3.9 KiB
High pass filters let through only signals with high frequencies and attenuate low frequency one.
Because high pass filters work exactly like low pass filters but in reverse, lets only do one example here:
Example:
Lets first calculate the cutoff frequency of this filter:
Resources/electricity/formulas
cards-deck: electricity
Ohms Law
Solve for voltage: #card
$\displaystyle V = I*R$ ^1654598090369
Solve for resistance: #card
$\displaystyle R = \frac{V}{I}$ ^1654598090389
Solve for current #card
$\displaystyle I = \frac{V}{R}$ ^1654598090398
Resistors in Series
#card
$R = R1 + R2 + R3 ...$ ^1654598090404
Resistors in Parallel
#card
\begin{flalign}
&\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &\\
\\
&\textit{For two resistors in parallel:} &\\
\\
&R = \frac{R1 * R2}{R1 + R2} &\\\
\end{flalign}
Tip: If resistors of the same value are in parallel the total resistance is a single resistor divided by the amount if resistors.
Voltage Divider
#card ^1654598090410
V_{out} = V_{in}(\frac{R_{1}}{R_1+R_2})
Thevenin’s Theorem
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.
Conservation of Charge (First Law)
#card
All current entering a node must also leave that node
\begin{flalign}
\sum{I_{IN}} = \sum{I_{OUT}}&&
\end{flalign}
Example: ^1654598090415
For this circuit kirchhoffs law states that:
\displaystyle i1 = i2 + i3 + i4
Conservation of Energy (Second Law)
All the potential differences around the loop must sum to zero.
\displaystyle \sum{V} = 0
Capacitors in Series
#card
$\displaystyle \frac{1}{C_{t}} = \frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} ...$ ^1654598090421
Impedance in a Circuit
#card
\begin{flalign}
&Z = \sqrt{R^2 + X^2} &\\\
\\
&X = X_{L} - X_{C} \\
\end{flalign}
Capacitive Reactance
#card ^1654598090426
\displaystyle X_{c} = \frac{1}{2 \pi fC}
Inductive Reactance
#card
$\displaystyle X_{l} = 2\pi fL$ ^1654598090432
Analog Filters
Cutoff Frequency for RC Filters
#card
$\displaystyle f_{c} = \frac{1}{2\pi RC}$ ^1654598090437
Cutoff Frequency for RL Filters
#card
$\displaystyle f_{c} = \frac{R}{2\pi L}$ ^1654598090445
Cutoff Frequency for multiple Low Pass Filters
\displaystyle f_{(-3db)} = f_{c}\sqrt{2^{(\frac{1}{n})}-1}
Where n = Number if identical filters
Resonance Frequency for RLC Low Pass Filter
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$\displaystyle f_{o} = \frac{1}{2\pi \sqrt{LC}}$ ^1654598090452
Center Frequency with Fc and Fh
#card
$f_{c} = \sqrt{f_{h}*f_{l}}$ ^1654598090459
Filter Response for RC Filters
#card
$V_{out} = V_{in}(\frac{X_c}{\sqrt{R_{1}^2+X_{c}^2}})$ ^1654598090466
Cutoff Frequency \pi Topology Filter
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When the two capacitors have the same capacitance, it can be calculated like this: ^1654598090479
\displaystyle f_c = \frac{1}{4\pi\sqrt{LC}}
Angular Frequency (\omega)
#card
$\omega = 2\pi f = \frac{2\pi}{T}$ ^1654598090492
RLC Series Response
This is basically Ohms Law:
\displaystyle V = IZ
Where Z is the impedance:
Z = \sqrt{R^2 + (X_L - X_C)^2}
X_L = Reactive Inductance
X_C = Reactive Capacativw
Current through a transistor
\displaystyle I_{EQ} = \frac{V_{BB}-{V_{BE}}}{\frac{R_B}{(\beta+1)}+R_E}
Gain Bandwidth Product
#card
$GBP = A_V * f_c$ ^1654598090498
\displaystyle f_c = \frac{GBP}{A_V}
Bandwidth of Multiple OpAmps
Where n = number of stages
and BW = Bandwidth of single op-amp
BW_E = BW\sqrt{2^\frac{1}{n}-1}
Power lost in a Resistor
#card
$P = IV = I^2R = \frac{V^2}{R}$ ^1654598090504
\displaystyle f_{c} = \frac{1}{2\pi 100 * 0.00000001}
\displaystyle f_{c} = 159154.94 \approx 159.1kHz