This question already has an answer here:

I know I am wrong, this question is just to understand logically why, I am not an engineer but an hobbist

While designing a pcb for a mcu I should put a decoupling capacitor near each power lines. We do it because the current draw of the ic could change and the capacitor will supply it immediately

Until here i hope we all agree

but why can't i put this cap only near the Vcc?

this is a hyperbole/example of what i am trying to say: enter image description here

I heard that the answer could be: "Because the electrons moves from - to +" but i don't think that this is correct logic, because this could bring me to make the board in the opposite situation: the cap near gnd and a very long track until vcc

I hope that i m not going to "shock" anyone with this dumb question, please it is just a curiosity


marked as duplicate by Olin Lathrop capacitor Dec 19 '17 at 19:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ What do you mean "only near the VCC"? Does your MCU only have one VCC pin? \$\endgroup\$ – The Photon Dec 19 '17 at 19:03
  • \$\begingroup\$ it is just an example, in this situation yes, immagine that the mcu just have one vcc/vss couple of pin, by the way i am not going to power it with a 9V battery \$\endgroup\$ – aster94 Dec 19 '17 at 19:08
  • \$\begingroup\$ If the question is, "why near Vcc rather than near the Ground line?" then nowadays, most layouts have a layer dedicated to a ground plane, while not all have another layer devoted to a power plane. \$\endgroup\$ – Brian Drummond Dec 19 '17 at 19:15

There is no difference between VCC and GND, both provide a voltage rail, and they are both defined only by the difference in voltage between them.

When you decouple VCC, the IC cannot pull the VCC pin down with a hard transient.

When you decouple GND, the IC cannot pull the GND pin up with a hard transient.

Either way would reduce the difference between VCC and GND.

An alternative view is to ignore the DC voltage, and treat the IC as the source of high-frequency noise that you want to stop from reaching the other components or the power supply. Having a capacitor on a short path between the supply pins shorts the high-frequency noise (remember that capacitors pass high frequency components and block lower frequencies), however the inductance of the traces to the capacitors form a low pass filter, and together with the capacitor, you get a band pass. Ideally you want a fairly wide band here, so you want high capacitance and low inductance on this path.

  • \$\begingroup\$ I think that this perfectly answer to my question, thank you! \$\endgroup\$ – aster94 Dec 19 '17 at 19:19

Having a long ground connection to the decoupling capacitor will increase its effective inductance, making it less effective at doing what it is prescribed for: handling sudden power bursts. The inductance slows down the current change, starving the chip and potentially causing it to misbehave.

  • \$\begingroup\$ Read the first sentence of the question again . \$\endgroup\$ – Harry Svensson Dec 19 '17 at 19:13
  • \$\begingroup\$ @HarrySvensson: I'm not sure that I'm picking up what you're getting at. The asker posted a layout with a degenerate example of what he meant, and my answer attempts to explain what is wrong with said layout. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 19 '17 at 19:19
  • \$\begingroup\$ "a degenerate example" ? Not sure what you mean there \$\endgroup\$ – TonyM Dec 19 '17 at 22:11
  • \$\begingroup\$ @TonyM: Look at the ground trace in the layout. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 20 '17 at 0:31
  • \$\begingroup\$ I did look but it didn't tell me what 'degenerate' means when referring to an example. \$\endgroup\$ – TonyM Dec 20 '17 at 6:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.