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I was told time and time again to always put decoupling capacitors as close to the IC pin's VCC as possible.

Sometimes when I make complex boards, I try my best to put the capacitor lead so close to the VCC pin, that the distance between the inside of the capacitor (where the pin runs into) and the VCC pin is less than 3mm.

There are times where routing the single-sided board gets so complex that I have to use 10 or 12 mil wide traces and make a track length of about 1/2 inch to even 1 inch in rare instances between VCC and the capacitor.

Whats the longest length I can go with traces between IC VCC and nearest capacitor before the decoupling effects of the capacitor become useless if the width is 10mils wide?

Mike -- No longer here
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    The actual ingress and egress noise suppressed or amplified by choices of decoupling cap(s) and long traces can be very significant high Q resonances at the remote cap affecting other IC's and visa versa. This can be reduced by adding ESR to small caps or using larger e-caps with inherently larger ESR in parallel with small ceramic caps. ( best case). But analysis becomes a complex LC filter analysis knowing the source noise current and dynamic Z with supply Z. As Cap SRF is very low Z, if supply is also very low Z with series trace L, supply noise can increase >40 dB at SRF from X/ESR ratio. – Tony Stewart EE75 Aug 06 '18 at 00:10

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The answer is unhelpfully "as short as possible", and/or "it depends".

Whether it needs to be 1mm, 1cm, or even 1m will depend on factors such as switching frequency, signalling standards, whether sensitive analogue components are involved, how much EMI you can get away with etc.

The purpose of the capacitor is to provide a nice low impedance path between supply rails. The longer and thinner the trace, the larger the inductance of said trace, and the higher the resulting impedance (remember inductors are short at DC, and increase in impedance with frequency).

If you have a signal that toggles very slowly, and/or can cope with noise generated on the signal due to poor decoupling, then you can get away with longer narrower traces.

If however you have a signal running at hundreds of MHz, or into the GHz range, even a few millimetres of trace might be too high of an inductance. At this point even a via might add a lot of unwanted inductance.

In circuits with analogue electronics, things also get critical because supply noise and ground noise have a nasty habit of finding their way into analogue front ends.

The simplest way to find out is to build a prototype, and test it. If things work as expected and performance is within parameters, then your traces are short enough. If you find issues with noise or signal problems, poor decoupling may be a culprit.

Tom Carpenter
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  • I guess I can get away with some length because my circuit in question is driven by a parallel port from the PC in Standard Parallel port mode and I believe the max data speed for the parallel port is 150kbps. – Mike -- No longer here Aug 06 '18 at 02:29
  • If your data protocol provides a wide dataeye (the safe time during which the incoming bit stream can be sampled without error), then many types of fidelity crimes get overlooked. – analogsystemsrf Aug 06 '18 at 04:40
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Here are some resonant-capacitor-inductor-Rdampen screen shots.

Rdampen is 1 milliOhm in the first screen-shot (the R in each of the 4 LRC is 1 milliOhm).

enter image description here

The following screen shot has all the dampening Resistors be 10 milliOHms. Notice the higher frequency peaks are NOT AFFECTED. You'll need more than 10 milliOhms (either lumped or ESR inside the cap or trace) for 10nH and 100pF.

Notice that resonance at 150MHz brings up the energy from -25dBc to +5dBc

enter image description here

Your Question was "What is the longest thinnest trace"? Depends on the inductance you can tolerate.

Traces over air (no underlying plane) have about 1nH/millimeter or 25 nH/inch. There is mild dependency upon width. [its a log(1 + length/width) ]

Traces over planes have about 0.1nH/millimeter, depending on distance from trace to plane. Thus an underlying plane (GND or VDD or otherwise) will raise the resonant frequency by sqrt(10) or 3.16X. Be aware of that.

What can you get away with? Clearly, as you are already aware, you have at least 3 degrees of Freedom: capacitor value, trace inductance, and losses due to ESR or trace. A mere 10 squares of trace (0.08" long, 0.008" wide) has, at 500uOhm per square, an Rdampen value of 5 milliOhms.

Thus many crimes of this nature are hidden under the Rdampen of the traces and Vias.

How to pick Rdampen? Use the (tiny) formula Rdampen = sqrt( L / C ). Thus 100uF and 10nH, which is the left-most LRC shunt circuit in each screen-shot, needs Rdampen of sqrt( 10nH / 100uF) = sqrt(1e-8 / 1e-4) = sqrt(1e-4)

and for that left-most LRC, use Rdampen of 1e-2 = 0.01 ohm = 10 milliohm.

Does this value of 10milliOhm suffice? At 1milliOhm, the dip at 160Khz is -93dB.

At 10 milliohm, that dip has risen (been dampened) to -74dB and is quite broad.

analogsystemsrf
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It is dependent on the capacitor, the capacitor packaging, trace geometry, and PCB stackup. Your best bet is to do an FEM simulation with the s-parameters of the capacitor, if you have access to the tools.

The effect of the trace length can be quite dramatic. capacitance response

user110971
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