# How can I determine the safe width of the PCB traces so that they withstand the momentary short circuit current before the circuit breaker trips?

Many lamps are supplied by 220v and a circuit breaker through a PCB as in the picture. The total current of this circuit is 1A. For this implemintation I am going to use 2A circuit breaker. I am afraid that a short circuit could happen at the load and the traces of the PCB may explode. How can I determine the safe width of the PCB traces so that they withstand the momentary short circuit current before the circuit breaker trips? Any tips would be appreciated.

• Use one of the many online ampacity calculators? – PlasmaHH Mar 26 '17 at 20:26
• Another possible approach to this is inrush current limiting (ICL). – Nick Alexeev Mar 26 '17 at 20:28
• @PlasmaHH Actually I could do that but I dont know how much the short circuit current would be. It could be 10A or 30A. It could be more or less. – Macit Mar 26 '17 at 20:47
• @Majid_L A short circuit more likely would be 6 kA (kilo amperes) more or less. – Marko Buršič Mar 26 '17 at 21:07
• @Majid_L The circuit breaker itself disconnects the section, but it doesn't limit the current. At SC it will disconnect for example in 5ms. The I^2*t value is now 6kA^2*5ms=180,000A^2*s. Whenever it will melt or not, you should calculate with some calculation: ultracad.com/articles/fusingr.pdf – Marko Buršič Mar 28 '17 at 11:02

A fuse might be more suitable.

SMD Fuse : Current Rating 1A , Voltage Rating - 350Vac , $0.18 @ 500pc or a Polyfuse 1A$0.40 (500pc) 10A max radial

The whole idea of a fuse or a breaker or a Polyfuse to choose protection that reacts quicker than the fragile path. If a short thin track heats up faster than a breaker or a polyfuse, then the only choice is a fast blow SMT fuse. got it? Which was my 1st answer. Ensure the track resistance is much much less than the SMT resistance, by design of the tracks and additional solder if necessary or busbar. A 2A fuse is 41 mOhm cold.

Otherwise if you don't then the track becomes the fuse. Beware of high resistance The whole idea of a fuse or a breaker or a Polyfuse to choose protection that reacts quicker than the fragile path. If a short thin track heats up faster than a breaker or a polyfuse, then the only choice is a fast blow SMT fuse. got it? Which was my 1st answer. Ensure the track resistance is much much less than the SMT resistance, by design of the tracks and additional solder if necessary or busbar. A 2A fuse is 41 mOhm cold. Otherwise if you don't then the track becomes the fuse. Beware of high resistance contacts.

• While it seems a good solution but I think it is not practical. Because if a lamp burned out a huge amount of current would go through the fuse and blow it. For sure in my approach the circuit breaker is going to trip too but it is easier to turn the circuit breaker back on instead of changing the fuse. – Macit Mar 26 '17 at 21:02
• Polyfuse could be a possible solution though. Can it handle 220v dead short current? – Macit Mar 26 '17 at 21:08
• An NTC (ICL) will protect a short cct on cold start but not hot and the Polyfuse is only rated for 10A so looks like a fuse is safest – Sunnyskyguy EE75 Mar 26 '17 at 21:55
• @Majid_L Wait... Are you concerned about a lamp burning out or a dead short? Those are two very different things. – marcelm Apr 30 '17 at 10:45
• @marcelm My main concern is the dead short. – Macit Apr 30 '17 at 12:17

A cube of copper 35 microns on a side (35 microns is the standard thickness of CU foil at 1ounce/foot^2) has 0.0005 (500 microOhms) resistance. I'll use a specific heat I know well ---- that of silicon, 2 picoJoules/cubic micron/deg C ---- to run through the math. In that 35micron cube, there are ~~ 40,000 cubes of size 1micron. Thus the energy storage is 40,000 * 2pJ/micron^2/deg C or 80 nanoJoules/35microncube CU. (again, I'm using silicon, not CU value).

What is the energy generated in that 35 micron cube, at 1 amp? P = I^2*R, or 1*1*0.0005, or 500 microJoules/second.

Lets just divide 500 microJoules/second by 80 nanoJoules/deg C, and we find the dimensions are "degC/second" and the scale factor 500,000nJ/80nJ = 6,000.

Thus the rate of heating is 6,000 degree C/second.

Now you need to compute the specific heat of a cubic micron of copper, and tweak that 6,000 to a more accurate number.

edit If you want 60 degree C rise, over 1 second, you need 100X the width I used. That 35micron cube is 1.4 mils, or 0.0014 inches. Thus 0.14 inches seems safe.

• Just to clarify, you have made your calculations for 1 amp current which is the steady state of the circuit. The problem is that I can not find a way to estimate the circuit's short current which I believe, with a type B circuit breaker, could last for 40ms. – Macit Mar 27 '17 at 8:52
• Thus a 40 millisecond short, at 25 amps, would also produce 60 degree Centigrade temperature rise. And 250 amps would produce 600 degree Centigrade rise; how fast will the breaker trip, at 250 Amps? – analogsystemsrf Mar 27 '17 at 16:19
• The breaker trips at currents higher than 10 Amps after about 40ms. – Macit Mar 27 '17 at 17:51
• What is the wiring resistance? – analogsystemsrf Mar 28 '17 at 2:18
• It is about 40 meters of 21 AWG cable. This cable has a resistance of 42 mΩ/m thus total resistance is going to be about 1.68Ω – Macit Mar 28 '17 at 10:35

I would not worry about that small amount of time. The goal is to have a trace/wire that has enough ampacity for the standard/max normal current load. In a short circuit situation with a circuit breaker, yes you would get a TON of amps, almost nothing would be able to handle that, however you only get them for a VERY short amount of time.

When you take time into consideration, your board wont blow up when something shorts, as long as the circuit breaker trips. Nothing will have enough time to heat up to a point if the current is cut off almost immediately.