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    EEVblog #1056 – Digilent Open Scope MZ Review

    Dave looks at the Digilent Open Scope MZ Review, an $89 open source oscilloscope, logic ...

    • David

      Whether adding solder to a copper trace improves current handling capability or not depends on how much solder you add (cross-sectional area of the solder compared to the cross-sectional area of the copper trace).

      For example, at room temperature a 1oz/sqft copper trace (0.034mm thick) will have a resistance that is 1.34 times greater than 63%-tin, 37%-lead (Eutectic point) solder tinning that is a semi-circle in cross section with a diameter equal to the trace width. This is regardless of trace length and/or width.

      This factor doubles for 0.5 Oz/sqft traces (better), and halves for 2.0 Oz/sqft traces (worse). So for 2.0 Oz/sqft, the tinning actually has a higher resistance than the copper trace.

      This assumes the resistivity of copper is 1.68X10-5 Ohms*mm, the resistivity of 63/37 solder is 1.44X10-4 Ohms*mm and the density of copper is 8.96 g/cm^3.

      As for heat dissipation, I have not seen a similar analysis of this so-far (interesting exercise though). But it should be noted that the thermal conductivity of copper is approximately ten times greater than that of 63/37 Sn/Pb solder [401 W/(m*k) for cu vs. 41 W/(m*k) for 63/37 Sn/Pb].

      I did this analysis some time ago (citing from lab notebook). Perhaps someone can check my findings.

      I do not have resistivity and thermal conductivity figures for RoHS compliant solder.

      • In response to the poster above:
        The solder and copper trace are effectively parallel resistances and the total resistance will always be smaller than the smallest resistance in this configuration (we’re talking DC). I don’t think it makes sense to argue about the resistance of the solder itself being higher than that of the trace in this case (you can also use silver plating if you really want to).

        You will use proper cable anyway, once you need to transport high currents; tinning a track can extend the range a little though (without much trouble in the production process, it has to run through the wave solder anyway)

        Good points about the thickness of the copper layer.

    • f4eru

      for the end resistance being a bit higher : that`s normal, when you solder, a bit of copper dissolves into the solder, just at the edge zone between the two metallic compounds.

    • Kedas

      Rule of thump (well, just math)
      A: Ratio resistivity solder tin/copper
      X: Ratio of thickness solder tin/copper
      Pr: Percent remaining resistance (%)
      (100% is without solder)

      Pr=A/(A+X) * 100%

      adding tin solder 8.4 times thicker than copper (X=8.4)
      8.4/(8.4+8.4)=0.5 –> 50% * R remaining.

    • Niklas

      The resistance is not the real issue with the current handling of solder plated traces. Check what happens with the thermal resistance, ie the ability to release the heat energy into the air. FR4 is really bad at transporting heat so most of the heat energy must leave via the surrounding air. Solder mask will increase the thermal resistance a bit.

      For non repetitive pulses the trace will handle higher currents, but it is not a viable option for steady state.

    • Robert Stever

      You could alternatively use flat-wind wire wire along with your slolder trace to increase current handling even further. I use this practice in high current (200+ amp 15v) voltage regulators, otherwise I need to use 16oz copper board for 10% of the tracing.

      • For sure.

        • Robert Stever

          The great one commented on me, my day is complete.

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