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PLL-Prob&SolsPLL Problems and Solutions (9/6/03) PLL PROBLEMS AND SOLUTIONS
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Problem 1 Rs = 5.3 Cs = 30fF Solve for and evaluate the series and parallel resonance Ls = 8.4mH frequencies of the crystal whose model is shown. It is Cp = 6pF suggested to make appropriate assumptions as the exact Crystal SU03H01P1 frequencies are difficult to achieve. Solution Solving the exact frequencies for this problem is very challenging. It is better to assume that series resonance (minimum impedance) will occur approximately when the impedance of Cs cancels the impedance of Ls. This gives series resonance as 1 ω s2 = L C s s → fs = 1 ≈ 10.026MHz 2 π L sC s
The parallel resonance can be approximated by assuming that it will occur close to the frequency when the impedance of the series branch equals the negati
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