How to calculate Deoupling CAP for removing High Frequency Noise ?

[vikasveshan]

Hi all ,

I wanted to understand the reason and the logic behind useing low value CAP for removing High Frequency Noise .
And why not high value CAP .


Thanks & Regards !!
V V

 

[momokochan]

It's not because of the value of the capacitance, but the low value caps tend to be physically smaller. Higher frequency means smaller wavelength so the smaller caps actually respond better at these frequencies.

 

[Orson Cart]

It does depend somewhat on the energy level of the noise, source impedance and its main frequency components, high energy levels require more capacitance, a low source impedance of the noise also requires more capacitance, smaller caps tend to have lower ESL and fairly low-ish ESR so they filter HF noise better because they present more pure capaciatnce at the frequencies of interest, larger caps may be very high impedance and inductive at frequencies above those intended by the makers.
Regards, Orson Cart.

 

[Milad-D]

high value capacitors like electrolyte capacitors suffer from series inductors and therefore if you plot the impedance of it, it starts to rise up at high frequency due to this inductance. Therefore they are are good for rejecting the low frequency noises. Small capacitors, like tantalum caps have smaller values of capacitance but they don't suffer from series inductor and they can efficiently reject the high frequency noise.
Therefore in good design you see that always a big electrolyte capacitor is in parallel with small tantalum capacitor to reject both low and high frequency noises

 

[naderi]

It is all about the Q-factor of the capacitors. Low-Q cap cannot pass the RF due to the series resistance.