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I have found and read many papers about bypass capacitors, but there is still one thing I am really struggling with. Let's say I have an IC and I place 3 bypass capacitors (eg. 10µ, 1µ and 100n). I am familiar with the idea that the smallest should be placed nearest, but I am still not certain why. From the papers I caught these 2 reasons:

  1. The smallest capacitors are faster; thus, they can react fastest.
  2. The goal of the smallest capacitor is to "filter" higher frequency noise. (This one is the one where I struggle.)

From what I've read, the reason to place the smallest closest is that high frequencies are affected by the length of the trace more than smaller frequencies. Is there someone who could explain this part to me?

Edit:
When I consider the typical graph of capacitor impedance based on frequency (the V shape graph), does it mean that the point of lowest impedance will be due to the inductance of LONG traces shifted left thus not filtering the high frequencies?

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    \$\begingroup\$ Smallest one needs smallest trace inductance, to drive its self-resonant frequency right up. \$\endgroup\$ – Brian Drummond Jun 3 '20 at 15:50
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Because inductive impedance increases with frequency $$Z=j\cdot2\cdot\pi\cdot f\cdot L$$ you need to have lowest inductance (=shortest trace) for highest frequency filtering.

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The traces are the reason why and the parasitic inductance, copper adds inductance, and it can be calculated with the equations below OR you can find a PCB trace calculator.

You then can take the calculated value of the inductance of the copper and plug it into a RLC calculator and find the filter pole

Another factor is the small amount of resistance that traces have, but it doesn't affect the pole of the filter as much as the inductance because it is so small.

The inductance of the copper (usually in nH for small traces around 8mil wide and short) makes a filter with the capacitor which will cause resonance and block the current source from the capacitor, this happens at high frequencies, but with IC's that run in the 100MHz or GHZ it can be a problem.

An easier thing than calculating is not calculate the filter pole and just place capacitor with the smallest value close to the IC.

enter image description here
Source: https://www.eetimes.com/power-tip-56-estimate-pwb-interconnect-inductance/#

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  • \$\begingroup\$ Yeah, I am not planing to calculate much, but out of curiosity I wanted to know the main reason, why the smallest capacitor is always closest to the IC. \$\endgroup\$ – Tomáš Havel Jun 3 '20 at 16:07
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enter image description here The DC power supply is suddenly applied to IV and this produces the spike ,which is high frequency signal and this may damage the IV. To protect IC from spike we connect capacitor which bypasses the spike to ground.

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    \$\begingroup\$ Please don't answer with scribbles in images, see this meta discussion \$\endgroup\$ – pipe Jun 5 '20 at 6:57

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