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I created a failed PCB. One of the reasons could be that I had tight track widths.

I was advised to create larger track widths, where high {voltages, current} exists.

I placed the two values in brackets, because I am not sure which of the two values are important for track widths.

However, how exactly do I know what the ideal track width should be? The max voltage is 36 V and max current 2 amps.

Is there a specific methodology or rules of thumb that I should stick to?

JYelton
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user1584421
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    How/what failed? – Rodo Jul 23 '23 at 16:29
  • @Rodo I had an erroneoulsy picked inductor. However, the traces were too small as well. I think this sums it up, however, if you want, you can read more on my original post here: https://electronics.stackexchange.com/questions/674675/step-up-circuit-does-not-work – user1584421 Jul 23 '23 at 16:31
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    the question should stand on its own ... please add the necessary information into this question – jsotola Jul 23 '23 at 16:57
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    I think all the important elements are here. The other question is the bigger picture, with data that is irrelevant. I used tight traces for high currents. – user1584421 Jul 23 '23 at 17:09
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    Googling for "pcb trace width calculator" will lead you to some useful resources. – Andrew Morton Jul 23 '23 at 17:54
  • @AndrewMorton I actually found this site: https://www.7pcb.com/trace-width-calculator However there are parameters unknown to me such as `Cu thickness`. That's why I asked if there are rules fo thumb. I generally use jlcpcb by the way. – user1584421 Jul 23 '23 at 22:17

3 Answers3

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Trace width matters for current. Spacing matters for voltage. The trace widths used on your PCB for 2 A are not wide enough. I think a bigger problem in your design is track length. For switching supplies you need to consider track inductance for the tracks carrying the switching currents. Track lengths must be minimized.

Besides using an inductor unsuitable for the task (mentioned in the previous post), the layout of the PCB can be better. Unfortunately, the datasheet doesn't give a sample layout like many switching I.C.s do. See if their Web site has a demo board layout. If they don't, pull up some data sheets from more main stream parts and read about what is important in the layout and the example layouts.

You should be using copper pours for the parts that carry the switching currents and minimize distances between parts that carry the switching currents. Also, you need ceramic bypass (around 100 nF) capacitors in addition to the electrolytic capacitors to take care of the fast switching edges. The ground connection needs to be a small area, not spread about with a maze of serpentine traces.

Schematics are generally drawn with the input on the left and output on the right. This makes it easier for experienced practitioners to read and understand the schematic.

qrk
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  • I know the layout can be better. I will group elements close together, especially decoupling capacitors... As for not having ceramic bypass capacitors, I just copied the circuit supplied by the StepUp IC datasheet. The front side copper is GND and the back side copper is Vcc (5V from the Step Down). For track length, which tracks should be small in length? The ones that supply the StepUp IC? Or the ones from the output of the IC? Finally, what is the ideal tack width I should use for 2A currents? Thank a lot! – user1584421 Jul 23 '23 at 17:41
  • @user1584421 It's less about track width, more about track length. Print out the schematic of the power supply and with a yellow highlight marker pen trace out the path of high current switching currents. Input and output capacitors are in this path. This will tell you where to tighten up your layout. The switching signals should be on the same layer. I have seen better layout designs fail due to track length. There is no "ideal" track width, just compromises. You can refer to IPC-2221 which suggests a 0.031 inch (0.8 mm) track width for 1 oz copper and a 10°C temperature rise. – qrk Jul 23 '23 at 18:16
  • Ok i will just make them too large to be sure. however, if they are too large, am I in danger of other effects? Like parasitic effects, RF effects and who knows what else? – user1584421 Jul 23 '23 at 22:21
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    @user1584421 At random, I pulled up the data sheet for the [LT7101](https://www.analog.com/media/en/technical-documentation/data-sheets/LT7101.pdf). Look on pages 33-34 to see good layout practices. Layout for switching power supplies can be complicated. I've seen experienced engineers do questionable layouts - some didn't work very well. – qrk Jul 24 '23 at 01:57
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The minimum spacing between conductors for a given voltage is covered in IPC-2221 "Generic standard on printed board design" (https://shop.ipc.org/ipc-2221/ipc-2221-standard-only/Revision-b/english). Trace width for a given current is discussed in IPC-2152 "Standard for Determining Current Carrying Capacity in Printed Board Design" (https://shop.ipc.org/ipc-2152/ipc-2152-standard-only/Revision-0/english). Both of these standard are excellent resources and are used for industrial, space, and military applications, but you need to pay for them.

There are free calculators, such as the "Saturn PCB Design Toolkit" (https://saturnpcb.com/saturn-pcb-toolkit/) that are based upon the above IPC standards.

For example, for a PCB with solder mask, the minimum spacing between traces with a 36 V difference is 0.13 mm (5.1 thou).

enter image description here

When determining the minimum conductor width, there are more things to consider, such as the presence of planes, are the conductors on an external or internal layer, the copper weight, the maximum permissible conductor temperature rise, etc. The following is an example of a PCB with 35 µm thick copper (1 oz) with external traces carrying 2 A of current.

enter image description here

C. Dunn
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  • Thank you! So for 2A traces, the conductor width should be only `1.2mm` wide? Also, the `conductor length` means the maximum length that the conductor that carries the 2A current should have, until it reaches another component? – user1584421 Jul 26 '23 at 14:40
  • @user1584421 Assuming your PCB matches the other inputs selected (conductor layer, copper weight, etc.) then yes a trace only needs to be 1.2 mm wide to carry 2 A with a 10 °C temperature rise. Please note, in the above scenario you are losing approximately 1 mV per mm of conductor length, which might be unacceptable for long traces. If so, make the trace wider. Also note, conductor length is only used to calculate power dissipation, resistance, and voltage drop. It is not an input to the conductor current equation. – C. Dunn Jul 27 '23 at 15:34
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The trace width of the power and ground should be as wide as possible. The ground trace is wider than the power trace. Ground trace > power trace> signal trace. Usually the width of the signal trace is 0.2~0.3mm, the thinest width is 0.05~0.07mm, the power trace is 1.2~2.5mm, the width of the power trace is best above 40mil, the minimum should also be above 25mil, if conditions permit, as wide as possible.

You also need consider the actual current, generally 10mil can withstand the maximum current 1a. You need choose the appropriate trace width according to the actual current.

lindayin
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  • Thank you! Should the traces from the chips to the GND be wide for ALL the components? Or only the chips which handle high currents? For example besides the chips with the high currents, I have a microcontroller with low currents. Should the microcontroller's trace from the GND pin, until it touches with the ground plane be wide? Also, unrelated, but how can I contact your company for pcb manufacturing and assembly? – user1584421 Jul 26 '23 at 14:38