Just a simple question: what exactly stands behind the need for placing the capacitors as close as possible to the current consuming device's pins? Is that the inductance, resistance or maybe impedance of the PCB track or wire that affects the electric charge?

  • \$\begingroup\$ This is a valid enough question, I don't understand the downvote: Not everyone is born with innate knowledge of the mysteries of decoupling capacitors, and there is way too much misinformation out there to be able to conclude a definitive answer merely by searching the web. \$\endgroup\$ – Anindo Ghosh Jan 19 '14 at 12:29
  • \$\begingroup\$ I discuss decoupling caps at length here: electronics.stackexchange.com/a/15143/4512 \$\endgroup\$ – Olin Lathrop Jan 19 '14 at 14:43

Is that the inductance,




or maybe impedance of the PCB track


or wire


that affects the electric charge?

hmm .. it affects the electric current, not so much the charge. The current from capacitor to decoupled device must meet as little "obstruction" as possible.

Devices can have huge inrush currents when switching and without decoupling this inrush current, together with resistance/inductance of the wiring can cause the power supply voltage to drop below the minimum operational power supply voltage. The decoupling cap is there to prevent this situation. By keeping the loop small, low inductance, low resistance, the capacitor can isolate the inrush current from the actual power supply which has much longer traces/leads and with that higher impedance.

  • 2
    \$\begingroup\$ This is how a tech kid asks questions from a geek father. I'd enjoyed this. Yes Yes Yes ! \$\endgroup\$ – Standard Sandun Jan 19 '14 at 12:13
  • \$\begingroup\$ Maybe you should add something about how you think track impedance (I think the correct term here is "characteristic impedance") matters? Length of a trace does not change the characteristic impedance. \$\endgroup\$ – Rolf Ostergaard Jan 19 '14 at 14:40
  • \$\begingroup\$ @Rolf: Unless your termination matches the track characteristic impedance (termination with a decoupling capacitor won't), then length matters a great deal. The effective impedance including reflections becomes frequency-dependent, and at some frequencies your capacitor+track turns into an inductor. The longer the track, the lower the frequency this begins to occur. A Smith chart is designed for visualizing this effect. \$\endgroup\$ – Ben Voigt Jan 19 '14 at 21:46

This is a BS specification (assuming you are talking about bypass caps for a modern digital IC). "As close as possible" is simply nonsense. Who defines "possible"?

We should all protest when we see stuff like that in a datasheet.

What we need to see is actual requirements. Like max impedance from DC to a max frequency - or something like that (I wrote about that here).

Assuming you are using two closely coupled solid power planes (which by far is the easiest way to do decent power distribution on a PCB for modern digital parts), the distance does not really matter in the typical case.

Surprised? This is actually old news. Well documented 20 years ago or so.

Look at the closely coupled power plane pair as a very wide transmission line (very low impedance). Remember a discrete capacitor has a resonance frequency around 100MHz or less.

enter image description here

If you recall the formula for going from bandwidth to rise-time: BW = 0.35/t_r it is obvious that a discrete capacitor will have a "rise-time" in the order of 3.5ns or more. That corresponds to more than 50cm on a board. Most boards are about that size or smaller, so pretty much anywhere on the board will be okay.

Inductance of the planes are virtually zero compared to the inductance of the capacitor and its mounting.

Resistance of a solid Cu plane is also very low, but something you have to consider not only for bypass, but at DC as well if you use very low voltage parts (1.2V as an example) with very high power consumption (10A as an example).

Feel free to detail your question, if you don't feel I covered the answer you were looking for? I can talk about this for hours. But the bottom line is:

Distance does NOT matter in the typical case.

  • 2
    \$\begingroup\$ If by typical case, you mean 4 layer boards with professional layout. I suspect this is in fact not the typical case for the person who asked this question. More likely, the boards are 1 layer and fabricated at home, or not even printed, but stripboard, or breadboard. In these cases, the inductance of the supply rails is way more than inductance of any capacitor. \$\endgroup\$ – Phil Frost Jan 19 '14 at 14:17
  • 2
    \$\begingroup\$ There is really no way to know that. That is why I was very careful to add the assumption I made in my answer: "Assuming you are using two closely coupled solid power planes". \$\endgroup\$ – Rolf Ostergaard Jan 19 '14 at 14:21

It's worth mentioning that on some occasions, the current taken down a relatively long PCB track can cause "other" chips to receive interference i.e. the main chip that takes the big surges might still be OK with a cap at some distance but, other (possibly more sensitive) circuitry on the same power lines may not be.

Radiated and conducted emissions can also be a problem when a capacitor is not placed as close as possible to the device that is taking the current surges.

There is also a small/rarer down-side and that occurs (as an example), on voltage regulators when "copper" feeding the chip has quite significant inductance. On power-up situations, the line inductance and very-local capacitor can formed a resonant tuned circuit and, the voltage across the capacitor may, for a short instant in time, rise well-above the maximum voltage rating of the device (despite the normal feeding voltage levels being perfectly acceptable). This can somewhat be alleviated by not having the capacitor so close or having a distributed capacitance that is able to befuddle the main peak of resonance. It's rare like I said.

  • \$\begingroup\$ Is this voltage spike mitigated by putting the capacitor beyond the chip, instead of in the track between the chip and its power supply? \$\endgroup\$ – Ben Voigt Jan 19 '14 at 21:50
  • \$\begingroup\$ that is a tricky one \$\endgroup\$ – Andy aka Jan 19 '14 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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