Choosing capacitor for LM6100 ethernet PHY XTAL


I want to use a crystal with the following characteristics for the LM3s6100 PHY

Load capacitance = 18pF

Freq tolerance       = 50ppm

Freq                         = 25Mhz

Link                         =

The Datasheet of the MCU shows there is a load capacitance of  CL = 50 pF

By using the equation CL = (C + (CP+CI)) / 2   assuming Cp = 5pF

Therefore 50pF = (C + 55pF)/2       

Hence C = 10pF Approx?

Did I get it right ?Should I connect a 10pF capacitor to the external crystal?

Please let me know your views



3 Replies

  • Hansen,

    I'm not familiar with the equation you posted: CL = (C + (CP + CI)) / 2.  Essentially, we use a parallel oscillating circuit which can be modeled by the following equation:

    CL = ((CL1 * CL2)/(CL1 + CL2)) + Cs

    Where Cs is the stray capacitance inherent from the PCB and pin (usually, we assume this is about 3 pf).  

    We do have a few kits that make use of the PHY that I can recommend using as reference designs.  Specifically, the 6965 and 8962 have evaluation kits ( and that are specifically designed to exemplify the proper usage of the Ethernet capability.  The 2965 ( is also of the same device family as the 6100 you're using, so you could look at its evaluation kit for reference as well.  I should note, though, that the 2965 kit was not specifically designed to showcase the Ethernet, so I would refer to the 6965 and 8962 first.


  • In reply to Stellaris Jordan:

    Thanks Im gonna get my calculator out to calculate the values.One quick question is it necessary to connect all the supply pins to VCC 3.3 volts and similarly ground?

    Or is it meant for decoupling purpose?

    Since I'm not gonna decouple all the pins can I skip a few pins to save board space?



  • In reply to Hansen Dsouza:

    Hi again,

    According to the equation CL = ((CL1 * CL2)/(CL1 + CL2)) + Cs

    So if my crystal CL is x farads CL1 and CL2 capacitors connected to the crystal should be approximately double 2x each ?