Example
The parameters of equivalent circuit of the crystal are given as follows:
L = 0.4 H, CS = 0.06 pF, R = 5 kΩ, Cm = 1.0 pF.
Determine series and parallel resonant frequencies of crystal.
Solution:
With reference to figure 2, the admittance of crystal Y can be given by
where, 
and 
The resonant frequencies are obtained by putting B = 0. Therefore,
Consider the term CS R2 / LS = CS R / [L R]. In the crystal, time constant (L / R) is very much greater than CS R. Thus the ratio is much less than 1. For the values given, this ratio is of the order of10-6. Neglecting this term in the comparison with 2, we get 2 roots as
here, ωs and ωp are the series and parallel resonant frequencies respectively. By substituting values, we get
ωs = 6.45 M Hz. and ωp = 6.64 MHz
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