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73M1903-IVT/F 参数 Datasheet PDF下载

73M1903-IVT/F图片预览
型号: 73M1903-IVT/F
PDF下载: 下载PDF文件 查看货源
内容描述: 调制解调器模拟前端 [Modem Analog Front End]
分类和应用: 调制解调器消费电路商用集成电路光电二极管
文件页数/大小: 46 页 / 530 K
品牌: TERIDIAN [ TERIDIAN SEMICONDUCTOR CORPORATION ]
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73M1903 Data Sheet  
DS_1903_032  
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:  
2 2304 2.4kHz 8/7 4  
.
Fvco / Fxtal  
=
24.576MHz  
After a few rounds of simplification this ratio reduces to:  
4
18  
1
Fvco/ Fxtal = (  
) (  
)
35  
Nnco1  
4
Dnco1  
Nnco2  
35  
1
=
=
Dnco2  
18  
, 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.  
44  
Rev. 2.0  
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