參數(shù)資料
型號: 73M1903C-IM
廠商: MAXIM INTEGRATED PRODUCTS INC
元件分類: 消費(fèi)家電
英文描述: SPECIALTY CONSUMER CIRCUIT, QCC32
封裝: LEAD FREE, QFN-32
文件頁數(shù): 38/45頁
文件大?。?/td> 459K
代理商: 73M1903C-IM
73M1903C
Modem Analog Front End
DATA SHEET
Page: 43 of 45
2005 TERIDIAN Semiconductor Corporation
Rev 3.4
After a few rounds of simplification this ratio reduces to:
18
1
35
4
Dnco2
Nnco2
Dnco1
Nnco1
)
1
18
(
)
35
4
(
/
=
=
Fxtal
Fvco
, where Nnco1 and Nnco2 must be < or equal to 8.
The ratio, Nnco1/Dnco1 = 4/35, is used to form a divide ratio for the NCO in pre-scaler and Nnco2/Dnco2
=1/18 for the NCO in the PLL.
Pre-scaler NCO: From Nnco1/Dnco1 = 4/35,
Pdvsr = Integer [ Dnco1/Nnco1 ] = 8;
Prst[2:0] = Nnco1 – 1 = 3;
Dnco1/Nnco1 = 35/4 = 8.75 suggests a divide sequence of {
÷9,÷9,÷9,÷8}, or
Pseq = {x,x,x,x,1,1,1,0} = xDh.
PLL NCO: From Nnco2/Dnco2 = 1/18,
Ndvsr = Integer [ Dnco2/Nnco2 ] = 18;
Nrst[2:0] = Nnco2 – 1 = 0; this means NO fractional divide. It always does ÷18. Thus Pseq becomes
“don’t care”.
Nseq = {x,x,x,x,x,x,x,x} = xxh.
Example3:
Crystal Frequency = 27MHz; Desired Sampling Rate, Fs = 7.2kHz
Step 1. First compute the required VCO frequency, Fvco, corresponding to
Fs = 2.4kHz x 3 = 7.2kHz.
Fvco = 2 x 2304 x Fs = 2 x 2304 x 2.4kHz x 3 = 33.1776MHz.
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.)
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