Electronic Devices and Amplifier Circuits

Crystal OscillatorsC 1 » C 2.ElectrodeC 2LCrystalC 1ElectrodeFigure 10.8. Symbol and equivalent circuit for a crystalRThe impedance of the equivalent circuit of Figure 10.8 in the s – domainisZs ( ) = 1 ⁄ sC 1|| ( R + sL + 1 ⁄ sC 2 )We now recall that in a series resonant circuit the quality factor QωQ 0S L0S = -----------Rat resonance is(10.16)(10.17)and since the Q of a crystal oscillator is very high, the value of the resistance R in (10.17) must bevery low and thus it can be omitted in relation (10.16) which can now be expressed asor( sL + 1 ⁄ sCZs ( ) 1⁄ sC 1|| ( sL + 1 ⁄ sC 2 ) ( 1⁄sC 1 )2 )= = ⋅ --------------------------------------------------1⁄ sC 1 + sL + 1 ⁄ sC 2s 2 + 1⁄LCZs ( ) = ( 1⁄sC 1 ) ⋅ ---------------------------------------------------------------2s 2 + [( C 1 + C 2 ) ⁄ ( LC 1 C 2 )](10.18)The denominator of (10.18) is a quadratic and it implies the presence of two resonant frequencieswhich can be found by inspection of the equivalent circuit of Figure 10.10. The resonance of theseries branch occurs when the imaginary part of the impedance is equal to zero. Thus, lettings = jω we obtain Zjω ( ) = jωL + 1⁄jωC 2 = 0 and denoting this frequency as ω 0S we obtain1ω 0S = --------------LC 2(10.19)We also can prove * that the resonance of the parallel combination occurs whenCω + 1 C 20P = -----------------LC 1 C 2(10.20)As stated above,(10.19).C 1 » C 2and under this condition relation (10.20) reduces to that of relation* The proof is left as an exercise for the reader at the end of this chapter.Electronic Devices and Amplifier Circuits with MATLAB® / Simulink® / SimElectronics® Examples, Third EditionCopyright © Orchard Publications10−9