PCchips M571LMR core voltage regulator
--------------------------------------

WARNING: The following information has not been properly tested and may
contain errors.

The following schematic diagram is of the core voltage regulator on the
M571LMR motherboard. Transistors Q17, Q15, Q19, and Q20 provide binary 
weighted contributions of 0.1V, 0.2V, 0.4V, and 0.8V, respectively, to the 
overall core voltage setting.

Undocumented header J9 appears to be intended for manufacturing tests. By 
means of jumper wires, this header can be used to override the BIOS core 
voltage settings. Pin 6 can also be used as a convenient Vcore test point.
Note that the transistors are NPN types, part number ZS94 (?), and that 
their emitters are connected to ground. Furthermore, when turned on their 
VCE is of the order of 8 or 9mV.

Note that the minimum Vcore is set by turning off all four transistors. This 
minimum is calculated from the formula:

 Vcore (min) = (8K06 + 4K99) / 8K06 x 1.25
             = 2.02V

                                     _____  Vcore
                                       |
                                      _|_ 
      85B     18C     47C     75C     | | 68B
      7K45    14K85   30K1    60K4    |_| 4K99   |\
       ________________________________|________5| \
       |       |       |       |       |         | E\ KA34063
      _|_     _|_     _|_     _|_      |  1.25V--|  / error amp
      | |     | |     | |     | |      |   Vref  | /
      |_|     |_|     |_|     |_|      |         |/
       |       |       |       |       | 
       |       |       |       |C     _|_ 
     |/      |/      |/     B|/       | | 88B
  |--|    |--|    |--|    |--|        |_| 8K06
  |  |\   |  |\   |  |\   |  |\        |
  |    |  |    |  |    |  |    |E      |
  |    |__|____|__|____|__|____|_______|
  |       |       |       |            |
  |   Q20 |   Q19 |   Q15 |   Q17     _|_
  |       |       |       |            =
  |       |       |       |    __
  |       |       |       |---|__|---->
  |       |       |       |    __
  |       |       |-----------|__|---->
  |       |       |       |    __        to chipset
  |       |-------------------|__|---->
  |       |       |       |    __
  |---------------------------|__|---->
  |       |       |       |    10K resistor
  |       |       |       |    pack
  |       |       |       |
 J9-4    J9-3    J9-2    J9-1
 0.8V    0.4V    0.2V    0.1V


J9 pinout
---------

        Q20  -----|  |------ Vcore
        Q15  --|  |  |  |--- Key
               |  |  |  |
              |--|--|--|--|
              |2 |4 |6 |x |
              |--|--|--|--|
              |1 |3 |5 |7 |
              |--|--|--|--|
               |  |  |  |
               |  |  |  |
        Q17  --|  |  |  |--- +5V
        Q19  -----|  |------ Ground


How to obtain a Vcore of 1.9V for certain AMD CPUs
--------------------------------------------------

This can be done by forcing all transistors off, and by placing a suitable 
resistor in parallel with the 4K99 resistor. The value of this resistor, R, 
can be calculated from the formulae:

 1/R = 1.25 / 8.06 / (Vcore - 1.25) - 1 / 4.99

 Vcore = 1.25 + 1.25 / 8.06 x R x 4.99 / (R + 4.99)

A value of 27K gives a Vcore of 1.90V, while 33K results in 1.92V.

To turn all transistors off, make up an 8-pin double row connector for J9 
by wiring pins 1,2,3,4, and 5 together. Run a separate wire from J9-6 to 
your new resistor, and wire the other end of this resistor to pin 5 of IC 
KA34063. Alternatively, ensure that a core voltage of 2.0V is selected in 
the BIOS setup.