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Old November 9th 13, 07:44 AM posted to alt.comp.periphs.mainboard.gigabyte
Paul
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Default GA-Z87X-UD4H: No USB Power?

(PeteCresswell) wrote:
In the process of building up a GA-Z87X-UD4H system, I notice that there
does not seem to be any power available on either the USB2 or USB3
outlets. I have not tried the internal USB3 headers pending purchase
of a 3.5" front panel to use them.

OTOH, on my EP45-UD3L, there is power on all the USB receptacles.

Am I missing something?


I don't know what to suggest.

The advertising for your board says there is a fuse per port.

http://www.gigabyte.com/products/pro...px?pid=4488#ov

http://www.gigabyte.com/fileupload/p...8/8122_big.jpg

The advertising shows a picture of a Polyfuse (the silhouette
has the distinctive feature on one end of the device). Yet,
looking at the motherboard picture, while I see a couple of
what look like Polyfuses, they're pretty tiny. Usually, a
Polyfuse is a bit bigger as components go. The Polyfuse
is polycrystalline, it melts on overcurrent, it crystallizes
when it cools off. That means the fuse recovers, and never
needs replacement. It needs a certain size, for that chemistry
to take place. So for so many amps of current, that
dictates the X-Y size. While they could make the device
thicker, I don't really see any taller SMT components
in the region.

So I can't say for sure, what current flow protection the
thing is using exactly. There seem to be an awful lot of
transistors in the area, but again, I can't be sure whether
they go to those headers or elsewhere.

It should have power, unless something is being switched off
for some reason.

Traditionally, at least for Southbridge ports, there
is one OC# input for each USB port. That's "Overcurrent
Active Low". When a logic zero appears on that signal,
the Southbridge generates an interrupt stating that
the measured current on the port is too high. But the
method of current measurement is left to the motherboard
designer. (On laptops, an 8 pin DIP per dual USB stack,
measures the current and generates OC# flags. On a desktop,
OC# is connected downstream of the Polyfuse, and when the
Polyfuse opens, the OC# drops to zero volts along with the
VBus level. Which is a cheap and clever trick.)

Sorry I can't be of more inspiration. I can't really
tell for sure what they're doing.

Other weirdness includes, only one inrush capacitor per
header. Implying the inrush is upstream of the dual
fuses, instead of the more normal downstream position.
Otherwise, their claim of a fuse per port would be a lie.
If the inrush cap was downstream, we'd need to see two
beefy caps per header. And I only see one. Intel
recommends something like 100uF per port, and that
holds up the VBus rail when a USB device is plugged
in hot.

Your board has some "iPad charger" settings, but as
far as I know, what that does is set resistor biasing
on D+ and D- of a USB port, so the iPad thinks it's
connected to a legit charger. Something like that.
I don't think that feature actually affects the
power path. A charger port presents a different
bias pattern, than a regular working USB port.

The USB powering these days, is normally done from
+5VSB. Your power supply has a 2 to 3 amp limit on
+5VSB, so it's a relatively weak rail. Yet, your
motherboard has ten USB3 ports (dual Intel, plus
two quad port external chips?). If drawing 900mA
from each port (the max), ten of those would draw
9 amps, or three times the power supply rating.
I don't think this is an issue in your case, but
it is a bit weird. While they could if they wanted,
connect some of the USB3 ports to +5V (only powered
when fans spin), that would then limit the "wake on"
features, and also limit which ports could be
used for "sleep charging". I assume in this case,
when you say the ports aren't powered, this was with
the system booted, and not a statement about iPod
charging status when the computer is asleep.

Paul