On the (two-layer) board I'm designing, I have a relatively large unused area. Instead of just pouring it with ground on both sides, I'm considering filling it with Vcc on one side and ground on the other, to create a small capacitance between the ground and Vcc. (I will still add adequate decoupling capacitance from regular capacitors, of course.)

The board is not exactly high speed (16 MHz microcontroller, doing only digital IO). And I'd be hard pressed to produced even 1 nF of capacitance from the available board area, I think. So you could argue that this extra capacitance is not doing much of a difference. But is there any reason why it might actually be a bad idea and should be avoided?


4 Answers 4


As a rule, the ground of the circuit serve as a reference point for all signals. That is why it must have very low voltage difference across the circuit. It is generally not true for the Vcc - as long as no signals are measured relative to the Vcc, some bigger voltage differences are acceptable.

The total current in the Vcc line is equal to the current of the ground line. But because of the above paragraph reasons, the ground line must have much lower resistance (actually impedance) than the Vcc line, where high inductance together with decoupling capacitors is even good.

That is why, the ground is often routed as an area, but Vcc simply as thick enough tracks.

This all was in theory. If your PCB is not dealing with high currents and/or very high frequencies, you probably can route it in whatever manner you want and everything will be OK. So, don't be paranoid.

Footnote: If you have so big empty areas, isn't it better to change the design in order to make the PCB smaller?

  • 1
    \$\begingroup\$ Re footnote: A big connector, as well as case mounting expands the width of the board. IC's and other components expand the height of the board. But it's really not that much space. The total area of the board is about 30*75 mm. \$\endgroup\$
    – nitro2k01
    Commented Oct 22, 2013 at 17:53
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    \$\begingroup\$ @nitro2k01 - well, I supposed something similar. But the last paragraph before the footnote is still valuable. :) \$\endgroup\$
    – johnfound
    Commented Oct 22, 2013 at 17:56
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    \$\begingroup\$ Yeah, I pretty much know the answer already, it won't make any detectable difference either way. The question is less "I have no clue" and more "surprise me". Of course, it can still generate interesting replies that are generally useful to the community. \$\endgroup\$
    – nitro2k01
    Commented Oct 22, 2013 at 18:10
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    \$\begingroup\$ "where high inductance together with decoupling capacitors is even good". Why? Because they form a low-pass filter? \$\endgroup\$
    – Rafael
    Commented Sep 25, 2015 at 18:09
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    \$\begingroup\$ @Rafael Indeed. Or put another away, series inductance gives a high impedance for high frequency content (current spikes) and prevents interference from propagating out to the PSU or the power bus. The capacitors (preferably low ESR and placed near the device of interest) on the other hand give a low impedance for frequency content, and makes sure the power rail can respond quickly enough. Ie, the low pass filter formed, if seen that way, concerns smoothing out the current draw that the PSU sees. \$\endgroup\$
    – nitro2k01
    Commented Sep 25, 2015 at 18:24

The tiny extra capacitance you will get on the power net will be pretty much irrelevant. It will be low ESR, but it won't be right where you need that, so it won't help much.

In your situation, I would not do the power plane. There is little drawback electrically, but it adds just a little more chance you will mess up something. It will also make the board harder to edit in case you need to make a few changes after testing. If you do use the power plane, make sure that any pads for that signal embedded in the plane have thermal releafs, else they will be hard to solder.

In short, it doesn't matter much either way, so I'd leave it out for simplicity and less chance of error.


Well there are both positives and negatives.

While you may not get a lot of capacitance it will be very high quality capacitance. Meaning that it's self resonance will be at a lot higher frequency that say a ceramic cap, because of the low inductance. So it would work as decoupling as a lot higher frequencies.

The negative is that you are now potentially moving an aggressor signal (your power rail) closer to other signals which then may pick this up because the coupling is increased because it is occupying more board area.

Do keep in mind that power is also a "ground" in terms of small signal analysis. While it could be dangerous to decouple to it, that might in some cases be advantageous.

  • \$\begingroup\$ As for what you're saying about moving closer, I don't see how this is the case. This part of the board is mostly unpopulated, and the only place where Vcc meets other signals (with or without the fill) is along one side of the trace or would-be plane. I'm under the impression that the only major added capacitive coupling is between ground and Vcc. \$\endgroup\$
    – nitro2k01
    Commented Oct 22, 2013 at 19:29

I second Olin's answer. It will make little difference because the capacitance is so far away and its so small. Its better to put actual capacitors near the pins.

However, as noted previously, you should add "Copper thieving" To those sections of the board to ensure faster and better handling of the PCB so that etching is better.


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