210 lines
3.9 KiB
Markdown
210 lines
3.9 KiB
Markdown
High pass filters let through only signals with high frequencies and attenuate low frequency one.
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Because high pass filters work exactly like low pass filters but in reverse, lets only do one example here:
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**Example:**
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![[rc-high-pass-example.png|300]]
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Lets first calculate the cutoff frequency of this filter:
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[[Resources/electricity/formulas|Formulas]]
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<!-- #include [[Resources/electricity/formulas]] -->
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---
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cards-deck: electricity
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---
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## Ohms Law
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*Solve for voltage:*
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#card
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$\displaystyle V = I*R$
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^1654598090369
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*Solve for resistance:*
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#card
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$\displaystyle R = \frac{V}{I}$
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^1654598090389
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*Solve for current*
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#card
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$\displaystyle I = \frac{V}{R}$
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^1654598090398
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## Resistors in Series
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#card
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$R = R1 + R2 + R3 ...$
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^1654598090404
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## Resistors in Parallel
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#card
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$$
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\begin{flalign}
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&\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &\\
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\\
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&\textit{For two resistors in parallel:} &\\
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\\
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&R = \frac{R1 * R2}{R1 + R2} &\\\
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\end{flalign}
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$$
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***Tip:***
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If resistors of the same value are in parallel the total resistance is a single resistor divided by the amount if resistors.
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## Voltage Divider
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#card
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^1654598090410
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$V_{out} = V_{in}(\frac{R_{1}}{R_1+R_2})$
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## Thevenin’s Theorem
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States that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.
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## Conservation of Charge (First Law)
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#card
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All current entering a node must also leave that node
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$$
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\begin{flalign}
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\sum{I_{IN}} = \sum{I_{OUT}}&&
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\end{flalign}
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$$
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**Example:**
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^1654598090415
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For this circuit kirchhoffs law states that:
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$\displaystyle i1 = i2 + i3 + i4$
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## Conservation of Energy (Second Law)
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All the potential differences around the loop must sum to zero.
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$\displaystyle \sum{V} = 0$
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## Capacitors in Series
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#card
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$\displaystyle \frac{1}{C_{t}} = \frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} ...$
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^1654598090421
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## Impedance in a Circuit
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#card
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$$
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\begin{flalign}
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&Z = \sqrt{R^2 + X^2} &\\\
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\\
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&X = X_{L} - X_{C} \\
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\end{flalign}
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$$
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## Capacitive Reactance
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#card
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^1654598090426
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$\displaystyle X_{c} = \frac{1}{2 \pi fC}$
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## Inductive Reactance
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#card
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$\displaystyle X_{l} = 2\pi fL$
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^1654598090432
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## Analog Filters
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## Cutoff Frequency for RC Filters
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#card
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$\displaystyle f_{c} = \frac{1}{2\pi RC}$
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^1654598090437
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## Cutoff Frequency for RL Filters
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#card
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$\displaystyle f_{c} = \frac{R}{2\pi L}$
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^1654598090445
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## Cutoff Frequency for multiple Low Pass Filters
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$\displaystyle f_{(-3db)} = f_{c}\sqrt{2^{(\frac{1}{n})}-1}$
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Where $n$ = Number if **identical** filters
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## Resonance Frequency for RLC Low Pass Filter
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#card
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$\displaystyle f_{o} = \frac{1}{2\pi \sqrt{LC}}$
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^1654598090452
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## Center Frequency with Fc and Fh
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#card
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$f_{c} = \sqrt{f_{h}*f_{l}}$
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^1654598090459
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## Filter Response for RC Filters
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#card
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$V_{out} = V_{in}(\frac{X_c}{\sqrt{R_{1}^2+X_{c}^2}})$
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^1654598090466
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## Cutoff Frequency $\pi$ Topology Filter
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#card
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When the two capacitors have the same capacitance, it can be calculated like this:
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^1654598090479
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$\displaystyle f_c = \frac{1}{4\pi\sqrt{LC}}$
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## Angular Frequency ($\omega$)
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#card
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$\omega = 2\pi f = \frac{2\pi}{T}$
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^1654598090492
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## RLC Series Response
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This is basically Ohms Law:
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$\displaystyle V = IZ$
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Where $Z$ is the impedance:
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$Z = \sqrt{R^2 + (X_L - X_C)^2}$
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$X_L$ = Reactive Inductance
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$X_C$ = Reactive Capacativw
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## Current through a transistor
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$\displaystyle I_{EQ} = \frac{V_{BB}-{V_{BE}}}{\frac{R_B}{(\beta+1)}+R_E}$
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## Gain Bandwidth Product
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#card
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$GBP = A_V * f_c$
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^1654598090498
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$\displaystyle f_c = \frac{GBP}{A_V}$
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## Bandwidth of Multiple OpAmps
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Where $n$ = number of stages
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and $BW$ = Bandwidth of single op-amp
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$BW_E = BW\sqrt{2^\frac{1}{n}-1}$
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## Power lost in a Resistor
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#card
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$P = IV = I^2R = \frac{V^2}{R}$
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^1654598090504
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<!-- /include -->
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```latex
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\displaystyle f_{c} = \frac{1}{2\pi 100 * 0.00000001}
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\displaystyle f_{c} = 159154.94 \approx 159.1kHz
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``` |