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?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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.
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.
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.
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.