| Electronic Components Datasheet Search |
|
AD9545 Datasheet(PDF) 125 Page - Analog Devices |
|
|
|||||||||||||||||||||||||||||
AD9545 Datasheet(HTML) 125 Page - Analog Devices |
|
125 / 157 page ![]() Data Sheet AD9545 Rev. A | Page 125 of 157 SYSTEM CLOCK COMPENSATION SYSTEM CLOCK COMPENSATION OVERVIEW The NCOs and the TDCs of the AD9545 derive their timekeeping from the system clock (see the System Clock PLL section). Therefore, the frequency accuracy of any of the NCOs relates directly to the accuracy of the system clock. Likewise, an inferred frequency based on the difference between successive TDC time stamps is subject to the accuracy of the system clock. Therefore, the stability of the system clock is crucial to the accuracy of the NCOs and TDCs within the AD9545. The stability of the system clock, in turn, relates directly to the stability of system clock source. The system clock source is the frequency source driving the XOA and XOB pins. As such, an ideally stable system clock source is desirable. In practice, however, the system clock source suffers from frequency instability caused by aging, variations in temperature, and similar physical factors. Any frequency instability introduced by the system clock source translates to a frequency instability in the NCOs and TDCs. Because the NCOs and TDCs are fundamentally numeric (digital) in nature, it is possible to tune the NCOs and TDCs numerically to counteract the system clock instability. That is, with a known frequency error associated with the system clock source, the user can apply a corresponding correction (numerically) to the NCOs and TDCs, which is the underlying concept of system clock compensation. The system clock compensation operates on fractional frequency error (FFE) rather than absolute frequency error. For a given nominal frequency, f0, the FFE of a deviated frequency, f, is FFE = (f − fO)/fO or FFE = f/fO − 1 In the case of TDCs, the difference between successive time stamps is the period of the underlying frequency. The difference between successive time stampes gives rise to the concept of fractional period error (FPE): FPE = (p − p0 )/p0 or FPE = p/p0 − 1 where: p0 is the nominal period. p is the deviated period. Because frequency relates to period as f = 1/p, FFE and FPE relate as follows: FFE = −FPE/(FPE + 1) or FPE = −FFE/(FFE + 1) In the context of system clock compensation, consider a given system clock frequency error expressed in terms of FFE. Applying an FFE correction factor to the NCOs and TDCs compensates for the FFE error of the system clock source. FFE as a correction factor to the NCOs and TDCs compensates for the FFE of system clock source. The AD9545 has the option of applying system clock compensation via two methods. • Open-loop method • Closed-loop method Open-Loop Method Figure 97 shows the open-loop method, with the system clock source driving the system clock PLL, which in turn serves as the clock source for a representative NCO within the AD9545. Although Figure 97 shows an NCO, this section also applies to a TDC. fNCO XOA XOB fSRC fS NCO AD9545 SYSTEM CLOCK PLL COMPENSATION CALCULATOR FTW 1 + FFECOMP CORRECTED FTW SYSTEM CLOCK SOURCE Figure 97. Open-Loop Method The system clock source provides the primary frequency, fSRC, with a nominal value of f0. The system clock PLL, which is the clock source to the NCO, multiplies fSRC by a constant, KPLL (that is, fS = KPLL × fSRC). Assuming FFECOMP = 0 in Figure 97, the NCO produces an output frequency, fNCO, proportional to the applied numeric frequency tuning word, FTW (that is fNCO = fS × FTW × KNCO, where KNCO is the proportionality constant). Therefore, express fNCO in terms of fSRC as fNCO = fSRC × KPLL × FTW × KNCO The ideal (error free) fNCO is then fNCO_IDEAL = f0 × KPLL × FTW × KNCO where f0 is the nominal frequency of the system clock source. If the system clock source experiences a fractional error, FFE, fNCO = f0 × (1 + FFE) × KPLL × FTW × KNCO Apply a fractional correction, FFECOMP, to compensate for FFE as follows: fNCO = f0 × (1 + FFE) × (1 + FFECOMP) × KPLL × FTW × KNCO If FFECOMP = –FFE, fNCO = f0 × (1 + FFE) × (1 – FFE) × KPLL × FTW × KNCO (23) |
|
Link URL |
| Does ALLDATASHEET help your business so far? [ DONATE ] |
About Alldatasheet | Advertisement | Contact us | Privacy Policy | Link to Datasheet | Link Exchange | Manufacturer List All Rights Reserved©Alldatasheet.com |
| Russian : Alldatasheetru.com | Korean : Alldatasheet.co.kr | Spanish : Alldatasheet.es | French : Alldatasheet.fr | Italian : Alldatasheetit.com Portuguese : Alldatasheetpt.com | Polish : Alldatasheet.pl | Vietnamese : Alldatasheet.vn Indian : Alldatasheet.in | Mexican : Alldatasheet.com.mx | British : Alldatasheet.co.uk | New Zealand : Alldatasheet.co.nz |
|
Family Site : ic2ic.com |
icmetro.com |