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Im designing a led driver for a friend of mine. The total consumption would be 0.5A (per Meter of led Strip) * 9 Meter = 4.5A, or around 5A. The strips are connected via screw terminals. The problem is, that if I use a online width calculator, which says that the thickness should be ~7mm. If I would connect the Voltage Source with a Screw Terminal, the trace would interfere with the other pins of it. Could I just use 3mm (which fits) and then thicker the traces with solder or even solder wires to the points? Thanks :)

Notes: -PCB thickness is 35um -aiming for single sided, to reduce amount of work.

~Straw

Pictures: Overview

7mm Trace width

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  • \$\begingroup\$ If jumpers are acceptable, terminate the tracks to a through hole via. From through hole take a guage wire and connect it to screw terminal. Feasible? \$\endgroup\$ – User323693 Jul 3 '15 at 0:50
  • \$\begingroup\$ Before you go into track widths and logistics, have you checked the datasheet for the screw-terminal connectors? Most molex-type wire-to-board connectors are only rated up to 3A. While the hole on the screw terminal's PCB footprint looks big enough to suggest that it is... those DC jack pin holes look suspiciously tiny. \$\endgroup\$ – CharlieHanson Jul 3 '15 at 1:02
  • \$\begingroup\$ Also, are you absolutely sure your driver IC will handle this? 5 amps through a single pin does not sound like a good idea. What IC are you using? \$\endgroup\$ – WhatRoughBeast Jul 3 '15 at 1:09
  • \$\begingroup\$ @chaaarlie2 screw terminals are fine, rated at 8A, the footprint was just made fast by me, so that i atleast have one. bad that i dont have any datasheet for the barrel jack, as i bought them from china someday. they seem to be the same as on a generic led strip driver which is rated for 5A. \$\endgroup\$ – Matze Strawberrymaker Jul 3 '15 at 1:12
  • \$\begingroup\$ @WhatRoughBeast . IC is ok. Im driving one Color (aka. ~0.15A per meter and color) with the uln2803, which is capable of 0.5A per pin. IC is Arduino. Its thought as an alternative led driver for stairs. when a PIR sensor outputs movement, the stairs will light up one by one, while still being a "normal" led strip driver \$\endgroup\$ – Matze Strawberrymaker Jul 3 '15 at 1:14
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Solder is 6 to 10 times more resistive than copper, so that's not a particularly good idea, unless you are able to put on so much solder that surface tension causes a domed cross sectional profile of substantial thickness.

Can you keep the trace wide, except right in the area of the connector?

Can you have the board house use thicker copper than normal, or plate it up (many start with exceptionally thin copper and do that anyway to achieve a normal thickness)?

If you are doing a handmade prototype (implicitly without solder mask), you could always solder a piece of copper wire along the length of the trace.

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  • \$\begingroup\$ sounds like the best option to me to just use copper wire, thanks :) \$\endgroup\$ – Matze Strawberrymaker Jul 3 '15 at 4:27
  • \$\begingroup\$ @MatzeStrawberrymaker Looks like you can get 50%+ decrease in resistance by just tinning the trace: youtube.com/watch?v=L9q5vwCESEQ \$\endgroup\$ – Golaž Jul 3 '15 at 13:50
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    \$\begingroup\$ No, you can't. That video does not show a trace that has been "tinned" but rather one that has been built up with a lot of extra solder to substantial thickness. The later experiment with a thin layer was only 15%. Some of the following comments on youtube are rather ignorant - for example, someone tries to claim that lead free solder is "silver based" - no, it's tin, with only a few percent silver at most. Even apart from the facts, and the melting temperature, that should be obvious from economics - enough silver to carry current would cost more than making thicker traces to start with. \$\endgroup\$ – Chris Stratton Jul 3 '15 at 13:59
  • \$\begingroup\$ Related: This excellent reference - TI Analog Engineer’s Pocket Reference - 4th edition provides some useful information on PCB track current/ voltage drop / heat / fusing issues. Especially pages 55-68. \$\endgroup\$ – Russell McMahon May 5 '16 at 5:31
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Track current ratings are for pcb temperature. As long as you keep the PCB temperature low, you can use thinner tracks and higher current. So...

1) What will be the ambient temperature?

2) As soon as you get away from the pin-out, broaden the track.

3) Make it a 4-layer board, (middle layers untracked copper as heat spreader)

4) Are you trying for 2oz copper? Make it 4oz or 6oz.

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  • \$\begingroup\$ It's not only about heating. Substantial voltage drop when high current must flow through small resistive traces can prevent many circuits from operating as designed. \$\endgroup\$ – Chris Stratton Jul 3 '15 at 16:14
  • \$\begingroup\$ ambient temperature goes from 15°C up to 30°C. 4 Layer board isnt easy to do at home. \$\endgroup\$ – Matze Strawberrymaker Jul 3 '15 at 18:00
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There's no reason not to dump as much solder onto that track as you like. I've seen plenty of PCBs use that technique, although the majority of them tend to do it on AC/mains parts of the boards. Nonetheless it does the trick for power lines.

Don't do it for signal/data lines with frequencies venturing into ultrasonic ranges and above; good old "60/40" solder is not designed for transmitting that sort of thing: serious signal degredation will result! Lead-free solder might be an improvement, but there will still be impurities galore once you've melted it and stuck it to your copper.

That is, of course, unless you're assembling this in a clean room and you've scrubbed the copper track, blitzed it with some nitrogen to ward off oxidation, and then used some gold solder heated by infrared to melt and bond it to the copper. No? You amateur.

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    \$\begingroup\$ "Ultrasonic" signals, let's say 100 kHz, have a wavelength much longer than the length of any reasonable trace. For example, 200 m. So, no effect will result from that because transmission-line effects are not important until one's signals get into the tens or hundreds of MHz, with wavelengths something close to the track length. The type of solder is irrelevant; all solders have more or less the same conductivity. \$\endgroup\$ – Oleksandr R. Jul 3 '15 at 2:01
  • \$\begingroup\$ As this is steady 12V only, guess it should be alright, but i think i'll go with direct copper wiring. \$\endgroup\$ – Matze Strawberrymaker Jul 3 '15 at 4:25

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