參數(shù)資料
型號： 73M1903-IVTR/F
廠商： Maxim Integrated Products
文件頁數(shù)： 40/47頁
文件大小： 0K
描述： IC MODEM AFE V.22BIS 20-TSSOP
產(chǎn)品培訓模塊： Lead (SnPb) Finish for COTS
Obsolescence Mitigation Program
標準包裝： 2,000
通道數(shù)： 2
電壓 - 電源，模擬： 3 V ~ 3.6 V
電壓 - 電源，數(shù)字： 3 V ~ 3.6 V
封裝/外殼： 20-TSSOP（0.173"，4.40mm 寬）
供應商設備封裝： 20-TSSOP
包裝： 帶卷 (TR)
DS_1903_032
73M1903 Data Sheet
Rev. 2.1
45
where Nnco1 and Nnco2 must be < or equal to 8.
The ratio, Nnco1/Dnco1 = 1/7, is used to form a divide ratio for the NCO in prescaler and Nnco2/Dnco2 =
1/18 for the NCO in the PLL.
Prescaler NCO: From Nnco1/Dnco1 = 1/7,
Pdvsr = Integer [ Dnco1/Nnco1 ] = 7;
Prst[2:0] = Nnco1 – 1 = 0; this means NO fractional divide. It always does ÷7. Thus Pseq becomes
“don’t care” and is ignored.
Pseq = {x,x,x,x,x,x,x,x} = xxh.
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” and is ignored.
Nseq = {x,x,x,x,x,x,x,x} = xxh.
Example 2:
Crystal Frequency = 24.576 MHz; Desired Sampling Rate, Fs = 10.971 kHz=2.4 kHz x 8/7 x4
Step 1. First compute the required VCO frequency, Fvco, corresponding to
Fs = 2.4 kHz x 8/7 x 4 =10.971 kHz.
Fvco = 2 x 2304 x Fs = 2 x 2304 x 2.4 kHz x 8/7 x 4 = 50.55634 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
576
.
24
4
8/7
2.4kHz
2304
2
/
=
.
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.
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