Question 7.16: QUARTZ CRYSTAL AND ITS INDUCTANCE A typical 1 MHz quartz cry...
QUARTZ CRYSTAL AND ITS INDUCTANCE A typical 1 MHz quartz crystal has the following properties:
f_{s} = 1 MHz f_{a} = 1.0025 MHz C_{o} = 5 pF R = 20 Ω
What are C and L in the equivalent circuit of the crystal? What is the quality factor Q of the crystal, given that
Q=\frac{1}{2 \pi f_{s} R C}
The expression for f_{s} is
f_{s}=\frac{1}{2 \pi \sqrt{L C}}
From the expression for f_{a}, we have
f_{a}=\frac{1}{2 \pi \sqrt{L C^{\prime}}}=\frac{1}{2 \pi \sqrt{\frac{C C_{o}}{C+C_{o}}}}
Dividing f_{a} by f_{s} eliminates L, and we get
\frac{f_{a}}{f_{s}}=\sqrt{\frac{C+C_{o}}{C_{o}}}
so that C is
C=C_{o}\left[\left(\frac{f_{a}}{f_{s}}\right)^{2}-1\right]=(5 pF )\left(1.0025^{2}-1\right)=0.025 pF
Thus
L=\frac{1}{C\left(2 \pi f_{s}\right)^{2}}=\frac{1}{\left(0.025 \times 10^{-12}\right)\left(2 \pi 10^{6}\right)^{2}}=1.01 H
This is a substantial inductance, and the enormous increase in the inductive reactance above f_{s} is intuitively apparent. The quality factor
Q=\frac{1}{2 \pi f_{s} R C}=3.18 \times 10^{5}
is very large.