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Bypass network

sert , 07-17-2017, 06:51 AM
Bypassing supply at analog devices is to reduce existing noise in power supply. But how about the positioning of the caps if two ICs are close to each other? Weren`t 50% of the caps redundant?
I place 2 ICs and their baypass caps close to each other. Would i get the same result if i only use 3 caps instead of 6?

The 2 ICs are close to each other again, but the bypass networks are placed on the opposite. Do i need in that case both bypass networks because of the spatial seperation?

mairomaster , 07-18-2017, 01:40 AM
The primary role of the decoupling capacitors is to be a close, low impedance current source, which keeps the voltage at the IC stable.

If you have 2 ICs you will have twice the current draw. Optimally you will want to use twice the capacitors. If you know what you are doing you might be able to make a compromise and save some of the capacitors, but just removing half of them is not optimal and might cause problems if the circuit is decoupling sensitive.
sert , 07-18-2017, 03:59 AM
Ok that makes sense. I thought about using bypass caps at analog components as filter for the power supply. And thus "clean" it and supply nearby ICs.
But since my analog signals have also a square shape, the current drawn by the devices plays a crucial role especially at the fast edges.

I will give it a try and see how good theory matches up with the real assembly.
mairomaster , 07-18-2017, 05:38 AM
The power supply output needs to be filtered as close to the output itself as possible - doing it differently is normally wrong. At the supplied components end you should already have a nice clean rail.
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