PC Chips M571 (V7.0A) undocumented voltage settings - Part 2
 ------------------------------------------------------------
 
 Here are the details of my core voltage calculations.
 
 The following schematic diagram shows the relevant section of the
 on-board power supply (reverse engineered):
 
 
                                  6K04    9K53    20K0 Resistor
                              12K4    8K20    15K8      values
 
                              2V2 3V5 3V3 3V2 2V9 2V8  Jumper ID
     From  _____  _____
     Load  |    \/    |14     R88 R89 R90 R91 R92 R93  Circuit ref
      |    |          |--------|---|---|---|---|---|
      |   1| |\Error  | Vref  |-| |-| |-| |-| |-| |-|
 Vout |----|-| \_Amp  | +5V   | | | | | | | | | | | | Resistors
       ____|_| /      |       |-| |-| |-| |-| |-| |-|
     V |  2| |/       |        |   |   |   |   |   |
       |   |          |        |   O   O   O   O   O  Jumpers
       |   | KA7500B  |        |   O   O   O   O   O  JP6
       |   |__________|        |   |   |   |   |   |
       |                       |---|---|---|---|---|
       |_______________________|   
                               |
                         R79  |-|
                         10K2 | |
                              |-|
                               |
                              _|_
 
 
 
 The Samsung KA7500B (=TL494) is a PWM regulator. The error amp senses
 the output voltage and compares it with a setpoint derived from an
 internal 5.0V reference and an external potential divider.
 
 Various undocumented voltages can be obtained by:
 
 (1) using multiple jumpers
 (2) using a resistor in place of a jumper
 (3) a combination of jumper and resistor
 
 In the following diagram R1 represents one of R89,R90,R91,R92,R93.
 R2 is the resistor that must be added to achieve a particular voltage.
 
     +5V ref
       ___
        |
        |---|
        |   |
       |-| |-|
 12K4  | | | | R1
       |-| |-|
        |   |   ______
        |   O---| R2 |----|
        |       |----|    |    R = R1 + R2
        |   O-------------|
        |   |
        |---|
        |
       |-|
 10K2  | |
       |-|
        |
       _|_
 
 The following formulae determine values for V and R (=R1+R2).
 
                5
 Vout =  ------------------
 (volt)         12.4/10.2
          1  +  ----------
                1 + 12.4/R
 
 
                  12.4
 R (kohm) = ----------------
             12.4/10.2
             ---------  - 1
              5/V - 1
 
 
 Example calculation:
 --------------------
 
 A CPU core voltage of 2.5V is desired. Allow for 0.07V voltage drop from
 regulator to CPU (as in original design). Solving the second equation
 for V=2.57 gives R = 43.4K ohm. We can approximate this resistance by
 placing a 22K resistor in the 2V8 jumper position, ie 20K + 22K = 42K.
 Solving the first equation for R=42 gives 2.58V, ie 2.51V at the CPU.
 
 The following table lists the voltages that are available for the
 various multiple jumper combinations (0=no jumper, 1=jumper installed).
 
 Jumpers       nominal V     V - 0.07
 3V5-2V8     (at regulator) (at CPU socket)
 
 00000          2.26          2.19
 OOOO1          2.86          2.79 
 OOO1O          2.97          2.9 
 OO1OO          3.27          3.2 
 OOO11          3.32          3.25 
 O1OOO          3.37          3.3 
 OO1O1          3.53          3.46 
 1OOOO          3.58          3.51 
 OO11O          3.59          3.52 
 O1OO1          3.6           3.53 
 O1O1O          3.65          3.58 
 OO111          3.76          3.69 
 1OOO1          3.76          3.69 
 O11OO          3.79          3.72 
 1OO1O          3.8           3.73 
 O1O11          3.82          3.75 
 1O1OO          3.91          3.84 
 O11O1          3.92          3.85 
 1OO11          3.93          3.86 
 O111O          3.95          3.88 
 11OOO          3.95          3.88 
 1O1O1          4.02          3.95 
 1O11O          4.04          3.97 
 11OO1          4.05          3.98 
 O1111          4.06          3.99 
 11O1O          4.07          4 
 1O111          4.13          4.06 
 111OO          4.14          4.07 
 11O11          4.15          4.08 
 111O1          4.21          4.14 
 1111O          4.23          4.16 
 11111          4.28          4.21 
 
 The following program was used to print the previous table. It may be
 useful for analysing similar configurations on other boards.
 
  FOR i = 1 TO &H1F
  r = 1 / 12.4
  jump$ = "OOOOO"
 
  IF (i AND &H1) = &H1 THEN r = r + 1 / 6.04 : MID$(jump$, 1) = "1"
  IF (i AND &H2) = &H2 THEN r = r + 1 / 8.2  : MID$(jump$, 2) = "1"
  IF (i AND &H4) = &H4 THEN r = r + 1 / 9.53 : MID$(jump$, 3) = "1"
  IF (i AND &H8) = &H8 THEN r = r + 1 / 15.8 : MID$(jump$, 4) = "1"
  IF (i AND &H10) = &H10 THEN r = r + 1 / 20 : MID$(jump$, 5) = "1"
 
  r = 1 / r
  v = 5 / (r / 10.2 + 1)
  v = CINT(v * 100) / 100
  PRINT jump$, v, v - .07
  NEXT i
 

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 This page was last modified on 17 July 2002