Test Your EQ (Engineering Quotient)

EQ #71

An engineer wanted to build an 8-MHz filter with a very narrow bandwidth, so he used a crystal lattice filter like what is shown in Figure 1:

Figure 1

However, when he built and tested his filter, he discovered that while it worked fine around 8 MHz, the attenuation at very high frequencies (e.g., more than 80 MHz) was very much reduced. What caused this?

The equivalent circuit for a quartz crystal is something like this:

The components across the bottom represent the crystal’s mechanical resonance, while the capacitor at the top represents the capacitance of the electrodes and holder. Typical values are:

• CSER: 10s of fF (yes, femtofarads, 10–15 F)
• L: 10 s of mH
• R: 10 s of ohms
• CPAR: 10 s of pF

The crystal has a series-resonant frequency based on just CSER and L. It has relatively low impedance (basically just R) at this frequency.

It also has a parallel-resonant (sometimes called “antiresonant”) frequency when you consider the entire loop, including CPAR. Since CSER and CPAR are essentially in series, together they have a slightly lower capacitance than CSER alone, so the parallel-resonant frequency is slightly higher. The crystal’s impedance is very high at this frequency.

But at frequencies much higher than either of the resonant frequencies, you can see that the impedance of CPAR alone dominates and it continues decreasing with increasing frequency. This reduces the crystal lattice filter to a simple capacitive divider, which passes high frequencies with little attenuation.

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EQ #71

by Circuit Cellar Staff time to read: 1 min