1
\$\begingroup\$

I am currently laying out the board for an audio amplifier I am working on. The power requirements aren't very high - there won't be much more than 500mA flowing through any given trace. However, I have ample space on my board, so I was thinking about using wide rectangular fills (40-50mil wide), like the following: enter image description here

Having wider traces minimizes resistance, an in addition to being much wider than they need to be, I find a layout like this to be aesthetically pleasing. However, could any issues arise from routing like this instead of using thinner traces? In other words, is there a reason why I should only make traces as wide as they need to be?

\$\endgroup\$
5
\$\begingroup\$

Usually that's just fine.

The parasitic capacitance to your ground plane will be a lot higher. In many cases that won't cause any problems (and may be better in the case of supply rails) but occasionally it can cause issues such as oscillation.

It can also cause excessive unwanted coupling affecting whatever is on the other side of the PCB if the voltage changes quickly on the trace.

\$\endgroup\$
4
\$\begingroup\$

If the traces going to a part are very large you may have trouble getting the pad hot enough to melt solder. Consider adding thermal relief

\$\endgroup\$
  • \$\begingroup\$ +1 Or use a reflow oven for soldering. \$\endgroup\$ – awjlogan Mar 1 '18 at 8:10
0
\$\begingroup\$

Wider traces produce larger loop areas, making the PCB more vulnerable to external (and internal) magnetic fields.

Suppose there are 10 amp surges in a wire, with 10 uS risetime, thus 1e6 amp/second slewrate. We'll assume the RETURN path is infinitely far away. What will be the induced voltage, in a 1cm by 10cm loop area? Assume the wire is 10cm away from the PCB.

Vinduce = [MUo * MUr * Area / (2 * pi * Distance)] * dI/dT

with MUo= 4 * pi * 1e-7, and MUr = 1 (air, copper, FR-4, vacuum)

Vinduce = 2e-7 * Area/Distance * dI/dT

Vinduce = 2e-7 * 1cm * 10cm/10cm * 1e6

Vinduce = 2e-7 * 1cm * 1e6 = 2 milliVolts trash floor.

If your average signal is 0.1 volts, your SNR = 50:1 or 34dB SNR

\$\endgroup\$
  • \$\begingroup\$ Thank you for the insight! These are power traces however, and they are all carrying low currents on them and there are plenty of decoupling capacitors to reduce ripple. I'll definitely keep this in mind for the audio signal traces though! \$\endgroup\$ – Billy Kalfus Mar 1 '18 at 4:15
  • \$\begingroup\$ The 1mV will be imposed on the VDD lines. Your audio amplifier likely cannot reject 10uS edges. Consider using a GND plane, which will dramatically reduce the loop area. \$\endgroup\$ – analogsystemsrf Mar 1 '18 at 4:19
  • \$\begingroup\$ +1 (I don't know who downvoted you, this is a valid point). \$\endgroup\$ – Spehro Pefhany Mar 1 '18 at 10:02
  • \$\begingroup\$ Perhaps a couple paragraphs of introduction was expected. \$\endgroup\$ – analogsystemsrf Mar 1 '18 at 23:07

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.