lab_EE1C1_4_microphone_amplifier_2025-10-22

🏠 Homework

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✏️ Execises

1.

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2. see above. the $e^{-t/\tau }$ with tau being small the discharging is slow
3. also above but curves the other way

2.

step 1: see previous step 2: the voltage stays the save for a few seconds, then it goes to 0.75 peak for 0.5 seconds to peak and then to 30mV peak to peak

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step3: with 2C peak to peak is 60mV after

3

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step 1: starts at 23peak to peak = 1.5 * 6 + 15V goes to 200mV when 15 is off step 2: 2L with 15V power supply

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without

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3

$$\begin{array}{rl}{\tau }_1& =\frac{L}{R_{th}}\\ & =\frac{1\text{mH}}{2.2\text{k}\Omega }\\ & =0.\overline{45}\mu \text{s}\\ {\tau }_2& =C{R}_{th}\\ & =150\text{nF}\cdot 150\Omega \\ & =22.5\mu \text{s}\end{array}$$

Here are the equations for the transient voltage behavior in RC and RL circuits, presented in the requested LaTeX format.

RC Circuit: Voltage Across the Capacitor (V_C)

Charging Process:

$$\begin{array}{rl}{V}_C(t)& =V_s\left(1-e^{-\frac{t}{\tau }}\right)\\ \text{where, }\tau & =RC\end{array}$$

Discharging Process:

$$\begin{array}{rl}{V}_C(t)& =V_0{e}^{-\frac{t}{\tau }}\\ \text{where, }\tau & =RC\end{array}$$

RL circuit

Charging Process (Current Rise):

$$\begin{array}{rl}{V}_L(t)& =V_s{e}^{-\frac{t}{\tau }}\\ \text{where, }\tau & =\frac{L}{R}\end{array}$$

“Discharging” Process (Current Decay):

$$\begin{array}{rl}{V}_L(t)& =-V_0{e}^{-\frac{t}{\tau }}\\ \text{where, }\tau & =\frac{L}{R}\end{array}$$

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📄Assignment