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AndrewT 29th January 2007 07:54 PM

ripple current calculation?
I have assembled a timer with relay.

The supply is a nominal 9Vac bridge rectified and minimal smoothing.
The scope shows ripple on the DC feed of 12V to 14V (flat topped saw tooth) when driving the 555 timer and it's few components.
The supply drops to 3V to 13.5V (10.5V of ripple) when the relay pulls in.

The relay is 400r and the timer draws about 5mA.

How does one calculate the ripple current in the first and only smoothing capacitor? Iripple=Vpk/[relay resistance] + timer current?

bays_9a5tt 29th January 2007 09:59 PM

hi Andrew, you might find Power Supply Designer good for your calculations.

AKN 29th January 2007 10:41 PM

Hi Andrew,

Wondered how to calculate ripple current myself, however found this:

Under calculation examples there is formulas that seems comprehensive.

ifrythings 30th January 2007 09:39 AM

C=Iload / 8 x F x Vout

Iload is in Amps

F is in Hz

Vout is desired ripple in V

So, C= 1Amp / 8 x 60Hz x 1Vripple = 2 083uF

richie00boy 30th January 2007 09:42 AM

Calculating ripple current is not easy. In the end we used a current clamp on the big power amps to measure it.

AndrewT 30th January 2007 04:24 PM

the 22uF allows the circuit to work correctly and after running for an hour still felt cold.

10uF does not work, so I will adopt 33uF or 47uF (I have a few in stock).
the ripple current for a range of 22uF varies between 37mA and 50mA and that seems very close to the running current that the circuit consumes.

4fun's link shows a calculation method, but, it is more complicated than I expected so I will try a few numbers and see what pops out.

Thanks for the replies.

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