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HMC832LP6GETR bảng dữ liệu(PDF) 27 Page - Analog Devices |
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HMC832LP6GETR bảng dữ liệu(HTML) 27 Page - Analog Devices |
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27 / 49 page ![]() HMC832 Data Sheet translates to a 1.2 Hz error. The exact frequency mode of the HMC832 addresses this issue and can eliminate quantization error by programming the channel step size to fPD/10 in Register 0x0C to 10 (in this example). More generally, this feature can be used whenever the desired frequency, fVCO, can be exactly represented on a step plan where there are an integer number of steps (<214) across integer-N boundaries. Mathematically, this situation is satisfied if ≥ = = 14 1 2 and ) , ( where 0 ) mod( PD GCD PD VCO GCD GCD VCOk f f f f GCD f f f (16) where: GCD means greatest common divisor. fPD = frequency of the phase detector. fVCOk is the channel step frequency where 0 < k < 224−1, as shown in Figure 46. Some fractional PLLs are able to achieve these exact frequencies by adjusting (shortening) the length of the phase accumulator (the denominator or the modulus of the Δ-Σ modulator) so that the Δ-Σ modulator phase accumulator repeats at an exact period related to the interval frequency (fVCOk − fVCO(k−1)) in Figure 46. Consequently, the shortened accumulator results in more frequent repeating patterns and as a result often leads to spurious emissions at multiples of the repeating pattern period, or at harmonic frequencies of fVCOk − fVCO(k−1). For example, in some applications, these intervals might represent the spacing between radio channels, with the spurious occurring at multiples of the channel spacing. In comparison, the Analog Devices method is able to generate exact frequencies between adjacent integer-N boundaries while still using the full 24-bit phase accumulator modulus, thus achieving exact frequency steps with a high phase detector comparison rate, which allows Analog Devices PLLs to maintain excellent phase noise and spurious performance in the exact frequency mode. Figure 46. Exact Frequency Tuning INTEGER BOUNDARY INTEGER BOUNDARY fN + 1 – fN = fPD fN fVCO1 fVCO2 fVCO3 fVCO = fVCO2 fVCO4 fN + 1 fVCO14 fVCO14 – 1 fVCO14 – 2 Rev. A | Page 26 of 48 |
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