"N´far" said:
(What I could find so far...)
"For the P4B533 remove the resistor between
GHI# and DPSLP#, get the system into standby mode and put a ground signal
toGHI# and the CPU will boot in Maximum Performance Mode (max. multipier)."
Mayby ít´s true...
What to do with an P4P800SE or an P4B266?
BTW: I tried CPUMSR, it doesn´t work. I only have DDR400 and the P4B266
doesn´t have a AGP/PCI lock - so I´m not interested in having a high FSB.
You are referring to the status signal that a mobile chipset
would use, to cause the processor to change speeds. The
mobile Intel processor has two speeds, and defaults to the
slower speed at startup. If a mobile processor is connected
to a desktop chipset, then the Speedstep control signal is
not available, to command a speed change.
As I understand the process, the processor is put to sleep, then
the GHI# signal changes state, to signal to the processor that
the higher speed is desired.
I guess my solution would be to get a worthy motherboard, like
a P4C800-E, as the FSB clock can be run up to 300MHz, while the
3:2 memory divider would allow DDR400 memory to run in spec
when the CPU is heavily overclocked. People take the mobiles,
which run at default 12x multiplier at CLK=100MHz, and run
them at 3.6GHz by raising the CPU clock to 300MHz. If the memory
is run at 1:1, then DDR600 memory is needed. If the memory
divider is set to 3:2, then only DDR400 memory is required.
Perhaps with a copy of a mobile processor datasheet, you can find
the location of the GHI# signal. Then measure to see if it is
connected to some other signal. Page 74 explains how the
chipset uses the GHI# signal. (Note: This kind of reverse
engineering is not easy to do. It will require a lot of
interpretation of the readings you get with an ohmmeter,
to figure out what is going on with the pins.) Page 42 shows
the pinout for a mobile CPU.
http://www.intel.com/design/mobile/datashts/25302804.pdf
If I look at the pinout of a desktop P4, pin A6 is labelled
"TESTHI11", a test signal. So, TESTHI11 <=> GHI# is the difference
between a desktop and a mobile. Similarly TESTHI12 <=> DPSLP#
on pin AD25. The explanation on page 20 shows that TESTHI11 and
TESTHI12 are connected via a resistor to VCC, as part of proper
treatment of those unused test inputs. If you just ground pin
A6, then both GHI# and DPSLP# will be logic 0, and the current
flowing through the pullup resistor will be 1.5V/60ohm = 25ma (OK).
The question would be, whether the processor can be awakened
again, if the DPSLP is asserted. You may need to break the
connection between GHI# and DPSLP#. So, you will need to use
the ohmmeter between GHI# and VCC, to see if there is a resistor.
Then, check between pin A6 and pin AD25, to see if GHI# is
shorted to DPSLP#. Then, lift the motherboard, and see if there
is a visible track between the two pins that can be cut.
Before cutting, you could just ground A6, and see if the speed
changes when you bring the computer out of sleep. If the
computer will not wake up, then you'd need to cut the trace
between A6 and AD25, wherever it happens to be.
http://developer.intel.com/design/pentium4/datashts/298643.htm
The Intel reference schematic for the 875 chipset, has a separate
resistor for TESTHI11 and TESTHI12. Which means, if you had an
Intel motherboard, you would not need to cut anything. It is
possible the Asus board works that way, but using the ohmmeter
will tell you for sure (whether A6 and AD25 are tied together).
It is even possible that Asus used some other crazy wiring
scheme - you never know.
In any case, you are going to need to use the pinouts in both
those documents, to see what pins a Mobile uses, and what they
connect to on a desktop version of the S478 socket. This is
not a project I would take on, unless I had zero budget for
a new motherboard, and a lot of time on my hands.
HTH,
Paul
