參數(shù)資料
型號: 73M1903-EVM-ETSI
廠商: Maxim Integrated Products
文件頁數(shù): 41/47頁
文件大?。?/td> 0K
描述: BOARD DEMO 73M1903 ETSI 203IMPED
產(chǎn)品培訓(xùn)模塊: Lead (SnPb) Finish for COTS
Obsolescence Mitigation Program
標(biāo)準(zhǔn)包裝: 1
主要目的: 接口,模擬前端(AFE)
已用 IC / 零件: *
主要屬性: *
次要屬性: *
已供物品: *
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
Fxtal
Fvco
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
(
/
=
=
Fxtal
Fvco
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.
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