67 lines
1.7 KiB
Markdown
67 lines
1.7 KiB
Markdown
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---
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author: Akbar Rahman
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date: \today
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title: MMME2051 // Piezoelectrics
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tags: [ piezoelectrics, op_amps ]
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uuid: ed7d0899-478d-4f0d-b0e9-634cdbb5b48a
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lecture_slides: [ ./lecture_slides/MMME2051EMD_Lecture7.pdf ]
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lecture_notes: []
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exercise_sheets: [ ./seminar_worksheets/MMME2051_Lec7_Top1_Quiz.pdf, ./seminar_worksheets/StrainGaugeHomework.pdf ]
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---
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Piezoelectricity is the charge that gets accumulated in some materials upon application of
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mechanical stress
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$$Q \propto F$$
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This relation allows the measurement of force using electric signals.
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\begin{align*}
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Q &\propto F \\
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Q &= k_1F \\
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&= k_1Ma \\
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\frac{\mathrm d Q}{\mathrm dt} &= i = k_1M\frac{\mathrm da}{\mathrm dt}
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\end{align*}
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# Integrating Amplifier
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Measuring current is expensive and difficult.
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Integrating the current helps to measure a voltage instead, which is easier.
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This is done using the following amplifier:
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\begin{align*}
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V_\text{out} &= A_{OL}(V^+-V__) = -A_{OL}V__ \\
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V__ &= V_\text{out} - V_C
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\end{align*}
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As input resistance of op amp is infinite:
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$$i_f = -i_n = -k_1M\frac{\mathrm da}{\mathrm dt}$$
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From the capacitor equation:
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$$i_f = C_f \frac{\mathrm dV_C}{\mathrm dt} = -k_1M\frac{\mathrm da}{\mathrm dt}$$
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Integrating both sides gives
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$$V_C = -\frac{k_1M}{C_f}a$$
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And it can be found that
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\begin{align*}
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V_\text{out} &= -A_{OL}(V_\text{out} - V_C) \\
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V_C &= -V\text{out} \frac{1+A_\text{OL}}{A_{OL}}
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\end{align*}
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To get
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$$V_\text{out} = \frac{k_1M}{C_f}a$$
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This circuit can be stacked to get velocity and displacement:
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