THE QUARTZ CRYSTAL AND ITS EQUIVALENT CIRCUIT From the following equivalent definition of the coupling coefficient, [latex] k^{2}=\frac{\text { Mechanical energy stored }}{\text { Total energy stored }} [/latex] show that [latex] k^{2}=1-\frac{f_{s}^{2}}{f_{a}^{2}} [/latex] Given that typically for an X-cut quartz crystal, [latex]k[/latex] = 0.1, what is [latex]f_{a}[/latex] for [latex]f_{s}[/latex] = 1 MHz? What is your conclusion?

[latex]C[/latex] represents the mechanical mass where the mechanical energy is stored, whereas [latex]C_{o}[/latex] is where the electrical energy is stored. If [latex]V[/latex] is the applied voltage, then [latex] k^{2}=\frac{\text { Mechanical energy stored }}{\text { Total energy stored }}=\frac{\frac{1}{2} C V^{2}}{\frac{1}{2} C V^{2}+\frac{1}{2} C_{o} V^{2}}=\frac{C}{C+C_{o}}=1-\frac{f_{s}^{2}}{f_{a}^{2}} [/latex] Rearranging this equation, we find [latex] f_{a}=\frac{f_{s}}{\sqrt{1-k^{2}}}=\frac{1 MHz }{\sqrt{1-(0.1)^{2}}}=1.005 MHz [/latex] Thus, [latex]f_{a} − f_{s}[/latex] is only 5 kHz. The two frequencies [latex]f_{s}[/latex] and [latex]f_{a}[/latex] in Figure 7.45d are very close. An oscillator designed to oscillate at [latex]f_{s}[/latex], that is, at 1 MHz, therefore, cannot drift far (for example, a few kHz) because that would change the reactance enormously, which would upset the oscillation conditions.

Question 7.15: THE QUARTZ CRYSTAL AND ITS EQUIVALENT CIRCUIT From the follo...

Principles of Electronic Materials and Devices

THE QUARTZ CRYSTAL AND ITS EQUIVALENT CIRCUIT From the following equivalent definition of the coupling coefficient,

$k^{2}=\frac{\text { Mechanical energy stored }}{\text { Total energy stored }}$

show that

$k^{2}=1-\frac{f_{s}^{2}}{f_{a}^{2}}$

Given that typically for an X-cut quartz crystal, $k$ = 0.1, what is $f_{a}$ for $f_{s}$ = 1 MHz? What is your conclusion?

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

$C$ represents the mechanical mass where the mechanical energy is stored, whereas $C_{o}$ is where the electrical energy is stored. If $V$ is the applied voltage, then

$k^{2}=\frac{\text { Mechanical energy stored }}{\text { Total energy stored }}=\frac{\frac{1}{2} C V^{2}}{\frac{1}{2} C V^{2}+\frac{1}{2} C_{o} V^{2}}=\frac{C}{C+C_{o}}=1-\frac{f_{s}^{2}}{f_{a}^{2}}$

Rearranging this equation, we find

$f_{a}=\frac{f_{s}}{\sqrt{1-k^{2}}}=\frac{1 MHz }{\sqrt{1-(0.1)^{2}}}=1.005 MHz$

Thus, $f_{a} − f_{s}$ is only 5 kHz. The two frequencies $f_{s}$ and $f_{a}$ in Figure 7.45d are very close. An oscillator designed to oscillate at $f_{s}$ , that is, at 1 MHz, therefore, cannot drift far (for example, a few kHz) because that would change the reactance enormously, which would upset the oscillation conditions.