Ripple Voltage -- The mystery

In the article on how to build a Power Supply, there's this statement: "If the input voltage is 9V AC, you will be able to draw 1A from the power supply. For the maximum input voltage of 35V you will be able to draw 0.1A."

If I want to calculate the ripple using c=470uF, f=100Hz and Vr = I/(C*f). I get Vripple_35v = 2v (0.1A) witch is seems fine, but I get Vripple_9v = 21v (1A) witch doesn't make sense.

What am I thinking wrong??
Thanks in advance.

Posted by André (evofaisca@hotmail.com) on 01/05/2008, at 19:16 GMT

Ripple voltage

The formula is an approximation which is valid for small ripple voltages. The capacitor must be sufficiently large for the current flowing. 470uF is too small to smooth the supply to a small ripple for a current of 1A so the result is a silly answer in this case. I suggest a much larger capacitor, such as 10,000 uF which gives about 1V ripple.

You may prefer to look at the problem another way: calculating the capacitor required to give 10% ripple (a good starting point for many circuits). There is a formula for this on my website: www.kpsec.freeuk.com/powersup.htm#smoothing

Posted by John Hewes on 07/05/2008, at 15:38 GMT

Ripple Voltage -- The mystery

Thanks.
It's problably better to view the matter from the point of view of the 10% ripple. But I'm seeing that for 1A of current I need it big, very big.

Posted by André on 15/05/2008, at 22:50 GMT