current capacity of a solder loaded track

I'm designing a PCB that is to carry 20 amps now and may go up to 50 amps in future designs, due to the pitch's of part pins the biggest tracks I can use are 0.15" which will carry up to 6 amps, just under a 3rd of my requirement, how much extra capacity can I get from loading the PCB tracks with 0.5mm of solder ? I guess it is down to the conductivity of lead and tin

Not much at all and not good practice. SnPb solder has less than 10% the electrical conductivity of copper.

**broken link removed** a calculator for solder resistance.

How many layers does the board have? You can place traces on more than one layer to increase the current capacity of a pin.

crutschow said:

**broken link removed** a calculator for solder resistance.

How many layers does the board have? You can place traces on more than one layer to increase the current capacity of a pin.

Click to expand...

single layer, I could use double I supose but that would only double the capacity so I might as well layout the board differently, is there any substance I can gest specificly for building up the thickness and carrying more current ?, I suppose another option is to solder a lump of wire on the tracks

according to your calculator a 2 cm length of 0.15" track with 0.5mm of solder loaded on is 0.00016 ohms, thats 0.4 W on 50 amps dissipated: bearable, only 64 mW at 20 amps: just fine

I've seen some audio circuits that use solder covered copper traces for extra but I don't think they're rated for as much current as you're talking about, you probably are much better off simply skipping the trace completely and soldering wire between the high current parts, or use terminal blocks. You may want to do this anyways as running that much power on a PCB is going to have some kind of cross talk effects on nearby or parallel traces.

I can't see how it doesn't make a difference.

Suppose solder only has 1/10th of the conductivity of copper.

The copper track is very thin compared to the solder laid on top of it.

If the solder has 5 times the crossectional area for the copper than the resistance would be reduced by a third.

If the solder has 10 times the crossectional area as the copper than the resistance would be halved.

Sceadwian said:

I've seen some audio circuits that use solder covered copper traces for extra but I don't think they're rated for as much current as you're talking about, you probably are much better off simply skipping the trace completely and soldering wire between the high current parts, or use terminal blocks. You may want to do this anyways as running that much power on a PCB is going to have some kind of cross talk effects on nearby or parallel traces.

Click to expand...

I'm thinking of a secondry PCB for the heavy current as basically it is just for a series of parallel blocking diodes

what about silver solder ? what is the resistance of silver solder ?

It'll obviously better, the resistance should be specified on the datasheet.

What's your budget?

You need to work out the crossectional area of the solder on the track.

You could polish the board, clean the soldering iron tip and measure how much solder you apply to a given length of track. If you know the thickness of the solder before melted and the length you've used you should be able to calculate the approximate crossectional area and perhaps knock off 10% to account for the rosin core and the tiny bit left on the tip afterwards.

Anyway, I'm pretty confident there will be enough solder so make a difference, even when the poor conductivity of tin/lead is taken into consideration.

well calculating for 20 amps I'm ok if I went higher I'd have to redo the board design to allow for wider main tracks, I'm not sure of the cost of silver solder (could maybe scrounge some at work for the little I'd need) but it could be worth the cost (if the thing ever gets sold it's down to how much profit I'd want to make but i put quality and safety first)

I don't see how silver could be cost effective, surely wider track or a heavier PCB would be cheaper?

What about lying thin tinned copper wire over the PCB tracks?

well when you need a track 0.5" wide with solder on top it kind of means you going to have to increase the board size considerably or use a seperate board altogether, if silver solder will greatly enhance conductivitry at 20 amps I'll consider it, if i wanted to carry more current and it allowed me to do that withoput redesigning the whole thing that would help

The problem with silver solder is the heat that it takes to melt it. I don't think the board will take it. At the least the copper will peel of the board.

What about using a jumper wire soldered as close to the components as you can get? you would use insulated hook-up wire of the right gage of course.

well if silver solder needs that much heat then its a no go, probably better to use wire soldered on top of the track to help carry the current

And it's cheaper too.

I don't know how gooder conductor silver solder is compared to copper. I know silver is a better conductor than copper but silver solder isn't pure silver.

Skip the track completly, just turn the end points into through hole parts and solder terminal blocks or heavy duty jumper wire, thick solid core jumper wire will stay in place quiet well on PCBs if it's bent neatly, and you can route them above and away from other traces to avoid interference.

As pointed out by Sceadwian the proper solution to this problem is to use point to point bus bar type wiring, or insulated wiring if the voltage requires it. Or just a straight up actual copper bus bar on the PCB. Up to 50 amps in future applications? Solder over a copper trace ain't gonna get it there, period. Think vaporized traces. I don't care what the initial calculations say, I've seen **** like it happen. I've been doing electronics R&D for 3 years now, including power stuff. Solder over a trace is considered a half-assed fix for when you ****ed up your board design and need to get the prototpye working until you can do another board spin. I don't know if it even meets IPC standards. I doubt it, because I've never seen it done in professional electronics.

I don't know how gooder conductor silver solder is compared to copper. I know silver is a better conductor than copper but silver solder isn't pure silver.

Click to expand...

Unless you're talking about the jewelry solder, silver bearing solder for electronics is only like 2% silver, with the balance being either pure tin and a smidge of copper (abbreviated as SAC for SnAgCu and used for RoHS stuff) or Sn and Pb for good old American stuff. The silver solder in jewelry making is mostly silver but is way high temperature stuff, would probably ruin a PCB in no time flat trying to solder with it.

Hi there,


It's true that regular solder has much less conductivity than copper, so that a copper trace of a certain width and thickness covered with a layer of regular solder that has the same width and thickness would carry only about 10 percent more current, but only if the electrical conductivity was the only thing to consider here, but there is another consideration. The electrical conductivity is about 10 times less than copper, and unfortunately the thermal conductivity is also about 10 times less than copper. This means when we coat a trace of copper with solder in addition to lowering the resistance we are also decreasing the trace's ability to dissipate heat by covering it with a (somewhat) heat insulator.
The ability of the trace to dissipate heat has a lot to do with it's current carrying capability, so when we reduce it's ability to dissipate heat we necessarily decrease it's ability to conduct current. Since the copper trace has a fiber or glass backing, when already one side of the trace is covered by a pseudo heat insulator, and the edges are not thick enough to have much impact on the heat dissipation, so when we cover the top with another pseudo heat insulator (albeit a somewhat poorer one though) we are again reducing the ability of the trace to dissipate heat and thus lowering it's ability to conduct current.

Just what is the overall impact then, having one parameter lower the resistance and the other decrease the heat dissipation factor?
The conductance of each trace (one being copper and the other solder) would be combined and then the temperature rise of the copper and solder combination would have to be calculated knowing the thermal conductivity is about 10 times less for solder, and that could be compared to copper alone.
With only one 'layer' of solder (a 'layer' being the same thickness as the copper) the current carrying capacity will most likely go DOWN due to the temperature rise, and multiple layers may or may not help the situation until some break point where the sides of the trace begin to have a more significant effect by adding significantly to the total dissipation surface area. It's hard to predict this action exactly though.

So how about a simpler way to deal with all these variables? Luckily, there is a way...
One way for example would be to etch a trace say 6 to 12 inches long on a scrap piece of PC board material that is the same type to be used with the final circuit. Next we would attach terminals to each end and run a DC test current through the trace. We would then measure the current through the trace and measure voltage drop. Knowing the voltage drop and current, we would calculate resistance.
Next we would cover the trace with a layer of solder, then test again and compare results.
A third test might be to try apply a heavy load of solder and then measure again.
Note however that in each case enough time has to pass for the trace to heat up to equilibrium before a measurement is made, and that it would be best to use the actual max current for the test that will be found in the actual application. It will certainly not work to use a small test current like 1ma, but the full 10 or 20 amps if that is what is needed for the finished circuit. A DC meter able to measure down to 1uv would be a good idea too, and it that isnt good enough than the test trace will have to be lengthened.

Silver solder comes in something like 2 percent to 4 percent for electrical use i think, and im not sure that that will decrease the resistance very much. One of the reasons for using that solder is because it flows better than regular solder and probably adheres better too...better conductivity is probably over stated for this kind of product.

Another variable is the surface condition of the added solder. If the solder is left to be very shiny, that will cause less dissipation too.

Ampacity of common wire, even magnet wire is well known, bugger the math, use more solid core jumper wire than you need and be done with it, I should hope that there are more pressing technical problems in your project.