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AD9552PCBZ Datasheet(PDF) 17 Page - Analog Devices |
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AD9552PCBZ Datasheet(HTML) 17 Page - Analog Devices |
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17 / 32 page ![]() AD9552 Rev. C | Page 17 of 32 PART INITIALIZATION AND AUTOMATIC POWER- ON RESET The AD9552 has an internal power-on reset circuit. At power-up, internal logic relies on the internal reference monitor to select either the crystal oscillator or the reference input and then initiates VCO calibration using whichever is found. If both are present, the external reference path is chosen. VCO calibration is required in order for the device to lock. If the input reference signal is not present, VCO calibration waits until a valid input reference is present. As soon as an input reference signal is present, VCO calibration starts. If the user wishes to use the crystal oscillator input even if the reference input is present, the user needs to set Bit 0 (use crystal resonator) in Register 0x1D. Any change to the preset frequency selection pins or the PLL divide ratios requires the user to recalibrate the VCO. OUTPUT/INPUT FREQUENCY RELATIONSHIP The frequency at OUT1 and OUT2 is a function of the PLL feedback divider values (N, FRAC, and MOD) and the output divider values (P0 and P1). The equations that define the frequency at OUT1 and OUT2 (fOUT1 and fOUT2, respectively) are as follows. ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + × = 1 0 1 P P N K f f MOD FRAC REF OUT fOUT2 = fOUT1 where: fREF is the input reference or crystal resonator frequency. K is the input mode scale factor. N is the integer feedback divider value. FRAC and MOD are the fractional feedback divider values. P0 and P1 are the OUT1 divider values. The numerator of the fOUT1 equation contains the feedback division factor, which has an integer part (N) due to an integer divider along with an optional fractional part (FRAC/MOD) associated with the feedback SDM. The following constraints apply: { } 47 , 36 ∈ MIN N { } 255 , , 1 , L + ∈ MIN MIN N N N { } 575 , 048 , 1 , , 1 , 0 L ∈ FRAC { } 575 , 048 , 1 , , 2 , 1 L ∈ MOD { }2 , 1 ∈ K { } 11 , , 5 , 4 0 L ∈ P { } 63 , , 2 , 1 1 L ∈ P Note that NMIN and K can each be one of two values. The value of NMIN depends on the state of the SDM. NMIN = 36 when the SDM is disabled or NMIN = 47 when it is enabled. The value of K depends on the 2× frequency multiplier. K = 1 when the 2× frequency multiplier is bypassed, or K = 2 when it is enabled. The frequency at the input to the PFD (fPFD) is calculated as follows: fPFD = K × fREF The operating range of the VCO (3.35 GHz ≤ fVCO ≤ 4.05 GHz) places the following constraint on fPFD: MHz 4050 MHz 3350 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ≤ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + MOD FRAC PFD MOD FRAC N f N CALCULATING DIVIDER VALUES This section provides a three-step procedure for calculating the divider values when given a specific fOUT1/fREF ratio (fREF is the frequency of either the REF input signal source or the external crystal resonator). The computation process is described in general terms, but a specific example is provided for clarity. The example is based on a frequency control pin setting of A[2:0] = 111 (see Table 9) and Y[5:0] = 101000 (see Table 10), yielding the following: fREF = 26 MHz fOUT1 = 625 × (66/64) MHz 1. Determine the output divide factor (ODF). Note that the VCO frequency (fVCO) spans 3350 MHz to 4050 MHz. The ratio, fVCO/fOUT1, indicates the required ODF. Given the specified value of fOUT1 (~644.53 MHz) and the range of fVCO, the ODF spans a range of 5.2 to 6.3. The ODF must be an integer, which means that ODF = 6 (because 6 is the only integer between 5.2 and 6.3). 2. Determine suitable values for P0 and P1. The ODF is the product of the two output dividers, so ODF = P0P1. It has already been determined that ODF = 6 for the given example. Therefore, P0P1 = 6 with the constraints that P0 and P1 are both integers and that 4 ≤ P0 ≤ 11 (see the Output/Input Frequency Relationship section). These constraints lead to the single solution: P0 = 6 and P1 = 1. Although this particular example yields a single solution for the output divider values with fOUT1 ≈ 644.53 MHz, some fOUT1 frequencies result in multiple ODFs rather than just one. For example, if fOUT1 = 100 MHz the ODF ranges from 34 to 40. This leads to an assortment of possible values for P0 and P1, as shown in Table 12. Table 12. Combinations for P0 and P1 P0 P1 ODF (P0 × P1) 4 9 36 4 10 40 5 7 35 5 8 40 6 6 36 7 5 35 8 5 40 9 4 36 10 4 40 |
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