Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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circuit_design:5_filter_circuits_i [2023/03/28 14:31] mexleadmin |
circuit_design:5_filter_circuits_i [2023/07/17 11:59] (aktuell) mexleadmin |
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| Zeile 11: | Zeile 11: | ||
| === Introductory example === | === Introductory example === | ||
| - | < | + | < |
| Various applications work in harsh environments, | Various applications work in harsh environments, | ||
| Zeile 54: | Zeile 54: | ||
| 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: | ||
| - | $\boxed{A_{\rm V}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{U_2}{U_1}\right)=20 dB \cdot \log_{10} A_\rm V}$(nbsp)(nbsp)(nbsp)(nbsp) | + | $\boxed{A_{\rm V}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{U_2}{U_1}\right)=20 |
| \\ | \\ | ||
| Zeile 93: | Zeile 93: | ||
| - For $A_{\rm V}= \color{green}{1} $ we get < | - For $A_{\rm V}= \color{green}{1} $ we get < | ||
| - | $ 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}}$ \\ | + | $ 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}}$ \\ |
| Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, | 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 < | - 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}}$ \\ | + | $ 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 $10^\color{blue}{x} = \color{green}{0.01}$, | 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 < | - 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}}$ \\ | + | $ 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, $\color{blue}{x}\approx 0.30103$. </ | Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$. </ | ||
| Zeile 119: | Zeile 119: | ||
| 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. \\ | 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: $A_{\rm V, eq}=\prod A_{\rm i} = A_1 \cdot A_2 \cdot A_3$. \\ | + | The total gain here is the product of the individual gains: $A_{\rm V, eq}=\prod A_{i} = A_1 \cdot A_2 \cdot A_3$. \\ |
| The determination of the total gain was rather laborious before the time of the pocket calculator due to the multiplications. | 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: $A_{\rm V,eq}^{\rm dB}=\sum A_i^{\rm dB} = A_1^{\rm dB} + A_2^{\rm dB} + A_3^{\rm dB}$. \\ | For the levels, the result is an addition: $A_{\rm V,eq}^{\rm dB}=\sum A_i^{\rm dB} = A_1^{\rm dB} + A_2^{\rm dB} + A_3^{\rm dB}$. \\ | ||
| Zeile 130: | Zeile 130: | ||
| - A linear factor of $\color{green}{\times 10}$ results in level $+ 20~{\rm dB}$. | - 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 $+ 6~{\rm dB}$. | - A linear factor of $\color{green}{\times 2} $ results in a level of $+ 6~{\rm dB}$. | ||
| - | - The linear value $A_{\rm V} = 1$ corresponds to $0 {\rm dB}$. | + | - 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 | ||
| Zeile 147: | Zeile 147: | ||
| Examples: | Examples: | ||
| - | - $A_{\rm V}^{\rm dB}=58~{\rm dB}$ \\ with interpolation points: $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$ \\ This becomes linear | + | - < |
| - | - $A_{\rm V}^{\rm dB}=56~{\rm dB}$ \\ with interpolation points: $A_{\rm V}^{dB}=56~{\rm dB} = 80~{\rm dB} - 24~{\rm dB} = \color{blue}{4}\cdot 20~{\rm dB} + \color{magenta}{-4}\cdot 6~{\rm dB}$ \\ This becomes linear $ \qquad \qquad \qquad \qquad \qquad \ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad | + | $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$ \\ |
| - | - $A_{\rm V}^{\rm dB}=55~{\rm dB}$ \\ with interpolation points: $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB} + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\ This becomes linear | + | 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 | ||
| + | - < | ||
| + | $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB} + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\ | ||
| + | This becomes linear $ \qquad \qquad | ||
| \\ | \\ | ||
| Zeile 204: | 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** | ||
| - | - $|A_{\rm V}(f)| = \mathcal{C} \cdot f$: 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. | + | - $|A_{\rm V}(f)| = \mathcal{C} \cdot f$: \\ 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. |
| - | - $|A_{\rm V}(f)| = \mathcal{C} / f$: If a function is considered that is reciprocal to $f$, then a tenfold increase in frequency results in a decrease in voltage gain to one-tenth. This results in a drop of $-20~{\rm dB}$ per decade (cf. <imgref picB> at high frequencies). | + | - $|A_{\rm V}(f)| = \mathcal{C} / f$: |
| - | - $|A_{\rm V}(f)| = \mathcal{C} / f^n$: Considering a function reciprocal to $f^n$, a tenfold increase in frequency results in a drop in voltage gain to one $1/10^n$th. This results in a drop of $-20~{\rm dB} \cdot n$ per decade. | + | - $|A_{\rm V}(f)| = \mathcal{C} / f^n$: \\ Considering a function reciprocal to $f^n$, a tenfold increase in frequency results in a drop in voltage gain to one $1/ |
| As an alternative to the actual course, $|A_{\rm V}(f)|$ 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. | ||
| Zeile 217: | 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 $A(\omega) \sim \frac{1}{\omega}$, | ||
| + | </ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 254: | Zeile 262: | ||
| The circuit thus created is called an **inverse integrator** or inverting integrator. | The circuit thus created is called an **inverse integrator** or inverting integrator. | ||
| - | 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~~ | ||
| Zeile 272: | Zeile 280: | ||
| Given the following equations: | Given the following equations: | ||
| - | |I. |Basic equation| $ | + | |I. |Basic equation| $ |
| - | |II. |Mesh 1 | $-U_{\rm I} + U_R - U_{\rm D} = 0$ | | + | |II. |Mesh 1 | $-U_{\rm I} + U_R - U_{\rm D} = 0$ | |
| - | |III. |Mesh 2 | $ U_{\rm D} + U_C + U_{\rm O} = 0$ | | + | |III. |Mesh 2 | $ U_{\rm D} + U_C + U_{\rm O} = 0$ | |
| - | |IV. | + | |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 = { Q \over U_C } = { 1 \over U_C }\cdot(\int_{t_0}^{t_1} I_C {\rm d}t+ Q_0(t_0)) $ | |
| - | |VI. | + | |VI. |
| === Calculation === | === Calculation === | ||
| Zeile 359: | Zeile 367: | ||
| $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\ | $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\ | ||
| - | $U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ dt+ Q_0(t_0)) \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**: |
| - **Steady state**: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since the switch-on. This takes out disturbances that are generated by switching on. | - **Steady state**: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since the switch-on. This takes out disturbances that are generated by switching on. | ||
| Zeile 421: | Zeile 429: | ||
| The phase $\varphi$ is calculated via | The phase $\varphi$ is calculated via | ||
| - | $\varphi = arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right) = arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = arctan \left( \infty \right) = +90°$ | + | $\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right) = \arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = \arctan \left( \infty \right) = +90°$ |
| 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 469: | Zeile 477: | ||
| < | < | ||
| - | In the simulation above, the circuit from <imgref pic7_1> is shown again. In addition, two switches $S1$ and $S2$ are built into the circuit by which the various feedback paths can be disabled: | + | In the simulation above, the circuit from <imgref pic7_1> is shown again. In addition, two switches $\rm S1$ and $\rm S2$ are built into the circuit by which the various feedback paths can be disabled: |
| - | - 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 ${\rm dB}$, 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. | ||
| Zeile 541: | 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 a $90°$ phase shift from the 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 Bode diagram as seen in <imgref pic10>. | From this knowledge, we get an expected Bode diagram as seen in <imgref pic10>. | ||
| Zeile 590: | Zeile 598: | ||
| 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. | 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. | ||
| - | $\varphi = arctan(\frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})})$ | + | $\varphi = \arctan(\frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})})$ |
| But here, too, there is a " | But here, too, there is a " | ||
| Zeile 600: | Zeile 608: | ||
| $\underline{A}_{\rm V}= \color{blue}{\mathcal{C}} \cdot (1 - {\rm j} \omega \cdot R_2 \cdot C)$ | $\underline{A}_{\rm V}= \color{blue}{\mathcal{C}} \cdot (1 - {\rm j} \omega \cdot R_2 \cdot C)$ | ||
| - | Thus, the phase $\varphi = arctan\left(\frac{\color{teal}{\Im(\underline{A}_{\rm V})}}{\color{brown}{\Re(\underline{A}_{\rm V})}}\right)$ 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. |
| $\underline{A}_{\rm V}= \mathcal{C} \cdot (\color{brown}{1} + {\rm j} \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$ | $\underline{A}_{\rm V}= \mathcal{C} \cdot (\color{brown}{1} + {\rm j} \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$ | ||
| - | $\boxed{\varphi = arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$ | + | $\boxed{\varphi = \arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$ |
| \\ | \\ | ||
| Zeile 624: | 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 $-0$. | ||
| Zeile 652: | Zeile 660: | ||
| The phase at the cut-off frequency is: | The phase at the cut-off frequency is: | ||
| - | $\varphi_{c} = arctan\left(-\omega_{c} \cdot R_2 \cdot C\right) = arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = arctan\left(-1 \right)$ | + | $\varphi_{c} = \arctan\left(-\omega_{c} \cdot R_2 \cdot C\right) = \arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = \arctan\left(-1 \right)$ |
| $\boxed{\varphi_{c} = \frac{3}{4} \pi =135°}$ | $\boxed{\varphi_{c} = \frac{3}{4} \pi =135°}$ | ||
| Zeile 667: | Zeile 675: | ||
| </ | </ | ||
| - | < | + | < |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 686: | Zeile 694: | ||
| Circuit analysis via differential equation yields: | Circuit analysis via differential equation yields: | ||
| - | $\boxed{U_{\rm O} = - R \cdot C \frac{d}{dt}U_{\rm I}}$ | + | $\boxed{U_{\rm O} = - R \cdot C \frac{\rm d}{{\rm d}t}U_{\rm I}}$ |
| With complex calculation, | With complex calculation, | ||
| Zeile 716: | Zeile 724: | ||
| ===== 5.4 High Pass Filter ===== | ===== 5.4 High Pass Filter ===== | ||
| - | A high pass filter can be created from the inverting differentiator, | + | A high pass filter can be created from the inverting differentiator, |
| < | < | ||
| Zeile 755: | Zeile 763: | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ===== 5.5 Overview high pass filter | + | ===== 5.5 Overview high pass Filter |
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| \\ {{drawio> | \\ {{drawio> | ||
| Zeile 800: | Zeile 808: | ||
| The amplification factors are: | The amplification factors are: | ||
| - | * $A_{V1} =80$ | + | * $A_{\rm V1} =80$ |
| - | * $A_{V2} =0.0125$ | + | * $A_{\rm V2} =0.0125$ |
| - | * $A_{V3} =250' | + | * $A_{\rm V3} =250' |
| Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$). | Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$). | ||
| Zeile 817: | Zeile 825: | ||
| < | < | ||
| ^ Amp. Name $\rightarrow$ ^ lin. Factor $\rightarrow$^ re-arrange $\rightarrow$ ^ as exponents of $2$ and $10$ $\rightarrow$^ | ^ Amp. Name $\rightarrow$ ^ lin. Factor $\rightarrow$^ re-arrange $\rightarrow$ ^ as exponents of $2$ and $10$ $\rightarrow$^ | ||
| - | | $A_{V1}$ | + | | $A_{\rm V1}$ | $80$ | $8 \cdot 10$ | $2^3 \cdot 10^1$ | $ 3 \cdot 6~{\rm dB} + |
| - | | $A_{V2}$ | + | | $A_{\rm V2}$ | $0.0125$ |
| - | | $A_{V3}$ | + | | $A_{\rm V3}$ | $250' |
| </ | </ | ||
| Zeile 843: | Zeile 851: | ||
| - Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$. | - Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$. | ||
| - 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. | - 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 $R_2 = 10 \rm k\Omega$ 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? | ||