feat: finished ac started capacitors

This commit is contained in:
max_richter 2022-03-13 19:17:20 +01:00
parent 9966ab367f
commit cbb0463352
24 changed files with 1302 additions and 53 deletions

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# AC (Alternating Current)
![](./assets/wave.gif)
A few important characteristics of an AC Signal:
## Time Period
How much time passes during one cycle of the signal (until it reaches the same point it started twice)
## Frequency
How many cycles the signal completes in a specified time frame.
## Amplitute
The maximum voltage of the signal compared to 0
## Peak-Peak Voltage
The difference between the highest and lowest peak.
## Root Mean Square Values
This helps calculating the current an equivalent DC Signal would need to provide the same amount of power.
![](./assets/rms.gif)
$$
\begin{flalign}
V_{RMS} = 0.7 * V_{PEAK} &&\\
V_{PEAK} = 1.4 * V_{RMS}
\end{flalign}
$$

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@ -32,27 +32,26 @@ $$
$$ $$
\begin{flalign} \begin{flalign}
&\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &\\ \frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &&\\
\\ \\
&\textit{For two resistors in parallel:} &\\ \textit{For two resistors in parallel:} &&\\
\\ \\
&R = \frac{R1 * R2}{R1 + R2} R = \frac{R1 * R2}{R1 + R2} &&\\
\end{flalign} \end{flalign}
$$ $$
*Tip:* ***Tip:***
If two resistors of the same value are in parallel the total resistance is half the value of a single resistor If resistors of the same value are in parallel the total resistance is a single resistor divided by the amount if resistors.
# Kirchhoff's Law # Kirchhoff's Law
### Conservation of Charge ## Conservation of Charge (First Law)
All current entering a node in a circuit must also leave that node All current entering a node must also leave that node
$$ $$
\begin{flalign} \begin{flalign}
& \sum{I_{IN}} = \sum{I_{OUT}} & \sum{I_{IN}} = \sum{I_{OUT}}&&
\end{flalign} \end{flalign}
$$ $$
@ -63,54 +62,15 @@ $$
For this circuit kirchhoffs law states that: For this circuit kirchhoffs law states that:
$$ $$
\begin{flalign} \begin{flalign}
&i1 = i2 + i3 + i4 & i1 = i2 + i3 + i4 &&
\end{flalign} \end{flalign}
$$ $$
### Conservation of Energy ## Conservation of Energy (Second Law)
All the potential differences around the loop must sum to zero.
$$ $$
\begin{flalign} \begin{flalign}
& \sum{V} = 0 & \sum{V} = 0 &&
\end{flalign} \end{flalign}
$$ $$
**Example:**
![](./assets/kirchhoffs-law-02.svg)
For this circuit this means.
$$
\begin{flalign}
V_{R1} = -(I*R1) && \text{Voltage Drop across R}_{1} \\
V_{R2} = -(I*R2) && \text{Voltage Drop across R}_{2}\\
\\
V_{S} - IR_{1} - IR_{2} &= 0 & | + IR_{1} + IR_{2}\\
V_{S} &= IR_{1} + IR_{2} \\
\\
V_{S} &= I(R_{1}+R_{2}) \\
V_{S} &= I*R_{t} \\
\end{flalign}
$$
We can change that formula around to find out the current of the entire circuit, aswell as current through single resistors
$$
\begin{flalign}
& V_{S} = I*R_{t} & | \div R_{t} \\
\\
& I = \frac{V_{S}}{R_{t}} = \frac{V_{S}}{R_{1}+R_{2}} \\
\\
& VR_{1} = V_{S}(\frac{R1}{R1+R2}) \\
& VR_{2} = V_{S}(\frac{R2}{R1+R2}) \\
\\
& VR_{1} = 5(\frac{500}{500+1500}) \\
& VR_{1} = 1.25v
\\
& VR_{2} = 5(\frac{1500}{500+1500}) \\
& VR_{2} = 3.75v
\end{flalign}
$$

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# Kirchhoffs Law
### Example 1
**Example:**
![](../assets/kirchhoffs-law-02.svg)
For this circuit this means.
$$
\begin{flalign}
V_{R1} = -(I*R1) && \text{Voltage Drop across R}_{1} \\
V_{R2} = -(I*R2) && \text{Voltage Drop across R}_{2}\\
\\
V_{S} - IR_{1} - IR_{2} &= 0 & | + IR_{1} + IR_{2}\\
V_{S} &= IR_{1} + IR_{2} \\
\\
V_{S} &= I(R_{1}+R_{2}) \\
V_{S} &= I*R_{t} \\
\end{flalign}
$$
We can change that formula around to find out the current of the entire circuit, aswell as current through single resistors
$$
\begin{flalign}
I &= I*R_{t} = \frac{V_{S}}{R_{t}} = \frac{V_{S}}{R_{1}+R_{2}}\\
\\
VR_{1} &= V_{S}(\frac{R1}{R1+R2})\\
VR_{2} &= V_{S}(\frac{R2}{R1+R2})\\
\\
VR_{1} &= 5(\frac{500}{500+1500})\\
VR_{1} &= 1.25v&\\
\\
VR_{2} &= 5(\frac{1500}{500+1500})\\
VR_{2} &= 3.75v
\end{flalign}
$$
### Example 2
```circuitjs
$ 1 0.000005 10.20027730826997 50 5 43 5e-11
v 192 288 192 112 0 0 40 10 0 0 1
x 125 208 167 211 4 24 10V
r 192 112 336 112 0 10
r 336 112 480 112 0 20
w 480 112 480 288 0
r 336 112 336 288 0 50
w 480 288 336 288 0
w 336 288 192 288 0
x 304 204 319 207 4 12 R3
x 256 140 271 143 4 12 R1
x 400 138 415 141 4 12 R2
x 332 104 340 107 4 12 A
x 332 304 340 307 4 12 B
x 205 102 215 105 4 12 I1
x 467 105 477 108 4 12 I2
x 346 276 356 279 4 12 I3
```
In this circuit we have three major loops we can apply [[formulas#Conservation of Energy Second Law|Kirchhoffs Second Law]] to, the one on the left, the one on the right and the most outer one. We can also use [[formulas#Conservation of Charge First Law | Kirchhoffs First Law]] for the node title **A**.
$$
\begin{flalign}
&\textit{Node A:} \\\
& I_{1} = I_{2}+I_{3} \\
\\
&\textit{Left Loop:} \circlearrowright &\\
& 10_{v} - I_{1}*R_{1} - I_{3}*R_{3} = 0& \\
\\
&\textit{Right Loop} \circlearrowleft &\\
&I_{2}*R_{2} - I_{3}*R_{3} = 0; &\\
\\
&\textit{Outer Loop:} \circlearrowright &\\
& 10_{v} - I_{1}*R_{1} - I_{2}*R_{2} = 0& \\
\\
\end{flalign}
$$
Now if we would like to find out I2 for example we can use the Right Loop Formula to do so:
$$
\begin{flalign}
& \textit{Right Loop} \circlearrowleft &\\
& I_{2}*R_{2} - I_{3}*R_{3} = 0 &&| + I_{3}*R_{3}\\\
\\
& I_{2}*R_{2} = I_{3}*R_{3} &&| \div R_{2} \\
\\
& I_{2} = \frac{I_{3}*R_{3}}{R_{2}} &&| \text{ Input Numbers } \\
\\
& I_{2} = \frac{I_{3}*5}{2} &&| \text{ Simplify} \\
\\
& I_{2} = \frac{5I_{3}}{2} &&| \text{ Simplify} \\
\\
&\textit{LeftLoop:} \\
& 10 - I_{1}*R_{1} - I_{3}*R_{3} = 0&&| \text{ Replace I1 with node a} \\
& 10 - (I_{2}+I_{3})*R_{1} - I_{3}*R_{3} = 0 &&| \text{ Replace I2 with previous} \\
\\
& 10 - (\frac{5I_{3}}{2}+I_{3})*R_{1} - I_{2}*R_{2} = 0
\end{flalign}
$$
On the last formula we only have one unknown variable, which is $I_{3}$ so lets solve for that.
$$
\begin{flalign}
& 10 - (\frac{5I_{3}}{2}+I_{3})R_{1} - I_{3}R_{3} = 0 &&| \text{ Replace Variables}\\\
\\
& 10 - (2.5I_{3}+I_{3})10 - 50I_{3} = 0 && \\
\\
& 10 - 35I_{3} - 50I_{3} = 0 &&\\
& 10 - 85I_{3} = 0 && | +85I_{3}\\
& 10 = 85I_{3} && | \div 85 \\
& 0.117647059 \approx I_{3} \\
& 117.64mA = I_{3}
\end{flalign}
$$
Now lets found out $I_{2}$ aswell
$$
\begin{flalign}
I_{2} = \frac{I_{3}*5}{2} &&\\
I_{2} = \frac{0.11764 * 5}{2} &&\\
\\
I_{2} \approx 0.2941 && \\
\\
I_2 \approx 294mA && \\
\end{flalign}
$$

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@ -9,6 +9,31 @@ Current
## Ohms ## Ohms
Resistance Resistance
## Watt (Power)
$Power = V * I = \frac{V^{2}}{R} = I^{2}R$
Joules per Second
Term | Symbol | Weight
-----------|----|------
Nanowatt | nW | 10-9
Microwatt | $\micro$W | $10^{-6}$
Milliwatt | mW | $10^{-3}$
Watt | W | $10^{0}$
Kilowatt | kW | $10^{3}$
Megawatt | MW | $10^{6}$
Gigawatt | GW | $10^{9}$
**Examples:**
Device | Power
-------|----------
Arduino| 167mW
Laptop | 1.5W
House | 2.2kW
## Ohms Law ## Ohms Law
$$ $$
V = \frac{I}{R} V = \frac{I}{R}
@ -25,6 +50,16 @@ Means if a component is symmetric or not
Polarised means that a component is not symmetric Polarised means that a component is not symmetric
## Voltage Divider ## Voltage Divider
## Farad
Term | Symbol | Weight
-----------|----|------
Picofarad | pW | $10^{-12}$
Nanofarad | nF | $10^{-9}$
Microfarad | $\micro$F | $10^{-6}$
Milifarad | mF | $10^{-3}$
Kilofarad | kF | $10^{3}$
## LED ## LED
Anode - The shorter Leg Anode - The shorter Leg

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@ -0,0 +1,65 @@
# Capacitors
Capacity is measured in [[glossary#Farad|Farads]].
Capacity is calculated as follows:
$$
\begin{flalign}
& C = \epsilon r \frac{A}{4\pi d} &&\\\
\\
& \epsilon r = \text{Dielectrics relative permittivity} &&\\
& A = \text{Amount of Area the plates overlap} &&\\
& d = \text{Distance between plates} &&\\
\end{flalign}
$$
![](../assets/Parallel_plate_capacitor.svg)
### Important Metrics
**Size:**
Larger Capacity $\approx$ Larger Size
**Maximum Voltage**
Each capacitor has a maximum voltage that can be dropped across it.
**Leakage Current**
Capacitors are not perfect, and leak some current across the terminals.
**Equivalent series Resistance (ESR)**
The terminals are not 100% conductive, so the will have some very small resistance, (usually less than $0.01\ohm$)
**Tolerance**
The capacity is not always exact, the tolerance describes how much it could vary, usually about $\mp 1\%$ to $\mp 20\%$
## Ceramic Capacitors
- least expansive
- relative small usually $< 10\micro F$
- low current leakage and ESR
- best for high frequency coupling
![](../assets/ceramic-capacitor.webp)
## Aluminium and Tantalum Electrolytic
- Usually polarized
- Capacity usuially $1\micro F - 1mF$
- Good for high voltage
![](../assets/tantalum-capacitor.jpg)
## Super Capacitors
- Usually can handle only low voltage
- Capacity in the range of farads
## Film Capacitor
- usually low ESR
## Mica Capacitor
- Can work in hot environments > $200\deg$
- Low ESR
- High Precision
- High Cost

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@ -30,7 +30,7 @@ Our power source must have more voltage than the voltage drop, otherwise we cant
First we need to find the datasheet of the specific component, it can be easily found by googling it First we need to find the datasheet of the specific component, it can be easily found by googling it
![TLUR DataSheet](./datasheets/tlur6400.pdf) ![TLUR DataSheet](tlur6400.pdf)
Now lets put that LED into a test circuit and calculate the resistance for it: Now lets put that LED into a test circuit and calculate the resistance for it:

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@ -0,0 +1,3 @@
# Resistors
Resistors transform voltage into heat.

0
a.md Normal file
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