Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
|
circuit_design:5_filter_circuits_i [2021/11/17 14:34] slinn |
circuit_design:5_filter_circuits_i [2023/07/17 11:59] (aktuell) mexleadmin |
||
|---|---|---|---|
| Zeile 3: | Zeile 3: | ||
| < | < | ||
| - | * Also recommended for basic circuits II is the nice introduction and textbook found [[https://wiki.mexle.org/circuit_design/3_opamp_basic_circuits_i|here]]. | + | * Also recommended for basic circuits II is the nice introduction and textbook found [[https://link.springer.com/book/10.1007%2F978-3-319-28127-8|here]]. |
| </ | </ | ||
| Zeile 11: | Zeile 11: | ||
| === Introductory example === | === Introductory example === | ||
| - | < | + | < |
| - | In various | + | Various |
| - | One possibility to process such signals is the use of filters. Filters have already been described in [[:electrical_engineering_2:networks_at_variable_frequency|electrical_engineering 2]]. The classical $R C$-filters are passive. This means that although the voltage value is filtered, the output current of the filter is always lower than the current measured at the input. To enable better filtering and subsequent | + | One possibility to process such noisy signals is the use of filters. Filters have already been described in [[:electrical_engineering_1:circuits_under_different_frequencies|Electrical Engineering 1]]. The classical $R C$-filters are passive. This means that although the voltage value is filtered, the output current of the filter is always lower than the current measured at the input. To enable better filtering and subsequently optimize the use of the signal, active filters can be applied. These are often built up by operational amplifiers. |
| - | Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{A1}$ after the first filter stage already shows significantly less noise. In the signal $U_{A2}$, even less noise is visible, but the rising and falling edges are also no longer displayed sharply. | + | Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{O1}$ after the first filter stage already shows significantly less noise. In the signal $U_{O2}$, even less noise is visible, but the rising and falling edges are also no longer displayed sharply. |
| - | With the switch (left in the simulation) a frequency-variable test signal ([[https://de.wikipedia.org/ | + | With the switch (left in the simulation) a frequency-variable test signal ([[https://en.wikipedia.org/ |
| This chapter will explain the basics of active high and low pass filters from operational amplifiers. | This chapter will explain the basics of active high and low pass filters from operational amplifiers. | ||
| Zeile 31: | Zeile 31: | ||
| After this lesson, you should: | After this lesson, you should: | ||
| - | - Be able to apply the superpostition | + | - Be able to apply the superposition |
| - | - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, | + | - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, |
| - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like. | - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like. | ||
| - | - Be able to name applications for the inverse-radiator, voltage-to-current converter, and current-to-voltage converter. | + | - Be able to name applications for the inverse |
| </ | </ | ||
| Zeile 40: | Zeile 40: | ||
| ===== 5.0 Representation of numerical values ===== | ===== 5.0 Representation of numerical values ===== | ||
| - | In order to be able to analyse | + | To be able to analyze |
| < | < | ||
| Zeile 48: | Zeile 48: | ||
| </ | </ | ||
| - | ==== 5.0.1 The dB measure | + | ==== 5.0.1 The dB scale ==== |
| - | The decibel measure | + | The Decibel scale is an auxiliary unit of measure that facilitates handling with ratios (e.g. $U_2/U_1$). These ratios are called **level** in engineering |
| The conversion to a level in $dB$ is defined for current or voltage ratios by the following equation: | The conversion to a level in $dB$ is defined for current or voltage ratios by the following equation: | ||
| - | $\boxed{A_V^{dB}=20 dB \cdot log_{10}\left(\frac{U_2}{U_1}\right)=20 dB \cdot log_{10}A_V}$ resp. | + | $\boxed{A_{\rm V}^{\rm dB}=20 |
| - | <WRAP right> | + | \\ |
| - | + | === Technical | |
| - | ^Name^Symbol^Formula^Reference value for 0dB| | + | <WRAP> |
| - | |[[wpde> | + | |
| - | |[[wpde> | + | |
| - | |[[wpde> | + | |
| - | |[[wpde> | + | |
| - | |[[wpde>sound pressure level]] | + | |
| + | ^ Name ^ Symbol | ||
| + | | [[wpde> | ||
| + | | [[wp> | ||
| + | | [[wp> | ||
| + | | [[wp> | ||
| + | | [[wp> | ||
| </ | </ | ||
| - | Note that this equation changes somewhat for __power__ | + | Note that this equation changes somewhat for __power__ |
| + | Since $P \sim U^2$ or $U \sim P^\frac{1}{2}$, | ||
| + | $A_P^{\rm dB}= | ||
| + | (nbsp)(nbsp)(nbsp)(nbsp)(nbsp)(nbsp) | ||
| + | $ =\color{blue}{20 | ||
| - | To the right of the table are various levels frequently used in engineering. In the following, only the voltage level is used and indicated with the symbol $dB$. | + | The table above shows various |
| - | + | === Simple examples of voltage levels in dB === | |
| - | <WRAP right> | + | |
| - | + | ||
| - | ^linear factor^level [$dB$]| | + | |
| - | |$\times 0.01$|$-40dB$| | + | |
| - | |$\times 0.1$|$-20dB$| | + | |
| - | |$\times 1$|$0dB$| | + | |
| - | |$\times 2$|$\approx +6dB$| | + | |
| - | |$\times 10$|$+20dB$| | + | |
| - | |$\times 100$|$+40dB$| | + | |
| + | < | ||
| + | ^ linear factor | ||
| + | |$\times 0.01$ |$-40 ~{\rm dB}$ | | ||
| + | |$\times 0.1$ |$-20 ~{\rm dB}$ | | ||
| + | |$\times 1$ |$ 0 ~{\rm dB}$ | | ||
| + | |$\times 2$ |$\approx +6~{\rm dB}$| | ||
| + | |$\times 10$ |$+20 ~{\rm dB}$ | | ||
| + | |$\times 100$ |$+40 ~{\rm dB}$ | | ||
| </ | </ | ||
| - | By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{dB}$ in $dB$. Occasionally, | + | By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{\rm dB}$ in $\rm dB$. |
| + | Occasionally, | ||
| Examples: | Examples: | ||
| - | - For $A_V= \color{green}{1} $ we get $ A_V^{dB}(\color{green}{1}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{1}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{1})} \quad = 20 dB \cdot 0 \quad \ \boldsymbol{= 0 dB}$ Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, | + | - For $A_{\rm V}= \color{green}{1} $ we get < |
| - | - For $A_V= \color{green}{0,01} $ we get $ A_V^{dB}(\color{green}{0,01}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{0,01}\right)}_\color{blue}{x} \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{0,01})} = 20 dB \cdot (-2) \boldsymbol{= -40 dB}$ Here $\color{blue}{x}$ has the value, that gives $10^\color{blue}{x} = \color{green}{0.01}$, | + | $ A_{\rm V}^{\rm dB}(\color{green}{1}) = 20 {\rm dB} \cdot \underbrace{\log_{10}\left(\color{green}{1}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{1})} \quad = 20 {\rm dB} \cdot 0 \quad \ \boldsymbol{= 0 {\rm dB}}$ \\ |
| - | - For $A_V= \color{green}{2} $, we get $ A_V^{dB}(\color{green}{2}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{2}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{2})} \quad\approx 20 dB \cdot 0.30103 \boldsymbol{\approx 6 dB}$ Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$. \\ \\ | + | Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, |
| + | - For $A_{\rm V}= \color{green}{0.01} $ we get < | ||
| + | $ A_{\rm V}^{\rm dB}(\color{green}{0.01}) = 20 {\rm dB} \cdot \underbrace{\log_{10}\left(\color{green}{0.01}\right)}_\color{blue}{x} \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{0.01})} = 20 {\rm dB} \cdot (-2) \boldsymbol{= -40 {\rm dB}}$ \\ | ||
| + | Here $\color{blue}{x}$ has the value, that gives $10^\color{blue}{x} = \color{green}{0.01}$, | ||
| + | - For $A_{\rm V}= \color{green}{2} $, we get < | ||
| + | $ A_{\rm V}^{\rm dB}(\color{green}{2}) = 20 {\rm dB} \cdot \underbrace{\log_{10}\left(\color{green}{2}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{2})} \quad\approx 20 {\rm dB} \cdot 0.30103 \boldsymbol{\approx 6 {\rm dB}}$ \\ | ||
| + | Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$. | ||
| === Use of the dB measure === | === Use of the dB measure === | ||
| Zeile 95: | Zeile 106: | ||
| The decibel offers some advantages, which are used in the filter elements considered below: | The decibel offers some advantages, which are used in the filter elements considered below: | ||
| - | * **handier numerical values**: If very large or very small linear values are needed, the number of the resulting level has fewer digits. Example: $A_V = 10000000 \rightarrow | + | * **handier numerical values**: If very large or very small linear values are needed, the number of the resulting level has fewer digits. Example: $A_{\rm V} = 10000000 \rightarrow |
| - | * **Relationship to Sensory Perceptions**: | + | * **Relationship to Sensory Perceptions**: |
| - | * **easier math**: The logarithm in the defining equation turns any multiplication of linear factors into an addition of levels: $A_V^{dB}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1) + 20dB \cdot log_{10}(A_2) = A_V^{dB}(A_1) + A_V^{dB}(A_2)$ | + | * **easier math**: The logarithm in the defining equation turns any multiplication of linear factors into addition of levels: $A_{\rm V}^{\rm dB}(A_1 \cdot A_2) = 20~{\rm dB} \cdot \log_{10}(A_1 \cdot A_2) = 20~{\rm dB} \cdot \log_{10}(A_1) + 20~{\rm dB} \cdot \log_{10}(A_2) = A_{\rm V}^{\rm dB}(A_1) + A_{\rm V}^{\rm dB}(A_2)$ |
| - | ~~PAGEBREAK~~ | + | |
| - | <WRAP right>< | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | Especially the last point of the calculation should be considered again. In <imgref pic0> several amplifiers connected in series can be seen with exemplary voltage gain values. \\ The total gain here is the product of the individual gains: $A_{V,ges}=\prod | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| + | |||
| + | Especially the last point of the calculation should be considered again. In <imgref pic0> several amplifiers connected in series can be seen with exemplary voltage gain values. \\ | ||
| + | The total gain here is the product of the individual gains: $A_{\rm V, eq}=\prod | ||
| + | The determination of the total gain was rather laborious before the time of the pocket calculator due to the multiplications. | ||
| + | For the levels, the result is an addition: $A_{\rm V,eq}^{\rm dB}=\sum A_i^{\rm dB} = A_1^{\rm dB} + A_2^{\rm dB} + A_3^{\rm dB}$. \\ | ||
| + | Here this would be: $A_{\rm V,eq}^{\rm dB} = \sum A^{\rm dB} = 88~{\rm dB} + (-58~{\rm dB}) + 14~{\rm dB} = 44~{\rm dB}$. | ||
| <WRAP column 100%> <panel type=" | <WRAP column 100%> <panel type=" | ||
| Zeile 108: | Zeile 128: | ||
| For current and voltage levels: | For current and voltage levels: | ||
| - | - A linear factor of $\color{green}{\times 10}$ results in level $+ 20dB$. | + | - A linear factor of $\color{green}{\times 10}$ results in level $+ 20~{\rm dB}$. |
| - | - A linear factor of $\color{green}{\times 2}$ results in a level of $+ 6dB$. | + | - A linear factor of $\color{green}{\times 2} $ results in a level of $+ 6~{\rm dB}$. |
| - | - The linear value $A_V = 1$ corresponds to $0 dB$. | + | - The linear value $A_{\rm V} = 1$ corresponds to $0 ~{\rm dB}$. |
| For systems connected in series, to determine the amplification | For systems connected in series, to determine the amplification | ||
| - | - multiply the linear measure $A_V$ and | + | - multiply the linear measure $A_\rm V$ and |
| - | - add the level $A_V^{dB}$. | + | - add the level $A_{\rm V}^{\rm dB}$. |
| </ | </ | ||
| - | ~~PAGEBREAK~~ | + | ~~PAGEBREAK~~ |
| + | |||
| + | \\ | ||
| + | === More difficult examples of voltage levels in dB=== | ||
| - | With this knowledge, the interpolation points $\color{green}{\times 10} \rightarrow + 20dB$ and $\color{green}{\times 2} \rightarrow + 6dB$ the linear values can easily be determined from a level in $dB$ without a calculator. | + | With this knowledge, the interpolation points $\color{green}{\times 10} \rightarrow + 20~{\rm dB}$ and $\color{green}{\times 2} \rightarrow + 6~{\rm dB}$ the linear values can easily be determined from a level in $dB$ without a calculator. |
| Examples: | Examples: | ||
| + | - < | ||
| + | $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$ \\ | ||
| + | This becomes linear | ||
| + | - < | ||
| + | $A_{\rm V}^{dB}=56~{\rm dB} = 80~{\rm dB} - 24~{\rm dB} = \color{blue}{4}\cdot 20~{\rm dB} + \color{magenta}{-4}\cdot 6~{\rm dB}$ \\ | ||
| + | This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \quad\ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad | ||
| + | - < | ||
| + | $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB} + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\ | ||
| + | This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad | ||
| - | - $A_V^{dB}=58dB$ | + | \\ |
| - | | + | The value $-3~{\rm dB}$ will still be used in the following examples. |
| - | - $A_V^{dB}=55dB$ | + | |
| - | ~~PAGEBREAK~~ | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| ==== 5.0.2 The Bode diagram ==== | ==== 5.0.2 The Bode diagram ==== | ||
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | The aim of the floor diagram | + | The Bode diagram |
| === Preliminary consideration: | === Preliminary consideration: | ||
| - | A complex number can always be reduced to two real number values. For the exact definition of these numerical values there are different possibilities (<imgref picA>): | + | A complex number can always be reduced to two real number values. For the exact definition of these numerical values, there are different possibilities (<imgref picA>): |
| - | - Definition over real part $\Re(\underline{A}_V)=A_V \cdot cos(\varphi)$ and imaginary part $\Im(\underline{A}_V)=A_V \cdot sin(\varphi)$ in $\underline{A}_V= \Re(\underline{A}_V) + j \cdot \Im(\underline{A}_V)$ | + | - Definition over real part $\Re(\underline{A}_{\rm V})=A_{\rm V} \cdot \cos(\varphi)$ and imaginary part $\Im(\underline{A}_{\rm V})=A_{\rm V} \cdot \sin(\varphi)$ in $\underline{A}_{\rm V}= \Re(\underline{A}_{\rm V}) + {\rm j} \cdot \Im(\underline{A}_{\rm V})$ |
| - | - Definition over magnitude | + | - Definition over absolute value $A_{\rm V} = |\underline{A}_{\rm V}|$ and phase $\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right)$ in $\underline{A}_{\rm V}=A_{\rm V} \cdot {\rm e}^{{\rm j} \varphi}$ |
| - | The 2nd definition is more appropriate when considering frequency-dependent voltage gain, since it allows the "time shift" (phase) to be separated from the gain. | + | The 2nd definition is more appropriate when considering frequency-dependent voltage gain since it allows the "time shift" (phase) to be separated from the gain. |
| ~~PAGEBREAK~~ | ~~PAGEBREAK~~ | ||
| - | === path to the Bode diagram | + | === Developing |
| - | < | + | < |
| - | To better understand the frequency dependence of the voltage gain, it can be plotted as $|A_V(f)|$ as a function of frequency. It is useful to represent the voltage gain as level $|A_V^{dB}(f)|$. The simulation | + | To better understand the frequency dependence of the voltage gain, it can be plotted |
| - | In this type of display, it is difficult to determine values such as the cut-off frequency from the frequency curve. However, if the frequency axis is also logarithmized, | + | In this type of display, it is difficult to determine values such as the cut-off frequency from the frequency curve. However, if the frequency axis is also logarithmized, |
| The phase can be made visible via '' | The phase can be made visible via '' | ||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| === Description of the Bode diagram === | === Description of the Bode diagram === | ||
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| The bottom diagram allows the **representation of a complex-valued, | The bottom diagram allows the **representation of a complex-valued, | ||
| Zeile 169: | Zeile 210: | ||
| A jump in frequency by a factor of $\times 10$ is called a **decade** | A jump in frequency by a factor of $\times 10$ is called a **decade** | ||
| - | - $|A_V(f)| = \mathcal{C} \cdot f$: If a function is considered that increases linearly with $f$, then a tenfold increase in frequency results in a tenfold increase in voltage gain. This results in an increase of $+20dB$ per decade. | + | - $|A_{\rm V}(f)| = \mathcal{C} \cdot f$: \\ If a function is considered that increases linearly with $f$, then a tenfold increase in frequency results in a tenfold increase in voltage gain. \\ This results in an increase of $+20~{\rm dB}$ per decade. |
| + | | ||
| + | | ||
| - | As an alternative to the actual course, $|A_V(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments. | + | As an alternative to the actual course, $|A_{\rm V}(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments. |
| ~~PAGEBREAK~~ <WRAP column 100%> <panel type=" | ~~PAGEBREAK~~ <WRAP column 100%> <panel type=" | ||
| Zeile 180: | Zeile 223: | ||
| - Phase response: linear phase over logarithmic frequency (i.e. single logarithmic representation) | - Phase response: linear phase over logarithmic frequency (i.e. single logarithmic representation) | ||
| - | This gives a straight line in the amplitude response for functions of the form $A(\omega) \sim \omega^n$. | + | This gives a straight line in the amplitude response for functions of the form $A(\omega) \sim \omega^n$. |
| + | In particular, this is true for $A(\omega) \sim \omega$, i.e., a slope of $+ ~\rm 20dB/decade$, and for $A(\omega) \sim \frac{1}{\omega}$, | ||
| + | </ | ||
| - | ===== 5.1 Reverse integrator | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| + | ===== 5.1 Inverse Integrator | ||
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| Up to now, operational amplifier circuits have been considered in which negative feedback took place via ohmic resistors. In the following, operational amplifier circuits with storing components ($C$, $L$) will be analyzed. | Up to now, operational amplifier circuits have been considered in which negative feedback took place via ohmic resistors. In the following, operational amplifier circuits with storing components ($C$, $L$) will be analyzed. | ||
| Zeile 190: | Zeile 240: | ||
| In electronics, | In electronics, | ||
| - | - Inductors are possible in integrated circuits, but are somewhat more difficult to calculate as such an element. | + | - Inductors are possible in integrated circuits but are somewhat more difficult to calculate as such an element. |
| - Inductors require a current source as current storage. The internal resistance results in a continuous power loss. | - Inductors require a current source as current storage. The internal resistance results in a continuous power loss. | ||
| Instead of inductors, capacitors are used in microelectronics and filter technology for sensor signals. The passive circuits formed are correspondingly also called $R$-$C$-links and were analyzed in [[: | Instead of inductors, capacitors are used in microelectronics and filter technology for sensor signals. The passive circuits formed are correspondingly also called $R$-$C$-links and were analyzed in [[: | ||
| - | The following basic circuit is a modified, [[: | + | The following basic circuit is a modified, [[: |
| - | For first circuit only the part between output voltage $U_A$ and virtual ground should be replaced by a capacitor (<imgref pic2>). | + | For the first circuit, only the part between output voltage $U_{\rm O}$ and virtual ground should be replaced by a capacitor (<imgref pic2>). |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 203: | Zeile 253: | ||
| ==== 5.1.1 Circuit analysis with differential equations ==== | ==== 5.1.1 Circuit analysis with differential equations ==== | ||
| - | < | + | < |
| - | The first active filter circuit can be seen on the right side of the simulation. The superimposed square-wave voltage sources result in a step function as input voltage. This generates a current via the RC element. If we now look at the output voltage in comparison to the input voltage, we can see that: | + | The first active filter circuit can be seen in the simulation |
| - | - for each constant input value $U_I \neq 0$ an output value with a fixed slope results and | + | - for each constant input value $U_{\rm I} \neq 0$ an output value with a fixed slope results and |
| - | - for each positive input value $U_I > 0$ a negative slope results, for a negative input value a positive slope results. | + | - for each positive input value $U_{\rm I} > 0$ a negative slope results, |
| - | The circuit thus created is called | + | The circuit thus created is called |
| - | If you look at the circuit, you can see that the node $K_1$ is at virtual ground. With a constant input voltage, the input current is therefore constant and defined only by the resistance. Thus, the capacitor charges with a constant current; the charge increases linearly. The voltage across the capacitor therefore also increases linearly. | + | If you look at the circuit, you can see that the node $\rm K_1$ is at virtual ground. With a constant input voltage, the input current is therefore constant and defined only by the resistance. Thus, the capacitor charges with a constant current; the charge increases linearly. The voltage across the capacitor therefore also increases linearly. |
| - | ~~PAGEBREAK~~ ~CLEARFIX~~ < | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | Just as in the basic circuits, the relationship between the output value and the input value is now to be determined mathematically. The meshes are drawn in <imgref pic2> for this purpose. The **transfer function $U_O = f(U_I)$** is now to be determined. | + | Just as in the basic circuits, the relationship between the output value and the input value is now to be determined mathematically. The meshes are drawn in <imgref pic2> for this purpose. The **transfer function $U_{\rm O} = f(U_{\rm I})$** is now to be determined. |
| - | $A_V = ? \quad -> \quad U_O = f(U_I) $ | + | $A_{\rm V} = ? \quad \rightarrow |
| === given equations === | === given equations === | ||
| Zeile 224: | Zeile 280: | ||
| Given the following equations: | Given the following equations: | ||
| - | |I.|Basic equation|$U_A = A_D \cdot U_D$| | + | |I. |Basic equation| $ U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$ | |
| - | |II.|mesh 1|$ -U_I+U_R-U_D=0 $| | + | |II. |
| - | |III.|mesh 2|$U_D+U_C+U_O=0$| | + | |III. |Mesh 2 | $ U_{\rm D} + U_C + U_{\rm O} = 0$ | |
| - | |IV.|Mesh|$I_R=I_C$| | + | |IV. |
| - | |V.|Capacity C|$C= { Q \over U_C } = { 1 \over U_C }\cdot(\int_{t_0}^{t_1} I_C dt+ Q_0(t_0)) $| | + | |V. |Capacity C | $C |
| - | |VI.|Resistance R|$R = { U_R \over I_R }$| | + | |VI. |
| - | === Derivation | + | === Calculation |
| - | The calculation is performed once in detail here (clicking on right arrow " | + | The calculation is performed once in detail here (clicking on the right arrow " |
| + | {{url> | ||
| - | ~~PAGEBREAK~~ | + | ~~PAGEBREAK~~ |
| + | \\ | ||
| - | <WRAP right>< | + | ==== 5.1.2 Signal-Time Curve ==== |
| - | By means of an example, the signal-time curve at the reverse integrator shall be explained. | + | < |
| + | < | ||
| + | </ | ||
| + | \\ | ||
| + | </ | ||
| - | | + | Using an example, the signal-time curve at the inverting integrator shall be explained. |
| - | Solution: | + | |
| + | - We are looking for the output voltage | ||
| - | | + | Solution: |
| - | - With the formula derived in 5.1.1 $U_O$ can be composed section by section: | + | - Over the given values of $R$ and $C$, the time constant $\tau$ is determined. |
| + | | ||
| + | - With the formula derived in 5.1.1 $U_{\rm O}$ can be composed section by section: | ||
| - | The calculation is performed once in detail here (clicking on right arrow " | + | The calculation is performed once in detail here (clicking on the right arrow " |
| - | <WRAP column 100%> <panel type=" | ||
| - | If a constant input voltage $U_E$ is applied to the reverse | + | <WRAP column 100%> <panel type=" |
| + | |||
| + | If a constant input voltage $U_{\rm I}$ is applied to the inverting | ||
| </ | </ | ||
| - | ~~PAGEBREAK~~ | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | ==== 5.1.3 Determination of amount | + | ==== 5.1.3 Determination of absolute value and phase ==== |
| - | In order to be able to determine | + | To be able to determine |
| - | $ U_I(t)= \hat{U}_I \cdot sin(\omega \cdot t)$ | + | $ U_{\rm I}(t)= \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$ |
| - | This definition of the input voltage can now be substituted into the above equation for $U_O$: | + | This definition of the input voltage can now be substituted into the above equation for $U_{\rm O}$: |
| - | The calculation is performed once in detail here (clicking on right arrow " | + | The calculation is performed once in detail here (clicking on the right arrow " |
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | The **amount** | ||
| - | The **phase** | + | The **absolute value** |
| - | $U_I = + \hat{U}_I \cdot sin(\omega \cdot t)$ $U_O = + \hat{U}_O \cdot cos(\omega \cdot t) = + \hat{U}_O \cdot sin(\omega \cdot t + 90°)$ $ \rightarrow | + | The **phase** |
| - | In order to be able to sketch the course in the bottom diagram, the **behavior of the transfer function | + | $U_{\rm I} = + \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$\\ |
| + | $U_{\rm O} = + \hat{U}_{\rm O} \cdot \cos(\omega | ||
| + | $ \rightarrow | ||
| + | \\ | ||
| + | === Consideration of extreme frequencies === | ||
| - | $ |A_V({\omega}\rightarrow 0 ; )| \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow 0}\quad \infty$ . $ |A_V({\omega}\rightarrow\infty)| | + | To be able to sketch the course in the bottom diagram, the **behavior of the transfer function |
| - | $\varphi = +90° \qquad \forall | + | $ |A_{\rm V}({\omega}\rightarrow 0 \ \; ) | \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow 0}\quad \infty$ \\ |
| + | $ |A_{\rm V}({\omega}\rightarrow \infty) | ||
| + | $\varphi = +90° \qquad \forall | ||
| - | From these boundary conditions the frequency response can already be sketched, see <imgref pic4>. | + | From these boundary conditions, the frequency response can already be sketched, see <imgref pic4>. |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ==== 5.1.4 Circuit | + | ==== 5.1.4 Circuit |
| - | In the previous chapters it could be seen that the circuit analysis with differential equations is already very tough and computationally intensive for a simple circuit like the reverse | + | In the previous chapters, it could be seen that the circuit analysis with differential equations is already very tough and computationally intensive for a simple circuit like the inverting |
| - | $U_R=R\cdot I \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ | + | $U_R=R\cdot I \qquad \qquad |
| + | $U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ {\rm d}t+ Q_0(t_0)) \qquad | ||
| However, this consideration can only be implemented under certain boundary conditions: | However, this consideration can only be implemented under certain boundary conditions: | ||
| - | - **sinusoidal quantities**: | + | - **sinusoidal quantities**: |
| - | - **Swinged | + | - **Steady |
| - | < | + | |
| + | < | ||
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| This can now be used to calculate the circuit (<imgref pic5>): | This can now be used to calculate the circuit (<imgref pic5>): | ||
| - | $\underline{Z}_1=R$ | + | $\underline{Z}_1=R$ |
| - | From the basic circuit of the [[: | + | $\underline{Z}_2=\frac{1}{{\rm j} \cdot \omega \cdot C} = \frac{\rm -j}{\omega \cdot C}$ |
| - | $A_V = \frac{U_A}{U_E}=-\frac{R_2}{R_1}$ | + | From the basic circuit of the [[: |
| + | |||
| + | $A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}=-\frac{R_2}{R_1}$ | ||
| This results in the complex: | This results in the complex: | ||
| - | $\underline{A}_V = \frac{\underline{U}_A}{\underline{U}_E}=-\frac{\underline{Z}_2}{\underline{Z}_1} = \frac{j}{\omega \cdot R \cdot C} $ | + | $\underline{A}_{\rm V} = \frac{\underline{U}_{\rm O}}{\underline{U}_{\rm I}}=-\frac{\underline{Z}_2}{\underline{Z}_1} = \frac{\rm j}{\omega \cdot R \cdot C} $ |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | === Plausibility check via extreme | + | === Plausibility check via extreme |
| ^From the formula^From the circuit| | ^From the formula^From the circuit| | ||
| - | |$\underline{A}_V \xrightarrow{\omega \rightarrow 0} \infty$|For $\omega \rightarrow 0$, the capacitor | + | |$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow 0} \infty$ |
| - | |$\underline{A}_V \xrightarrow{\omega \rightarrow \infty} 0$|For $\omega \rightarrow \infty$ the capacitor | + | For $\omega \rightarrow 0$, the capacitor |
| + | Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of | ||
| + | $\underline{U}_{\rm O} \rightarrow \infty$. | ||
| + | |$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow \infty} 0$ | < | ||
| + | For $\omega \rightarrow \infty$ the capacitor | ||
| + | Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of | ||
| + | $\underline{U}_{\rm O} \rightarrow 0$. </ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | === magnitude | + | === Absolute Value and Phase === |
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | <WRAP right>< | ||
| - | The amount | + | The absolute value $A_{\rm V}$ is given by: |
| - | $|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$ | + | $|\underline{A}_{\rm V}|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$ |
| - | Specifically, | + | Specifically, |
| - | $|\underline{A}_V(0dB)|\overset{!}{=} 1 \widehat{=} | + | $|\underline{A}_{\rm V}(0~{\rm dB})|\overset{!}{=} 1 \widehat{=} |
| The phase $\varphi$ is calculated via | The phase $\varphi$ is calculated via | ||
| - | $\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right) = arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = arctan \left( \infty \right) = +90°$ | + | $\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right) = \arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = \arctan \left( \infty \right) = +90°$ |
| The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (<imgref pic6>) for $x \rightarrow \infty$. | The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (<imgref pic6>) for $x \rightarrow \infty$. | ||
| Zeile 342: | Zeile 437: | ||
| ==== 5.1.5 Frequency response ==== | ==== 5.1.5 Frequency response ==== | ||
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | The frequency response is to be illustrated | + | The frequency response is to be illustrated |
| - | - Let $R=1 k\Omega$, $C=16 nF$ be given. | + | - Let $R=1 ~\rm k\Omega$, $C=16 ~\rm nF$ be given. |
| - We are looking for the Bode diagram | - We are looking for the Bode diagram | ||
| Solution | Solution | ||
| - | - Determining the time constant: \\ $\tau = R \cdot C = 16 \mu s$ \\ \\. | + | - Determining the time constant: \\ $\tau = R \cdot C = 16 ~\rm µs$ \\ \\ |
| - | - Determining the frequency $f$ for $|\underline{A}_V(0dB)|$: $\omega(0dB)= \frac{1}{\tau} = 2\pi \cdot f(0dB)$ \ This gives $f(0dB)$ via: \f(0dB)=\frac{1}{2\pi} \cdot \frac{1}{16 \mu s} \approx | + | - Determining the frequency $f$ for $|\underline{A}_{\rm V}(0~{\rm dB})|$: $\omega(0~{\rm dB})= \frac{1}{\tau} = 2\pi \cdot f(0~{\rm dB})$ \\ This gives $f(0~{\rm dB})$ via: \\ $f(0~{\rm dB})=\frac{1}{2\pi} \cdot \frac{1}{16 |
| - | - Consideration of the slope: \ $|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$ \ From this, a tenfold increase in $f$ results in one-tenth the amount | + | - Consideration of the slope: |
| - From this information, | - From this information, | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ===== 5.2 Lowpass | + | ===== 5.2 Low Pass Filter |
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | Another circuit can be derived from the reverse | + | Another circuit can be derived from the inverting |
| * first be considered practically with a simulation, | * first be considered practically with a simulation, | ||
| - | * then a picture of the system' | + | * then a picture of the system' |
| - | * be checked by a circuit analysis with complex | + | * be checked by a circuit analysis with complex |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | === Lowpass | + | === Low Pass Filter |
| - | < | + | < |
| - | In the simulation | + | In the simulation |
| - | - If only switch $S1$ is closed, the circuit is an inverting amplifier. | + | - If only switch $\rm S1$ is closed, the circuit is an inverting amplifier. |
| - | - If only switch $S2$ is closed, the circuit is an inverting integrator. | + | - If only switch $\rm S2$ is closed, the circuit is an inverting integrator. |
| - | In the simulation, the Bode diagram is sketched below. By clicking on the Bode diagram, the distribution of the current in the circuit corresponding to the frequency is displayed and - in addition to the Bode diagram - also the gain in $dB$, or the phase in degrees. | + | In the simulation, the Bode diagram is sketched below. By clicking on the Bode diagram, the distribution of the current in the circuit corresponding to the frequency is displayed and - in addition to the Bode diagram - also the gain in ${\rm dB}$, or the phase in degrees. |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | <panel type=" | + | <panel type=" |
| - First set up an inverting amplifier and read off the gain and phase (click on the Bode diagram). | - First set up an inverting amplifier and read off the gain and phase (click on the Bode diagram). | ||
| - Now change the circuit to an inverting integrator and read off the gain and phase there as well. | - Now change the circuit to an inverting integrator and read off the gain and phase there as well. | ||
| - Now both switches should be closed. | - Now both switches should be closed. | ||
| - | - In which frequency ranges does the inverting amplifier or the reverse | + | - In which frequency ranges does the inverting amplifier or the inverting |
| - | - Consider the distribution of the currents in the circuit at different frequencies. At what frequency does the current divide approximately equally between the $1k\Omega$ resistor and the capacitor? What is the value of the gain and phase here? | + | - Consider the distribution of the currents in the circuit at different frequencies. At what frequency does the current divide approximately equally between the $1~\rm k\Omega$ resistor and the capacitor? What is the value of the gain and phase here? |
| - After reading through the following analyses, the gain and phase at the "kink point" can be determined. Do these deviate from your measurement? | - After reading through the following analyses, the gain and phase at the "kink point" can be determined. Do these deviate from your measurement? | ||
| Zeile 396: | Zeile 499: | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ==== 5.2.1 Consideration | + | ==== 5.2.1 First Consideration ==== |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | === Extreme | + | === Extreme |
| - | From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of magnitude | + | From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of the absolute value of voltage gain and phase can be analyzed. |
| - $\omega \rightarrow 0$: | - $\omega \rightarrow 0$: | ||
| Zeile 408: | Zeile 511: | ||
| - Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$ | - Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$ | ||
| - Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially | - Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially | ||
| - | - Thus the circuit behaves like an inverting amplifier: $|\underline{A}_V|=|-\frac{R_2}{R_1}|$ | + | - Thus the circuit behaves like an inverting amplifier: $|\underline{A}_{\rm V}|=|-\frac{R_2}{R_1}|$ |
| - For the inverting amplifier, a sinusoidal input signal results in a negated sinusoidal signal. This corresponds to a phase of $\varphi=\pm 180°$. The sign is not defined by the inverting amplifier. | - For the inverting amplifier, a sinusoidal input signal results in a negated sinusoidal signal. This corresponds to a phase of $\varphi=\pm 180°$. The sign is not defined by the inverting amplifier. | ||
| - $\omega \rightarrow \infty$: | - $\omega \rightarrow \infty$: | ||
| Zeile 414: | Zeile 517: | ||
| - Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$ | - Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$ | ||
| - Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts | - Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts | ||
| - | - Thus the circuit behaves like a reverse | + | - Thus the circuit behaves like a inverting |
| - | From this it can be seen that | + | From this, it can be seen that |
| * for low frequencies a constant gain is expected and | * for low frequencies a constant gain is expected and | ||
| - | * for high frequencies a drop as known from the reverse | + | * for high frequencies a drop is known from the inverting |
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| === expected Bode diagram === | === expected Bode diagram === | ||
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | A floor diagram can be estimated from the extreme | + | A Bode diagram can be estimated from the extreme |
| **Frequency Response:** | **Frequency Response:** | ||
| - | * For low frequencies, | + | * For low frequencies, |
| - | * For higher frequencies, | + | * For higher frequencies, |
| * There is a frequency where both situations seem to occur simultaneously | * There is a frequency where both situations seem to occur simultaneously | ||
| Zeile 441: | Zeile 549: | ||
| For the intermediate area, there must be a transition between the two extremal situations. | For the intermediate area, there must be a transition between the two extremal situations. | ||
| - | One problem still seems to be that for the inverting amplifier it is not clear whether the phase is now $+180°$ or $-180°$. In the mathematical consideration of the inverting integrator, it turned out that for a capacitor an integration step ($U=\frac{1}{C} \int I_C \ dt$) must be performed. In the inverting amplifier, no integration step was necessary. Thus, a sinusoidal input signal is shifted by 90° at most. So there must be just $90°$ phase shift from inverting amplifier to high frequencies at low frequencies. | + | One problem still seems to be that for the inverting amplifier, it is not clear whether the phase is now $+180°$ or $-180°$. In the mathematical consideration of the inverting integrator, it turned out that for a capacitor an integration step ($U=\frac{1}{C} \int I_C \ {\rm d}t$) must be performed. In the inverting amplifier, no integration step was necessary. Thus, a sinusoidal input signal is shifted by 90° at most. So there must be just a $90°$ phase shift from the inverting amplifier to high frequencies at low frequencies. |
| - | From this knowledge, we get an expected | + | From this knowledge, we get an expected |
| <WRAP column 100%> <panel type=" | <WRAP column 100%> <panel type=" | ||
| Zeile 449: | Zeile 557: | ||
| The following rules apply to filters: | The following rules apply to filters: | ||
| - | * for each energy | + | * for each energy |
| - | * for each energy | + | * for each energy |
| - | * each energy | + | * each energy |
| * The phase response is monotonically decreasing. | * The phase response is monotonically decreasing. | ||
| Zeile 458: | Zeile 566: | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | === RC element and cutoff | + | === RC element and cut-off |
| - | In the circuit, the parallel circuit $R_2$ and $C$ behaves | + | In the circuit, the parallel circuit $R_2$ and $C$ behave |
| - | For this: \\ $\underline{X}_C|=R_2$ \\ $\frac{1}{\omega_{Gr} \cdot C}=R_2 \rightarrow \omega_{Gr} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{Gr}$ \\ $\boxed{f_{Gr} = \frac{1}{2\pi | + | For this: \\ |
| + | $|\underline{X}_C|=R_2$ \\ | ||
| + | $\frac{1}{\omega_{c} \cdot C}=R_2 \rightarrow \omega_{c} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{c}$\\ | ||
| + | $\boxed{f_{c} = \frac{1}{2\pi \cdot R_2 \cdot C}} $\\ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 468: | Zeile 579: | ||
| ==== 5.2.2 Circuit Analysis with Complex Calculus ==== | ==== 5.2.2 Circuit Analysis with Complex Calculus ==== | ||
| - | Now the circuit is to be analyzed again by means of complex | + | Now the circuit is to be analyzed again using complex |
| - | $\underline{A}_V=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{j \cdot \omega \cdot C}}{R_2 + \frac{1}{j \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\ | + | $\underline{A}_{\rm V}=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{{\rm j} \cdot \omega \cdot C}}{R_2 + \frac{1}{{\rm j} \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\ |
| + | $\boxed{\underline{A}_{\rm V}= - \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j} \omega \cdot R_2 \cdot C}}$ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | === Calculation of magnitude | + | === Calculation of Absolute Value and Phase === |
| - | For the calculation of the amount | + | For the calculation of the absolute value $A_{\rm V}$ a " |
| - | $|\underline{A}_V| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{\mathcal{c}|}$ | + | $|\underline{A}_{\rm V}| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{|\mathcal{c}|}$ |
| - | This results in for the amount: | + | This results in the absolute value: |
| - | $\boxed{|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}}$ | + | $\boxed{|\underline{A}_{\rm V}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}}$ |
| - | For the phase $\varphi$ real value $\Re(\underline{A}_V)$ and imaginary value $\Im(\underline{A}_V)$ must be determined by multiplication with the conjugate complex value. | + | For the phase, $\varphi$ real value $\Re(\underline{A}_{\rm V})$ and imaginary value $\Im(\underline{A}_{\rm V})$ must be determined by multiplication with the conjugate complex value. |
| - | $\varphi = arctan(\frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)})$ | + | $\varphi = \arctan(\frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})})$ |
| But here, too, there is a " | But here, too, there is a " | ||
| - | $\underline{A}_V= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}} \cdot \frac{1 - j \omega \cdot R_2 \cdot C}{\color{blue}{1 - j \omega \cdot R_2 \cdot C}}$ | + | $\underline{A}_{\rm V}= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j} \omega \cdot R_2 \cdot C}} \cdot \frac{1 - {\rm j} \omega \cdot R_2 \cdot C}{\color{blue}{1 - {\rm j} \omega \cdot R_2 \cdot C}}$ |
| - | After all, it is just the conjugate complex value that is multiplied to get a real denominator. Thus the blue marked part is a real constant $\mathcal{C}$, because all factors of the constant are real: | + | After all, it is just the conjugate complex value that is multiplied to get a real denominator. Thus the blue marked part is a real constant $\mathcal{C}$ because all factors of the constant are real: |
| - | $\underline{A}_V= \color{blue}{\mathcal{C}} \cdot (1 - j \omega \cdot R_2 \cdot C)$ | + | $\underline{A}_{\rm V}= \color{blue}{\mathcal{C}} \cdot (1 - {\rm j} \omega \cdot R_2 \cdot C)$ |
| - | Thus, the phase $\varphi = arctan\left(\frac{\color{teal}{\Im(\underline{A}_V)}}{\color{brown}{\Re(\underline{A}_V)}}\right)$ is obtained as. | + | Thus, the phase $\varphi = \arctan\left(\frac{\color{teal}{\Im(\underline{A}_{\rm V})}}{\color{brown}{\Re(\underline{A}_{\rm V})}}\right)$ is obtained as. |
| - | $\underline{A}_V= \mathcal{C} \cdot (\color{brown}{1} + j \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$ | + | $\underline{A}_{\rm V}= \mathcal{C} \cdot (\color{brown}{1} + {\rm j} \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$ |
| - | $\boxed{\varphi = arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$ | + | $\boxed{\varphi = \arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$ |
| - | For the __amount__ | + | \\ |
| + | === Consideration of extreme frequencies === | ||
| - | | + | For the __absolute value__ |
| + | | ||
| + | - at $\omega \rightarrow \infty$: | ||
| - | < | + | < |
| + | <panel type=" | ||
| + | < | ||
| + | </ | ||
| < | < | ||
| Zeile 514: | Zeile 632: | ||
| </ | </ | ||
| - | For finding the phase $\color{red}{\varphi} = arctan\left(\color{teal}{-\omega \cdot R_2 \cdot C}\right)$ it helps to look at the arctangent in the diagram. To do this, the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is plotted against the phase $\color{red}{\varphi}$ (<imgref pic20>). For the extremal values $\omega$ of results in: | + | For finding the phase $\color{red}{\varphi} = \arctan\left(\color{teal}{-\omega \cdot R_2 \cdot C}\right)$ it helps to look at the arctangent in the diagram. To do this, the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is plotted against the phase $\color{red}{\varphi}$ (<imgref pic20>). For the extremal values $\omega$ of results in: |
| - at $\omega \rightarrow 0$: the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$. | - at $\omega \rightarrow 0$: the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$. | ||
| - at $\omega \rightarrow \infty$: $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ argument goes towards $-\infty$. | - at $\omega \rightarrow \infty$: $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ argument goes towards $-\infty$. | ||
| - | In the diagram the argument $Arg$ is changeable by the slider in the upper left corner. The course in the diagram must be continuous, because also the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is continuous between $-0$ and $-\infty$. This is only possible on the __upper branch__: the point $Arg \rightarrow -0$ then just corresponds to approaching the y-axis (here $\color{red}{\varphi}$-axis) from the left in <imgref pic20>. The point $Arg \rightarrow -\infty$ corresponds to the path to the left in the diagram. | + | In the diagram, the argument $Arg$ is changeable by the slider in the upper left corner. The course in the diagram must be continuous because also the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is continuous between $-0$ and $-\infty$. This is only possible on the __upper branch__: the point $Arg \rightarrow -0$ then just corresponds to approaching the y-axis (here $\color{red}{\varphi}$-axis) from the left in <imgref pic20>. The point $Arg \rightarrow -\infty$ corresponds to the path to the left in the diagram. |
| This results in a progression of the phase $\color{red}{\varphi}$ from $\color{red}{\varphi}(\omega \rightarrow -0)=\pi = 180°$ to $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$. | This results in a progression of the phase $\color{red}{\varphi}$ from $\color{red}{\varphi}(\omega \rightarrow -0)=\pi = 180°$ to $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$. | ||
| - | == PAGEBREAK | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | === Calculation of the cutoff | + | === Calculation of the cut-off |
| - | The cut-off frequency can also be understood as the transition from the inverting amplifier to the reverse | + | The cut-off frequency can also be understood as the transition from the inverting amplifier to the inverting |
| - | Thus, for the cut-off frequency $f_{Gr}$ we get | + | Thus, for the cut-off frequency $f_{c}$ we get |
| - | $\frac{R_2}{R_1} = \frac{1}{\omega_{Gr} R_1 \cdot C}$ $\omega_{Gr} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{Gr}$ | + | $\frac{R_2}{R_1} = \frac{1}{\omega_{c} R_1 \cdot C}$ \\ |
| + | $\omega_{c} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{c}$ | ||
| - | At the cut-off frequency, the result is an amount | + | At the cut-off frequency, the result is an absolute value of: |
| - | $|\underline{A}_{V, | + | $|\underline{A}_{V, |
| - | $\boxed{|\underline{A}_{V, | + | $\boxed{|\underline{A}_{V, |
| - | The phase at the cutoff | + | The phase at the cut-off |
| - | $\varphi_{Gr} = arctan\left(-\omega_{Gr} \cdot R_2 \cdot C\right) = arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = arctan\left(-1 \right)$ | + | $\varphi_{c} = \arctan\left(-\omega_{c} \cdot R_2 \cdot C\right) = \arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = \arctan\left(-1 \right)$ |
| - | $\boxed{\varphi_{Gr} = \frac{3}{4} \pi =135°}$ | + | $\boxed{\varphi_{c} = \frac{3}{4} \pi =135°}$ |
| - | Because of the $-3dB$ attenuation of the low-frequency gain at the cutoff | + | Because of the $-3~{\rm dB}$ attenuation of the low-frequency gain at the cut-off |
| - | ~PAGEBREAK~ ~CLEARFIX~~ | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| + | ===== 5.3 Inverting Differentiator ===== | ||
| - | ===== 5.3 Reverse Differential ===== | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | < | + | < |
| - | <WRAP right> | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | ~~PAGEBREAK~~~CLEARFIX~~~ | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | <WRAP right><panel type=" | + | In <imgref pic11_1> an inverting differentiator is shown. Compared to the integrator here just the resistor and the capacitor are swapped. |
| - | In <imgref pic11_1> an inverse differentiator | + | In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal |
| - | In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal is output at the output. The derivative at the reversal points (" | ||
| - | In the following, only the results will be discussed without | + | In the following, only the results will be discussed without |
| Circuit analysis via differential equation yields: | Circuit analysis via differential equation yields: | ||
| - | $\boxed{U_A = - R \cdot C \frac{d}{dt}U_E}$ | + | $\boxed{U_{\rm O} = - R \cdot C \frac{\rm d}{{\rm d}t}U_{\rm I}}$ |
| With complex calculation, | With complex calculation, | ||
| - | $\boxed{\underline{A}_V=-j \cdot \omega \cdot R \cdot C}$ | + | $\boxed{\underline{A}_{\rm V}=-{\rm j} \cdot \omega \cdot R \cdot C}$ |
| From this, the Bode diagram shown in <imgref pic11> can be determined. | From this, the Bode diagram shown in <imgref pic11> can be determined. | ||
| - | ~~PAGEBREAK~~~CLEARFIX~~ | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | < | + | <panel type=" |
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| - | For the inverse | + | For the inverting |
| - | - Circuit analysis | + | - Circuit analysis |
| - | - Determination of magnitude | + | - Determination of absolute value and phase from differential equation (incl. consideration of extreme cases) |
| - | - Example of a signal-time-curve with: $R = 10 k\Omega$ and $C = 2µF$ and $U_E$ as right shown | + | - Example of a signal-time-curve with $R = 10 ~\rm k\Omega$ and $C = 2 ~\rm µF$ and $U_{\rm I}$ - as shown in the diagram. |
| - | - Circuit analysis | + | - Circuit analysis |
| - | - Consideration of magnitude | + | - Consideration of absolute value and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$ |
| - | - Frequency response (Bode plot) for circuit with: $R = 10 k\Omega$ and $C = 16nF$ | + | - Frequency response (Bode plot) for a circuit with $R = 10 ~\rm k\Omega$ and $C = 16 ~\rm nF$. |
| - | </ | + | </ |
| - | ===== 5.4 High Pass ===== | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | <WRAP left> | + | ===== 5.4 High Pass Filter ===== |
| - | <WRAP right>< | + | A high pass filter |
| - | ~~PAGEBREAK~~~CLEARFIX~~~ | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | <WRAP right>< | + | With complex calculation this results in: $\boxed{\underline{A}_{\rm V} = - \frac{R_2}{R_1} \cdot \frac{{\rm j} \cdot \omega \cdot R_1 \cdot C}{1 + {\rm j} \cdot \omega \cdot R_1 \cdot C}} $ |
| - | A high-pass filter can be created from the reversal differentiator if the purely capacitive impedance for $Z_1$ is suitably extended via a resistive component (< | + | From this, the Bode diagram shown in < |
| - | With complex calculation this results in: $\boxed{\underline{A}_V | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | From this, the Bode diagram shown in <imgref pic12> can be determined. | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | ~~PAGEBREAK~~~CLEARFIX~~ | + | <WRAP> |
| - | In [[# | ||
| - | | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| + | |||
| + | <panel type=" | ||
| + | |||
| + | In [[# | ||
| + | |||
| + | | ||
| - Expected Bode diagram | - Expected Bode diagram | ||
| - | - RC element and cutoff | + | - RC element and cut-off |
| - Circuit analysis with complex calculation | - Circuit analysis with complex calculation | ||
| - | - Calculation of magnitude | + | - Calculation of absolute value and phase |
| </ | </ | ||
| - | ~~PAGEBREAK~~ ~CLEARFIX~~ | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | ===== 5.5 Overview high pass / low pass ===== | + | ===== 5.5 Overview high pass Filter |
| - | < | + | < |
| + | < | ||
| + | </ | ||
| + | \\ {{drawio> | ||
| + | </ | ||
| - | ~PAGEBREAK~ ~CLEARFIX~~ | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| - | ====== | + | ====== |
| - | <panel type=" | + | <panel type=" |
| - | Derive the linear factor for the following levels in dB. State how this factor can be determined in each case using the interpolation points 20 dB ≙ factor 10 and 6 dB ≙ factor 2. | + | Derive the linear factor for the following levels in dB. State how this factor can be determined in each case using the interpolation points |
| Solve without a calculator | Solve without a calculator | ||
| - | As an example, the derivation | + | As an example, the calculation |
| - | ^level^over interpolation points in dB^over interpolation points linear^linear factor| | + | ^level^over interpolation points in $\rm dB$^over interpolation points linear^linear factor| |
| - | |$10 dB$|$5 \cdot 6 dB - 20 dB$|$2^5 \cdot {1\over 10}$|$3, | + | |$10 |
| - | |$2 dB$| | | | | + | |$2 ~{\rm dB}$| | | | |
| - | |$4 dB$| | | | | + | |$4 ~{\rm dB}$| | | | |
| - | |$6 dB$| | | | | + | |$6 ~{\rm dB}$| | | | |
| - | |$8 dB$| | | | | + | |$8 ~{\rm dB}$| | | | |
| - | |$12 dB$| | | | | + | |$12 |
| - | |$14 dB$| | | | | + | |$14 |
| - | |$16 dB$| | | | | + | |$16 |
| - | |$18 dB$| | | | | + | |$18 |
| - | |$15 dB$| | | | | + | |$15 |
| - | |$79 dB$| | | | | + | |$79 |
| - | |$128 dB$| | | | | + | |$128 ~{\rm dB}$| | | | |
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | < | + | The following image shows a series of amplifiers, which shall be analyzed. |
| + | |||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | The amplification factors are: | ||
| + | * $A_{\rm V1} =80$ | ||
| + | * $A_{\rm V2} =0.0125$ | ||
| + | * $A_{\rm V3} =250' | ||
| + | |||
| + | Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$). | ||
| + | |||
| + | <button size=" | ||
| + | < | ||
| + | |||
| + | - Rearrange the given linear factors as exponents of $2$ and $10$, e.g. $2^6 \cdot 10^7$ | ||
| + | - Use the exponent values to transfer it into $\rm dB$, in this example: $6 \cdot 6~{\rm dB} + 7 \cdot 20~{\rm dB} $ | ||
| + | - Calculate the $\rm dB$ value, in this example: $36~{\rm dB} + 140~{\rm dB} = 176~{\rm dB}$ | ||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | < | ||
| + | ^ Amp. Name $\rightarrow$ ^ lin. Factor $\rightarrow$^ re-arrange $\rightarrow$ ^ as exponents of $2$ and $10$ $\rightarrow$^ | ||
| + | | $A_{\rm V1}$ | $80$ | $8 \cdot 10$ | $2^3 \cdot 10^1$ | $ 3 \cdot 6~{\rm dB} + | ||
| + | | $A_{\rm V2}$ | $0.0125$ | ||
| + | | $A_{\rm V3}$ | $250' | ||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | < | ||
| + | | ||
| + | | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | The following circuit shall be given with $R= 10 ~\rm k\Omega$, $C = 1.6 ~\rm µF$ and a sinusoidal input voltage $U_{\rm I} = 1 ~\rm V $ with $f = 1 ~\rm kHz$ | ||
| + | |||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| - | Let the circuit shown opposite with $R= 10 k\Omega$, $C = 1.6 uF$ and a sinusoidal input voltage $U_E = 1 V $ with $f = 1 kHz$ be given. | + | As described in the course, the Bode diagram can be displayed in Tina TI via '' |
| - Simulate this circuit in Tina TI \\ - with an ideal operational amplifier \\ - using the operational amplifiers uA776, LM301A and LM318. | - Simulate this circuit in Tina TI \\ - with an ideal operational amplifier \\ - using the operational amplifiers uA776, LM301A and LM318. | ||
| - Attach the Bode diagram. | - Attach the Bode diagram. | ||
| - | - Briefly describe the differences in the amplitude response of the gain $A_V$. | + | - Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$. |
| - | - What happens if instead of $R= 10 k\Omega$, $C = 1.6 uF$ the same time constant is implemented with $R= 10 M\Omega$, $C = 1.6 nF$? Verify this by comparing the Bode plot of the LM318 operational amplifier. | + | - What happens if instead of $R= 10 ~\rm k\Omega$, $C = 1.6 ~\rm µF$ the same time constant is implemented with $R= 10 ~\rm M\Omega$, $C = 1.6 ~\rm nF$? Verify this by comparing the Bode plot of the LM318 operational amplifier. |
| - | - Simulate an inverting amplifier in Tina TI. To do this, replace the capacitor in the circuit with a resistor $R_2 = 10 k\Omega$ and use the LM318 op-amp. | + | - Simulate an inverting amplifier in Tina TI. To do this, replace the capacitor in the circuit with a resistor $R_2 = 10 ~\rm k\Omega$ and use the LM318 op-amp. |
| - Attach the Bode diagram. | - Attach the Bode diagram. | ||
| - What should the Bode diagram (amplitude and phase response) look like for an ideal inverting amplifier? What deviation do you notice in the real setup? | - What should the Bode diagram (amplitude and phase response) look like for an ideal inverting amplifier? What deviation do you notice in the real setup? | ||
| - | - Up to what frequency can the circuit be operated as an inverting amplifier (maximum deviation of 1dB)? Use zoom and/or cursor to determine. | + | - Up to what frequency can the circuit be operated as an inverting amplifier (maximum deviation of $1~{\rm dB}$)? Use zoom and/or cursor to determine. |
| - | - Simulate an open-loop operational amplifier LM318 (i.e. without feedback network). The non-inverting input should be connected to ground. The inverting input should have the above sinusoidal input voltage. | + | - Simulate an open-loop operational amplifier LM318 (i.e. without feedback network). The non-inverting input should be connected to the ground. The inverting input should have the above sinusoidal input voltage. |
| - Attach the Bode diagram. | - Attach the Bode diagram. | ||
| - What is the cut-off frequency? | - What is the cut-off frequency? | ||
| - | - How many dB per decade does the amplitude response drop at high frequencies? | + | - How many $\rm dB$ per decade does the amplitude response drop at high frequencies? |
| </ | </ | ||