Sorry all. I was hoping to do this back over a month ago.
Since there are many threads discussing how to get to and remove the ficm I will skip this. I'm starting with the assumption that you have the ficm apart and the power supply board is sitting in front of you.
This will be very straight forward because almost everything needed is in the pictures. I know I will forget something (because I'm distracted) so feel free to PM me if I don't respond to the thread in a timely fashion.
To start, here's a basic schematic showing the circuit you are working with. The actual power path (inductor, capacitors, fet switch and diode) are repeated 4 times on the board. However, there is only one feedback path which sets the output voltage.
The idea is simple... lower the value of the 2.15KOhm resistor by placing another resistor in parallel with it.
Here's a picture of the 7-screw ficm power board (If someone with the 4-screw model can post a picture that would be great). Please note, part values and names are the same for both variants and so is the general location of the parts (IIRC). I've crudely labeled the points you need to find and what I see as the best spot to make the modification.
The screw hole potentials (from left to right, listed as one row) are:
48V, 5V, (Not Determined), Ground, Ground, 12V.
That's about it! Install the adjustment resistor and you'll have your new voltage.
If you want something other than 58V (10K resistor) or 54V (20K resistor), you can find the necessary resistor value using the following equation:
R2 = 48.7 KOhm
R1 = 2.15 KOhm
Vout = Desired Output Voltage
Radd = Parallel resistor to R1 necessary to achieve Vout
Radd = (2.0295 * R2 * R1) / ((Vout * R1) - 2.0295 * (R1 + R2))