Datasheet

73M1903 Data Sheet DS_1903_032
46 Rev. 2.1
Example 3:
Crystal Frequency = 27 MHz; Desired Sampling Rate, Fs = 7.2 kHz
Step 1. First compute the required VCO frequency, Fvco, corresponding to
Fs = 2.4 kHz x 3 = 7.2 kHz.
Fvco = 2 x 2304 x Fs = 2 x 2304 x 2.4 kHz x 3 = 33.1776 MHz.
Step 2. Express the required VCO frequency divided by the Crystal Frequency as a ratio of two integers.
This is initially given by:
MHz
FxtalFvco
27
3 2.4kHz 2304 2
/
•••
=
.
After a few rounds of simplification this reduces to:
96
5
125
8
Dnco2
Nnco2
Dnco1
Nnco1
)
5
96
()
125
8
(/
==
•=FxtalFvco
The two ratios are not unique and many other possibilities exist. But for this particular application, they
are found to be the best set of choices within the constraints of Prst and Nrst allowed. (Nnco1, Nnco2
must be less than or equal to 8.)
The ratio, Nnco1/Dnco1 = 8/125, is used to form a divide ratio for the NCO in prescaler and Nnco2/Dnco2
=5/96 for the NCO in the PLL.
Pre-scaler NCO: From Nnco1/Dnco1 = 8/125,
Pdvsr = Integer [ Dnco1/Nnco1 ] = 15;
Prst[2:0] = Nnco1 – 1 = 7;
Dnco1/Nnco1 = 125/8 = 15.625 suggests a divide sequence of {÷16,÷16,÷15,÷16,÷16,÷15,÷16,÷15}, or
Pseq = {1,1,0,1,1,0,1,0} = DAh.
PLL NCO: From Nnco2/Dnco2 = 5/96,
Ndvsr = Integer [ Dnco2/Nnco2 ] = 19;
Nrst[2:0] = Nnco2 – 1 = 4;
Dnco2/Nnco2 = 19.2 suggests a divide sequence of {÷19, ÷19, ÷19, ÷19, ÷20}, or
Nseq = {x,x,x,0,0,0,0,1} = x1h.