by 2. Its already a 100 (supply frequency by 2)???
Thanks.
 

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t06afre

Joined May 11, 2009
5,934
The frequency is the number of occurrences of a repeating event per unit time. I took this drawing from Wiki to illustrate. After the rectifier the waveform has changed and frequency has doubled, compared to the AC input frequency. It is that simple
 

Thread Starter

cdennis414

Joined Dec 19, 2012
15
Ok I understand that but if you look at the calculation from the book the 50 Hz original AC frequency been doubled which makes a 100Hz and then doubled again. The part i cant understand why they doubled it twice. Sorry if i didn't explain it properly. I attached the first page from the book. As i understand the original frequency is 50Hz.
Thanks.
 

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t06afre

Joined May 11, 2009
5,934
The number 2 has nothing to do with the frequency.
The formula is for half-wave rectification:
Vripple=I/fC
And for full-wave rectification
Vripple=I/2fC
Remember that the frequency for half-wave rectification will be the AC voltage frequency. So the ripple in a full-wave rectification. Will be four times less compared to half-wave rectification
 
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