<<snip>>
It could be stressed two ways. Motherboards now, tend to run the USB
ports off +5VSB. Which is a poor choice from the perspective of getting
enough amps. If you had 14 ports, and actually supported all of them,
that would be 14*500mA or 7 amps (this assumes every port draws as much
current as the standard would suggest is possible). And power supplies
tend to top out at around 5VSB @ 3A perhaps. So if there really were
14 ports, all in heavy usage, the supply would shut off on overcurrent.
(But you could easily run 14 optical mice, and the supply would stay up. :-) )
The motherboard itself, has ampacity limits on conductors. The average
board, over time, was probably four layers thick. A few manufacturers use
2 ounce copper, versus say 0.5 ounce that was more typical for signal
layers. And the heavy copper, means the tracks don't have to be as wide,
for a given max current flow. Tracks are sized according to target
delta_T (the motherboard heats up, over top of the track). There are
tables you can consult, for picking copper track thickness, for a
desired delta_T.
Near the main power connector, there is a lot of need to move amps of
current away from the connector. And not a lot of layers to do it with.
That's the puzzling part of motherboard design. How do they manage to
do that ?
So that's the motherboard stress part. Figuring out how to route
enough wide copper, so there isn't too much delta_T on any
individual conductor.
The motherboard has a Tg or glass transition temperature, which defines
how much heat the motherboard can take. And the Tg provides the
budget for allowing the motherboard to "get hot". The price of the
bare motherboard varies with the Tg of the material selected, and
with motherboards, you'd assume not the best material got used
(for pricing reasons).
If you scroll half way down this page, you can see such a curve for
sizing conductors in a PCB.
http://wiki.xtronics.com/index.php/Wire-Gauge_Ampacity
If you actually were to carry 7 amps in the motherboard (which the
power supply would not allow on +5VSB), and were willing to accept
a delta_T of 20C, then the track is about 0.060" wide. Which sounds
quite easy to do, assuming there are no other heavy loads on the
motherboard.
http://wiki.xtronics.com/images/7/74/PCB_current.png
All those ports would have to be fused as well, and there is a
Polyfuse per pair of ports, typically. When picking the distribution
track size, you'd assume one port goes shorted at a time, not all of
them :-) So you'd engineer for 12*500ma + approx 2A on the shorted
port pair (to open the Polyfuse), or about 8 amps. Which again,
is stopped by the power supply.
Given how stupid this design idea is (USB +5VSB-only operation), the
track never has to be sized for more than about 3 amps. Just enough,
to hit the supply limit, on the "best" power supply. So now the
track is down to around 0.030" wide, to carry +5VSB without significant
heating.
Now, if they'd allowed it to be powered from +5V, the end-users
would like that, but the track to distribute USB power ends up
being wider. And then, you can actually use all the ports, without
worrying about it (run 14 iPhone chargers or whatever).
Paul