0
\$\begingroup\$

enter image description here 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?

enter image description here

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 enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ Could you add some information on the specific IC? Yo provided none. \$\endgroup\$
    – Edgar Brown
    Nov 29, 2018 at 4:44
  • \$\begingroup\$ @EdgarBrown Ok,I've added some information to describe this problem. \$\endgroup\$
    – Ross
    Nov 29, 2018 at 4:54
  • 2
    \$\begingroup\$ Why do you think it's useless to widen the trace as soon as possible? Is using 2oz copper out of the question? Any particular reason you chose 10°C rise? \$\endgroup\$
    – Spehro Pefhany
    Nov 29, 2018 at 4:59
  • 2
    \$\begingroup\$ Make the trace as wide as possible. Also, as the trace leads away from the pin, widen it as soon as there is room. A short narrow section is no big deal. Some of the answers have good advice for you. You are mistaken to think that since one section is only 12 mils, the whole thing might as well be 12 mils. Make as much of it as wide as you can. Also, TI would not be so stupid as to make a part that can't be routed out. \$\endgroup\$
    – user57037
    Nov 29, 2018 at 7:56
  • 1
    \$\begingroup\$ You can also push the two traces at the top left inwards a long way, allowing you to make your VCC trace much thicker closer to the chip. \$\endgroup\$
    – Tom Carpenter
    Nov 29, 2018 at 8:09

3 Answers 3

Reset to default
6
\$\begingroup\$

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?

suggested PCB layout

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Regarding bond wires, double-bonding and multiple bond pads are rather common. Although it increases the complexity of the package, bond pads don’t have to have a 1:1 correspondence with IC pins. \$\endgroup\$
    – Edgar Brown
    Nov 29, 2018 at 5:30
  • \$\begingroup\$ Another way to look at it is that an infinitely long trace can only lose heat to the board or the air. A wide trace which narrows down to a short, narrow trace will allow heat to flow from the narrow portion to the cooler, wider section (and copper is a great conductor of heat). Within limits, of course. \$\endgroup\$
    – WhatRoughBeast
    Nov 29, 2018 at 23:12
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.