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5 Filter Circuits

5.0 Representation of numerical values

In order to be able to analyse filter circuits, various possibilities for representing the numerical values should be explained beforehand.

5.0.1 The dB measure

The decibel measure 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 level makes it possible to refer to a reference quantity. In electronic circuitry, the decibel is used as a dimensionless unit for current or voltage ratios. In the future, this will be particularly interesting for the amplification $A_V = \frac{U_O}{U_I}$ and factors.

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. $A_C^{dB}=20 dB \cdot log_{10}\left(\frac{I_2}{I_1}\right)$

Note that this equation changes somewhat for power quantities, i.e. ratios of $P$. If $P \sim U^2$ or $U \sim P^\frac{1}{2}$ is considered, then we get: \cdot $A_P^{dB}=20 dB \cdot log_{10}\left(\frac{P_2^\frac{1}{2}}{P_1^\frac{1}{2}}\right)$\ $ =\color{blue}{20 dB \cdot \frac{1}{2}} \cdot log_{10}\left(\frac{P_2}{P_1}\right) =\color{blue}{10 dB} \cdot log_{10}\left(\frac{P_2}{P_1}\right) $

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 $dB$.

By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{dB}$ in $dB$. Occasionally, the superscript $\boxed{}^{dB}$ is omitted below, in which case the level is denoted by the unit after the numerical value.

Examples:

1. 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 $10^\color{blue}{x} = \color{green}{1}$, so just $\color{blue}{x}=0$.
   
   
2. 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 $10^\color{blue}{x} = \color{green}{0.01}$, so even $\color{blue}{x}=-2$.
   
   
3. 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, $\color{blue}{x}\approx 0.30103$.
   
   

Use of the dB measure

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: $A_V = 10000000 \rightarrow A_V^{dB}= 140dB$. This also results in less „zero counting“.
• Relationship to Sensory Perceptions: Sensory perceptions such as brightness and loudness have an almost exponential effect. That is, any tenfold increase in the underlying physical quantity (number of photons or sound pressure) does not have ten times the effect, but seems to have an additive effect.
• easier math: The logarithm in the defining equation turns any multiplication of linear factors into an addition of levels: $A_V^{dB}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1) + 20dB \cdot log_{10}(A_2) = A_V^{dB}(A_1) + A_V^{dB}(A_2)$

Especially the last point of the calculation should be considered again. In Abbildung 1 several amplifiers connected in series can be seen with exemplary voltage gain values.
The total gain here is the product of the individual gains: $A_{V,ges}=\prod A_i = A_1 \cdot A_2 \cdot A_3$.
The determination of the total gain was rather laborious before the times of the pocket calculator due to the multiplications. For the levels, the result is an addition: $A_{V,ges}^{dB}=\sum A_i^{dB} = A_1^{dB} + A_2^{dB} + A_3^{dB}$.
Here this would be: $A_{V,ges}^{dB}=\sum A^{dB} = 88dB + (-58dB) + 14dB = 44dB$.

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.

Examples:

1. $A_V^{dB}=58dB$ \color{blue}{2}\cdot: $A_V^{dB}=58dB = 40dB + 18dB = \color{blue}{2}\cdot 20dB + \color{magenta}{3}\cdot 6dB$ This becomes linear to $ \qquad \qquad \qquad \qquad A_V = 10^\color{blue}{2} \qquad \cdot \qquad 2^\color{magenta}{3} \qquad = 100 \cdot 8 = 800$
2. $A_V^{dB}=56dB$ with interpolation points: $A_V^{dB}=56dB = 80dB - 24dB = \color{blue}{4}\cdot 20dB + \color{magenta}{-4}\cdot 6dB$ This becomes linear to $ \qquad \qquad \qquad \qquad A_V = 10^\color{blue}{4} \qquad \cdot \qquad 2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$ or $A_V^{dB} = 20dB + 36dB \rightarrow A_V= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$ \rightarrow
3. $A_V^{dB}=55dB$ with interpolation points: $A_V^{dB}=56dB = 40dB + 18dB - 3dB = \color{blue}{2}\cdot 20dB + \color{magenta}{3}\cdot 6dB + \color{teal}{-\frac{1}{2}}\cdot 6dB$ This becomes linear to $ \qquad \qquad \qquad \qquad \qquad A_V = 10^\color{blue}{2} \qquad \cdot \qquad 2^\color{magenta}{3} \qquad \cdot \qquad 2^\color{teal}{-\frac{1}{2}} \approx 100 \cdot 8 \cdot 0,707 = 560$

5.0.2 The Bode diagram

The aim of the floor diagram is to show the transmission behaviour of systems clearly and concisely.

Preliminary consideration: complex numbers

A complex number can always be reduced to two real number values. For the exact definition of these numerical values there are different possibilities (Abbildung 2):

1. 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 $\underline{A}_V= \Re(\underline{A}_V) + j \cdot \Im(\underline{A}_V)$
2. Definition over magnitude $A_V = |\underline{A}_V|$ and phase $\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right)$ in $\underline{A}_V=A_V \cdot e^{j \varphi}$

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.

path to the Bode diagram

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 on the right shows the $|A_V^{dB}(f)|$ curve for a lowpass filter in the lower half of the image. The curve starts at $0dB$ on the left and drops to about $-45dB$. If the mouse is dragged over the diagram, the gain value in $dB$ can be displayed for each frequency. Clicking on the curve displays current flow and voltage ratios. Only clicking on the leftmost frequency range results in a situation where $U_A$ assumes noticeably high voltages (visible via the color of the line).

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, a different picture results. This is done via Options » Linear Scale. Now the initially flat course of the voltage gain for low frequencies and the drop for higher frequencies becomes very clear. Also the cut-off frequency can be read in the „kink“.

The phase can be made visible via Options » Show Phase.

Description of the Bode diagram

The bottom diagram allows the representation of a complex-valued, frequency-dependent quantity in logarithmic form. It is also referred to as the frequency response and is divided into (cf. Abbildung 3):

• the amplitude response which represents amplitude in double logarithmic form (dB level is a logarithmic representation); and
• the phase response which represents the phase single logarithmically.

A jump in frequency by a factor of $\times 10$ is called a decade (abbreviated Dec.). The amplitude response of various functions will be briefly discussed ($\mathcal{C}$ is an arbitrary frequency-independent factor):

1. $|A_V(f)| = \mathcal{C} \cdot f$: 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. - $|A_V(f)| = \mathcal{C} / f$: 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 $-20dB$ per decade (cf. Abbildung 3 at high frequencies). - $|A_V(f)| = \mathcal{C} / f^n$: 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 $-20dB \cdot n$ per decade.

As an alternative to the actual course, $|A_V(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments.

5.1 Reverse integrator

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.

In electronics, inductors are rarely used for this purpose. This has several reasons:

1. Inductors are possible in integrated circuits, but are somewhat more difficult to calculate as such an element.
2. 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 Elektrotechnik 2.

The following basic circuit is a modified, inverted amplifier (Abbildung 4) in which one or both ohmic resistors are replaced by (complex-valued) impedances.

For first circuit only the part between output voltage $U_A$ and virtual ground should be replaced by a capacitor (Abbildung 5).

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:

1. for each constant input value $U_I \neq 0$ an output value with a fixed slope results and
2. for each positive input value $U_I > 0$ a negative slope results, for a negative input value a positive slope results.

The circuit thus created is called a reverse 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.

~CLEARFIX~~

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 Abbildung 5 for this purpose. The transfer function $U_O = f(U_I)$ is now to be determined.

$A_V = ? \quad -> \quad U_O = f(U_I) $

given equations

Given the following equations:

Derivation

The calculation is performed once in detail here (clicking on right arrow „►“ leads to the next step, alternative representation):

By means of an example, the signal-time curve at the reverse integrator shall be explained. - Let $R=5 c\Omega$, $C=1 \mu F$, and the input voltage waveform $U_E$ shown in Abbildung 6 be given.

1. We are looking for the output voltage $U_O$.

Solution: - Over the given values of $R$ and $C$, the time constant $\tau$ is determined.

1. With the inverse integrator the input value is integrated and inverted. For the given course of the input voltage the calculation of interpolation points is sufficient.
2. With the formula derived in 5.1.1 $U_O$ can be composed section by section:

The calculation is performed once in detail here (clicking on right arrow „►“ leads to the next step, alternative representation):

5.1.3 Determination of amount and phase

In order to be able to determine magnitude and phase, purely sinusoidal input and output variables are to be considered first. The following function is used as input voltage $U_E$:

$ U_I(t)= \hat{U}_I \cdot sin(\omega \cdot t)$

This definition of the input voltage can now be substituted into the above equation for $U_O$:

The calculation is performed once in detail here (clicking on right arrow „►“ leads to the next step, alternative representation):

~CLEARFIX~~

The amount $|A_V|$ is given by the amplitude ratio of $\hat{U}_O \over \hat{U}_I$: $$|A_V|={\hat{U}_O \over \hat{U}_I} = {1 \over {\omega \cdot R\cdot C}} $$

The phase can be determined from the „time offset“ of the peak input voltage $U_E = \hat{U}_I \cdot sin(\omega \cdot t)$ and output voltage $U_O = { {\hat{U}_I } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t)$ can be determined. The phase is given by considering the trigonometric functions and the sign:

$U_I = + \hat{U}_I \cdot sin(\omega \cdot t)$ $U_O = + \hat{U}_O \cdot cos(\omega \cdot t) = + \hat{U}_O \cdot sin(\omega \cdot t + 90°)$ $ \rightarrow \varphi = 90°$

In order to be able to sketch the course in the bottom diagram, the behavior of the transfer function $U_A=f(U_E)$ in the extreme cases for low (${\omega}\rightarrow 0$) and high frequencies (${\omega}\rightarrow \infty$) shall be considered. For magnitude $|A_V|$ and phase $\varphi$ we obtain:

$ |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)| \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow\infty}\quad 0$ .

$\varphi = +90° \qquad \forall \omega$

From these boundary conditions the frequency response can already be sketched, see Abbildung 7.

5.1.4 Circuit analysis with complex calculation

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 integrator. Now the complex calculation will be considered as a method which simplifies the analysis of such circuits. For the complex calculation, the resistances and capacitances are replaced by complex impedances:

$U_R=R\cdot I \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ $U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C dt+ Q_0(t_0)) \qquad \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ with $\quad \underline{Z}_C= \frac{1}{j \cdot \omega \cdot C}$ .

However, this consideration can only be implemented under certain boundary conditions:

1. sinusoidal quantities: complex current or voltage pointers (cf. ET2 Wechselstromtechnik) can only represent sinusoidal quantities.
2. Swinged state: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since switch-on. This takes out disturbances that are generated by switching on.

This can now be used to calculate the circuit (Abbildung 8):

$\underline{Z}_1=R$ $\underline{Z}_2=\frac{1}{j \cdot \omega \cdot C} = \frac{-j}{\omega \cdot C}$

From the basic circuit of the inverting amplifier, its voltage gain is known:

$A_V = \frac{U_A}{U_E}=-\frac{R_2}{R_1}$

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} $

Plausibility check via extreme value consideration

magnitude and phase

The amount $A_V$ is given by:

$|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$

Specifically, for a magnitude of $|\underline{A}_V(0dB)|$ at $0dB$, the result is:

$|\underline{A}_V(0dB)|\overset{!}{=} 1 \widehat{=} 0dB \rightarrow \omega(0dB) = \frac{1}{R \cdot C}$

The phase $\varphi$ is calculated via

$\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right) = arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = arctan \left( \infty \right) = +90°$

The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (Abbildung 9) for $x \rightarrow \infty$.

5.1.5 Frequency response

The frequency response is to be illustrated by means of an example.

1. Let $R=1 k\Omega$, $C=16 nF$ be given.
2. We are looking for the Bode diagram

Solution

1. Determining the time constant:
   $\tau = R \cdot C = 16 \mu s$
   \\.
2. 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 10kHz$
   \
3. 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 $|\underline{A}_V|$, i.e., a slope of $-20dB$ per decade \ \.
4. From this information, the full Bode diagram can be determined (Abbildung 10)

5.2 Lowpass

Another circuit can be derived from the reverse integrator. For this purpose, the hitherto purely capacitive value of the impedance between the output side and virtual ground is to be supplemented by a resistive component. This circuit can be seen in Abbildung 11. In the following, this circuit

• first be considered practically with a simulation,
• then a picture of the system's effect will be drawn up without detailed calculation, and finally *
• be checked by a circuit analysis with complex calculation.

Lowpass in simulation

In the simulation on the right, the circuit from Abbildung 11 is shown again. In addition, two switches $S1$ and $S2$ are built into the circuit by which the various feedback paths can be disabled:

1. If only switch $S1$ is closed, the circuit is an inverting amplifier.
2. If only switch $S2$ 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.

5.2.1 Consideration without detailed account

Extreme value consideration

From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of magnitude of voltage gain and phase can be analyzed.

1. $\omega \rightarrow 0$:
   1. The capacitor acts like an open circuit
   2. Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$
   3. Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially
   4. Thus the circuit behaves like an inverting amplifier: $|\underline{A}_V|=|-\frac{R_2}{R_1}|$
   5. 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.
2. $\omega \rightarrow \infty$:
   1. The capacitor acts like a short circuit
   2. Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$
   3. Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts
   4. Thus the circuit behaves like a reverse integrator: $|\underline{A}_V|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$ and $\varphi=+ 90°$

From this it can be seen that

• for low frequencies a constant gain is expected and
• for high frequencies a drop as known from the reverse integrator.

expected Bode diagram

A floor diagram can be estimated from the extreme value consideration.

Frequency Response:

• For low frequencies, the filter behaves like an inverting amplifier with $|\underline{A}_V|=|-\frac{R_2}{R_1}|$
• For higher frequencies, the filter behaves like an inverting integrator with $|\underline{A}_V|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$
• There is a frequency where both situations seem to occur simultaneously

Phase response:

• For low frequencies, the filter behaves like an inverting amplifier with $\varphi=\pm 180°$.
• For higher frequencies the filter behaves like an inverting integrator with $\varphi=+90°$
• There is a frequency where both situations seem to occur simultaneously

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.

From this knowledge, we get an expected floor diagram as seen in Abbildung 12.

RC element and cutoff frequency

In the circuit, the parallel circuit $R_2$ and $C$ behaves like a passive RC element. That is, it affects the frequency response. At a certain frequency, the circuit behaves just in such a way that the current runs half over $R_2$ and $C$, thus acting „half“ like an inverting amplifier and „half“ like a reverse integrator. This frequency is the cutoff frequency $f_{Gr}$:

For this:
$\underline{X}_C|=R_2$
$\frac{1}{\omega_{Gr} \cdot C}=R_2 \rightarrow \omega_{Gr} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{Gr}$
$\boxed{f_{Gr} = \frac{1}{2\pi \cdot R_2 \cdot C}}$
$\boxed{\underline{A}_V=-\frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}}$

5.2.2 Circuit Analysis with Complex Calculus

Now the circuit is to be analyzed again by means of complex calculation. The impedances are again understood as complex numbers. The starting point is again the voltage gain of the (complex) inverting amplifier. The impedances from Abbildung 11 are taken into account:

$\underline{A}_V=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{j \cdot \omega \cdot C}}{R_2 + \frac{1}{j \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$

Calculation of magnitude and phase

For the calculation of the amount $A_V$ a „trick“ can be used. In principle, the amount can always be determined by multiplication with the conjugate complex value. But here it is easier to calculate the amount of the fraction via the amount of numerator and amount of denominator:

$|\underline{A}_V| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{\mathcal{c}|}$

This results in for the amount:

$\boxed{|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}}$

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.

$\varphi = arctan(\frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)})$

But here, too, there is a „trick“:

$\underline{A}_V= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}} \cdot \frac{1 - j \omega \cdot R_2 \cdot C}{\color{blue}{1 - j \omega \cdot R_2 \cdot C}}$

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:

$\underline{A}_V= \color{blue}{\mathcal{C}} \cdot (1 - j \omega \cdot R_2 \cdot C)$

Thus, the phase $\varphi = arctan\left(\frac{\color{teal}{\Im(\underline{A}_V)}}{\color{brown}{\Re(\underline{A}_V)}}\right)$ is obtained as.

$\underline{A}_V= \mathcal{C} \cdot (\color{brown}{1} + j \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$

$\boxed{\varphi = arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$

For the amount we get

1. for $\omega \rightarrow 0$: $|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 }}$,     since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \gg 1$ So the magnitude of the gain goes towards $|\underline{A}_V| = \frac{R_2}{R_1}$. The effect is similar to the inverting amplifier - at $\omega \rightarrow \infty$: $|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{\color{blue}{(\omega \cdot R_2 \cdot C)^2}} $,     da $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \ll 1$ So the amount of gain goes towards $|\underline{A}_V| = \frac{1}{\color{blue}{\omega \cdot} R_1 \color{blue}{\cdot C}}$. The action resembles the inverse integrator

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 $\color{red}{\varphi}$ (Abbildung 13). For the extremal values $\omega$ of results in:

1. at $\omega \rightarrow 0$: \the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$.
2. 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 Abbildung 13. 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 $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$.

PAGEBREAK ==CLEARFIX

Calculation of the cutoff frequency

The cut-off frequency can also be understood as the transition from the inverting amplifier to the reverse integrator. In the Bode diagram (Abbildung 12), the cut-off frequency can be found at the intersection of the straight lines for the inverting amplifier and the reverse integrator.

Thus, for the cut-off frequency $f_{Gr}$ we get

$\frac{R_2}{R_1} = \frac{1}{\omega_{Gr} R_1 \cdot C}$ $\omega_{Gr} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{Gr}$

At the cut-off frequency, the result is an amount of:

$|\underline{A}_{V,Gr}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega_{Gr} \cdot R_2 \cdot C)^2}}= \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C)^2}} = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + 1^2}} = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{2}}$

$\boxed{|\underline{A}_{V,Gr}| = \frac{1}{2} \sqrt{2} \cdot \frac{R_2}{R_1} = -3dB + |\underline{A}^{dB}_V(\omega \rightarrow 0)|}$

The phase at the cutoff frequency is:

$\varphi_{Gr} = arctan\left(-\omega_{Gr} \cdot R_2 \cdot C\right) = arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = arctan\left(-1 \right)$

$\boxed{\varphi_{Gr} = \frac{3}{4} \pi =135°}$

Because of the $-3dB$ attenuation of the low-frequency gain at the cutoff frequency, it is also called the $-3dB$ cutoff frequency.

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5.3 Reverse Differential

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In Abbildung 14 an inverse differentiator is shown. Compared to the integrator here just the resistor and the capacitor are swapped.

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 („peaks“) of the signal cannot be determined (see Differentiability of the sum function). This leads to problems in the calculation during the simulation and can be seen as overshoot or „deflection“ at $U_A$. To reduce this, a small resistor (relative to the feedback resistor) is inserted after the capacitor.

In the following, only the results will be discussed without derivation.

Circuit analysis via differential equation yields:

$\boxed{U_A = - R \cdot C \frac{d}{dt}U_E}$

With complex calculation, the transfer function becomes:

$\boxed{\underline{A}_V=-j \cdot \omega \cdot R \cdot C}$

From this, the Bode diagram shown in Abbildung 15 can be determined.

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5.4 High Pass

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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 (Abbildung 16). The simulation above shows this high pass. A reverse integrator forms from this with switch $S1$ closed and switch $S2$ open. When the switch is inverted, an inverting amplifier is formed. In the simulation, clicking on a frequency point again shows the distribution of the currents.

With complex calculation this results in: $\boxed{\underline{A}_V = - \frac{R_2}{R_1} \cdot \frac{j \cdot \omega \cdot R_1 \cdot C}{1 + j \cdot \omega \cdot R_1 \cdot C}} $

From this, the Bode diagram shown in Abbildung 17 can be determined.

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5.5 Overview high pass / low pass

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tasks

Learning questions

• What affects the rise time of an amplifier circuit?
• What is the difference between an ideal and a real operational amplifier?

References

References to the media used

Element	License	Link
Abbildung 9: Arc tangent course	CC-BY-SA 4.0	https://de.m.wikipedia.org/wiki/Datei:Arctangent.svg