I was wondering if a PCB trace can support peak currents, larger than the nominal current. E.g., using a PCB trace width calculator, a 7mm trace supports 8A, but can it support larger peak currents? If so, what is the duration of the peak and the value of current supported and how is it calculated?
\$\begingroup\$
\$\endgroup\$
3
-
\$\begingroup\$ There are a lot of online calculators but an important parameter is maximum allowable temperature rise. I have no idea about the duration though. \$\endgroup\$– Rohat KılıçMay 4, 2017 at 9:12
-
\$\begingroup\$ You can fudge on the temp rise over duty cycle, but the voltage drop will spike with the current. This may not excuse you from any applicable safety requirements if that applies to you though. \$\endgroup\$– DanielMay 4, 2017 at 9:18
-
2\$\begingroup\$ Define "peak". The shorter the time period, the higher the current that can be handed. \$\endgroup\$– PuffafishMay 4, 2017 at 9:37
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
There is a difference between "peak" and "sustained".
As others have pointed out, the time matters here.
When you define the "maximum sustained" current, you gennerally aproach it as steady state system - given a maximum allowed temperature of the copper, and knowing the thermal impedance from the conductor to the enviroment, what is the maximum steady state power we can dissipate in the conductor.
When we talk about peak currents, the key here is that we don't reach a steady state. The copper has a certain thermal mass. This means that we can pump more current through the conductor for a short period - as long as we don't surpass our maximum allowed temperature. The current has to heat up the copper first - if the current pulse lasts shorter than the time it would take to heat the copper past this maximum rated temperature, there is no issue.
This also means that the shorter the pulse, the higher it's value may be.
When using a resistive model to simulate temperatures, the thermal mass is represented by a capacitance, and the thermal impedance to the atmosphere is a resistor. The current pulse would be a spike in power (thermal current) through this RC filter.
\$\begingroup\$
\$\endgroup\$
Copper foil has 0.0005 ohms per square, at the standard 1 ounce/foot^2 thickness. That foil is 35 microns thick. A cube of that is ~~ 50,000 cubic microns.
Using the specific heat for silicon (quick answer) of 2 picoJoules per cubic micron per degree Centi, the above cube of copper has specific heat of
$$50,000 cubic microns * 2 picoJoules / cubic micron/ degree Centi$$
Or 100,000 picoJoules per degree Centi. or 0.1 microJoules per degree Centi.
How fast will 1,000 amps heat that 35 micron cube of copper? The resistance is 0.0005 ohms. Power is IIR = 1000 *1000 * 0.0005 = 500 watts energy rate.
At 500 watts, or 500 joules/second, the rate of rise is 5 billion degrees C per second. In about a microsecond, you have 5,000 degree C, thus plasma?
answered May 4, 2017 at 15:27