I am building a test probe PCB to test 75-115V. What is the minimum distance between traces that I should adhere to when building my PCB? I will have ground, 3v and 115V traces running next to each other, there will be low amperage, (1ma max), and the board is single sided.
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\$\begingroup\$ What's limiting the current from the 75V-115V source? \$\endgroup\$– Jason SApr 21, 2012 at 18:58
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\$\begingroup\$ It is 4 battery packs. I don't know the limiting current. \$\endgroup\$– ReidApr 23, 2012 at 2:05
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\$\begingroup\$ rough size per battery pack = ??? \$\endgroup\$– Jason SApr 23, 2012 at 2:35
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\$\begingroup\$ 18*6*6in, 30 batteries per pack. \$\endgroup\$– ReidApr 23, 2012 at 18:43
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2\$\begingroup\$ @Reid I think what Jason S is saying is that battery packs can generally supply far more than 1 mA. Your intended load might be 1 mA, but if something goes wrong (say you tie the output wires together) your circuit board might see a lot more than that. You should generally put some sort of fusing between the battery and anything else. Otherwise, your PCB traces might themselves fuse in a fault situation. \$\endgroup\$– Stephen CollingsSep 6, 2013 at 12:51
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2 Answers
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Is it AC or DC that will be running on the trace? If DC then don't worry because that low of voltage won't arc. If it's AC then try to run AC perpendicular to the DC (opposite sides of the board would be ideal) so your collecting as little EMF as possible. Otherwise if your going to run parallel you'll need to do math to find the radius of the magnetic field produced by the AC lines and that would be your gap.
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1\$\begingroup\$ Note: electric fields extend to infinity (photons are massless outside plasmas). \$\endgroup\$– tybluApr 21, 2012 at 4:59
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\$\begingroup\$ I was wondering about that actually. After i wrote that i thought i remembered that they extend infinitely like gravity. \$\endgroup\$ Apr 21, 2012 at 5:28
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\$\begingroup\$ To be precise: 75V may not be enough to initiate an arc (I'm trying to find this), but all you need is a momentary fault that causes current to flow and get interrupted, and trace inductance will initiate an arc, and it will be maintained until the arc is extinguished because the source of energy is gone or the traces burn up far apart. \$\endgroup\$– Jason SApr 21, 2012 at 14:42
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1\$\begingroup\$ Thanks @KalleMP for bringing back this post that I forgot I wrote back in 2012. UPDATE: The first version of the board caught on fire multiple times. Not due to trace spacing, but due to poor isolation protection (Ahh, to be young, dumb, and working on a 38.4KW battery pack). Subsequent boards got significantly better as I progressed through my EE degree, and the final version ended up in commercial use =) \$\endgroup\$– ReidNov 26, 2018 at 6:08
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Obviously there are different standards for different environments and applications, so please don't take this as any sort of universal wisdom. But for the UL and CE clearance specs I work with, PCB trace clearance should be at least .2 mils per volt. So if you have 115V difference between two conductors, clearance should be at least 23 mils. Unless your board needs to be ridiculously small for some reason, you shouldn't have a problem quadrupling that.
If it still worries you, the PCB itself is an excellent insulator. Put the high voltage trace on the opposite side from the low voltage.
Remember, batteries can supply enough current to blow traces and components off your board, arc, and potentially explode the battery itself. If you're hooking stuff up to a battery, fuse the link.
answered May 25, 2012 at 21:36
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2\$\begingroup\$ Saw this after posting my answer: electronics.stackexchange.com/questions/22796/… Much more detail! \$\endgroup\$ May 25, 2012 at 21:47