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My colleague claims it is best practice to put a bypass capacitor "before the circuit" on the PCB, and even complicates the PCB design in order to ensure this rule is followed. I claim it doesn't matter, and as such it it silly to complicate the PCB design when a simpler design with the capacitor "after" would be available, but I however cannot bring any evidence to back up my claim.

(In the sketches, the integrated circuit is replaced by a current source for simplicity. The imperfection of write from power supply is modelled with a small resistor Rwire and a small inductor Lwire).

schematic

simulate this circuit – Schematic created using CircuitLab

But I do not see how it would make any difference if the bypass capacitor was "after" the circuit such as this :

schematic

simulate this circuit

Does it "before" vs "after" really makes any difference?

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    \$\begingroup\$ You could to 'simulate' the resistance of the (horizontal) connecting tracks between the components by adding 0.05 ohm resistors (an arbitary value), 4 per circuit, and see what happens. \$\endgroup\$
    – JIm Dearden
    Commented Dec 17, 2016 at 20:44
  • \$\begingroup\$ The answer depends on how you model the parasitic resistance and inductance of the supply and ground return traces. The simplified lumped-constant circuits shown in this question don't include any parasitic effects from PCB trace resistance and inductance, so there's nothing for the bypass capacitor to compensate. But in the real world, such parasitic effects do exist and sometimes need to be compensated. \$\endgroup\$
    – MarkU
    Commented Dec 17, 2016 at 21:42

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It makes no difference.

What matters is (mainly) the loop area from the positive terminal on the IC, through the IC to its ground pin, through copper to the negative terminal of the capacitor, through the capacitor, and then through copper to the positive terminal of the IC.

In both cases, assuming the capacitor is very close to the circuit, it will absorb the noise caused by the long supply lines ...

The noise isn't caused by the supply lines, it's caused by the switching currents from the IC (which you correctly model with a current source) and the impedance of the network providing power. You are correct that it doesn't matter "which side" of the IC the capacitor is located, if it provides a low-impedance path for the switching currents it will prevent them from producing voltage variation on the supply net.

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  • \$\begingroup\$ Your first statement is correct. The rest, not so much. Generally, at least for logic ICs, this just isn't true. What is important is the resistance and inductance of the traces connecting the capacitor pins to the ground and power pins of the IC. (Plus, of course, the internal parasitics in the capacitor) And, since a capacitor is an insulator, just what constitutes a loop? \$\endgroup\$
    – WhatRoughBeast
    Commented Dec 17, 2016 at 20:41
  • \$\begingroup\$ Then I need to use a concrete proof there's no difference to show my colleague I'm right - just saying "someone on stack overflow said so" won't convince him. \$\endgroup\$
    – Bregalad
    Commented Dec 17, 2016 at 21:23
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    \$\begingroup\$ @WhatRoughBeast, 1. The area of the loop I described is what will determine the inductance of the connection to the capacitor. 2. Displacement current flows through the dielectric of the capacitor. \$\endgroup\$
    – The Photon
    Commented Dec 17, 2016 at 21:35
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    \$\begingroup\$ @Bregalad, what is his reasoning for thinking the capacitor has to be where he wants it? \$\endgroup\$
    – The Photon
    Commented Dec 17, 2016 at 21:37
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    \$\begingroup\$ @WhatRoughBeast, the conductor dimensions have an effect but generally it's a smaller effect than from the loop area. \$\endgroup\$
    – The Photon
    Commented Dec 18, 2016 at 2:15

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