Onion said:
I was being simplistic. The 'charger' is a circuit powered from the
main power supply.
All DC voltage creation originates at the main transformer hence they
are all connected.
Drastically so in the case of a short.
I can explain futher, but I hope you see the point.
There's a picture of a power supply here, so you can understand how
it works. Transformers are used for isolation. The turns ratio
helps define the output voltages. Rectifiers convert the high frequency
AC from the output transformer, back into DC. It's a switch mode supply,
where the switching occurs at high frequencies, to help keep the
size of the transformer smaller, and make it easier to filter
on the output. A separate transformer, separates the controller
portion, from the bipolar transistor base drive, as the switching
transistors have high voltage on them. (This is one of my favorite
free schematics on the web.)
http://www.pavouk.org/hw/en_atxps.html
As for the DC blocking of the CMOS (powering) thing, you can see
that on PDF page 82 here. The upper left corner shows the diodes
selecting the battery or +5VSB derived source of power for the CMOS storage.
PDF page 85, shows the regulator that takes 5VSB and makes V_3P3_Standby
from it. The diodes only allow current to flow in one direction.
http://www.intel.com/design/chipsets/schematics/252812.htm
The 12V fan, is well removed from CMOS battery, 5VSB, 3P3_Standby
and so on. It has nothing to do with them. I don't see any
obvious backfeed paths to consider, either.
The CMOS battery is *not* to be charged. It says so right on the
datasheet for the CR2032. The spec to meet, is 1 microamp, which
means no more than 1 microamp may flow backwards into the
CMOS battery (which would charge it). Therefore, it cannot and
must not be charged. The diodes on page 82, help meet that spec.
The battery runs, until it runs out of juice. When the computer
is line powered, and +5VSB is available, the +5VSB supplies the
current required, which is why the battery lasts so long.
Even when the computer is completely unplugged, the load
on the CMOS battery is around 5-10 microamps or so. 2 microamps
is enough to run a digital watch circuit, and the SB well
uses a bit more than that.
To get back to the original symptoms, there is one additional
piece of information we can use. The CMOS well, powers both
256 bytes of CMOS RAM, as well as the RTC clock. If the CMOS
BIOS settings were lost, as well as the RTC clock being
set to the year 1980, that would be evidence for a failure
of the CMOS voltage (the output of the two diodes). My
suspicion is, Man-wai Chang saw the BIOS settings reset
to defaults, but the clock maintained the correct time.
And that is evidence of the Asus "overclocking failure,
reset to defaults" feature. It doesn't reset the clock
to the year 1980.
I don't see how the dead fan would have triggered that
feature, unless the computer crashed during the period in
question. Even switching off the power, without safely
shutting down the PC, could trigger that Asus feature.
Paul
