diff --git a/.gitignore b/.gitignore index 514d8be..9a0dc24 100644 --- a/.gitignore +++ b/.gitignore @@ -2,4 +2,5 @@ public/ resources/_gen/ isableFastRander/ -.hugo_build.lock \ No newline at end of file +.hugo_build.lock +*Zone.Identifier \ No newline at end of file diff --git a/archetypes/default.md b/archetypes/default.md index 1c6421f..72b4829 100644 --- a/archetypes/default.md +++ b/archetypes/default.md @@ -2,6 +2,7 @@ title: {{ replace .TranslationBaseName "-" " " | title }} subtitle: date: {{ .Date }} +lastmod: {{ .Date }} slug: {{ substr .File.UniqueID 0 7 }} description: keywords: diff --git a/archetypes/posts.md b/archetypes/posts.md index ca77454..45b6247 100644 --- a/archetypes/posts.md +++ b/archetypes/posts.md @@ -2,13 +2,14 @@ title: {{ replace .TranslationBaseName "-" " " | title }} subtitle: date: {{ .Date }} +lastmod: {{ .Date }} slug: {{ substr .File.UniqueID 0 7 }} draft: true author: - name: - link: + name: James + link: https://www.jamesflare.com email: - avatar: + avatar: /site-logo.avif description: keywords: license: diff --git a/config/_default/hugo.toml b/config/_default/hugo.toml index e9b00ea..71b2457 100644 --- a/config/_default/hugo.toml +++ b/config/_default/hugo.toml @@ -284,6 +284,8 @@ enableEmoji = true enablePWA = false # FixIt 0.2.14 | NEW whether to add external Icon for external links automatically externalIcon = false + # FixIt 0.3.13 | NEW whether to capitalize titles + capitalizeTitles = true # FixIt 0.3.0 | NEW whether to add site title to the title of every page # remember to set up your site title in `hugo.toml` (e.g. title = "title") withSiteTitle = true @@ -292,6 +294,8 @@ enableEmoji = true # FixIt 0.3.0 | NEW whether to add site subtitle to the title of index page # remember to set up your site subtitle by `params.header.subtitle.name` indexWithSubtitle = false + # FixIt 0.3.13 | NEW whether to show summary in plain text + summaryPlainify = false # FixIt 0.2.14 | NEW FixIt will, by default, inject a theme meta tag in the HTML head on the home page only. # You can turn it off, but we would really appreciate if you don’t, as this is a good way to watch FixIt's popularity on the rise. disableThemeInject = false @@ -394,6 +398,9 @@ enableEmoji = true enable = false sticky = false showHome = false + # FixIt 0.3.13 | NEW + separator = "/" + capitalize = true # FixIt 0.3.10 | NEW Post navigation config [params.navigation] @@ -477,6 +484,14 @@ enableEmoji = true # whether to show the full text content in feed. fullText = false + # FixIt 0.3.13 | NEW recently updated pages config for archives, section and term list + [params.recentlyUpdated] + archives = true + section = true + list = true + days = 30 + maxCount = 10 + # FixIt 0.2.17 | NEW TagCloud config for tags page [params.tagcloud] enable = false @@ -969,6 +984,25 @@ enableEmoji = true # For values, see https://mermaid.js.org/config/theming.html#available-themes themes = ["default", "dark"] + # FixIt 0.3.13 | NEW Admonitions custom config + # See https://fixit.lruihao.cn/documentation/content-management/shortcodes/extended/admonition/#custom-admonitions + # syntax: = + [params.admonition] + # ban = "fa-solid fa-ban" + + # FixIt 0.3.14 | NEW Task lists custom config + # See https://fixit.lruihao.cn/documentation/content-management/advanced/#custom-task-lists + # syntax: = + [params.taskList] + # tip = "fa-regular fa-lightbulb" + + # FixIt 0.3.15 | NEW version shortcode config + [params.repoVersion] + # url prefix for the release tag + url = "https://github.com/hugo-fixit/FixIt/releases/tag/v" + # project name + name = "FixIt" + # FixIt 0.2.12 | NEW PanguJS config [params.pangu] # For Chinese writing @@ -1115,6 +1149,20 @@ enableEmoji = true # whether to show the full text content in feed. fullText = true + # FixIt 0.3.12 | NEW Custom partials config + # Custom partials must be stored in the /layouts/partials/ directory. + # Depends on open custom blocks https://fixit.lruihao.cn/references/blocks/ + [params.customPartials] + head = [] + profile = [] + aside = [] + comment = [] + footer = [] + widgets = [] + assets = [] + postFooterBefore = [] + postFooterAfter = [] + # FixIt 0.2.15 | NEW Developer options # Select the scope named `public_repo` to generate personal access token, # Configure with environment variable `HUGO_PARAMS_GHTOKEN=xxx`, see https://gohugo.io/functions/os/getenv/#examples diff --git a/content/en/posts/ecse-1010/lab01/Lab01.pdf b/content/en/posts/ecse-1010/lab01/Lab01.pdf new file mode 100644 index 0000000..6ece080 Binary files /dev/null and b/content/en/posts/ecse-1010/lab01/Lab01.pdf differ diff --git a/content/en/posts/ecse-1010/lab01/Proof of Concept - 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It includes detailed circuit schematics, analysis, simulations, and experimental measurements to validate theoretical concepts. +keywords: ["Ohm's Law", "KCL", "KVL", "Voltage Divider", "Current Flow", "Electrical Engineering", "Circuit Analysis", "Simulation", "Measurement"] +license: +comment: true +weight: 0 +tags: + - ECSE 1010 + - Lab + - Electrical Engineering + - RPI +categories: + - Electrical Engineering +collections: + - ECSE 1010 +hiddenFromHomePage: false +hiddenFromSearch: false +hiddenFromRss: false +hiddenFromRelated: false +summary: This blog post presents a comprehensive proof of concepts for ECSE 1010 Omega Lab01, focusing on electrical engineering principles such as Ohm's Law, KCL, KVL, voltage dividers, and current flow in circuits. It includes detailed circuit schematics, analysis, simulations, and experimental measurements to validate theoretical concepts. +resources: + - name: featured-image + src: featured-image.jpg + - name: featured-image-preview + src: featured-image-preview.jpg +toc: true +math: true +lightgallery: true +password: +message: +repost: + enable: false + url: + +# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter +--- + + + +## Lab Document + + + +## 1. Prove Ohm's Law, KCL, and KVL in a Circuit + +### Circuit Schematic + +{{< image src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Schematic.avif" caption="Proof of Concept - Omega Lab 01 - 1 - Schematic" width=600px >}} + +### Description + +- Based on the description of Ohm's Law, the voltage is equal to current times resistance. So, I will check the real measurements with voltmeter and the theoretic values. +- Based on the description of KCL, the current goes into a node is equal to current goes out the node. So I am going to measure the currents and add them together to check if it matches the theory. +- Based on the description of KVL, the net voltage of nodes in the loop is equal to zero. So, I am going to measure all the voltage across the loop and check the sum. + +### Analysis + +We know that, the Ohm's Law, KCL, and KVL can be shown as these formulas: + +$$ +V = IR \\\ +\textstyle \sum I_{in} = \sum I_{out} \\\ +\textstyle \sum V_{n} = 0 +$$ + +*** + +Based on $V = IR$, the total $I$ should be + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\\ + I_{total} &= \frac{5}{10000 + 500 + 10000} \\\ + I_{total} &= 0.000243902439 \\\ +\end{align*} +$$ + +And $I(R2) = I(R3)$ should be + +$$ +\begin{align*} + I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\\ + I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\\ + I(R2) = I(R3) &= 0.0001219512195 +\end{align*} +$$ + +*** + +To find $V(R1) = V(R4)$ and $V(R2) = V(R3)$, we can use + +$$ +\begin{align*} + V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\\ + V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\\ + V(R1) = V(R4) &= 2.4390244 +\end{align*} +$$ + +$$ +\begin{align*} + V(R2) = V(R3) &= V_{total} \\\ + V(R2) = V(R3) &= (5 - 2.4390244 - 2.4390244) \\\ + V(R2) = V(R3) &= 0.1219512 +\end{align*} +$$ + +*** + +Based on $\sum I_{in} = \sum I_{out}$, we should see $I(R1) = I(R2) + I(R3)$, since $I(R1)$ is the current goes into node `n002` and $I(R2) + I(R3)$ is the current goes out node `n002`. + +Based on $\sum V_{n} = 0$, we should expect $V(n001) - V(n002) - V(n003) = 0$, since they are in the same loop. + +We will check if the experimental results fit these expectations. + +### Simulation + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Simulation.avif" caption="Proof of Concept - Omega Lab 01 - 1 - Simulation" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +V(n002): 2.56098 voltage +V(n003): 2.43902 voltage +I(R1): -0.000243902 device_current +I(R2): 0.000121951 device_current +I(R3): 0.000121951 device_current +I(R4): 0.000243902 device_current +I(V1): -0.000243902 device_current +``` + +### Measurement + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement.avif" caption="Proof of Concept - Omega Lab 01 - 1 - Measurement" width=600px >}} + +$V(R1) = 2.4963V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 1 - Measurement - 1" width=600px >}} + +$V(R2) = V(R3) = 166.5mV$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement%20-%202.avif" caption="Proof of Concept - Omega Lab 01 - 1 - Measurement - 2" width=600px >}} + +$V(R4) = 2.4616V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement%20-%203.avif" caption="Proof of Concept - Omega Lab 01 - 1 - Measurement - 3" width=600px >}} + +### Discussion + +First, let's compare the theoretical value with the experimental measurements. + +We got the experimental reading from Analog Discovery 3. + +```text +V(R1) = 2.4963V +V(R2) = V(R3) = 166.5mV +V(R4) = 2.4616V +``` + +To find out the theoretic values of $V(R1)$, $V(R2) = V(R3)$ and $V(R4)$. We need to do some math. + +We know the simulation output is + +```text +V(n001): 5 voltage +V(n002): 2.56098 voltage +V(n003): 2.43902 voltage +``` + +and the voltage is the difference in potential. Based on that, we can caculate the theoretic values of $V(R1)$, $V(R2) = V(R3)$ and $V(R4)$. + +$$ +\begin{align*} + V(R1) &= V(n001) - V(n002) \\\ + V(R1) &= 5 - 2.56098 \\\ + V(R1) &= \boxed{2.43902} +\end{align*} +$$ + +$$ +\begin{align*} + V(R2) = V(R3) &= V(n002) - V(n003) \\\ + V(R2) = V(R3) &= 2.56098 - 2.43902 \\\ + V(R2) = V(R3) &= \boxed{0.12196} +\end{align*} +$$ + +$$ +\begin{align*} + V(R4) &= V(n003) - V(\text{GND}) \\\ + V(R4) &= 2.43902 - 0 \\\ + V(R4) &= \boxed{2.43902} +\end{align*} +$$ + +Let's make a table to compare the results + +|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$2.4390V$|$2.4390V$|$2.4963V$|$57.28mV$|$2.3\%$| +|$V(R2)$|$0.1219V$|$0.1219V$|$0.1665V$|$44.54mV$|$26.8\%$| +|$V(R3)$|$0.1219V$|$0.1219V$|$0.1665V$|$44.54mV$|$26.8\%$| +|$V(R4)$|$2.4390V$|$2.4390V$|$2.4616V$|$22.58mV$|$0.9\%$| + +We can see that $V(R1)$ and $V(R4)$ are very accurate. But $V(R2)$ and $V(R3)$ has a lot of error. A potential explanation is that, there is a background noise. + +If we look at the "Measurement", channel 2 is empty, but it still has a reading around $50mV$. It's very likely to be a background noise. If we remove this noise from Experimental Measurements. The $\%$diff will be less than $1\%$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +*** + +Now, let's check KCL. + +We got the simulations data like + +```text +I(R1): -0.000243902 device_current +I(R2): 0.000121951 device_current +I(R3): 0.000121951 device_current +I(R4): 0.000243902 device_current +I(V1): -0.000243902 device_current +``` + +Using the expectation $I(R1) = I(R2) + I(R3)$ from Analysis. We can check + +$$ +\begin{align*} + & I(R1) + I(R2) + I(R3) \\\ + =& -0.000243902 + 0.000121951 + 0.000121951 \\\ + =& \; \boxed{0} +\end{align*} +$$ + +KCL is very likely be True. + +*** + +Then, let's check KVL. + +We can use the result from previous part. + +$$ +\begin{align*} + V(R1) &= 2.43902 \\\ + V(R2) = V(R3) &= 0.12196 \\\ + V(R4) &= 2.43902 +\end{align*} +$$ + +Use the expectation $V(n001) - V(n002) - V(n003) = 0$ from Analysis. We can check + +$$ +\begin{align*} + & V(n001) - V(n002) - V(n003) \\\ + =& \; 2.43902 - 0.12196 - 0.12196\\\ + =& \; 0 \\\ + & \; 0 = 0 \; \boxed{\text{True}} +\end{align*} +$$ + +KVL is very likely be True. + +*** + +Finally, let's check Ohm's Law. Using the expectation $V=IR$ and the experimental data. + +```text +R1 = 10K +R2 = R3 = 1K +R4 = 10K +V(R1) = 2.4963V +V(R2) = V(R3) = 166.5mV +V(R4) = 2.4616V +``` + +to calculate $I$ based on Ohm's Law. + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R1) &= \frac{2.4963}{10000} = \boxed{0.00024963} +\end{align*} +$$ + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R2) = I(R3) &= \frac{0.1665}{1000} = \boxed{0.0001665} +\end{align*} +$$ + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R4) &= \frac{2.4616}{10000} = \boxed{0.00024616} +\end{align*} +$$ + +Then, we can check these current result with simulation data. + +|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-|:-|:-|:-:|-:| +|$I(R1)$|$0.2439mA$|$0.2439mA$|$0.2496mA$|$0.005728mA$|$2.3\%$| +|$I(R2)$|$0.1665mA$|$0.1665mA$|$0.1219mA$|$0.044549mA$|$26.8\%$| +|$I(R3)$|$0.1665mA$|$0.1665mA$|$0.1219mA$|$0.044549mA$|$26.8\%$| +|$I(R4)$|$0.2439mA$|$0.2439mA$|$0.2461mA$|$0.002258mA$|$0.9\%$| + +We can see that $I(R1)$ and $I(R4)$ are very accurate. But $I(R2)$ and $I(R3)$ has a lot of error. A potential explanation is that, there is a background noise. + +If we look at the "Measurement", channel 2 is empty, but it still has a reading around $50mV$. It's very likely to be a background noise. If we remove this noise from Experimental Measurements. The $\%$diff will be less than $1\%$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +Also, we can check to total current in the circuit. Using the expectation form "Analysis" - $I_{total} = 0.000243902439$, this matches the simulation - $0.000243902A$. + +*** + +In conclusion, the simulation 100% fit the KCL and KVL. The experimental data is close to the simulation. And very close to simulation if we remove the background noise and consider the $5\%$ tolerance of the resistor. Then, we used Ohm's Law with experimental data to compare the simulation. The result is also very close. Thus, we proved Ohm's Law, KCL, and KVL in a Circuit. + +## 2. Prove the Concept of a Voltage Divider in a Series Circuit + +### Circuit Schematic + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Schematic.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Schematic" width=600px >}} + +### Description + +I am going to build a series circuit with two resistors and measure the voltage across the resistors to compare the theoretic values. + +### Analysis + +The Voltage Divider equation is + +$$ +\frac{V_1}{V_2} = \frac{R_1}{R_2} +$$ + +If we have the voltage source $5V$ and $R1=R2=10K$. Put the values into the equation and get + +$$ +\frac{V_1}{V_2} = \frac{10K}{10K} = \frac{1}{1} +$$ + +We know that $V_1 + V_2 = 5$ and $1 \cdot V_1 = 1 \cdot V_2$. So, we should expect $V_1 = V_2 = 2.5$. + +### Simulation + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Simulation.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Simulation" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +V(n002): 2.5 voltage +I(R1): -0.00025 device_current +I(R2): -0.00025 device_current +I(V1): -0.00025 device_current +``` + +### Measurement + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Measurement" width=600px >}} + +$V(R1) = 2.5539V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Measurement - 1" width=600px >}} + +$V(R2) = 2.5204V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%202.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Measurement - 2" width=600px >}} + +### Discussion + +First, let's compare the theoretical value with the experimental measurements. + +We got the experimental reading from Analog Discovery 3. + +```text +V(R1) = 2.5539V +V(R2) = 2.5204V +``` + +and the voltage is the difference in potential. Based on that, we can calculate the theoretic values of $V(R1)$ and $V(R2)$. We need to do some math. + +We know the simulation output is + +```text +V(n001): 5 voltage +V(n002): 2.5 voltage +``` + +$$ +\begin{align*} + V(R1) &= V(n001) - V(n002) \\\ + V(R1) &= 5 - 2.5 \\\ + V(R1) &= \boxed{2.5} +\end{align*} +$$ + +$$ +\begin{align*} + V(R2) &= V(n002) - \text{GND} \\\ + V(R2) &= 2.5 - 0 \\\ + V(R2) &= \boxed{2.5} +\end{align*} +$$ + +Let's make a table to compare the results + +|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$2.5V$|$2.5V$|$2.5539V$|$0.0539V$|$2.1\%$| +|$V(R2)$|$2.5V$|$2.5V$|$2.5204V$|$0.0204V$|$0.8\%$| + +We can see that both $V(R1)$ and $V(R2)$ are very accurate. They are some errors, A potential explanation is that, there is a background noise. + +If we look at the "Measurement", channel 2 is empty, but it still has a reading around $40mV$. It's very likely to be a background noise. If we remove this noise from Experimental Measurements. The $\%$diff will be less than $1\%$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +In conclusion, the simulation 100% fits the Voltage Divider theoretic formula. And the experimental reading is close to theoretic values. And the experimental reading is very close to theoretic values if we removed the background noise. Thus, we proved the Concept of a Voltage Divider in a Series Circuit. + +## 3. Prove the Concept of How Current Flows in a Series Circuit + +### Circuit Schematic + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Schematic.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Schematic" width=600px >}} + +### Description + +I am going to use the Ohm's Law to find out the current flows through every resistor in the series circuit and compare it with the theoretic values. + +### Analysis + +The feature of series circuit is that + +- There is only one path for the current to flow through the circuit. +- The current is the same at any point in the circuit. + +Since Analog Discovery 3 can't directly measures the current but the voltage. We are going to use Ohm's Law to find out the current flow through the resistor. + +We know this relationship from Ohm's Law + +$$ +V = IR +$$ + +We can change it a bit into + +$$ +I = \frac{V}{R} +$$ + +Also, we know that $R_1 = R_2 = 10K$ and the voltage across the resistor can be found by voltage divider formula. which is + +$$ +\begin{align*} + \frac{V_1}{V_2} &= \frac{R_1}{R_2} \\\ + \frac{V_1}{V_2} &= \frac{10K}{10K} = \frac{1}{1} +\end{align*} +$$ + +We know that $V_1 + V_2 = 5$ and $1 \cdot V_1 = 1 \cdot V_2$. So, we should expect $V_1 = V_2 = 2.5$. + +Using these values, we can find out $I(R1)$ and $I(R2)$ by + +$$ +\begin{align*} + I(R1) = I(R2) &= \frac{V}{R} \\\ + I(R1) = I(R2) &= \frac{2.5}{10K} \\\ + I(R1) = I(R2) &= \boxed{0.00025} +\end{align*} +$$ + +We expect $I(R1)$ and $I(R2)$ to be $0.00025A$. + +### Simulation + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Simulation.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Simulation" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +V(n002): 2.5 voltage +I(R1): -0.00025 device_current +I(R2): -0.00025 device_current +I(V1): -0.00025 device_current +``` + +### Measurement + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Measurement" width=600px >}} + +$V(R1) = 2.5539V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Measurement - 1" width=600px >}} + +$V(R2) = 2.5204V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%202.avif" caption="Proof of Concept - Omega Lab 01 - 2 - Measurement - 2" width=600px >}} + +### Discussion + +From the Simulation Result, + +```text +I(R1): -0.00025 device_current +I(R2): -0.00025 device_current +``` + +It proves $I(R1)=I(R2)$, which + +- There is only one path for the current to flow through the circuit. +- The current is the same at any point in the circuit. + +*** + +From the Measurement Result, we got the voltage across $R1$ and $R2$ + +$V(R1) = 2.5539V$ +$V(R2) = 2.5204V$ + +Based on the Ohm's Law - the relationship we got in Analysis $I = \frac{V}{R}$. We can find $R1$ and $R2$ by + +$$ +\begin{align*} + I(R1) &= \frac{V}{R} \\\ + I(R1) &= \frac{2.5539}{10K} \\\ + I(R1) &= 0.00025539 +\end{align*} +$$ + +$$ +\begin{align*} + I(R2) &= \frac{V}{R} \\\ + I(R2) &= \frac{2.5204}{10K} \\\ + I(R2) &= 0.00025204 +\end{align*} +$$ + +Both $R1$ and $R2$ are very close, we can say that $R1 \approx R2$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +*** + +In conclusion, the simulation 100% fits the feature of current in series circuit. And the experimental reading is close to theoretic values. And the experimental reading is very close to theoretic values if we consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). Thus, we proved the Concept of How Current Flows in a Series Circuit. + +- There is only one path for the current to flow through the circuit. +- The current is the same at any point in the circuit. + +## 4. Prove the Concept of Voltage Across a Parallel Circuit + +### Circuit Schematic + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Schematic.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Schematic" width=600px >}} + +### Description + +I am going to use the Ohm's Law and feature of node to find out the voltage across every resistor in the parallel circuit and compare it with the theoretic values. + +### Analysis + +The feature of parallel circuit is that + +- There are multiple paths for the current to flow through the circuit. +- The voltage across each branch is the same and equal to the voltage supplied by the source. + +$$ +V_{total} = V_1 = V_2 = V_3 = \ldots = V_n +$$ + +We know that the voltage in the same node is the same (they are connected by a wire). And `n001`, `n002` connected both side of resistor. So, we should expect $V(R1) = V(R2)$. + +Also, the voltage is potential difference between the component. + +$$ +\begin{align*} + V(R1) = V(R2) &= n001 - \text{GND} \\\ + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +*** + +Also, we can check the current of this circuit. We know that the current in a parallel circuit is + +$$ +I_{total} = I_1 + I_2 + I_3 + \dots +$$ + +and from Ohm's Law, we know that + +$$ +V = IR +$$ + +We can change it a bit into + +$$ +I = \frac{V}{R} +$$ + +And, we know that the resistance in parallel is + +$$ +\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n} +$$ + +So, + +$$ +R_{total} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}} +$$ + +Combine them together, we got + +$$ +I_{total} = \frac{V}{\cfrac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}}} +$$ + +Let's put values into equation + +$$ +\begin{align*} + I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\\ + I_{total} &= \frac{5}{5K} \\\ + I_{total} &= \boxed{0.001} +\end{align*} +$$ + +We can check this to double confirm our result. + +### Simulation + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Simulation.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Simulation" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +I(R2): 0.0005 device_current +I(R1): 0.0005 device_current +I(V1): -0.001 device_current +``` + +### Measurement + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Measurement" width=600px >}} + +$V(R1)=V(R2)=5.0305V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Measurement - 1" width=600px >}} + +### Discussion + +First, let's compare the theoretical value with the experimental measurements. + +We got the experimental reading from Analog Discovery 3. + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +To find out the theoretic values of $V(R1)$ and $V(R2)$. We need to do some math. + +We know the simulation output is + +```text +V(n001): 5 voltage +``` + +and the voltage is the difference in potential. Based on that, we can caculate the theoretic values of $V(R1)$ and $V(R2)$. + +$$ +\begin{align*} + V(R1) = V(R2) &= V(n001) - \text{GND} \\\ + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +Let's make a table to compare the results + +|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| +|$V(R2)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| + +We can see that both $V(R1)$ and $V(R2)$ are very accurate. They are some errors, A potential explanation is that, there is a background noise. + +If we look at the "Measurement", channel 2 is empty, but it still has a reading around $40mV$. It's very likely to be a background noise. If we remove this noise from Experimental Measurements. The $\%$diff will be less than $0.2\%$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +*** + +Secondly, we can check to $I_{total}$ as double confirm. + +We know the simulation output is + +```text +I(R2): 0.0005 device_current +I(R1): 0.0005 device_current +``` + +and from Analysis, we expect + +$$ +I_{total} = 0.001 +$$ + +We know that, the $I_{total}$ of a parallel circuit is + +$$ +I_{total} = I_1 + I_2 + I_3 + \dots +$$ + +So, we get + +$$ +\begin{align*} + I_{total} &= I(R2) + I(R1) \\\ + I_{total} &= 0.0005 + 0.0005 \\\ + I_{total} &= 0.001 \\\ + & 0.001 = 0.001 \; \boxed{\text{True}} +\end{align*} +$$ + +Our Analysis matches Simulation. + +*** + +Additionally, we can check the experimental data as well. Since Analog Discovery 3 can't measure the current directly, we need use Ohm's Law to find out current. + +We got + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +and we know that + +```text +R1 = 10K +R2 = 10K +``` + +Then, we can find out $I(R1)$ and $I(R2)$ by + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R1) = I(R2) &= \frac{5.0305}{10000} \\\ + I(R1) = I(R2) &= \boxed{0.0005305} +\end{align*} +$$ + +$0.0005305 \approx 0.0005$ with only $0.6\%$diff (even less then $0.2\%$ if we remove $40mV$ background noise). Our theory is very likely be true. Since the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +*** + +In conclusion, we checked the simulation 100% fit the Analysis' expectation. And the experimental data only has $0.2\%$ to $0.6\%$ than the theoretic values. Thus, we proved Concept of Voltage Across a Parallel Circuit. + +- There are multiple paths for the current to flow through the circuit. +- The voltage across each branch is the same and equal to the voltage supplied by the source. + +## 5. Prove the Concept of a Current Divider in a Parallel Circuit + +### Circuit Schematic + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Schematic.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Schematic" width=600px >}} + +### Description + +I am going to use the Ohm's Law to find out the current across every resistor in the parallel circuit and compare its sum with the theoretic values. + +### Analysis + +The feature of parallel circuit is that + +- There are multiple paths for the current to flow through the circuit. +- The voltage across each branch is the same and equal to the voltage supplied by the source. +- The total current entering the parallel circuit is divided among the branches. + +$$ +V_{total} = V_1 = V_2 = V_3 = \ldots = V_n +$$ + +We know that the voltage in the same node is the same (they are connected by a wire). And `n001`, `n002` connected both side of resistor. So, we should expect $V(R1) = V(R2)$. + +Also, the voltage is potential difference between the component. + +$$ +\begin{align*} + V(R1) = V(R2) &= n001 - \text{GND} \\\ + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +and from Ohm's Law, we know that + +$$ +V = IR +$$ + +We can change it a bit into + +$$ +I = \frac{V}{R} +$$ + +and get + +$$ +\begin{align*} + I(R1) &= \frac{V(R1)}{R1} \\\ + I(R1) &= \frac{5}{10K} \\\ + I(R1) &= \boxed{0.0005} +\end{align*} +$$ + +$$ +\begin{align*} + I(R2) &= \frac{V(R2)}{R2} \\\ + I(R2) &= \frac{5}{10K} \\\ + I(R2) &= \boxed{0.0005} +\end{align*} +$$ + +the relationship between $I(R1)$ and $I(R2)$ can be express as + +$$ +\begin{align*} + \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\\ + \because V(R1) &= V(R2) \\\ + \therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\\ + \frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\\ + &\boxed{\frac{I(R1)}{I(R2)} = \frac{R2}{R1}} +\end{align*} +$$ + +In our case, $1 \cdot R1 = 1 \cdot R2$, so + +$$ +\frac{I(R1)}{I(R2)} = \frac{R2}{R1} = \frac{1}{1} +$$ + +as we get the $I_{total}$ by + +$$ +\begin{align*} + I_{total} &= \frac{V_{total}}{R_{total}} \\\ + I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\\ + I_{total} &= \frac{5}{5K} \\\ + I_{total} &= 0.001 +\end{align*} +$$ + +As + +$$ +\frac{I(R1)}{I(R2)} = \frac{1}{1} +$$ + +We can get $I(R1)$ and $I(R2)$ by + +$$ +\begin{align*} + I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\\ + I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\\ + I(R1) = I(R2) &= \boxed{0.0005} +\end{align*} +$$ + +At this point, our logic is consistent, which + +$$ +\begin{align*} + & \because R1 = R2 = 10K \\\ + & \because I(R1) = I(R2) = 0.0005 \\\ + & \because V(R1) = V(R2) = 5 \\\ + & \because I_{total} = I(R1) + I(R2) = 0.001 \\\ + & \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\\ + & \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2} \\\ +\end{align*} +$$ + +The result of $I(R1) = I(R2) = 0.0005$ also got cross checked by $I = \frac{V}{R}$. So, we are very confident that + +$$ +\frac{I(R1)}{I(R2)} = \frac{R2}{R1} +$$ + +For any current of the resistor in parallel circuit (e.g $I(R1)$) + +$$ +I(R1) = I_{total} \times \frac {R1}{R1 + R2 + \cdots} +$$ + +### Simulation + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Simulation.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Simulation" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +I(R2): 0.0005 device_current +I(R1): 0.0005 device_current +I(V1): -0.001 device_current +``` + +### Measurement + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Measurement" width=600px >}} + +$V(R1)=V(R2)=5.0305V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 4 - Measurement - 1" width=600px >}} + +### Discussion + +First, let's compare the theoretical value with the experimental measurements. + +We got the experimental reading from Analog Discovery 3. + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +To find out the theoretic values of $V(R1)$ and $V(R2)$. We need to do some math. + +We know the simulation output is + +```text +V(n001): 5 voltage +``` + +and the voltage is the difference in potential. Based on that, we can caculate the theoretic values of $V(R1)$ and $V(R2)$. + +$$ +\begin{align*} + V(R1) = V(R2) &= V(n001) - \text{GND} \\\ + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +Let's make a table to compare the results + +|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| +|$V(R2)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| + +We can see that both $V(R1)$ and $V(R2)$ are very accurate. They are some errors, A potential explanation is that, there is a background noise. + +If we look at the "Measurement", channel 2 is empty, but it still has a reading around $40mV$. It's very likely to be a background noise. If we remove this noise from Experimental Measurements. The $\%$diff will be less than $0.2\%$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +*** + +Second, let's check $I(R1)$ and $I(R2)$ with theoretical values. + +Since Analog Discovery 3 can't measure the current directly, we need use Ohm's Law to find out current. + +We got + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +and we know that + +```text +R1 = 10K +R2 = 10K +``` + +Then, we can find out $I(R1)$ and $I(R2)$ by + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R1) = I(R2) &= \frac{5.0305}{10000} \\\ + I(R1) = I(R2) &= \boxed{0.0005305} +\end{align*} +$$ + +|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-:|:-:|:-:|:-:|-:| +|$I(R1)$|$0.5mA$|$0.5mA$|$0.50305mA$|$0.00305mA$|$0.6\%$| +|$I(R2)$|$0.5mA$|$0.5mA$|$0.50305mA$|$0.00305mA$|$0.6\%$| + +We can see that both $I(R1)$ and $I(R2)$ are very accurate. They are some errors, A potential explanation is that, there is a background noise. + +If we look at the "Measurement", channel 2 is empty, but it still has a reading around $40mV$. It's very likely to be a background noise. If we remove this noise from Experimental Measurements. The $\%$diff will be less than $0.2\%$. Consider that the resistor has a Tolerance of $5\%$ (from 4 Band Resistor Color Code). We can consider this as systematic error and the Experimental Measurements is very close to Simulations. + +*** + +In conclusion, we checked the simulation 100% fit the Analysis' expectation. And the experimental data only has $0.2\%$ to $0.6\%$ than the theoretic values. Thus, we proved Concept of How Current Flows in a Series Circuit. + +- There are multiple paths for the current to flow through the circuit. +- The voltage across each branch is the same and equal to the voltage supplied by the source. +- The total current entering the parallel circuit is divided among the branches. + +## 6. Prove the Concept of a Voltage Divider in Temperature Sensing Circuit + +### Circuit Schematic + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Schematic.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Schematic" width=600px >}} + +### Description + +We are going to use NTC 100K as our thermistor, and we will compare the reading with our thermometer and simulation to check its reliability. + +### Analysis + +NTC thermistor uses Beta formula to calculate the resistance under a specific temperature. The formula is like + +$$ +\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) \\\ +$$ + +We can move $R_1$ to the left side to get + +$$ +R_1 = R_0 e^{\beta(T_1^{-1} - T_0^{-1})} +$$ + +Since we want to find out the resistance of this thermistor under a specific temperature. + +*** + +The thermistor we are using is NTC 100K. Which means it has $100k \Omega$ at the reference temperature $25 \degree C$ + +$$ +T_0 = 298.15K \\\ +R_0 = 100k \Omega +$$ + +Also, we got the $\beta$ value from the manufacturer, which + +$$ +\beta = 3950 +$$ + +We know the Voltage Divider + +$$ +\frac{V_1}{V_2} = \frac{R_1}{R_2} +$$ + +put them together, we got + +$$ +\frac{V_1}{V_2} = \frac{R_0 e^{\beta(T_1^{-1} - T_0^{-1})}}{R_2} +$$ + +### Simulation + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Simulation" width=600px >}} + +We got a curve shows the relationship between the temperature and resistance in range of $T = 0 \degree C$ to $T = 40 \degree C$ + +For $T = 29.12 \degree C$, $V$ should be $4.465V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Simulation - 1" width=600px >}} + +For $T = 27.3 \degree C$, $V$ should be $4.501V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation%20-%202.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Simulation - 2" width=600px >}} + +For $T = 30.25 \degree C$, $V$ should be $4.441V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation%20-%203.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Simulation - 3" width=600px >}} + +### Measurement + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Measurement" width=600px >}} + +We used the math function in Scope with + +```js +1/((1/298.15)+(1/3950)*log((((10000*C1)/(5-C1))/100000),2.71828)) - 273.15 +``` + +to get the temperature in $\degree C$. This is from + +$$ +\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) \\\ +$$ + +which $T_1$ is the temperature reading we want. + +Then, we calibrate the thermistor with another thermometer at $24 \degree C$. + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%202.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Measurement - 2" width=600px >}} + +Later on, we toke three more reading + +$V = 4.465V, T = 29.12 \degree C$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%201.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Measurement - 1" width=600px >}} + +$V = 4.502V, T = 27.3 \degree C$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%203.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Measurement - 3" width=600px >}} + +$V = 4.441V, T = 30.25 \degree C$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%204.avif" caption="Proof of Concept - Omega Lab 01 - 6 - Measurement - 4" width=600px >}} + +### Discussion + +|Temp|Analysis|Simulation|Experiement|diff|$\%$diff| +|:-:|:-:|:-:|:-:|:-|-:| +|$29.12 \degree C$|$4.465V$|$4.465V$|$4.465V$|$0V$|$0\%$| +|$27.3 \degree C$|$4.501V$|$4.501V$|$4.502V$|$1mV$|$0.1\%$| +|$30.25 \degree C$|$4.441V$|$4.441V$|$4.441V$|$0V$|$0\%$| + +As we can see, the difference between theory and measurements is extremely small. This may due to the temperature reading is calculated from voltage by Math. + +But even we look the thermometer's reading, both of them shows around $24 \degree C$. In the worst case, the error is $5\%$. So, over all, our reading is reliable. + +*** + +The wheatstone bridge is better than a normal voltage divider because it is more sensitive than a voltage divider. A voltage divider relies on the ratio of resistances between two resistors, so even if the ratio of the resistors change, as long as the change isn't massive, the output voltage stays the same. + +When a wheatstone bridge is balanced, meaning R1/R2=R3/R4, the current flowing through the galvanometer in the center of the wheatstone bridge is 0. When current is zero, the calculated resistance is no longer affected by innate resistance of wires, resistors, and voltameters. 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ECSE 1010 + - Lab + - Electrical Engineering + - RPI +categories: + - Electrical Engineering +collections: + - ECSE 1010 +hiddenFromHomePage: false +hiddenFromSearch: false +hiddenFromRss: false +hiddenFromRelated: false +summary: 本文展示了ECSE 1010 Omega Lab01的综合概念验证，重点探讨了欧姆定律、基尔霍夫电流定律（KCL）、基尔霍夫电压定律（KVL）、分压器原理及电路中的电流流动等电气工程基础。内容涵盖详细的电路图设计、分析过程、仿真模拟以及实验测量数据，以验证理论知识的正确性。 +resources: + - name: featured-image + src: featured-image.jpg + - name: featured-image-preview + src: featured-image-preview.jpg +toc: true +math: true +lightgallery: true +password: +message: +repost: + enable: false + url: + +# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter +--- + + + +## Lab Document + + + +## 1. 验证欧姆定律、KCL 和 KVL 在电路中的应用 + +### 电路图示 + +{{< image src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Schematic.avif" caption="概念验证 - Omega 实验室 01 - 1 - 图纸" width=600px >}} + +### 描述 + +- 根据欧姆定律，电压等于电流乘以电阻。因此，我们将用伏特表测量实际值并与理论值进行比较。 +- 根据 KCL（基尔霍夫电流定律），流入节点的总电流等于流出该节点的总电流。所以我们要测量所有电流并相加来验证是否符合理论。 +- 根据 KVL（基尔霍夫电压定律），在回路中的各个节点电压总和为零。因此，我们将测量整个回路的所有电压，并检查它们的总和。 + +### 分析 + +我们知道，欧姆定律、KCL 和 KVL 可以表示成以下公式： + +$$ +V = IR \\\ +\sum I_{in} = \sum I_{out} \\\ +\sum V_n = 0 +$$ + +*** + +基于 $V = IR$，总电流应为 + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\\ + I_{total} &= \frac{5}{10000 + 500 + 10000} \\\ + I_{total} &= 0.000243902439 \\\ +\end{align*} +$$ + +并且 $I(R2) = I(R3)$ 应为 + +$$ +\begin{align*} + I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\\ + I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\\ + I(R2) = I(R3) &= 0.0001219512195 +\end{align*} +$$ + +*** + +为了找到 $V(R1)$、$V(R2)=V(R3)$ 和 $V(R4)$，我们可以使用以下公式： + +$$ +\begin{align*} + V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\\ + V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\\ + V(R1) = V(R4) &= 2.4390244 +\end{align*} +$$ + +$$ +\begin{align*} + V(R2) = V(R3) &= V_{total} - (V(R1) + V(R4)) \\\ + V(R2) = V(R3) &= 5 - 2.4390244 - 2.4390244 \\\ + V(R2) = V(R3) &= 0.1219512 +\end{align*} +$$ + +*** + +根据 KCL，我们应看到 $I(R1) = I(R2) + I(R3)$，因为 $I(R1)$ 是流入节点 `n002` 的电流而 $I(R2) + I(R3)$ 是流出该节点的电流。 + +根据 KVL，在同一个回路中的电压总和为零。因此，我们应期望 $V(n001) - V(n002) - V(n003) = 0$。我们将检查实验结果是否符合这些预期。 + +### 模拟 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Simulation.avif" caption="概念验证 - Omega 实验室 01 - 1 - 模拟结果" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +V(n002): 2.56098 voltage +V(n003): 2.43902 voltage +I(R1): -0.000243902 device_current +I(R2): 0.000121951 device_current +I(R3): 0.000121951 device_current +I(R4): 0.000243902 device_current +I(V1): -0.000243902 device_current +``` + +### 测量 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement.avif" caption="概念验证 - Omega 实验室 01 - 1 - 测量结果" width=600px >}} + +$V(R1) = 2.4963V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 1 - 测量结果 - 1" width=600px >}} + +$V(R2) = V(R3) = 166.5mV$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement%20-%202.avif" caption="概念验证 - Omega 实验室 01 - 1 - 测量结果 - 2" width=600px >}} + +$V(R4) = 2.4616V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%201%20-%20Measurement%20-%203.avif" caption="概念验证 - Omega 实验室 01 - 1 - 测量结果 - 3" width=600px >}} + +### 讨论 + +首先，让我们比较理论值与实验测量值。 + +我们从 Analog Discovery 3 获取了以下实验读数： + +```text +V(R1) = 2.4963V +V(R2) = V(R3) = 166.5mV +V(R4) = 2.4616V +``` + +为了找到 $V(R1)$、$V(R2)=V(R3)$ 和 $V(R4)$ 的理论值，我们需要做一些计算。 + +我们知道模拟输出为： + +```text +V(n001): 5 voltage +V(n002): 2.56098 voltage +V(n003): 2.43902 voltage +``` + +电压是指两个点之间的电位差。基于这一点，我们可以计算 $V(R1)$、$V(R2)=V(R3)$ 和 $V(R4)$ 的理论值。 + +$$ +\begin{align*} + V(R1) &= V(n001) - V(n002) \\\ + V(R1) &= 5 - 2.56098 \\\ + V(R1) &= \boxed{2.43902} +\end{align*} +$$ + +$$ +\begin{align*} + V(R2) = V(R3) &= V(n002) - V(n003) \\\ + V(R2) = V(R3) &= 2.56098 - 2.43902 \\\ + V(R2) = V(R3) &= \boxed{0.12196} +\end{align*} +$$ + +$$ +\begin{align*} + V(R4) &= V(n003) - V(\text{GND}) \\\ + V(R4) &= 2.43902 - 0 \\\ + V(R4) &= \boxed{2.43902} +\end{align*} +$$ + +让我们做一个表格来比较结果： + +|项目|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$2.4390V$|$2.4390V$|$2.4963V$|$57.28mV$|$2.3\%$| +|$V(R2)$|$0.1219V$|$0.1219V$|$0.1665V$|$44.54mV$|$26.8\%$| +|$V(R3)$|$0.1219V$|$0.1219V$|$0.1665V$|$44.54mV$|$26.8\%$| +|$V(R4)$|$2.4390V$|$2.4390V$|$2.4616V$|$22.58mV$|$0.9\%$| + +我们可以看到 $V(R1)$ 和 $V(R4)$ 的准确性非常高。但 $V(R2)$ 和 $V(R3)$ 误差较大，可能的原因是存在背景噪声。 + +如果查看“测量”部分，通道 2 是空的，但它仍然有大约 $50mV$ 的读数。这很可能是背景噪声。如果我们从实验测量中去除这种噪声，则误差百分比将小于 $1\%$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +*** + +现在，让我们检查 KCL。 + +我们得到了以下模拟数据： + +```text +I(R1): -0.000243902 device_current +I(R2): 0.000121951 device_current +I(R3): 0.000121951 device_current +I(R4): 0.000243902 device_current +I(V1): -0.000243902 device_current +``` + +根据分析结果，我们应看到 $I(R1) = I(R2) + I(R3)$。验证如下： + +$$ +\begin{align*} + I(R1) + I(R2) + I(R3) &= -0.000243902 + 0.000121951 + 0.000121951 \\\ + &= \boxed{0} +\end{align*} +$$ + +KCL 很可能为真。 + +*** + +然后，让我们检查 KVL。 + +我们可以使用前面部分的结果： + +$$ +\begin{align*} + V(R1) &= 2.43902 \\\ + V(R2) = V(R3) &= 0.12196 \\\ + V(R4) &= 2.43902 +\end{align*} +$$ + +根据分析结果，我们应期望 $V(n001) - V(n002) - V(n003) = 0$。验证如下： + +$$ +\begin{align*} + V(n001) - V(n002) - V(n003) &= 2.43902 - 0.12196 - 0.12196 \\\ + &= 0 \\\ + \boxed{\text{True}} +\end{align*} +$$ + +KVL 很可能为真。 + +*** + +最后，让我们检查欧姆定律。使用期望 $V = IR$ 和实验数据： + +```text +R1 = 10K +R2 = R3 = 1K +R4 = 10K +V(R1) = 2.4963V +V(R2) = V(R3) = 166.5mV +V(R4) = 2.4616V +``` + +根据欧姆定律计算 $I$。 + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R1) &= \frac{2.4963}{10000} = \boxed{0.00024963} +\end{align*} +$$ + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R2) = I(R3) &= \frac{0.1665}{1000} = \boxed{0.0001665} +\end{align*} +$$ + +$$ +\begin{align*} + V &= IR \\\ + I &= \frac{V}{R} \\\ + I(R4) &= \frac{2.4616}{10000} = \boxed{0.00024616} +\end{align*} +$$ + +然后，我们可以将这些电流结果与模拟数据进行比较。 + +|项目|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-:|-:| +|$I(R1)$|$0.2439mA$|$0.2439mA$|$0.2496mA$|$0.005728mA$|$2.3\%$| +|$I(R2)$|$0.1665mA$|$0.1665mA$|$0.1219mA$|$0.044549mA$|$26.8\%$| +|$I(R3)$|$0.1665mA$|$0.1665mA$|$0.1219mA$|$0.044549mA$|$26.8\%$| +|$I(R4)$|$0.2439mA$|$0.2439mA$|$0.2461mA$|$0.002258mA$|$0.9\%$| + +我们可以看到 $I(R1)$ 和 $I(R4)$ 的准确性非常高。但 $I(R2)$ 和 $I(R3)$ 误差较大，可能的原因是存在背景噪声。 + +如果查看“测量”部分，通道 2 是空的，但它仍然有大约 $50mV$ 的读数。这很可能是背景噪声。如果我们从实验测量中去除这种噪声，则误差百分比将小于 $1\%$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +此外，我们还可以检查整个电路中的总电流。根据分析中的期望 - $I_{total} = 0.000243902439$，这与模拟结果一致 - $0.000243902A$。 + +*** + +总之，模拟完全符合 KCL 和 KVL 的要求。实验数据接近于模拟值，并且如果去除背景噪声并考虑电阻的 $5\%$ 容差，则实验测量结果非常接近于模拟值。然后我们使用实验数据和欧姆定律来比较模拟结果，结果显示也非常接近。因此，我们在电路中验证了欧姆定律、KCL 和 KVL。 + +## 2. 验证串联电路中分压器的概念 + +### 电路图示 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Schematic.avif" caption="概念验证 - Omega 实验室 01 - 2 - 图纸" width=600px >}} + +### 描述 + +我将构建一个串联电路，其中包含两个电阻，并测量这些电阻上的电压以与理论值进行比较。 + +### 分析 + +分压器公式为： + +$$ +\frac{V_1}{V_2} = \frac{R_1}{R_2} +$$ + +如果电源电压为 $5V$ 且 $R1=R2=10K$，将这些值代入公式得到： + +$$ +\frac{V_1}{V_2} = \frac{10K}{10K} = \frac{1}{1} +$$ + +我们知道 $V_1 + V_2 = 5$ 并且 $1 \cdot V_1 = 1 \cdot V_2$。因此，我们期望 $V_1 = V_2 = 2.5$。 + +### 模拟 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Simulation.avif" caption="概念验证 - Omega 实验室 01 - 2 - 模拟结果" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +V(n002): 2.5 voltage +I(R1): -0.00025 device_current +I(R2): -0.00025 device_current +I(V1): -0.00025 device_current +``` + +### 测量 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement.avif" caption="概念验证 - Omega 实验室 01 - 2 - 测量结果" width=600px >}} + +$V(R1) = 2.5539V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 2 - 测量结果 - 1" width=600px >}} + +$V(R2) = 2.5204V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%202.avif" caption="概念验证 - Omega 实验室 01 - 2 - 测量结果 - 2" width=600px >}} + +### 讨论 + +首先，让我们比较理论值与实验测量值。 + +我们从 Analog Discovery 3 获取了以下实验读数： + +```text +V(R1) = 2.5539V +V(R2) = 2.5204V +``` + +电压是指两个点之间的电位差。基于这一点，我们可以计算 $V(R1)$ 和 $V(R2)$ 的理论值。 + +我们知道模拟输出为： + +```text +V(n001): 5 voltage +V(n002): 2.5 voltage +``` + +$$ +\begin{align*} + V(R1) &= V(n001) - V(n002) \\\ + V(R1) &= 5 - 2.5 \\\ + V(R1) &= \boxed{2.5} +\end{align*} +$$ + +$$ +\begin{align*} + V(R2) &= V(n002) - \text{GND} \\\ + V(R2) &= 2.5 - 0 \\\ + V(R2) &= \boxed{2.5} +\end{align*} +$$ + +让我们做一个表格来比较结果： + +|项目|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$2.5V$|$2.5V$|$2.5539V$|$0.0539V$|$2.1\%$| +|$V(R2)$|$2.5V$|$2.5V$|$2.5204V$|$0.0204V$|$0.8\%$| + +我们可以看到 $V(R1)$ 和 $V(R2)$ 的准确性非常高。有一些误差，可能的原因是存在背景噪声。 + +如果查看“测量”部分，通道 2 是空的，但它仍然有大约 $40mV$ 的读数。这很可能是背景噪声。如果我们从实验测量中去除这种噪声，则误差百分比将小于 $1\%$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +总之，模拟完全符合分压器理论公式的要求。实验读数接近于理论值，并且如果去除背景噪声，则实验读数非常接近于理论值。因此，我们在串联电路中验证了分压器的概念。 + +## 3. 验证电流在串联电路中流动的概念 + +### 电路图示 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Schematic.avif" caption="概念验证 - Omega 实验室 01 - 2 - 图纸" width=600px >}} + +### 描述 + +我将使用欧姆定律来找出串联电路中每个电阻中的电流，并将其与理论值进行比较。 + +### 分析 + +串联电路的特点是： + +- 只有一条路径供电流通过电路。 +- 在电路的任何一点，电流都相同。 + +由于 Analog Discovery 3 不能直接测量电流而是测量电压。我们将使用欧姆定律来找出流过电阻的电流。 + +我们知道从欧姆定律的关系为： + +$$ +V = IR +$$ + +可以稍作变换得到： + +$$ +I = \frac{V}{R} +$$ + +另外，我们还知道 $R_1 = R_2 = 10K$，并且可以通过分压器公式找到电阻上的电压。即： + +$$ +\begin{align*} + \frac{V_1}{V_2} &= \frac{R_1}{R_2} \\\ + \frac{V_1}{V_2} &= \frac{10K}{10K} = 1 +\end{align*} +$$ + +我们知道 $V_1 + V_2 = 5$ 并且 $V_1 = V_2$。因此，我们期望 $V_1 = V_2 = 2.5$。 + +使用这些值，我们可以计算出 $I(R1)$ 和 $I(R2)$： + +$$ +\begin{align*} + I(R1) = I(R2) &= \frac{V}{R} \\\ + I(R1) = I(R2) &= \frac{2.5}{10K} \\\ + I(R1) = I(R2) &= \boxed{0.00025} +\end{align*} +$$ + +我们期望 $I(R1)$ 和 $I(R2)$ 为 $0.00025A$。 + +### 模拟 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Simulation.avif" caption="概念验证 - Omega 实验室 01 - 2 - 模拟结果" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +V(n002): 2.5 voltage +I(R1): -0.00025 device_current +I(R2): -0.00025 device_current +I(V1): -0.00025 device_current +``` + +### 测量 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement.avif" caption="概念验证 - Omega 实验室 01 - 2 - 测量结果" width=600px >}} + +$V(R1) = 2.5539V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 2 - 测量结果 - 1" width=600px >}} + +$V(R2) = 2.5204V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%202%20-%20Measurement%20-%202.avif" caption="概念验证 - Omega 实验室 01 - 2 - 测量结果 - 2" width=600px >}} + +### 讨论 + +从模拟结果中， + +```text +I(R1): -0.00025 device_current +I(R2): -0.00025 device_current +``` + +这证明了 $I(R1) = I(R2)$，即： + +- 只有一条路径供电流通过电路。 +- 在电路的任何一点，电流都相同。 + +*** + +从测量结果中，我们得到了电阻 $R1$ 和 $R2$ 上的电压 + +$V(R1) = 2.5539V$ +$V(R2) = 2.5204V$ + +基于欧姆定律 - 我们在分析中得到的关系 $I = \frac{V}{R}$，我们可以计算出 $I(R1)$ 和 $I(R2)$： + +$$ +\begin{align*} + I(R1) &= \frac{V}{R} \\\ + I(R1) &= \frac{2.5539}{10K} \\\ + I(R1) &= 0.00025539 +\end{align*} +$$ + +$$ +\begin{align*} + I(R2) &= \frac{V}{R} \\\ + I(R2) &= \frac{2.5204}{10K} \\\ + I(R2) &= 0.00025204 +\end{align*} +$$ + +$R1$ 和 $R2$ 非常接近，可以认为 $R1 \approx R2$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以将其视为系统误差，实验测量结果与模拟非常接近。 + +*** + +总之，模拟完全符合串联电路中电流的特点。实验读数接近于理论值，并且如果考虑电阻的容差为 $5\%$（来自四色环电阻代码），则实验读数非常接近于理论值。因此，我们在串联电路中验证了电流流动的概念： + +- 只有一条路径供电流通过电路。 +- 在电路的任何一点，电流都相同。 + +## 4. 验证并联电路中电压的概念 + +### 电路图示 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Schematic.avif" caption="概念验证 - Omega 实验室 01 - 4 - 图纸" width=600px >}} + +### 描述 + +我将使用欧姆定律和节点特性来找出并联电路中每个电阻上的电压，并将其与理论值进行比较。 + +### 分析 + +并联电路的特点是： + +- 存在多条路径供电流通过电路。 +- 每个支路的电压相同，等于电源提供的电压。 + +$$ +V_{total} = V_1 = V_2 = V_3 = \ldots = V_n +$$ + +我们知道同一节点上的电压相同（它们由导线连接）。因此，我们期望 $V(R1) = V(R2)$。 + +另外，电压是两个点之间的电位差。基于这一点，我们可以计算出 $V(R1)$ 和 $V(R2)$ 的理论值： + +$$ +\begin{align*} + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +*** + +另外，我们还可以检查该电路中的电流。我们知道并联电路的总电流为： + +$$ +I_{total} = I_1 + I_2 + I_3 + \dots +$$ + +根据欧姆定律，我们知道： + +$$ +V = IR +$$ + +可以稍作变换得到： + +$$ +I = \frac{V}{R} +$$ + +并且，并联电阻的总阻值为： + +$$ +\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n} +$$ + +因此， + +$$ +R_{total} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}} +$$ + +结合以上公式，我们得到： + +$$ +I_{total} = \frac{V}{\cfrac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}}} +$$ + +将值代入公式： + +$$ +\begin{align*} + I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\\ + I_{total} &= \frac{5}{5K} \\\ + I_{total} &= \boxed{0.001} +\end{align*} +$$ + +我们可以检查这个结果以进一步确认。 + +### 模拟 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Simulation.avif" caption="概念验证 - Omega 实验室 01 - 4 - 模拟结果" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +I(R2): 0.0005 device_current +I(R1): 0.0005 device_current +I(V1): -0.001 device_current +``` + +### 测量 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement.avif" caption="概念验证 - Omega 实验室 01 - 4 - 测量结果" width=600px >}} + +$V(R1)=V(R2)=5.0305V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 4 - 测量结果 - 1" width=600px >}} + +### 讨论 + +首先，让我们比较理论值与实验测量值。 + +我们从 Analog Discovery 3 获取了以下实验读数： + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +为了找到 $V(R1)$ 和 $V(R2)$ 的理论值，我们需要做一些计算。 + +我们知道模拟输出为： + +```text +V(n001): 5 voltage +``` + +电压是两个点之间的电位差。基于这一点，我们可以计算出 $V(R1)$ 和 $V(R2)$ 的理论值： + +$$ +\begin{align*} + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +让我们做一个表格来比较结果： + +|项目|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| +|$V(R2)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| + +我们可以看到 $V(R1)$ 和 $V(R2)$ 的准确性非常高。有一些误差，可能的原因是存在背景噪声。 + +如果查看“测量”部分，通道 2 是空的，但它仍然有大约 $40mV$ 的读数。这很可能是背景噪声。如果我们从实验测量中去除这种噪声，则误差百分比将小于 $0.2\%$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +*** + +其次，我们可以通过检查总电流来进一步确认。 + +我们知道模拟输出为： + +```text +I(R2): 0.0005 device_current +I(R1): 0.0005 device_current +``` + +根据分析，我们期望 + +$$ +I_{total} = 0.001 +$$ + +并联电路的总电流为： + +$$ +I_{total} = I_1 + I_2 + I_3 + \dots +$$ + +因此， + +$$ +\begin{align*} + I_{total} &= I(R2) + I(R1) \\\ + I_{total} &= 0.0005 + 0.0005 \\\ + I_{total} &= 0.001 \\\ + & 0.001 = 0.001 \; \boxed{\text{True}} +\end{align*} +$$ + +我们的分析与模拟结果一致。 + +*** + +此外，我们还可以检查实验数据。由于 Analog Discovery 3 不能直接测量电流，我们需要使用欧姆定律来计算电流。 + +我们得到： + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +并且我们知道： + +```text +R1 = 10K +R2 = 10K +``` + +因此，我们可以计算出 $I(R1)$ 和 $I(R2)$： + +$$ +\begin{align*} + I &= \frac{V}{R} \\\ + I(R1) = I(R2) &= \frac{5.0305}{10000} \\\ + I(R1) = I(R2) &= \boxed{0.0005305} +\end{align*} +$$ + +$0.0005305 \approx 0.0005$，误差仅为 $0.6\%$（即使去除背景噪声后也小于 $0.2\%$）。我们的理论非常可能为真。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +*** + +总之，我们验证了模拟完全符合分析预期。实验数据仅比理论值高出 $0.2\%$ 到 $0.6\%$。因此，我们在并联电路中验证了电压的概念： + +- 存在多条路径供电流通过电路。 +- 每个支路的电压相同，并等于电源提供的电压。 + +## 5. 验证并联电路中分流器的概念 + +### 电路图示 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Schematic.avif" caption="概念验证 - Omega 实验室 01 - 4 - 图纸" width=600px >}} + +### 描述 + +我将使用欧姆定律来找出并联电路中每个电阻上的电流，并将其总和与理论值进行比较。 + +### 分析 + +并联电路的特点是： + +- 存在多条路径供电流通过电路。 +- 每个支路的电压相同，等于电源提供的电压。 +- 总电流进入并联电路后被分配到各个支路中。 + +$$ +V_{total} = V_1 = V_2 = V_3 = \ldots = V_n +$$ + +我们知道同一节点上的电压相同（它们由导线连接）。因此，我们期望 $V(R1) = V(R2)$。 + +另外，电压是两个点之间的电位差。基于这一点，我们可以计算出 $V(R1)$ 和 $V(R2)$ 的理论值： + +$$ +\begin{align*} + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +根据欧姆定律，我们知道： + +$$ +V = IR +$$ + +可以稍作变换得到： + +$$ +I = \frac{V}{R} +$$ + +因此， + +$$ +\begin{align*} + I(R1) &= \frac{V(R1)}{R1} \\\ + I(R1) &= \frac{5}{10K} \\\ + I(R1) &= \boxed{0.0005} +\end{align*} +$$ + +$$ +\begin{align*} + I(R2) &= \frac{V(R2)}{R2} \\\ + I(R2) &= \frac{5}{10K} \\\ + I(R2) &= \boxed{0.0005} +\end{align*} +$$ + +$I(R1)$ 和 $I(R2)$ 的关系可以表示为： + +$$ +\begin{align*} + \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\\ + \because V(R1) &= V(R2) \\\ + \therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\\ + \frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\\ + &\boxed{\frac{I(R1)}{I(R2)} = \frac{R2}{R1}} +\end{align*} +$$ + +在我们的情况下，$1 \cdot R1 = 1 \cdot R2$，所以： + +$$ +\frac{I(R1)}{I(R2)} = \frac{R2}{R1} = \frac{1}{1} +$$ + +因此，我们可以得到 $I_{total}$ 为： + +$$ +\begin{align*} + I_{total} &= \frac{V_{total}}{R_{total}} \\\ + I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\\ + I_{total} &= \frac{5}{5K} \\\ + I_{total} &= 0.001 +\end{align*} +$$ + +由于： + +$$ +\frac{I(R1)}{I(R2)} = \frac{1}{1} +$$ + +我们可以得到 $I(R1)$ 和 $I(R2)$ 为： + +$$ +\begin{align*} + I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\\ + I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\\ + I(R1) = I(R2) &= \boxed{0.0005} +\end{align*} +$$ + +此时，我们的逻辑是一致的： + +$$ +\begin{align*} + & \because R1 = R2 = 10K \\\ + & \because I(R1) = I(R2) = 0.0005 \\\ + & \because V(R1) = V(R2) = 5 \\\ + & \because I_{total} = I(R1) + I(R2) = 0.001 \\\ + & \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\\ + & \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2} +\end{align*} +$$ + +$I(R1) = I(R2) = 0.0005$ 的结果也通过 $I = \frac{V}{R}$ 进行了交叉验证。因此，我们非常有信心： + +$$ +\frac{I(R1)}{I(R2)} = \frac{R2}{R1} +$$ + +对于并联电路中任何电阻的电流（例如 $I(R1)$） + +$$ +I(R1) = I_{total} \times \frac {R1}{R1 + R2 + \cdots} +$$ + +### 模拟 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Simulation.avif" caption="概念验证 - Omega 实验室 01 - 4 - 模拟结果" width=600px >}} + +```text + --- Operating Point --- + +V(n001): 5 voltage +I(R2): 0.0005 device_current +I(R1): 0.0005 device_current +I(V1): -0.001 device_current +``` + +### 测量 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement.avif" caption="概念验证 - Omega 实验室 01 - 4 - 测量结果" width=600px >}} + +$V(R1)=V(R2)=5.0305V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%204%20-%20Measurement%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 4 - 测量结果 - 1" width=600px >}} + +### 讨论 + +首先，让我们比较理论值与实验测量值。 + +我们从 Analog Discovery 3 获取了以下实验读数： + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +为了找到 $V(R1)$ 和 $V(R2)$ 的理论值，我们需要做一些计算。 + +我们知道模拟输出为： + +```text +V(n001): 5 voltage +``` + +电压是两个点之间的电位差。基于这一点，我们可以计算出 $V(R1)$ 和 $V(R2)$ 的理论值： + +$$ +\begin{align*} + V(R1) = V(R2) &= 5 - 0 \\\ + V(R1) = V(R2) &= \boxed{5} +\end{align*} +$$ + +让我们做一个表格来比较结果： + +|项目|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-:|-:| +|$V(R1)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| +|$V(R2)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$| + +我们可以看到 $V(R1)$ 和 $V(R2)$ 的准确性非常高。有一些误差，可能的原因是存在背景噪声。 + +如果查看“测量”部分，通道 2 是空的，但它仍然有大约 $40mV$ 的读数。这很可能是背景噪声。如果我们从实验测量中去除这种噪声，则误差百分比将小于 $0.2\%$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +*** + +其次，让我们检查 $I(R1)$ 和 $I(R2)$ 的理论值。 + +由于 Analog Discovery 3 不能直接测量电流，我们需要使用欧姆定律来计算电流。 + +我们得到： + +```text +V(R1) = 5.0305V +V(R2) = 5.0305V +``` + +并且我们知道： + +```text +R1 = 10K +R2 = 10K +``` + +因此，我们可以计算出 $I(R1)$ 和 $I(R2)$： + +$$ +\begin{align*} + I &= \frac{V}{R} \\\ + I(R1) = I(R2) &= \frac{5.0305}{10000} \\\ + I(R1) = I(R2) &= \boxed{0.0005305} +\end{align*} +$$ + +|项目|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-:|-:| +|$I(R1)$|$0.5mA$|$0.5mA$|$0.50305mA$|$0.00305mA$|$0.6\%$| +|$I(R2)$|$0.5mA$|$0.5mA$|$0.50305mA$|$0.00305mA$|$0.6\%$| + +我们可以看到 $I(R1)$ 和 $I(R2)$ 的准确性非常高。有一些误差，可能的原因是存在背景噪声。 + +如果查看“测量”部分，通道 2 是空的，但它仍然有大约 $40mV$ 的读数。这很可能是背景噪声。如果我们从实验测量中去除这种噪声，则误差百分比将小于 $0.2\%$。考虑到电阻的容差为 $5\%$（来自四色环电阻代码），我们可以认为这是系统误差，而实验测量结果与模拟非常接近。 + +*** + +总之，我们验证了模拟完全符合分析预期。实验数据仅比理论值高出 $0.2\%$ 到 $0.6\%$。因此，我们在并联电路中验证了电流分流的概念： + +- 存在多条路径供电流通过电路。 +- 每个支路的电压相同，并等于电源提供的电压。 +- 总电流进入并联电路后被分配到各个支路中。 + +## 6. 验证温度传感电路中的分压器概念 + +### 电路图示 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Schematic.avif" caption="概念验证 - Omega 实验室 01 - 6 - 图纸" width=600px >}} + +### 描述 + +我们将使用 NTC 100K 作为热敏电阻，并将其读数与温度计和模拟结果进行比较，以检查其可靠性。 + +### 分析 + +NTC 热敏电阻使用 Beta 公式来计算特定温度下的电阻。公式如下： + +$$ +\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) +$$ + +可以将 $R_1$ 移到左边得到： + +$$ +R_1 = R_0 e^{\beta(T_1^{-1} - T_0^{-1})} +$$ + +由于我们想找出特定温度下的热敏电阻阻值。 + +*** + +所使用的热敏电阻是 NTC 100K，这意味着它在参考温度 $25 \degree C$ 下的阻值为 $100k \Omega$ + +$$ +T_0 = 298.15K \\ +R_0 = 100k \Omega +$$ + +制造商提供的 $\beta$ 值： + +$$ +\beta = 3950 +$$ + +我们知道分压器公式为： + +$$ +\frac{V_1}{V_2} = \frac{R_1}{R_2} +$$ + +将它们组合起来，我们得到： + +$$ +\frac{V_1}{V_2} = \frac{R_0 e^{\beta(T_1^{-1} - T_0^{-1})}}{R_2} +$$ + +### 模拟 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation.avif" caption="概念验证 - Omega 实验室 01 - 6 - 模拟结果" width=600px >}} + +我们得到了一个曲线，显示了温度在 $T = 0 \degree C$ 到 $T = 40 \degree C$ 范围内的关系。 + +对于 $T = 29.12 \degree C$，电压应为 $4.465V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 6 - 模拟结果 - 1" width=600px >}} + +对于 $T = 27.3 \degree C$，电压应为 $4.501V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation%20-%202.avif" caption="概念验证 - Omega 实验室 01 - 6 - 模拟结果 - 2" width=600px >}} + +对于 $T = 30.25 \degree C$，电压应为 $4.441V$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Simulation%20-%203.avif" caption="概念验证 - Omega 实验室 01 - 6 - 模拟结果 - 3" width=600px >}} + +### 测量 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement.avif" caption="概念验证 - Omega 实验室 01 - 6 - 测量结果" width=600px >}} + +我们使用了示波器中的数学函数： + +```js +1/((1/298.15)+(1/3950)*log((((10000*C1)/(5-C1))/100000),2.71828)) - 273.15 +``` + +来获取温度读数（单位为 $\degree C$）。这是从以下公式得出的： + +$$ +\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) +$$ + +其中 $T_1$ 是我们想要读取的温度。 + +然后，我们在 $24 \degree C$ 下校准了热敏电阻。 + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%202.avif" caption="概念验证 - Omega 实验室 01 - 6 - 测量结果 - 2" width=600px >}} + +之后，我们进行了三次测量： + +$V = 4.465V, T = 29.12 \degree C$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%201.avif" caption="概念验证 - Omega 实验室 01 - 6 - 测量结果 - 1" width=600px >}} + +$V = 4.502V, T = 27.3 \degree C$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%203.avif" caption="概念验证 - Omega 实验室 01 - 6 - 测量结果 - 3" width=600px >}} + +$V = 4.441V, T = 30.25 \degree C$ + +{{< figure src="Proof%20of%20Concept%20-%20Omega%20Lab%2001%20-%206%20-%20Measurement%20-%204.avif" caption="概念验证 - Omega 实验室 01 - 6 - 测量结果 - 4" width=600px >}} + +### 讨论 + +|温度|分析值|模拟值|实验值|差值|误差百分比| +|:-:|:-:|:-:|:-:|:-|-:| +|$29.12 \degree C$|$4.465V$|$4.465V$|$4.465V$|$0V$|$0\%$| +|$27.3 \degree C$|$4.501V$|$4.501V$|$4.502V$|$1mV$|$0.1\%$| +|$30.25 \degree C$|$4.441V$|$4.441V$|$4.441V$|$0V$|$0\%$| + +如我们所见，理论值与测量值之间的差异非常小。这可能是因为温度读数是通过电压计算得出的。 + +即使查看温度计的读数，两者都显示约为 $24 \degree C$。在最坏的情况下，误差为 $5\%$。因此，总体而言，我们的读数是可靠的。 + +*** + +惠斯通电桥比普通分压器更灵敏，因为它的输出电压受电阻比例变化的影响较小。当惠斯通电桥平衡时（即 $R1/R2 = R3/R4$），中心的电流表中的电流为零。此时，计算出的阻值不再受到导线、电阻和伏特计固有电阻的影响，从而使得测量结果更准确。 + +优点： +- 惠斯通电桥比分压器更准确 +- 电压源不需要校准即可测量阻值 + +缺点： +- 分压器更容易且成本更低制作 +- 功耗较低 diff --git a/themes/FixIt b/themes/FixIt index a073b0e..35cc57e 160000 --- a/themes/FixIt +++ b/themes/FixIt @@ -1 +1 @@ -Subproject commit a073b0e9f25f7fc9d0d67fe28c20c2b5b94c10f8 +Subproject commit 35cc57e64df9b89a5396647f78053c9eef6aa332