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參數資料
| 型號： | 73M1903C-IMR/F |
| 廠商： | MAXIM INTEGRATED PRODUCTS INC |
| 元件分類： | 消費家電 |
| 英文描述： | SPECIALTY CONSUMER CIRCUIT, PQCC32 |
| 封裝： | LEAD FREE, QFN-32 |
| 文件頁數： | 41/48頁 |
| 文件大小： | 487K |
| 代理商： | 73M1903C-IMR/F |
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73M1903C Data Sheet
DS_1903C_033
46
Rev. 5.0
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.
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. For this particular application, they are
found to be the best set of choices within the allowed constraints of Prst and Nrst. (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.
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