Let's solve each part step by step: --- ## **2) Charge Amplifier Output Voltage Calculation** ### **Given:** - MEMS capacitance (\(C_{MEMS}\)): - Nominal (\(C_{nom}\)) = 3 pF - Minimum (\(C_{min}\)) = 2 pF - Maximum (\(C_{max}\)) = 5 pF - Input Voltage (\(V_{in}\)) = 10 V - Feedback Capacitance (\(C_2\)) = 10 pF ### **Charge Amplifier Output Equation:** The output voltage of a charge amplifier at the end of the \(\phi_2\) cycle is: \[ V_{out} = \frac{C_{MEMS}}{C_2} \cdot V_{in} \] We'll calculate \(V_{out}\) for each value of \(C_{MEMS}\): --- #### **a) For \(C_{nom} = 3~\text{pF}\):** \[ V_{out,nom} = \frac{3~\text{pF}}{10~\text{pF}} \times 10~\text{V} = \frac{3}{10} \times 10 = 3~\text{V} \] --- #### **b) For \(C_{min} = 2~\text{pF}\):** \[ V_{out,min} = \frac{2~\text{pF}}{10~\text{pF}} \times 10~\text{V} = \frac{2}{10} \times 10 = 2~\text{V} \] --- #### **c) For \(C_{max} = 5~\text{pF}\):** \[ V_{out,max} = \frac{5~\text{pF}}{10~\text{pF}} \times 10~\text{V} = \frac{5}{10} \times 10 = 5~\text{V} \] --- ### **Summary Table for Output Voltage:** | Capacitance | Output Voltage (\(V_{out}\)) | |---------------------|------------------------------| | \(C_{min} = 2\) pF | 2 V | | \(C_{nom} = 3\) pF | 3 V | | \(C_{max} = 5\) pF | 5 V | --- ## **3) CMOS Relaxation Oscillator Frequency** ### **Given:** - Supply Voltage (\(V_{DD}\)) = 5 V - Resistors (\(R_1 = R_2 = R\)) = 100 k\(\Omega\) - Capacitance (\(C_{MEMS}\)): use \(C_{min}\), \(C_{nom}\), \(C_{max}\) ### **Frequency of CMOS Relaxation Oscillator:** The period (\(T\)) of a CMOS relaxation oscillator is: \[ T = 2R C \ln(2) \] So, the frequency (\(f\)) is: \[ f = \frac{1}{T} = \frac{1}{2RC \ln(2)} \] Plug in the values: - \(R = 100~\text{k}\Omega = 100 \times 10^3~\Omega\) - \(\ln(2) \approx 0.693\) \[ f = \frac{1}{2 \times 100 \times 10^3 \times C \times 0.693} \] \[ f = \frac{1}{138600~C} \] Now plug in each capacitance value: --- #### **a) \(C_{min} = 2~\text{pF} = 2 \times 10^{-12}~\text{F}\):** \[ f_{min} = \frac{1}{138600 \times 2 \times 10^{-12}} = \frac{1}{277200 \times 10^{-12}} = \frac{1}{2.772 \times 10^{-7}} = 3.61 \times 10^6~\text{Hz} = 3.61~\text{MHz} \] --- #### **b) \(C_{nom} = 3~\text{pF} = 3 \times 10^{-12}~\text{F}\):** \[ f_{nom} = \frac{1}{138600 \times 3 \times 10^{-12}} = \frac{1}{415800 \times 10^{-12}} = \frac{1}{4.158 \times 10^{-7}} = 2.41 \times 10^6~\text{Hz} = 2.41~\text{MHz} \] --- #### **c) \(C_{max} = 5~\text{pF} = 5 \times 10^{-12}~\text{F}\):** \[ f_{max} = \frac{1}{138600 \times 5 \times 10^{-12}} = \frac{1}{693000 \times 10^{-12}} = \frac{1}{6.93 \times 10^{-7}} = 1.44 \times 10^6~\text{Hz} = 1.44~\text{MHz} \] --- ### **Summary Table for Oscillator Frequency:** | Capacitance | Frequency (\(f\)) | |---------------------|---------------------| | \(C_{min} = 2\) pF | 3.61 MHz | | \(C_{nom} = 3\) pF | 2.41 MHz | | \(C_{max} = 5\) pF | 1.44 MHz | --- **Let me know if you need any more explanation of the steps!**

Let's carefully re-solve each part of the problem, paying close attention to the details and ensuring correctness. --- ## **Part 2: Charge Amplifier Output Voltage Calculation** ### **Given Data:** - MEMS capacitance (\(C_{MEMS}\)): - Nominal (\(C_{nom}\)) = 3 pF - Minimum (\(C_{min}\)) = 2 pF - Maximum (\(C_{max}\)) = 5 pF - Input voltage (\(V_{in}\)) = 10 V - Feedback capacitance (\(C_2\)) = 10 pF ### **Understanding the circuit:** In a charge amplifier configuration, the **charge** stored on the MEMS capacitor during the sensing phase is: \[ Q = C_{MEMS} \times V_{in} \] This charge is transferred to the feedback capacitor \(C_2\), producing an output voltage: \[ V_{out} = \frac{Q}{C_2} = \frac{C_{MEMS} \times V_{in}}{C_2} \] --- ### **Step-by-step calculations:** #### **a) Nominal capacitance \(C_{nom} = 3\,\text{pF}\):** \[ V_{out,nom} = \frac{3\,\text{pF}}{10\,\text{pF}} \times 10\,\text{V} = \frac{3}{10} \times 10 = 3\,\text{V} \] --- #### **b) Minimum capacitance \(C_{min} = 2\,\text{pF}\):** \[ V_{out,min} = \frac{2\,\text{pF}}{10\,\text{pF}} \times 10\,\text{V} = \frac{2}{10} \times 10 = 2\,\text{V} \] --- #### **c) Maximum capacitance \(C_{max} = 5\,\text{pF}\):** \[ V_{out,max} = \frac{5\,\text{pF}}{10\,\text{pF}} \times 10\,\text{V} = \frac{5}{10} \times 10 = 5\,\text{V} \] --- ### **Summary:** | Capacitance | Output Voltage \(V_{out}\) | |--------------|---------------------------| | 2 pF | 2 V | | 3 pF | 3 V | | 5 pF | 5 V | **Note:** Since the charge amplifies the voltage proportionally to \(C_{MEMS}\), the output voltage increases with capacitance. --- ## **Part 3: CMOS Relaxation Oscillator Frequency Calculation** ### **Given Data:** - Supply voltage \(V_{DD} = 5\,\text{V}\) - Resistors \(R_1 = R_2 = R = 100\,k\Omega\) - Capacitance values as above: \(C_{min} = 2\,pF\), \(C_{nom} = 3\,pF\), \(C_{max} = 5\,pF\) ### **Oscillator period and frequency:** For a CMOS relaxation oscillator, the **period \(T\)** is approximately: \[ T = 2 R C \ln(2) \] where \(\ln(2) \approx 0.693\). The **frequency**: \[ f = \frac{1}{T} = \frac{1}{2 R C \ln(2)} \] --- ### **Calculations:** Convert \(R\) to ohms: \(R= 100\,k\Omega = 100 \times 10^3\,\Omega\). Plugging in the constants: \[ f = \frac{1}{2 \times 100 \times 10^3 \times C \times 0.693} = \frac{1}{138600 \times C} \] where \(C\) is in farads. --- ### **Compute for each capacitance:** #### **a) \(C_{min} = 2\,pF = 2 \times 10^{-12}\,F\):** \[ f_{min} = \frac{1}{138600 \times 2 \times 10^{-12}} = \frac{1}{277200 \times 10^{-12}} = \frac{1}{2.772 \times 10^{-7}} \] \[ f_{min} \approx 3.605 \times 10^{6}\,\text{Hz} = 3.605\,\text{MHz} \] --- #### **b) \(C_{nom} = 3\,pF = 3 \times 10^{-12}\,F\):** \[ f_{nom} = \frac{1}{138600 \times 3 \times 10^{-12}} = \frac{1}{415800 \times 10^{-12}} = \frac{1}{4.158 \times 10^{-7}} \] \[ f_{nom} \approx 2.405 \times 10^{6}\,\text{Hz} = 2.405\,\text{MHz} \] --- #### **c) \(C_{max} = 5\,pF = 5 \times 10^{-12}\,F\):** \[ f_{max} = \frac{1}{138600 \times 5 \times 10^{-12}} = \frac{1}{693000 \times 10^{-12}} = \frac{1}{6.93 \times 10^{-7}} \] \[ f_{max} \approx 1.445 \times 10^{6}\,\text{Hz} = 1.445\,\text{MHz} \] --- ## **Final Results:** | Capacitance | Oscillator Frequency \(f\) | |--------------|----------------------------| | 2 pF | **3.605 MHz** | | 3 pF | **2.405 MHz** | | 5 pF | **1.445 MHz** | --- **Summary:** - The charge amplifier output voltage scales directly with the MEMS capacitance. - The relaxation oscillator frequency inversely scales with the capacitance. Let me know if you'd like further clarifications!

Certainly! Let's carefully work through both problems step by step, ensuring clarity and correctness. --- ## **Problem 2: Charge Amplifier Output Voltage** ### **Given:** - MEMS capacitance \( C_x \): - Nominal \( C_{nom} = 3\,\text{pF} \) - Minimum \( C_{min} = 2\,\text{pF} \) - Maximum \( C_{max} = 5\,\text{pF} \) - Input voltage \( V_{in} = 10\,\text{V} \) - Feedback capacitor \( C_2 = 10\,\text{pF} \) ### **Understanding the circuit:** In a charge amplifier: - The MEMS capacitor \( C_x \) is charged during the sensing phase: \[ Q = C_x \times V_{in} \] - This charge is transferred to the feedback capacitor \( C_2 \), producing an output voltage: \[ V_{out} = -\frac{Q}{C_2} = -\frac{C_x V_{in}}{C_2} \] (The negative sign indicates inversion, which is typical for a charge amplifier; for magnitude, we focus on the absolute value unless specified otherwise.) --- ### **Calculations:** #### **a) Nominal capacitance \( C_{nom} = 3\,\text{pF} \):** \[ V_{out} = -\frac{3\,\text{pF} \times 10\,\text{V}}{10\,\text{pF}} = -\frac{30\,\text{pF}\cdot\text{V}}{10\,\text{pF}} = -3\,\text{V} \] #### **b) Minimum capacitance \( C_{min} = 2\,\text{pF} \):** \[ V_{out} = -\frac{2\,\text{pF} \times 10\,\text{V}}{10\,\text{pF}} = -2\,\text{V} \] #### **c) Maximum capacitance \( C_{max} = 5\,\text{pF} \):** \[ V_{out} = -\frac{5\,\text{pF} \times 10\,\text{V}}{10\,\text{pF}} = -5\,\text{V} \] --- ### **Final Results (magnitude):** | Capacitance | Output Voltage \( V_{out} \) | |--------------|------------------------------| | 2 pF | **-2.0 V** | | 3 pF | **-3.0 V** | | 5 pF | **-5.0 V** | *Note:* The negative sign indicates inversion; if only magnitude is needed, just state the absolute value. --- ## **Problem 3: CMOS Relaxation Oscillator Frequency** ### **Given:** - Resistance \( R = 100\,\text{k}\Omega = 100 \times 10^3\,\Omega \) - Capacitance \( C \): - \( C_{min} = 2\,\text{pF} = 2 \times 10^{-12}\,\text{F} \) - \( C_{nom} = 3\,\text{pF} = 3 \times 10^{-12}\,\text{F} \) - \( C_{max} = 5\,\text{pF} = 5 \times 10^{-12}\,\text{F} \) --- ### **Oscillator frequency formula:** A standard CMOS relaxation oscillator period: \[ T = 2 R C \ln 3 \] where: - \(\ln 3 \approx 1.0986\), - The frequency: \[ f = \frac{1}{T} = \frac{1}{2 R C \ln 3} \] --- ### **Calculations:** \[ f = \frac{1}{2 \times 100 \times 10^3\,\Omega \times C \times 1.0986} \] Simplify denominator: \[ \text{Denominator} = 2 \times 100,000 \times C \times 1.0986 = 200,000 \times C \times 1.0986 \] Calculate for each \( C \): --- ### **a) For \( C_{min} = 2\,\text{pF} = 2 \times 10^{-12}\,\text{F} \):** \[ T_{min} = 2 \times 100,000 \times 2 \times 10^{-12} \times 1.0986 \approx 2 \times 10^5 \times 2 \times 10^{-12} \times 1.0986 \] \[ T_{min} = (2 \times 10^5) \times (2 \times 10^{-12}) \times 1.0986 = (4 \times 10^{-7}) \times 1.0986 \approx 4.394 \times 10^{-7}\,\text{s} \] Frequency: \[ f_{max} = \frac{1}{T_{min}} \approx \frac{1}{4.394 \times 10^{-7}} \approx 2.276\,\text{MHz} \] --- ### **b) For \( C_{nom} = 3\,\text{pF} = 3 \times 10^{-12}\,\text{F} \):** \[ T_{nom} \approx 2 \times 10^5 \times 3 \times 10^{-12} \times 1.0986 = 6 \times 10^{-7} \times 1.0986 \approx 6.5916 \times 10^{-7}\,\text{s} \] Frequency: \[ f_{nom} \approx \frac{1}{6.592 \times 10^{-7}} \approx 1.517\,\text{MHz} \] --- ### **c) For \( C_{max} = 5\,\text{pF} = 5 \times 10^{-12}\,\text{F} \):** \[ T_{max} \approx 2 \times 10^5 \times 5 \times 10^{-12} \times 1.0986 = 1 \times 10^{-6} \times 1.0986 \approx 1.0986 \times 10^{-6}\,\text{s} \] Frequency: \[ f_{max} \approx \frac{1}{1.0986 \times 10^{-6}} \approx 0.91\,\text{MHz} \] --- ## **Final Summary:** | Capacitance | Oscillation Frequency \(f\) | |--------------|------------------------------| | 2 pF | **~2.28 MHz** | | 3 pF | **~1.52 MHz** | | 5 pF | **~0.91 MHz** | --- **Summary:** - The **charge amplifier's output voltage** scales linearly with capacitance, with negative inversion. - The **oscillator frequency** inversely scales with capacitance; larger capacitance yields lower frequency. Feel free to ask if you'd like further clarifications!

Certainly! Here's the detailed solution formatted with the equations exactly as you provided, step-by-step: --- ### **Problem 2 — Charge Amplifier Output** 1. **Charge on the MEMS capacitor:** \[ Q = C_x V_{in} \] 2. **Output voltage of the charge amplifier:** \[ V_{out} = - \frac{Q}{C_2} = - \frac{C_x V_{in}}{C_2} \] *(The minus sign indicates inversion; if only magnitude is required, consider the absolute value.)* 3. **Plugging in the given values:** \[ V_{in} = 10\,\text{V}, \quad C_2 = 10\,\text{pF} \] --- **For \( C_{min} = 2\,\text{pF} \):** \[ V_{out} = - \frac{2\,\text{pF} \times 10\,\text{V}}{10\,\text{pF}} = - 2\,\text{V} \] **For \( C_{nom} = 3\,\text{pF} \):** \[ V_{out} = - \frac{3\,\text{pF} \times 10\,\text{V}}{10\,\text{pF}} = - 3\,\text{V} \] **For \( C_{max} = 5\,\text{pF} \):** \[ V_{out} = - \frac{5\,\text{pF} \times 10\,\text{V}}{10\,\text{pF}} = - 5\,\text{V} \] --- ### **Answer:** \[ \boxed{ \begin{aligned} & \textbf{For } C_{min} = 2\,\text{pF}:\quad V_{out} = -2\,\text{V} \\ & \textbf{For } C_{nom} = 3\,\text{pF}:\quad V_{out} = -3\,\text{V} \\ & \textbf{For } C_{max} = 5\,\text{pF}:\quad V_{out} = -5\,\text{V} \end{aligned} } \] --- ### **Problem 3 — CMOS Relaxation Oscillator Frequency** The period \( T \) of a CMOS relaxation oscillator is: \[ T = 2 R C \ln(3) \] where: \[ \ln(3) \approx 1.0986 \] The frequency: \[ f = \frac{1}{T} = \frac{1}{2 R C \ln(3)} \] Given: \[ R = 100\,\text{k}\Omega = 100 \times 10^3\,\Omega \] --- **Calculations for each \( C \):** \[ f = \frac{1}{2 \times 100 \times 10^3 \times C \times 1.0986} \] \[ f \approx \frac{1}{220\,000 \times C} \] --- **For \( C_{min} = 2\,\text{pF} = 2 \times 10^{-12}\,\text{F} \):** \[ f_{max} = \frac{1}{220\,000 \times 2 \times 10^{-12}} = \frac{1}{440\,000 \times 10^{-12}} = \frac{1}{4.4 \times 10^{-7}} \approx 2.27\,\text{MHz} \] **For \( C_{nom} = 3\,\text{pF} \):** \[ f_{nom} = \frac{1}{220\,000 \times 3 \times 10^{-12}} = \frac{1}{660\,000 \times 10^{-12}} = \frac{1}{6.6 \times 10^{-7}} \approx 1.52\,\text{MHz} \] **For \( C_{max} = 5\,\text{pF} \):** \[ f_{min} = \frac{1}{220\,000 \times 5 \times 10^{-12}} = \frac{1}{1.1 \times 10^{-6}} \approx 0.91\,\text{MHz} \] --- ### **Final Results:** \[ \boxed{ \begin{aligned} & \textbf{For } C_{min} = 2\,\text{pF}:\quad f \approx 2.27\,\text{MHz} \\ & \textbf{For } C_{nom} = 3\,\text{pF}:\quad f \approx 1.52\,\text{MHz} \\ & \textbf{For } C_{max} = 5\,\text{pF}:\quad f \approx 0.91\,\text{MHz} \end{aligned} } \] --- Let me know if you'd like any further modifications!

Certainly! Here's the detailed solution written exactly with your provided equations, step-by-step, without any modifications: --- ### **Problem 2 — Charge Amplifier Output** 1. When the MEMS capacitor \( C_x \) is sampled with input voltage \( V_{in} \), the charge stored is: \[ Q = C_x V_{in} \] 2. This charge is transferred onto the feedback capacitor \( C_2 \) of the charge amplifier. The output voltage at the end of the \(\phi_2\) cycle is: \[ V_{out} = - \frac{Q}{C_2} = - \frac{C_x V_{in}}{C_2} \] *(The minus sign indicates inversion, as typical for charge amplifiers.)* 3. Plugging in the given values: \( V_{in} = 10\,\mathrm{V} \), \( C_2 = 10\,\mathrm{pF} \): - For \( C_{min} = 2\,\mathrm{pF} \): \[ V_{out} = - \frac{2\,\mathrm{pF} \times 10\,\mathrm{V}}{10\,\mathrm{pF}} = - 2\,\mathrm{V} \] - For \( C_{nom} = 3\,\mathrm{pF} \): \[ V_{out} = - \frac{3\,\mathrm{pF} \times 10\,\mathrm{V}}{10\,\mathrm{pF}} = - 3\,\mathrm{V} \] - For \( C_{max} = 5\,\mathrm{pF} \): \[ V_{out} = - \frac{5\,\mathrm{pF} \times 10\,\mathrm{V}}{10\,\mathrm{pF}} = - 5\,\mathrm{V} \] --- ### **Answer (2):** \[ \boxed{ \begin{aligned} & \text{For } C_{min} = 2\,\mathrm{pF}:\quad V_{out} = -2\,\mathrm{V} \\ & \text{For } C_{nom} = 3\,\mathrm{pF}:\quad V_{out} = -3\,\mathrm{V} \\ & \text{For } C_{max} = 5\,\mathrm{pF}:\quad V_{out} = -5\,\mathrm{V} \end{aligned} } \] --- ### **Problem 3 — CMOS Relaxation Oscillator Frequency** The period \( T \) of a CMOS relaxation oscillator is: \[ T = 2 R C \ln(3) \] where: \[ \ln(3) \approx 1.098612 \] and the frequency: \[ f = \frac{1}{T} = \frac{1}{2 R C \ln(3)} \] Given: \[ R = 100\,\mathrm{k}\Omega = 100 \times 10^{3}\,\Omega \] Calculating the denominator: \[ 2 R C \ln(3) = 2 \times 100 \times 10^{3} \times C \times 1.098612 \] \[ = 200 \times 10^{3} \times C \times 1.098612 \] \[ = 220,000 \times C \] with \( C \) in farads. --- ### **Calculations for each capacitance:** - For \( C_{min} = 2\,\mathrm{pF} = 2 \times 10^{-12}\,\mathrm{F} \): \[ T_{max} = 220,000 \times 2 \times 10^{-12} = 4.4 \times 10^{-7}\,\mathrm{s} \] \[ f_{max} = \frac{1}{T_{max}} = \frac{1}{4.4 \times 10^{-7}} \approx 2.27\,\mathrm{MHz} \] - For \( C_{nom} = 3\,\mathrm{pF} = 3 \times 10^{-12}\,\mathrm{F} \): \[ T_{nom} = 220,000 \times 3 \times 10^{-12} = 6.6 \times 10^{-7}\,\mathrm{s} \] \[ f_{nom} = \frac{1}{T_{nom}} \approx 1.52\,\mathrm{MHz} \] - For \( C_{max} = 5\,\mathrm{pF} = 5 \times 10^{-12}\,\mathrm{F} \): \[ T_{min} = 220,000 \times 5 \times 10^{-12} = 1.1 \times 10^{-6}\,\mathrm{s} \] \[ f_{min} = \frac{1}{T_{min}} \approx 0.91\,\mathrm{MHz} \] --- ### **Final Results:** \[ \boxed{ \begin{aligned} & \text{For } C_{min} = 2\,\mathrm{pF}:\quad f \approx 2.27\,\mathrm{MHz} \\ & \text{For } C_{nom} = 3\,\mathrm{pF}:\quad f \approx 1.52\,\mathrm{MHz} \\ & \text{For } C_{max} = 5\,\mathrm{pF}:\quad f \approx 0.91\,\mathrm{MHz} \end{aligned} } \] --- Let me know if you'd like further clarifications!