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A simplified formula for ripple on a capacitor input power supply:
Vrip = I/F*C
Where Vrip is peak to peak ripple, I is the current in amps, F the frequency in Hz, and C the capacitor in farads. Note that a full wave rectifier will give you double the line frequency. This formula is only an approximation, but will suffice for most designs.
I never seen this formula before Vrip = I/FC can you tell me where it is derived from. Some people are saying to use RC or LC filter. SMPS is at 100khz.
how can i measure the frequency of the noise. Do you i do fourier analysis to figure out the harmonics or...?
Yes, you normally use a LC filter in a SMPS output stage. You would only use the capacitor input filter to rectify and filter the incoming AC line voltage into DC for the SMPS to chop at 100Khz. I can't remember exactly where it is derived from, but it probably starts with v(t)=q(t)/C, etc. Sleepy.
However on the mains side, this will be too small.
Because the current will increase as the voltage falls. An SMPS can be approximated as a constant power device so you need to look at the power drawn and the minimum input voltage.
I haven't worked out a formula for the primary side of SMPS, I'll give it a go if you like (assuming Google doesn't find anything) but my calculus isn't that good so someone else will probably beat me to it.
For the secondary side of the SMPS just use the formula I described above and add a 100nF ceramic capacitor in parallel because electrolytics aren't good at high frequencies.
Using inductors can be problematic if the circuit being driven draws a pulsed current because it could excite the resonance in the inductor and filter capacitors. Your probably better off with a ferrite bead which is lossy at high frequencies and shouldn't cause too much resonance.
In a SMPS the size of the output capacitor and inductor will depend on, amongst other things, the duty cycle... specifically the "off-time"... i.e. if I have say a 5V output with a10% duty cycle and another 5V supply that has a 90% duty cycle... the output inductors & cap's are completly different for the same pk-pk ripple with the same load level on both.
Thanks guys that was really good information. I did however approach a solution as to solving the ripple. I just used a basic LC filter now formula for it is f = 1/(2pi root(LC)) this will reduce the ripple. as for the output capacitor i tried the simplified formula but it doesn't seem to do the job in my simulator.
The C = 1 / (Vrip * F), i believe is for bypass capacitor formula. But never the less i will try to see its effects on my circuit simulation.