CDCE6214: Expected Output Frequency (FOUT)

Hello,

Are you referring to table 8?

In the description "PSA = 4, IOD = 25" -- And the equation is "FVCO = PSA × IOD × FOUT."
So the VCO frequency is 4 * 25 * 24.756 MHz = 2457.6 MHz.
And the output frequency can be solved as = FVCO / PSA / IOD.

The VCO frequency can be calculated from the reference frequency using:
VCO = xxxREF / PLL_R * (PLL_N + PLL_num / PLL_den)

If the doubler is used then:
VCO = xxxREF * 2 * (PLL_N + PLL_num / PLL_den)

In this example they don't give us the (PRI|SEC)REF input frequency.  But they state their phase detector frequency is 25 MHz.  25 MHz could be the reference frequency with PLL R = 1.

VCO = 25 MHz / 1 * (98 + 152000 / 500000)

Hopefully this gives you an idea of the equations used for the PLL.

73,
Timothy