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Hi,
Could someone check these calculations are correct. I'm unwilling to believe 1mm³ or 2mm³ of solder create such a high resistance value as 1.5Ω and 0.75Ω, respectively. Maybe I made a mistake somewhere doing the sums.
Must add first, after doing more research, found a table that says "solder" has a resistivity of 15 * (10^-8), the same 10^-8 as copper and tin, so not sure which to go with.
Anyway, let's say, solder has a resistivity of 15uOhm/cm, the solder blob could be 1mm³ or 2mm³.
For 3D shapes, Resistance is
ρ * (L/(W * t))
(Length, Width, thickness)
These are calculated in metres. e.g. 2mm is 0.002m, 15µΩ/cm * 100 is 0.0015Ω/m
2mm³ solder resistance = 0.0015Ω * (0.002m/(0.002m * 0.002m))
= 0.0015Ω * (0.002m/0.000004m)
= 0.0015Ω * 500
= 0.75 Ω
1mm³ solder resistance = 0.0015Ω * (0.001m/(0.001m * 0.001m))
= 0.0015Ω * (0.001m/0.000001m)
= 0.0015Ω * 1000
= 1.5 Ω
This seems very high to me, compared to the e.g. 0.00122Ω a 2cm * 1cm bit of PCB track causes, ...which is why it would be nice to know if the calculations are correct or the 10^-8 figure
may be the more accurate one..
Also, why does 1mm work out as having more resistance than 2mm, when it is smaller on every side and in volume?
Apart, but related: To choose the ppm/ºC for solder, with nothing much to go on for solder specifically, is it best to use the worst ppm value of Sn Pb Cu, which is the tin that has a ppm/ºC of +4,700 (copper is +3,930; lead is +3,400)?
Thanks.
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Also, why does 1mm work out as having more resistance than 2mm, when it is smaller on every side and in volume?
I really asked that question... Slight mental disconnect between increasing wiring gauge for greater current and little sums about blobs of solder... Embarrassing.