PCB trace width can't meet the rated current

Question

The rated current for VCC5_AC from AC adaptor is 2A. According to the calculation on http://circuitcalculator.com, the trace width for 2A should be 30.3mil.

However, as indicated above, the width of AC input pin of the BQ24030 is only around 12mil and I think it's useless to connect a wider trace to this pin, so the actual trace width of VCC5_AC connected to the pin is just 12mil.

My question is:

1. Does this mean the regulator cannot drain the rated current 2A?
2. If this is the case, is there any solution to solve this problem?

Edit: Thanks to everyone's suggestion. I realize my mistakes are

1. The uncertainty about changing the trace length in the middle, even a short part, which is not very significant in this case;

2. The misconfiguration of trace width calculation tools;

   I've changed my layout as below which widen the power line

Answer 1

Some considerations:

1. pins and solder have a much larger sectional area than a PCB trace. So the pin and pad can handle much more current than a trace of the same width (even if they are made of lower conductivity materials).
2. The estimated temperature rise assumes a relatively long trace. A short trace segment, that is sandwiched between large metal areas will dissipate heat much faster.
3. A 10ºC temperature rise is rather conservative.
4. You can rearrange adjacent traces to maximize the wide trace run to within a few millimeters of the pad.

But, is there any reason why you didn’t use the datasheet’s suggested layout as a guidance?

Answer 2

You did not consider that the temperature rise of a trace also has a dependency on trace length. (The calculator assumes a infinitely long trace).

Let’s take a trace with total length of 1m with thickness of 30mils. This is rated with 2A with 10°C temperature rise. Now imagine a very tiny region/spot of that trace with length of 10mils where the 30mil trace with is reduced to 2mils. Of course this „spot“ would increase the resistance a little and it would absorb a little more energy. But the trace would still be able to carry that 2A current with only 10°C temperature rise.

Also consider that an IC has internal bond wires that usually have very small wire diameters. But because they are very short, this is not a problem for the same reasons.

Answer 3

According to the calculation on http://circuitcalculator.com, the trace width for 2A should be 30.3mil.

*With a 10 degree temperature rise.

Everything is a resistor, even board traces. The maximum current trough a resistor is defined by it's power rating, which is the maximum temperature. If you want more, you need to dissipate more heat.
This means that for the a typical board trace it can run 2 Amps with a 10 degree temperature rise for the specified width and thickness, but 15 mil for a 30 degree temperature rise.
You did not specify the thermal resistance in C/W. Which is also difficult to calculate.

The small trace of 12mil is connected on one side to the regulator part, and the other side a SMD component.
These are both parts that allow heat to be transferred away from the trace, increasing the current capacity of the trace.

1. Does this mean the regulator cannot drain the rated current 2A?

The regulator can do 2A without problems, so can this short trace.