Does anyone know which "chip" exactly they talking about? It doesn't
seem that we have the replace the entire motherboard just because of one
chip. I have experience removing and replacing surface mount chips, but
not quite sure where to look on this particular MB
Any help or advice is appreciated
You would have to trace backwards from the PS_ON# pin of the main
power supply connector on the motherboard. The PCB is multi-layer,
which means the signal could snake just about anywhere. On one motherboard
here, it took me two hours, with the motherboard in hand, just to
trace some signals from one side of the motherboard to the other.
An ohmmeter helps in this regard (when you have a theory as to where
a signal goes, you can use the ohm meter to verify there is zero ohms
from one point to the next).
If you have a schematic for the motherboard, then the job is much easier.
As you can instantly determine the origin of the signal, and identify the
culprit visually.
PS_ON# is open collector. The amount of current the driver has to sink
is relatively low. The designer has the option of using something like
a high power driver (7407), or if they wanted, they could use a switching
transistor, something like a 2N2222 perhaps, in a small package. They would
not necessarily drive PS_ON# directly from one of the larger chipset
chips, choosing instead to buffer the signal, to make repair easier.
This is the basic logic flow. Contrary to intuition, the front power
switch is not directly involved in turning the computer on and off.
The front power switch is momentary contact, and some other chip latches
a pulse coming from that chip, to make a steady logic level.
Front
Power ----------- Southbridge ---- SuperIO ---- OC driver -----> PS_ON#
Button RTC wakeup (Wake on transistor Pin 14
Alarm key press) or chip
Computers have multiple reasons to "wake up". All the waking reasons,
must be connected somewhere in the logic flow. For example, if the
PC is scheduled to wake at 5 A.M. by the Windows Task Scheduler, the
RTC in the Southbridge asserts the "I want to wake up". And that
propagates to PS_ON#. Another example would be the support in the
SuperIO chip, for waking when a key is pressed on a PS/2 keyboard.
All of the waking reasons are logically OR'ed to make the final
PS_ON# signal. There are many reasons for the computer to awake, so
there are many terms in the digital logic equation.
To get the pin number for PS_ON#, you can look in the various ATX
power supply specs. On the 20 pin connector, it is pin 14. The
wire color is supposed to be green. Grounding pin 14, turns the
supply on. Grounding pin 14 would also ground the "OC driver" in the
diagram, but that is OK, because the logic type is open collector and
is safe to ground.
http://www.formfactors.org/developer/specs/atx/ATX12V_1_3dg.pdf
http://www.formfactors.org/developer/specs/ATX12V_PSDG_2_2_public_br2.pdf
To understand the logic, we can take as an example, the twelve year old
reference schematic for a 440BX based Intel motherboard.
http://www.intel.com/design/chipsets/designex/BXDPDG10.htm
http://www.intel.com/design/chipsets/designex/BXDPDG10.PDF
On PDF page 18, upper right hand corner, is an example of
a circuit to drive PS_ON#. The 74F07 is a Fairchild FAST
hex buffer with open collector output drive. It sinks
current but doesn't source current. It knows how to pull
PS_ON# to ground to turn on the power supply. When 74F07
"lets go" or goes open circuit, a pullup resistor (similar
in function to R413 (10K ohms) next to the chip) , ensures
the logic signal returns to +5V voltage.
If you then go to PDF page 32 of the schematic, in the bottom
center of the page you can see the 20 pin ATX main power connector.
The signal created on page 18 (B_SUSC) is used to drive pin 14
of the ATX connector. And that tells you that B_SUSC = PS_ON#.
The "#" convention is intended to indicate "active low"
- the signal is in the activated state when the signal
is at zero volts.
So that is the basic principle, but "anything goes" when it
comes to circuit design. Just because the Intel example
does it that way, does not mean any other companies have
to do it the same way. You may not find a 74F07 14 pin DIP
chip on the motherboard. Maybe it is one of the many three
legged transistors on the board, doing the final driving.
That is why you'd have to trace backwards from pin 14 on
the 20 pin main power connector, to find out what is driving
that signal - and that isn't exactly an easy thing to do without
product documentation that is not normally given to the public.
It is even possible for one of the larger chips to drive that
signal line directly, in which case, forget about repairing
it. On one motherboard, Asus actually repaired a board, by
replacing the Southbridge on a damaged board. So it is
physically possible, with a hot air station and the appropriate
adapter on the end of the tool, to remove a BGA. If I ask
a local company to perform such a repair here, they quoted
$1000 to change a chip like that and Xray to verify proper
soldering (it is how they pay their hired help to sit around
waiting for repair work). In Taiwan or China, done in bulk,
they can do that for peanuts. It is not a particularly
easy repair, to replace a BGA, as there is some probability
of permanently damaging the PCB by doing so. So there is
some risk associated with the attempt. The better quality the
PCB, the lower the risk when doing it. Where I worked, they
claimed you could safely change a BGA chip like that three times,
before the PCB would be ruined.
And that is why, a motherboard designer might use a buffer chip
or a transistor to drive PS_ON#. If the buffer fails, even a
hobbyist could attempt to repair it. Just a bunch of snipping
and soldering.
To quote some odds, if I was attempting the repair myself,
with no documentation, there is a 10% probability I'd find
the thing doing the driving. Plenty of guys are more
motivated than I am, and their odds will be higher of
fixing it. I've worked with some amazing repair people, and
they're better at that kind of stuff than I am.
Good luck,
Paul