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2024-12-18 02:06:38 -05:00
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content/en/posts/ecse-1010/lab02/index.md
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@@ -10,7 +10,7 @@ author:
link: https/www.jamesflare.com
email:
avatar: /site-logo.avif
description: This blog post discusses a detailed lab assignment focusing on proving various electrical concepts using resistors, diodes, op-amps, and nodal analysis. The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
description: The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
keywords: ["Electrical Engineering","Ohm's Law","IV curve","Nodal Analysis","Op-Amp"]
license:
comment: true
@@ -28,7 +28,7 @@ hiddenFromHomePage: false
hiddenFromSearch: false
hiddenFromRss: false
hiddenFromRelated: false
summary: This blog post discusses a detailed lab assignment focusing on proving various electrical concepts using resistors, diodes, op-amps, and nodal analysis. The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
summary: The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
resources:
- name: featured-image
src: featured-image.avif
@@ -68,7 +68,7 @@ Let's pick two resistor. The first one is
$$
\begin{align*}
33 \times (1\times10^1) = 330 \Omega \pm 5\%
33 \times (1\times10^1) = 330 \Omega \pm 5\%
\end{align*}
$$
@@ -84,7 +84,7 @@ The second one is
$$
\begin{align*}
11 \times (1\times10^2) = 1100 \Omega \pm 5\%
11 \times (1\times10^2) = 1100 \Omega \pm 5\%
\end{align*}
$$
@@ -350,7 +350,7 @@ we got
### Discussion
Our experimental matches the datasheet. Consider the datasheet said
Our experimental matches the datasheet. Consider the datasheet said
- $V_F = 1.7V$
- $I_F = 100 mA$
@@ -667,7 +667,7 @@ $$
\begin{equation*}
V_{out}=\begin{cases}
\text{if} \; V_{in} < V_{ref}, V_{out} = V_s - \\\
\text{if} \; V_{in} > V_{ref}, V_{out} = V_s + \\\
\text{if} \; V_{in} > V_{ref}, V_{out} = V_s + \\\
\end{cases}
\end{equation*}
$$
@@ -678,7 +678,7 @@ $$
\begin{equation*}
V_{out}=\begin{cases}
\text{if} \; V_{in} < 0V, V_{out} = -5V \\\
\text{if} \; V_{in} > 0V, V_{out} = 5V \\\
\text{if} \; V_{in} > 0V, V_{out} = 5V \\\
\end{cases}
\end{equation*}
$$
@@ -723,8 +723,8 @@ In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it c
$$
\begin{align*}
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
\end{align*}
$$
@@ -782,8 +782,8 @@ In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it c
$$
\begin{align*}
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
\end{align*}
$$