I am making a "pcb" with perfboard and solder traces. Unfortunately I need to run up to 8 amps through some parts of it. In the places where I need to use wires to jump over solder traces I plan on having two strands of 20 gauge copper wire each with their own solder pad.Teh best diagram ever!! Would normal solder traces be able to carry enough current or would I need to also double up the solder traces in order to prevent a lot of resistance and possible incineration of the perfboard. Note I will be using actual brand name veroboard and the solder traces that carry more than 2 amps of current will probably be about 5 inches long.

  • 1
    \$\begingroup\$ Theres an existing protoboard tag that seems appropriate. (Interestingly, I've always seen "veroboard" used synonymously with "stripboard" ... never realised that they make perfboard too) \$\endgroup\$ – Jules Jul 28 '18 at 18:52
  • \$\begingroup\$ ahh I didn't see that one. For anyone reading this I requested a perfboard tag in my original post then edited it out. \$\endgroup\$ – 5hutd0wn R35tart Jul 28 '18 at 18:54
  • \$\begingroup\$ What kind of series resitance you can accept? How much power can be dissipated in the tracks? I think this is more a problem of voltage drop than ampacity. \$\endgroup\$ – Vladimir Cravero Jul 28 '18 at 20:47
  • \$\begingroup\$ It would only be 5 inches long trace so I dont think their would be that much voltage drop \$\endgroup\$ – 5hutd0wn R35tart Jul 28 '18 at 21:42

Each square of copper foil has 1/2000 ohm (0.000498 ohms, at 25 degree C) if the weight per square foot is the default 1 ounce of copper. This produces a 35 micron (0.035 milliMeters) or 1.4 mils (0.00014 inches) of foil.

The resistance of any square of this foil (any size, measured from opposite edges) is that 1/2000 ohm, if the holes are not present. Lets ignore that fact.

What is the power dissipation of 8 amps in 0.0005 ohms? P = I^2 * R

P = 8 *8 / 2000 = 64/2000 = 32/1000 = 32 milliWatts.

The thermal resistance of this foil is also 70 degree Centigrade per square, ignoring the holes and the etched gaps between the squares.

Assuming these squares are 0.1 inches, then you'll have 50 squares (with the etched gaps and the holes), with thermal resistance end-to-end of 50 * 70 = 3,500 degrees per watt. Your power dissipation (again ignoring the holes and the gaps ) will be 50 squares * 0.032 watts = 1.6 watts.

Assuming that 1.6 watts flows from one end to the other end, you'll have a temperature rise of 1.6 * 3,500 = 3500 + 2100 = 5,600 degree Centigrade.

Given the heat is (likely) uniformly generated, and flows both direction, you'll have considerably less that 5,600 degree C rise in temperature.

This heat has 5 exit paths (1) into the air (maybe your best bet, with a fan if needed) (2) radiated out into space, if the squares become really really hot (3) conducted along the line of squares, thru the heavy copper wire you've promised to solder along the 5" line of squares (4) conducted THROUGH and DOWN and UP in the insulating board material (FR-4 ?) to the numerous adjacent squares that will serve as a poor-man's sloppy heatsink (5) conducted DOWN and OVER to adjacent squares connected to some large mechanical structures to dump that 1.6 watts to a large area of the case

Note that 1.6 watts in a 1cm^2 surface will likely BURN your fingers. Your surface here is 1" by 0.5" equivalent surface, and it is metal, so you should be safe.

However, any nearby temperature-sensitive components will not be at room temperature, and measurement errors become possible.

  • 1
    \$\begingroup\$ Ahh... well that was a very through answer... I was thinking that what I will do is solder two 20 gauge wires directly onto the copper trace and entomb them in a thick solder trace for any high current parts of the board. The copper should help reduce heat. I will have two 80 mm fans mounted to the top and back of the enclosure. One to pull in cool air from the back and one to push out warm air out the top. \$\endgroup\$ – 5hutd0wn R35tart Jul 29 '18 at 18:58
  • \$\begingroup\$ You might also embed tiny heatsinks into the 5" of 8amp trace. Have thick wires sticking up into the airflow, every inch or 0.5 inch. And can you provide a photo of the finished Vector board? \$\endgroup\$ – analogsystemsrf Jul 30 '18 at 5:40
  • \$\begingroup\$ Unlikely. I need to order about 50% of the materials still and I do not want to pre solder the vero until I have all the parts in hand incase I decide to change something. \$\endgroup\$ – 5hutd0wn R35tart Jul 31 '18 at 18:11

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.