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So i recently inherited a design as a new graduate, and am currently in the process of working on the next board revision. The design currently calls for a 24VAC line to come out of a G6k relay, leaves the board via a connector, and eventually turns a larger contactor on in the boards housing. The current design typically has 10mil signal lines, but for the 24VDC and 24VAC, the previous designer tried to use 20mil lines.

The problem is, in this case, he used 10mil lines for the 24VAC trace which eventually outputs to the contactor. The good news is we are just powering relays with the trace. I tested the amperage coming out of the relay on the board, and it is only about 230mA or so, which seems well within the thresholds for a 1oz 10mil trace.

I know this is a frequently asked question and I have been doing the reading, but would love an opinion from someone with a little more experience. 24VAC on a 10mil trace ok if we arent trying to pull too much current? Seems like I could pull 4x that before I had to start worrying if i can trust these calculators to any extent.

Thanks for any help or guidance!

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Using the graph of AnalogKid, with a mere 10 ° C rise, at 10 mils crosssection area, you are allowed 0.75 amps.

With standard foil being 1.4 mils thick, you need only 7 mils of width. So that graph/nomograph suggests you are quite OK.

As comments/answers state, its all about HEAT and TEMPERATURE RISE>

If you have a PLANE UNDER that trace, the heat will flow downward and greatly cool the trace.

If the trace Is SHORT, then the heat will flow out the ends, easily and greatly cool the trace.

Standard foil has 70 ° C thermal rise per watt per square of foil (1.4 mils thick, or 35 microns). Thus 10 mil wide trace, standard foil thickness, tha tis 1 inch long will have 70 ° C times 1,000 mils / 10 mils = 70 * 100 ==

  • 7,000 degree C heating if you dissipate a watt in the middle of the trace

But to dissipate a watt, with 100 squares times 0.000500 ohms per square, or 0.05 ohms, using Power = I * I * R, you need 1watt/0.05ohms = 20, thus 4.5 amps.

With 0.45 amps, the power is 0.01 watt (and is spread over all the trace, not just in the middle), the rise is 0.01watt * 7,000 degree C per watt, or

  • 70 degree rise, if all the hreat flows from one end of the trace to the other end

If you generate the heat uniformly, and let the heat flow out BOTH ends, the temperature rise is much less.

And that is without a PLANE underneath.

Notice that 4_layer PCBs can have the plane 2X closer to hot traces, giving you a FREE 2:1 reduction in temperature rise.

To develop your own instincts on this, get a quadrille pad, and sketch out the copper trace you care about.

Use these numbers:

  • thermal resistance of standard thickness foil if 70 ° C per watt per square of foil, for any size square.

  • electrical resistance of that same foil is 0.000500 (1/2,000) ohms per square, for any size square

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Without getting into constant impedances and RF effects, sizing a pc board trace for its power application is all about heat. Some online calculators give you trace temperature increase, with different values for internal traces on multi-layer boards and those on the two surface layers. For 1/4 A, I would increase the trace width to at least 50 mils, and use 100 mils if there is room. It costs zero.

A search for 'pc board trace width chart' brought up the classic chart. I see the 10 degC line as a boundary of desperation - don't ever cross it, and don't go near it unless absolutely necessary.

https://images.search.yahoo.com/search/images;_ylt=A0geKLrBwD5fg8IASA1XNyoA;_ylu=X3oDMTE0bHMzczh1BGNvbG8DYmYxBHBvcwMxBHZ0aWQDQzAxNjRfMQRzZWMDcGl2cw--?p=pc+board+trace+width+chart&fr2=piv-web#id=3&iurl=https%3A%2F%2Fi.stack.imgur.com%2Fy4p8b.png&action=click

enter image description here

My background: 50+ years in circuit and circuit board design, the last 16 in rugged MIL systems.

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  • \$\begingroup\$ Thanks for the quick response, but I am having a hard time squaring the graph with what you said. So if I had 10mil width trace, and a 1oz weight had a thickness of around 1.4mil. Wouldn't the cross section = 10*1.4=14sqmils. Wouldnt that mean I would be approaching the line more around the .75A mark? That would mean a 50mil width trace would be more for around 3A or so according to that graph. Where is my logic failing me here. Thanks! \$\endgroup\$ – Gus Aug 20 at 19:15
  • \$\begingroup\$ Double check the source of that plot.if it is IPC-2221 it has been superceeded by the IPC-2152 in 2009. My advice, download SaturnPCB and use the track width calculator \$\endgroup\$ – JonRB Aug 20 at 19:46
  • \$\begingroup\$ Yeah I was using Saturn 7.12 in the IPC-2152 mode to try and guesstimate, which was what was giving me hope we wouldn't have to do a huge redesign given our needs. \$\endgroup\$ – Gus Aug 20 at 20:06

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